Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: Tony Williams on March 21, 2006, 10:58:35 AM
-
I have posted on my website a new article concerning aircraft guns:
http://www.quarry.nildram.co.uk/CannonMGs.htm (Cannon or machine guns in WW2)
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Great article Sir.:aok
-
Thanx Tony nice read.
Bronk
-
Hi Tony,
Good article, but the conclusion surprised me:
"To return to the original question, were the Americans right to rely so heavily on the .50 M2 when all other combatant nations had a clear preference for cannon of at least 20 mm calibre? The answer has to be yes. It was adequate for its purpose, and was the only satisfactory aircraft gun in production in the USA."
If the .50 M2 was the only combat-worthy weapon anyhow, why even ask if using it was the right decision? As far as I can see, it was not a decision at all.
Here is a short firepower comparison for roughly equivalent batteries based on energy output as usual:
2x MG 151/20 - 155 rpg - 150 kg - 149% firepower - firepower per weight: 337%
2x MG-FF - 168 rpg - 170 kg - 92% firepower - firepower per weight: 183%
4x MG 151 - 224 rpg - 331 kg - 103% firepower - firepower per weight: 105%
6x ,50 Browning M2 - 250 rpg - 339 kg - 100% firepower - firepower per weight: 100%
Easy to see why the 0.60" MG151 copy was not introduced :-)
Regards,
Henning (HoHun)
-
IMHO the right question to ask is why the Americans utilized the Browning M2 at so slow rate of fire; the FN version of the Browning M2 did 1100 rounds/min at 1939 ie it had roughly 50% better fire power.
gripen
-
Originally posted by HoHun
Hi Tony,
Good article, but the conclusion surprised me:
"To return to the original question, were the Americans right to rely so heavily on the .50 M2 when all other combatant nations had a clear preference for cannon of at least 20 mm calibre? The answer has to be yes. It was adequate for its purpose, and was the only satisfactory aircraft gun in production in the USA."
If the .50 M2 was the only combat-worthy weapon anyhow, why even ask if using it was the right decision? As far as I can see, it was not a decision at all.
It wasn't the only choice they could have made: they could have decided to go with the Hispano (in which case they would probably have been forced to make alterations, and suffered reliability problems meanwhile), or they could have pressed on with the .60 MG 151 (not really worth it).
There were various less likely projects as well, such as the .90 (23mm) T1-4 series (heavy and slow-firing).
They could even have decided to follow the RAF's early example and fitted up to 12 .30 cals.
There are always alternatives - frequently, the ones chosen only seem obvious with hindsight.
-
2x MG 151/20 - 155 rpg - 150 kg - 149% firepower - firepower per weight: 337%
2x MG-FF - 168 rpg - 170 kg - 92% firepower - firepower per weight: 183%
4x MG 151 - 224 rpg - 331 kg - 103% firepower - firepower per weight: 105%
6x ,50 Browning M2 - 250 rpg - 339 kg - 100% firepower - firepower per weight: 100%
Why does 2x151 have the firepower figure of 149% and 4x151 only 103%?
What do you use to determine "firepower"?
-C+
-
Originally posted by Charge
2x MG 151/20 - 155 rpg - 150 kg - 149% firepower - firepower per weight: 337%
These are the 20mm's.....
2x MG-FF - 168 rpg - 170 kg - 92% firepower - firepower per weight: 183%
4x MG 151 - 224 rpg - 331 kg - 103% firepower - firepower per weight: 105%
These are only 15mm.....
6x ,50 Browning M2 - 250 rpg - 339 kg - 100% firepower - firepower per weight: 100%
Why does 2x151 have the firepower figure of 149% and 4x151 only 103%?
What do you use to determine "firepower"?
-C+
-
Ahh, okay. Didn't notice that missing /20... :D
What about "firepower", is that MP?
-C+
-
Originally posted by Charge
2x MG 151/20 - 155 rpg - 150 kg - 149% firepower - firepower per weight: 337%
2x MG-FF - 168 rpg - 170 kg - 92% firepower - firepower per weight: 183%
4x MG 151 - 224 rpg - 331 kg - 103% firepower - firepower per weight: 105%
6x ,50 Browning M2 - 250 rpg - 339 kg - 100% firepower - firepower per weight: 100%
Why does 2x151 have the firepower figure of 149% and 4x151 only 103%?
What do you use to determine "firepower"?
-C+
The two guns are 20mm and the four guns are 15mm.
-
Hi Charge,
>What about "firepower", is that MP?
It's total power of the projectiless (based on kinetic and chemical energy, the latter calculated from mass of the chemically active content at the energy density of TNT) measured at the muzzle.
The total energy of the ammunition supply of the different batteries is identical. Naturally, the lower-firepower batteries require a longer firing duration to expend ammunition of equal effectiveness.
Regards,
Henning (HoHun)
-
Hi Tony,
>It wasn't the only choice they could have made: they could have decided to go with the Hispano (in which case they would probably have been forced to make alterations, and suffered reliability problems meanwhile), or they could have pressed on with the .60 MG 151 (not really worth it).
>There were various less likely projects as well, such as the .90 (23mm) T1-4 series (heavy and slow-firing).
>They could even have decided to follow the RAF's early example and fitted up to 12 .30 cals.
Hm, I don't have the data for the 23 mm cannon unfortunately. I suppose the weapon itself is going to be listed your books, but for a complete comparison I'd need the weight of the belted ammunition, which can be hard to find.
Here is a comparison of the most obvious American options (using 1880 rounds for the 0.50" Browning to reflect the P-51D loadout):
2x Hispano II - 193 rpg - 195 kg - 125% firepower - firepower per weight: 244%
4x 0.60" MG 151 copy - 281 rpg - 373 kg - 103% firepower - firepower per weight: 105%
6x ,50 Browning M2 - 313 rpg - 381 kg - 100% firepower - firepower per weight: 100%
12x Browning ,303 - 782 rpg - 402 kg - 62% firepower - firepower per weight: 59%
To translate that into US terms, replacing the American 50 caliber Browning with British 20 mm cannon would have reduced the weight of the P-51D by 410 lbs (or 4.3% of its loaded weight of 9600 lbs). That's a lot of weight, and as the extra weight is combined with inferior firepower, I really doubt that it was the right decision not to make every effort to get the US production 20 mm cannon into service as quickly as possible.
Regards,
Henning (HoHun)
-
Well, increasing the rate of fíre of the M2 was an obivious option, it required some modifications and decreased life of the barrel but it was in production and it actually saw some use. For one reason or another it was not adobted by USAF nor NAVY during war; possibly yet another "not invented here" case again.
gripen
-
Originally posted by gripen
Well, increasing the rate of fíre of the M2 was an obivious option, it required some modifications and decreased life of the barrel but it was in production and it actually saw some use. For one reason or another it was not adobted by USAF nor NAVY during war; possibly yet another "not invented here" case again.
It wasn't perfected until quite late in the war, and required so many changes that many parts were not interchangeable. This would have been a pain in wartime.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Actually the FN version of the M2 with 1100 rpm was a production and combat ready gun allready 1939. Check your mail box.
gripen
-
"the latter calculated from mass of the chemically active content at the energy density of TNT"
Ok, but I think you need to use PETN figures to get correct energy for Minengeschoss if you use historical belting.
Exlosive velocities are 8,400 m/s for PETN (pure at dens. 1,7) and 6,900 m/s for TNT. (PETN was not used as such in ammo.)
-C+
-
Originally posted by gripen
Actually the FN version of the M2 with 1100 rpm was a production and combat ready gun allready 1939. Check your mail box.
gripen
I have FN's booklet on the 13.2mm Browning (no date, but obviously pre-1940) in which they claim 1,050 rpm.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi,
Nice article Tony,
the .50 cal was the better weapon for the US need, cause, while the gun was good enough to destroy all japanese planes rather fast and to keep the german fighters away from the Bombers, it did provide a much better hitprobability than the relative slow firing Hispano.
Better 6 x 12,5 rps than 4 x 10 rps, so around 35 rps more flying toward the target. Who need cannons, if the enemy dont have big tough bombers?
The much higher gunpower isnt a advantage, if the smaler weapon already is able to bring the enemy fast down, specialy if the smaler gun provide a much higher hitprobability.
The hitprobability is what i miss a bit in your articles, at least imho you should point more to this very important aspect of armament.
For example the MK108 was a nice powerfull weapon, maybe similar powerfull like 2 x MG151/20, but this only count vs very big strait flying targets(bombers). VS fighters, where some 20mm rounds already was enough, the MK108 was not nearly as effective as the 2 x MG151/20.
Without to point to the hitprobability in a realy drastical way, people run around and compare the armament performence by comparing the gunpower with 100% hitquote for all guns.
Anyway, here i wanna say a special thank you! Your work is the base to my damage model work in EAW. :)
Greetings, Knegel
-
Knegel,
To paraphrase an ace "If a man can't hit with two guns, he isn't going to hit with eight."
Basically the RoF of four Hispano's or six Browning .50s is more than enough to get you hits if your aim is right and the Hispano is much more likely to kill.
-
Hi Tony,
>I have FN's booklet on the 13.2mm Browning (no date, but obviously pre-1940) in which they claim 1,050 rpm.
Actually, rate of fire is strictly secondary in the machine gun vs. cannon issue.
The amount of energy released by cannon ammunition is much greater than that of machine gun bullets. This means the cannon win by the weight of the ammunition, not (primarily) by the lower weight of the guns themselves.
Here is a comparison of the ammunition mass alone required for the same total energy:
2x Hispano II - 193 rpg - 95 kg
4x 0.60" MG 151 copy - 281 rpg - 205 kg
6x ,50 Browning M2 - 313 rpg - 207 kg
12x Browning ,303 - 782 rpg - 282 kg
The total Hispano II reference battery - cannon and ammunition - had a mass of just 195 kg, about the same as the bare ammunition weight of the 0.50" Browning battery. Even if you had a Browning machine gun capable of 5850 rounds per minute to equal the firepower of the dual Hispanos, it would still be heavier than the cannon battery:
2x Hispano II - 193 rpg - 195 kg - 125% firepower - firepower per weight: 244%
1x Wonder Browning - 1880 rpg - 236 kg - 125% firepower - firepower per weight: 202%
Regards,
Henning (HoHun)
-
Hi Charge,
>Ok, but I think you need to use PETN figures to get correct energy for Minengeschoss if you use historical belting.
>Exlosive velocities are 8,400 m/s for PETN (pure at dens. 1,7) and 6,900 m/s for TNT. (PETN was not used as such in ammo.)
The problem is, destructiveness and energy are not directly tied in different explosives. The higher blast wave velocity you pointed out will increase damage beyond what you'd expect from the energy comparison alone.
The TNT assumption just serves as a rational and verifyable approximation, it's not perfect of course!
"The guide is always correct, reality is frequently inaccurate" ;-)
Regards,
Henning (HoHun)
-
Hi Knegel,
>Better 6 x 12,5 rps than 4 x 10 rps, so around 35 rps more flying toward the target.
Well, that's a common misconception. You don't just want to hit the enemy, you want to knock him down. To achieve that, you need heavy hits - it probability terms, you want to maximize the product of the rate of fire, the probility of achieving a hit and the probability that the hit kills.
Cannon are considerably superior in Pk, so ROF and Ph can be allowed to drop a bit without sacrificing overall superiority.
>Without to point to the hitprobability in a realy drastical way, people run around and compare the armament performence by comparing the gunpower with 100% hitquote for all guns.
You are confusing rate of fire and hit probability, by the way. I know what you mean, but hit probability is really just the ratio of hits to total shots fired and has nothing to do with rate of fire. For example, a modern air-to-air missile has a great hit probability, but a rather lousy rate of fire :-)
Regards,
Henning (HoHun)
-
Originally posted by HoHun
Actually, rate of fire is strictly secondary in the machine gun vs. cannon issue.
Hm... the point here is that there was 50% firepower increase available with existing technology and minor modifications to airframes.
gripen
-
Originally posted by Karnak
Knegel,
To paraphrase an ace "If a man can't hit with two guns, he isn't going to hit with eight."
Basically the RoF of four Hispano's or six Browning .50s is more than enough to get you hits if your aim is right and the Hispano is much more likely to kill.
This is only true if you have time and the distance to aim well!! While a fightercombat (not a suprise attack) you barely have this time. While a attack to a bomber you wanna stay as far away as possible.
Specialy while shooting with big deflection more bullets to the target = better hitprobability.
Better bring 3 emenys damaged downward, than 1 exploded, specialy if you 'only' wanna cover your bombers.
Greetings, Knegel
-
Originally posted by HoHun
Hi Knegel,
>Better 6 x 12,5 rps than 4 x 10 rps, so around 35 rps more flying toward the target.
Well, that's a common misconception. You don't just want to hit the enemy, you want to knock him down. To achieve that, you need heavy hits - it probability terms, you want to maximize the product of the rate of fire, the probility of achieving a hit and the probability that the hit kills.
Cannon are considerably superior in Pk, so ROF and Ph can be allowed to drop a bit without sacrificing overall superiority.
>Without to point to the hitprobability in a realy drastical way, people run around and compare the armament performence by comparing the gunpower with 100% hitquote for all guns.
You are confusing rate of fire and hit probability, by the way. I know what you mean, but hit probability is really just the ratio of hits to total shots fired and has nothing to do with rate of fire. For example, a modern air-to-air missile has a great hit probability, but a rather lousy rate of fire :-)
Regards,
Henning (HoHun)
Hi,
the hitprobability have much to do with rate of fire(of course its not the only aspect)! Not that much while shooting to a strait flying target on close distance or on big targets. But as longer the range to target and as more the target move, as more aiming is a weak point.
Not many pilots was able to aim like marsaille, not many had the skill to get always that close into a perfect attacking position.
Rate of fire = many bullets to target = more possible hits.
How whas the hitquote of a average WWII pilot, around 2% of the amoload??
If we consider this, i would say if almost twice as many bulltes fly to target the hitprobability will get increase a 'bit'.
Of course if we use Marsaille as representative WWII pilots, this dont count that much anymore.
As i wrote before: Cannons make sence on big tough targets.
Shooting with cannons to japanese planes simply wasnt needed, the .50 cal was more than enough.
Same count for german fighters.
Greetings, Knegel
-
Knegel,
You are very greatly exagerating the effecvt of rate of fire on landing rounds on the target.
If you are off target with eight .50s you will just miss with a lot of bullets. If you are on target with eight .50s or two 20mm cannon you will hit in either case. There is no significant difference in hitting with them.
All of the other combatant nations had no exceptional difficulty hitting fighters with 20mm cannon.
You are creating an issue where there is none.
Long range shooting is just dumb regardless of your armament. I am not going to say that because George Buerling shot down a Bf109 at 800+ yards with the Hispanos on his Spit that long rang fire was practical. In all cases you should get as close as you can.
Saburo Sakai certainly didn't feel threatened by USN pilots using their "long range fire" to try to hit him. He mentally urged them to continue throwing their ammo away.
-
Hi Knegel,
>the hitprobability have much to do with rate of fire(of course its not the only aspect)!
Nothing at all. You are confusing terms.
Nf = number of rounds fired
Nh = number of hits
Ph = hit probability
Ph = Nf / Nh
You are talking about Nh, not about Ph.
ROF = rate of fire
Nh = ROF * Ph
Ph only depends on the ballistic capabilities of the gun and the skill of the shooter, not on the rate of fire.
>As i wrote before: Cannons make sence on big tough targets.
You have been wrong before. Cannon are superior against every kind of target.
>Better bring 3 emenys damaged downward, than 1 exploded, specialy if you 'only' wanna cover your bombers.
Scaring doesn't win battles. You need to kill people to win.
If you look at the way the war was actually fought, the USAAF didn't achieve air superiority by merely scaring the Germans. They only managed to reduce the bomber losses when they unleashed the escort fighters to seek out and shoot down the Luftwaffe fighter pilots. With the production rate of the German aircraft industry, aircraft or spare parts hardly mattered.
Regards,
Henning (HoHun)
-
While Knegel is confusing the terms a bit, there is still point in the rate of fire. With 50% increase in rate of fire you get 50% more hits at same amount of time assuming that all other parameters remain the same.
And there is even point in the probability of the hit, as these are burst type weapons which means that in the case of the uncertainty (range, aiming error, dispersion etc.) there is better probability to get hits with faster rate of fire with one burst (ie larger concentration of the projectiles in the target area).
Besides, that comparison (FN M2 vs American M2) does not rely on questionable rating system of the HEI content of the projectile; in some targets might be easier to destroy with large HEI content while others might be easier with pure hitting power.
gripen
-
The hit probability question really depends on gun harmonisation.
If the guns are mounted in the nose, or are otherwise harmonised to group tightly at a particular distance, then it doesn't matter what the rate of fire is, if you're off-target all of your shots will miss.
You can only use RoF to increase the hit probability if you spread the aim of your guns to cover a wide area. You then stand a better chance of scoring hits if your aim is bad.
The problem with spreading the aim of the guns is, of course, that you will only achieve a few hits even if your aim is very good; so although you may score hits, you are less likely to shoot down your opponent.
If you're only going to score a few hits, then you're better off with cannon than HMGs, because a cannon hit is more likely to inflict serious damage.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Looking at the big dispersion of six wing-mounted 50"s, ROF will, certainly, increase the hit %.
-
Tony,
It does not matter how the guns are located or harmonized; if we assume that all other parameters are constant except the ROF and the guns are used as burst type weapons (short aimed bursts), then the higher ROF will give more hits and also higher probability of the hits regardless the source(s) of the uncertainty (harmonization, dispersion, aiming error and/or what ever).
gripen
-
Hi,
Hohun,
Nf = number of rounds fired
Nh = number of hits
Ph = hit probability
Ph = Nf / Nh
I dont know where you got this formula from, but NF / NH is = hitquote, not hitprobability.
A probability never is exact!!
It depends on many factors, like Pilot skill, dispersion, middle bullet velocity, distance to target, size of target, speed of the gun in relation to the target, attacking angle etc etc.
Many of this factors influence each other, even the weather(turbulences) is a factor.
The rate of fire increase the hitprobability as more the dispersion increase (longer range, turbulences, nervouse pilot, unstable plane at slowspeed etc) and as faster the target move while shooting with big deflection.
Compare the extreme to see the influence of the rof to the hitprobability.
Take the M4 37mm with 2,5 rps(900m/s Muzzvel) and a MG151/15 with 12rps (i often saw 15 rps, also around 900m/s).
You realy think the hitprobability of the slow firing gun while a turnfight was same high like with the MG151/15??
If yes, i wonder why you think the militair did introduce MG´s at all??
Of course a muzzle loader rifle( 1 shot) have the same hitprobability like a MG42, if the target is a elephant in 5m distance, but what if there is a kangaroo running/jumping around like mad in 100m distance??
Tony,
"If the guns are mounted in the nose, or are otherwise harmonised to group tightly at a particular distance, then it doesn't matter what the rate of fire is, if you're off-target all of your shots will miss."
Most WWII aimpoints was around the target most of the time, the pilot shot, failed in most cases and adjusted the aim by watching the tracers. Since a plane dont always react always nice to this adjustings the aim often was only for some milli sec on the target, specialy if the attacked one started to evade. Probably thats the reason why many planes only got some hits and 'smal' damages.
As higher the rof, as higher the hitprobability while this short moments on target is.
I also think wingmounted guns suffer more by a smaler rof, but even nosemounted guns can show a big pilot/plane related dispersion.
Greetings, Knegel
-
Hi Knegel,
>I dont know where you got this formula from, but NF / NH is = hitquote, not hitprobability.
It is the expectancy value of the hit probability and equivalent to the hit probability for sufficiently large values of Nf.
>The rate of fire increase the hitprobability as more the dispersion increase (longer range, turbulences, nervouse pilot, unstable plane at slowspeed etc) and as faster the target move while shooting with big deflection.
The rate of fire increases the number of rounds fired and the number of hits, but not the hit probability.
The only increase you get is the probability of scoring AT LEAST n hits, which is something completely different than the hit probability.
And is quite obvious that the kangaroo vs. elephant question depends on the lethality of the round you are using. Assume you've got two almost identical guns, just that one has five times the rate of fire and the other has five times the lethality per round:
Pkill1 = Pkill0
Pkill2 = 5 * Pkill0
ROF1 = 5 * ROF0
ROF2 = ROF0
Ph = const
Ptrophy1 = Pkill1 * ROF1 * Ph
Ptrophy2 = Pkill2 * ROF2 * Ph
Ptrophy1/Ptrophy2 = Pkill0/(5*Pkill0) * 5*ROF0/ROF0 * Ph/Ph = 1/5 * 5 * 1 = 1
=>
Ptrophy1 = Ptrophy2
In short, both guns have the same probability of turning your target into a trophy. There is no benefit to be gained by distributing the same combined lethality into more rounds unless you start with an overkilling round. (For the US gun question, a single 20 mm round can hardly be considered overkilling even against a fighter, so this is no concern.)
The confusion in this thread is due to not telling apart the basic random event ("Fire one bullet") and its basic probability space with the complex compound event ("Fire n bullets") and then failing to properly define the event to be analysed ("Plane is shot down"/"Plane is hit by at least one bullet").
What's the desired event? "Plane is shot down"? Nothing to gain from rate of fire if it comes at the cost of lethality of the individual rounds. Nothing to gain from an increase in dispersion either.
"Plane is hit by at least one bullet"? The probability for this event does in deed increase with the rate of fire, and it might be influenced both up or down by dispersion.
Now back to air combat: I have not seen any historical evidence to suggest that any air force ever actively pursued a "better three damaged than one shot down" policy. Most tactical guidelines provided by successful fighter pilots do indeed stress "Get in close and don't waste your ammunition for uncertain shots". I have yet to see a real-world pilot to advise "Your best option is to spray them from long range". Wing gun trajectory divergence, which is not the same as dispersion, is seldom commented on, but all the comments I have found are negative. Dispersion is especially mentioned in the German "Schießfibel": "Verlaß dich also nicht auf die Waffenstreuung - sie hilft dir nicht, wenn du schlecht gezielt hast! Du siehst hier klar, wie g e n a u Du den Vorhalt kennen und halten muß, wenn nicht der ganze Segen daneben gehen soll." ('Thus, don't rely on weapon dispersion - it won't help you if your aim was poor! You can clearly see here how a c c u r a t e you have to know and hold the lead if you don't want the entire 'blessing' to miss.")
I don't know where the "shotgun" romanticism comes from, but apparently the mistake of thinking that weapon dispersion will make it easier to get kills was common in WW2, too, as the Schießfibel directly addresses and rejects it.
Regards,
Henning (HoHun)
-
Originally posted by HoHun
Dispersion is especially mentioned in the German "Schießfibel": "Verlaß dich also nicht auf die Waffenstreuung - sie hilft dir nicht, wenn du schlecht gezielt hast! Du siehst hier klar, wie g e n a u Du den Vorhalt kennen und halten muß, wenn nicht der ganze Segen daneben gehen soll." ('Thus, don't rely on weapon dispersion - it won't help you if your aim was poor! You can clearly see here how a c c u r a t e you have to know and hold the lead if you don't want the entire 'blessing' to miss.")
Actually that part of the "Schießfibel" means that if you aim good enough, the some amount of dispersion will help you. The only case when some dispersion does not help you is the purely theoretical case of shooting without any kind of error.
Naturally that means that there is a connection between amount of error (which is practically allways systematical) and optimal amount of dispersion; the more error, the higher the optimal dispersion. The good point in dispersion (if compared to harmonization or other solutions to same problem) is that it's absolute amount increases with range just like the error in the aerial shooting (neglible at short range).
Originally posted by HoHun
I don't know where the "shotgun" romanticism comes from...
The shotgun analogy comes from the simple facts that these weapons are burst type and there is allways some error in the aerial shooting. As Galland noted: Flying targets are normally shot with shotgun.
Besides, the analogy with shotgun fits particularly well to the comparison between the FN M2 and the american M2 (the FN version having 50% higher ROF). Calculating probability of the hit for a single projectile does not account the time effect which is the main advantage of the higher ROF.
In fact the only things I see here as "romanticism" are assumtions based on extremely accurate shooting and avoiding burst nature of these weapons.
gripen
-
Hi,
the lethality dont have to do anything with hitprobability!!
It have to do with kill probability!
If your bullet with 5 times as high lethality simply dont hit, cause the much smaler hitprobability, its worthless, at least less worth than the smaler gun.
Specialy if the smaler gun is enough to strike the kangaroo down.
"I have not seen any historical evidence to suggest that any air force ever actively pursued a "better three damaged than one shot down" policy.
Me neighter,t hats why i wrote:"Better bring 3 emenys damaged downward, than 1 exploded"
But actually while a escort and while trying to help a wingi it count as more important to bring the enemy off target than to destroy him. Thats why the pilots didnt expected many kills while escort missions. Thats the reason for the germans and brits to keep the smal MG´s that long. They was good enough to "ring the bell" and did provide a big amoload(not unimportant while a escort).
Of course it was/is the wish of all countrys to have a gun which provide higherst firepower and highest hitprobability, but in WWII no nation had the perfect gun, so they had to decide between most firepower and hitprobability. Often the decission was based on the order.
And of course the pilots got told "Get in close and don't waste your ammunition for uncertain shots", but as we know this was more a wish than reality. While a turnfight there are not many certain shots.
"What's the desired event? "Plane is shot down"? Nothing to gain from rate of fire if it comes at the cost of lethality of the individual rounds. Nothing to gain from an increase in dispersion either."
If a .50cal is good enough to bring a target down with some lonly hits, cause missing selfsealing tanks or cause the bullet was strong enough to get through the plating( according to all i did read, this was the case vs Japanese planes and smal fighters), there is nothing to gain from lethality at the cost of rate of fire.
Only if you consider the big dispersion even while good aiming, and so the high quote of not hitting bullets, you can see in what way the hitprobability get increased by the rof.
Its simply more likely that a bullet hit, if more rounds fly toward the target.
I dont say that dispersion help aiming!!! Its the other way around, the dispersion(gun/pilot/plane/turbolences related) minimize the hitprobability. Specialy with the very smal archived hitprobability due to this dispersion in general, the rof increase the hitprobability much.
Its like shooting with rifle to a bird and a shotgun. The hitquote of the rifle may be similar and the destructive power is much more big, but the hitprobability of the shotgun is much more big on a moving target.
Only if we consider a absolut exact aiming, the rof dont count, but thats unrealistic in WWII, specialy on ranges above 100m and specialy while attacking a turning enemy.
This is what Tony wrote in one of his articles to this:
"It is sometimes argued that a projectile with a high muzzle velocity and a good ballistic shape (which reduces the rate at which the initial velocity is lost) provides a longer effective range. To some extent this is true, but the greatest limitation on range in air fighting in the Second World War was the difficulty in shooting accurately. The problem of hitting a target moving in three dimensions from another also moving in three dimensions (and probably at a different speed and on a different heading) requires a complex calculation of range, heading and relative speed, while bearing in mind the flight time and trajectory of the projectiles. Today, such a problem can easily be solved by a ballistic computer linked to a radar or laser rangefinder, but at the time we are examining, the "radar" was the human eyeball and the "ballistic computer" the human brain. The range, heading and speed judgements made by the great majority of pilots were notoriously poor, even in training. And this was without considering the effects of air turbulence, G-forces when manoeuvring, and the stress of combat. These factors limited the effective shooting range to around 400 m against bombers (longer in a frontal attack) and against fighters more like 250 m. "
As higher the rof, as longer the practical effective range was(the theoretical range of a .50cal got assumed by the USAF with 900yard).
Specialy in the PTO it was wise to carry 6 x M2 .50cal with enough amo to shoot 30sec(F4U) than 4 x hispano M2 with enough amo to shoot 12sec(also F4U).
The hitprobability while shooting in general was more big, the hitprobability per amoload was much bigger.
I think you mix up hitquote with hitprobability.
The hitquote is a backward calculation, based on results, describing a special moment of history. Although the hitquote is = the hitprobability for this exact circumstance, its only a smal part of the whole picture.
The hitprobability, so the hitquote of the same armament will increase with a better gunsight, better pilot, good weather(no turbolences), bigger target, poor enemy pilot skill etc. and the other way around.
The hitquote depends on the hitprobability, but the hitprobability dont depends on the hitquote.
To say the rof dont influence the hitprobability, cause if the pilot aim well he hit anyway, and if he dont aim well he fail anyway, is the same like to say a better gunsight dont improve the hitprobability, cause a good pilot always go very close and always know the needed lead.
Greetings, Knegel
-
Originally posted by gripen
Tony,
It does not matter how the guns are located or harmonized; if we assume that all other parameters are constant except the ROF and the guns are used as burst type weapons (short aimed bursts), then the higher ROF will give more hits and also higher probability of the hits regardless the source(s) of the uncertainty (harmonization, dispersion, aiming error and/or what ever).
I agree with you that a higher RoF (other things being equal) will give more hits - once you're on target. But the probability of getting into the 'hit zone' has more to do with the shot pattern of the guns - their dispersion - than their RoF.
I would basically summarise the situations like this:
1. Accurate shooting and concentrated fire = many hits (increased RoF will further increase the hits, and allow a shorter burst of fire to be effective)
2. Accurate shooting and dispersed fire = few hits (increased RoF will increase the number of hits)
3. Less accurate shooting and concentrated fire = no hits (increased RoF will have no effect, except wasting more ammo)
4. Less accurate shooting and dispersed fire = few hits (increased RoF will increase the number of hits)
In the first case, increased RoF is not so important because you'll probably shoot the target down anyway, so it's mainly important in the case of dispersed fire.
Even then, however, you have to be good enough to get your cone of fire to coincide with the position of the target; if you're very inaccurate then you're going to miss regardless of shot dispersion or RoF.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi,
if we believe the 2% hitquote estimation of the Luftwaffe while attacking strait flying 4Mots, the dispersion must have been a very big aspect, therfor i would say number 1. and 3. dont weight that much while aircombat.
With dispersion i dont mean only the gunsetting and gun related dispersion.
I guess the biggest dispersion is caused by the try of the pilot to keep the target in the gunsight. The planes wasnt fixed, the pilot had to adjust the aiming point all the time, while this he moved often for a very short period over the target.
So i would say:
As shorter the range or as bigger the target = as higher the hitprobability of a single round = as less important the rof is, regarding the hitprobability of the armament.
As longer the range or as smaler the target = as smaler the hitprobability of a single round = as more important the rof is, regarding the hitprobability of the armament.
Since on smal targets around 50m count as safe kill distance, but already 250m turn to be luck, the rof will be a not to smal factor at a range of around 150m.
Greetings,
-
Hi Knegel,
>the lethality dont have to do anything with hitprobability!!
>It have to do with kill probability!
We agree.
>If your bullet with 5 times as high lethality simply dont hit, cause the much smaler hitprobability, its worthless, at least less worth than the smaler gun.
The hit probability Ph is the same. It has nothing to do with rate of fire.
>Specialy if the smaler gun is enough to strike the kangaroo down.
To be exact: If the lethality of the greater gun is so high that it achieves a Pk of 100% and still has some lethality reserve against the kangaroo. That is the overkill phenomenon I wrote about, and it certainly does not apply to the 20 mm cannon.
>Me neighter,t hats why i wrote:"Better bring 3 emenys damaged downward, than 1 exploded"
Well, I misunderstood that as normal WW2 air force use was "kill" to describe a target that was probably shot down, "probable" a target that might have been shot down, and "damaged" a target that was not shot down. Anyway, then your point is overkill again, and the 20 mm cannon does not overkill.
>Often the decission was based on the order.
How about some clear historical evidence?
>>"What's the desired event? "Plane is shot down"? Nothing to gain from rate of fire if it comes at the cost of lethality of the individual rounds. Nothing to gain from an increase in dispersion either."
>If a .50cal is good enough to bring a target down with some lonly hits, cause missing selfsealing tanks or cause the bullet was strong enough to get through the plating( according to all i did read, this was the case vs Japanese planes and smal fighters), there is nothing to gain from lethality at the cost of rate of fire.
Not anything that could happen will happen. You could bring down a Zero with a single 12.7 mm hit, but even the Zero is not all unprotected fuel tank. Even the chance for hitting the fuel tank, never mind doing lethal damage to hit, might be as low as 15% even if you manage to hit the Zero. A 20 mm mine shell can do serious damage on 90% of the airframe, so it will end up with a higher Pk in the end, making cannon superior aginst Zeros as well. (Interesting side note: The un-armoured, un-protected, machine-gun-armed Curtiss CW-21 light-weight fighter fought some battles against cannon-armed Zeros in the Pacific, suffering much heavier losses than they could afflict to the Zeros.)
Anway, it's all rolled up in this formula:
Pdestruction = Pkill * Nf * Ph
>Only if you consider the big dispersion even while good aiming, and so the high quote of not hitting bullets, you can see in what way the hitprobability get increased by the rof.
Hit probability does not increase through rate of fire. Basic stochastics.
>Its simply more likely that a bullet hit, if more rounds fly toward the target.
That's the completely different event "at least one bullet hits the target". This is not connected to the probability of destruction.
>Its like shooting with rifle to a bird and a shotgun. The hitquote of the rifle may be similar and the destructive power is much more big, but the hitprobability of the shotgun is much more big on a moving target.
Actually, it's not. The trick about the shotgun is that it shoot a vast number of small projectiles, most of which miss. Basic stochastics.
What you are considering is the event "at least one of the shotgun pellets hits", which does indeed have a higher probability, but makes the example different from the MG vs. cannon example.
The probability for each shotgun pellet to hit is just the same as the probabilty for each rifle bullet to hit. It's just that the shotgun fires hundreds of projectiles in a fraction of second, but most of them miss just the same as hundreds of rifle projectiles would miss.
The shotgun analogy typically is completely misunderstood by simulator pilots.
>As higher the rof, as longer the practical effective range was(the theoretical range of a .50cal got assumed by the USAF with 900yard).
Best keep that out of the discussion since "effective range" is another can of worms. Just to illustrate this: What is the effective range of one 12.7 mm machine gun? And if rate of fire increases effective range, what is the effective range of eight 12.7 mm guns? And of course lethality enters the picture, too, since "effective" implies the capability not just to hit but to damage upon hitting. (If I remember correctly, when I asked Tony, he mentioned that there was no scientific definition of the term "effective range" anyway.)
>I think you mix up hitquote with hitprobability.
Read this first: http://de.wikipedia.org/wiki/Erwartungswert
(First paragraph could be enough.)
I regret that there seems to be no similar concise explanation in the English Wikipedia, it would be quite useful for our discussion here.
Regards,
Henning (HoHun)
-
Hi Knegel,
>Since on smal targets around 50m count as safe kill distance, but already 250m turn to be luck, the rof will be a not to smal factor at a range of around 150m.
This is the formula in which ROF is a factor:
Pdestruction = Pkill * ROF * Tf * Ph
Note that it is just one factor.
Using the approximation that Pkill is approximately proportional to the total energy of the projectiles, you get the following comparison:
2x Hispano II - 193 rpg - 195 kg - 125% firepower - firepower per weight: 244%
6x ,50 Browning M2 - 313 rpg - 381 kg - 100% firepower - firepower per weight: 100%
Which is exactly what I posted above. By the way, since you mentioned ammunition loads: If you examine the weight of the ammunition load required for equal total energy, you'll find that cannon are superior to machine guns and thus would have been a better choice for the USAAF even by your criterium.
Regards,
Henning (HoHun)
-
Originally posted by Tony Williams
I would basically summarise the situations like this:
1. Accurate shooting and concentrated fire = many hits (increased RoF will further increase the hits, and allow a shorter burst of fire to be effective)
Shorter bursts will result smaller dispersion due to wandering of the aim ie higher probability of the hit with given amount of rounds even in the case of the relatively accurate shooting. This can be seen from the gun camera films; even in the close range shooting most projectiles seem to miss due to wandering.
Note that the time is the key to understand the advantages of the increased ROF; there is basicly allways lack of time to aim and shoot. A good example of this is head on attack on bombers as the Germans did against American bombers, another example is shooting a maneuvering fighter.
Shortly the advantages of the increased ROF here are that shorter burst are needed for same amount of hits (same killing probability with better hit probability) or similar bursts will result more hits (better killing probability with same hitting probability).
Originally posted by Tony Williams
2. Accurate shooting and dispersed fire = few hits (increased RoF will increase the number of hits)
That is true only in the more or less theoretical case of shooting without any error.
In reality the sitution is pretty much same as in the first case because the dispersion (from the gun or mounting) is practically allways much smaller than the dispersion caused by even very small aiming errors (or other errors; wandering, harmonization etc.).
Notable thing here is that there is no reason to believe that the other errors (aiming or what ever) are some how evenly distributed ie the error is practically allways systematical. And that is the reason why some guns have some amount of purpose built dispersion.
Originally posted by Tony Williams
3. Less accurate shooting and concentrated fire = no hits (increased RoF will have no effect, except wasting more ammo)
Generally I understand less accurate shooting as more error in shooting.
In practice this case is similar with the first case, the probabilities of the killing and hitting are just lower due to larger error, the advantages of the ROF being the same.
Originally posted by Tony Williams
4. Less accurate shooting and dispersed fire = few hits (increased RoF will increase the number of hits)
Agreed, this is a typical shotgun case; the dispersion and the ROF help you.
gripen
-
Hi,
if you agree that a shotgun provide a higher hitprobability, while shooting to a smal fast moving target, than a normal rifle, you also must agree that a MG, which shoot 12 rps have a higher hitprobability than a rifle.
And of course also 6 x 13rps(78 rps) provide a higher hitprobability, while shooting to this smal target, than 4 x 10rps(40 rps).
If you combine this hitprobability advantage with the much bigger amoload and the fact that the .50cal brought suprising easy kills vs almost all japanese planes, the .50 cal turn out to be the better weapon.
If the rof is not important, why they did introduce AAA guns with 4 x 20mm instead of using the realy deadly but slow firing 37mm only??
While a flyby of a plane, which did last maybe 30sec, the 3,7-cm-Flak 43 could shoot max 30 rounds, while the 2-cm-Flakvierling 38 was able to shoot max 600 rounds, and you realy think the hitprobability was the same???
I dont think that the kill probability of a 20mm vs a unprotected plane, without selfsealing tanks was that much better than that of .50cal. At least the lot of snapshot kill storys while shooting down japanese planes make me believe it couldnt get much better.
Hitprobability, not expection is what we talking about, right?
http://de.wikipedia.org/wiki/Wahrscheinlichkeit
Are you kidding regarding the CW21?? Many Zeros with combat experienced pilots vs some splitted Dutch planes??
The question is: How it was possible to get a Zero down with this obvious disadvantage?
http://www.warbirdforum.com/cw215.htm
Later the Zeros and other japanese planes got shot down in a high sequence almost only by 4-6 x .50cal, until 1943 mainly carried by F4F´s and P40´s.
Specialy while shooting to the nimble japanese fighters, often from horrible positions(cause the japanese planes simply had a advanced turn climb performence) the increased hitprobability was a good advantage.
Greetings, Knegel
-
Hi Knegel,
>if you agree that a shotgun provide a higher hitprobability ...
The point is, I don't.
What the shotgun does it to provide a greatly increased number of rounds at the same hit probability as a rifle:
Pigeon-killing example:
Shotgun: Pkill = 2% per pellet, Nf = 250 pellets, Ph = 20%
Rifle: Pkill = 100% per bullet, Nf = 1 bullet, Ph = 20%
Shotgun: Pdestruction = Pkill * Nf * Ph = 2% * 250 pellets * 20% = 100%
Rifle: Pdestruction = Pkill * Nf * Ph = 100% * 1 bullet * 20% = 20%
So in this example, everything is different between shotgun and rifle except for the hit probability.
Of course, the shotgun example is a bad choice for comparing it to WW2 aerial guns since the firing of the individual pellets is not a series in independend random events. Still, if one insists on comparing them, one should be aware that a shotgun fire a large number of projectiles, most of which miss even if you manage to hit the pigeon.
>Hitprobability, not expection is what we talking about, right?
The link explains why it's perfectly fine to use Ph = Nf / Nh. If you don't follow, try to re-read the first paragraph in the Wikipedia article I gave you, and click the link "Gesetz der großen Zahlen", too.
>Are you kidding regarding the CW21?? Many Zeros with combat experienced pilots vs some splitted Dutch planes??
It should be pretty obvious that I am not kidding. The CW21 example shows the result of applying 20 mm fire to unprotected airframes. Study history with an open mind, it might have a lesson for you.
Regards,
Henning (HoHun)
-
Hi,
if the hitprobability while shooting with a shotgun to a smal fast moving target is the same like with a rifle, can you please explain why a hunter use a shotgun??
I always thought its cause the shotgun provide a better probability to hit a rabbit or bird, no??
I dont know your definiion of probability, according to the link i gave you its this:
Wahrscheinlichkeit ist
(probability is:)
-ein Maß für die Unsicherheit zukünftiger Ereignisse oder zweifelhafter
Aussagen(a measurement for the uncertainness of comming occurrences or problematic testimonys) ,
-ein Maß für die relative Häufigkeit des Auftretens von Ereignissen bei Auswahl aus mehreren Möglichkeiten(a measurement for the relative frequency of occurrence of occurrences by selection out of multiple possibilitys).
So can you please explain why you assume exact the same hitprobability in your formula?? The hitprobability is what we wanna calculate, better to say estimate!!
Actually i dont understand the corelation of the "Erwartungswert einer diskreten Zufallsvariablen" and "Gesetz der großen Zahlen" to the hitprobability.
The hitprobability of a gun is related to much more important factors than luck.
In our shotgun/rifle example the hitprobability of the shotgun is bigger, cause we dont need to aim that exact, cause the many bullets flying with big dispersion toward the bird(or rabbit).
Actually you already did agree to this:"What you are considering is the event "at least one of the shotgun pellets hits", which does indeed have a higher probability, but makes the example different from the MG vs. cannon example."
There is no different between this and the high rof vs low rof example. A shotgun represent a gun with high rof, it dont matter if the bullets get shot at same time or in one hour. Who shot more often simply have a higher probability to hit the target, of course this only is important as long the hitprobability of one bullet is very smal(very smal targets, fast moving targets, very long range, no time to aim etc).
As more the hitprobability of a single gun shot increase as less important is a high rof. Thats why they use rifles while hunting big, still standing animals and thats why they did use rifles with at least two barrel while hunting a tiger(two barrels = a double hitprobability if the 1st bullet fail) and also while hunting partridges and rabbits, cause even the shotgun dont provide nearly a 100% hitprobability.
While estimating the hitprobability it dont matter how many of my bullets hit the target, it only matter how likely it is to hit the target at all. The kill probability determine how likely it is to get enough hits to bring the target down, also here it dont matter how many bullets dont hit, it only matter that at least one bullet hit a critical area.
As long as a probability isnt 100% we cant calculate it exact, as lower the hitprobability get as more the calculation turn into a estimation.
Out of the hitquote while tests we can estimate the hitprobability in a better way. But we will get much different hitquotes with different testsetups, like different attacking angle, different stable gunplattform(due to turbolences or slow speed), different distances, strait flying or turning target and much more. Therfor its not possible to estimate a hitprobability without to determine the circumstances exact.
What we know is that the Luftwaffe did estimate the hitquote to around 2% when a average pilot did attack a B17 from the rear.
We also know some storys of Marsaille where he did need only 6 rounds to bring down a plane, how many did hit we dont know, but according to his wingis it looked like almost all did hit. So we have a differeence in hitprobability only due to pilot skill from 2% up to around 90%.
"It should be pretty obvious that I am not kidding. The CW21 example shows the result of applying 20 mm fire to unprotected airframes. Study history with an open mind, it might have a lesson for you."
After i did read the article in what way the CW21´s got shot down and that they was able to destroy some Zeros as well, i dont see any proofe for cannons beeing better vs light armored planes. I think noone told that 20mm´s cant bring down a light armored plane.
Do you think i dont study history with an open mind, while you be sure you do?? I think thats somewhat a contradiction.
Maybe you dont believe me, but the whole time i try to understand your thought´s, but i always come to the same conclusion that you mix up hitprobability with hitquote.
A hitquote can get calculated, a hitprobability is a estimation without a exact value. A hitquote is exact cause it get calculated out of tested results, a hitprobability is a estimation of how it will be in future under rather less than more exact circumstances.
Greetings, Knegel
-
Hi Knegel,
>if the hitprobability while shooting with a shotgun to a smal fast moving target is the same like with a rifle, can you please explain why a hunter use a shotgun??
I have already explained that above :-)
If a hunter fires his shotgun, the shot spreads out so much that maybe only 10% of them hit the rabbit, which drops dead. Hit probability is 10%, rabbit-kill probability is 100%.
If a hunter fires a rifle, he will miss the rabbit 90% of the time, but that still means he has a 10% hit probability. (And if he hits, the rabbit drops dead.) This translates into a rabbit-kill probability of 10%.
Two weapons with equal hit probability - but the hunter rightly chooses the shotgun.
>Actually i dont understand the corelation of the "Erwartungswert einer diskreten Zufallsvariablen" and "Gesetz der großen Zahlen" to the hitprobability.
As long as we talk about a very large number of rounds fired, we can treat expectancy value and probability as equivalent.
If a guy has fired 110 rounds in one fight, of which 11 hit, that doesn't mean that we can say with any kind of reliability that the hit probability for that weapon under those circumstances is 10%. However, if at the end of the war, we're looking at 1.1 million rounds fired and 110000 hits, we can say with great certainty that the hit probability was 10%.
>There is no different between this and the high rof vs low rof example.
The difference is that the rifle overkills a rabbit. The 20 mm cannon does not overkill a fighter.
>While estimating the hitprobability it dont matter how many of my bullets hit the target, it only matter how likely it is to hit the target at all.
>Maybe you dont believe me, but the whole time i try to understand your thought´s, but i always come to the same conclusion that you mix up hitprobability with hitquote.
That's exactly the confusion I pointed out in the beginning: The hit probability describes the likelihood of one shot fired at a target hitting that target. That's a basic random experiment.
You confuse that with the question "If n shots are fired at a target, how likely is it that at least one hit is achieved".
Using the relevant axioms, you can calculate that as:
P (at least one hit with n shots) = 1 - P (no hits with n shots) = 1 - P (no hit with one shot)^n
For example, firing 10 shots with a hit probability of 10%, how likely is it that we score at least one hit?
P (>=1 hit with 10 shots)
= 1 - P (no hit with one shot)^10
= 1 - (1 - P (one hit with one shot))^10
= 1 - (1 - 0,1)^10
= 1 - 0,9^10
= 1 - 0,349
= 65.1%
So we have an exact hit probability of 10%, but the chance of scoring at least one hit with 10 rounds is 65.1%.
These are the two aspects of probabilities that have lead to much confusion in this thread.
I hope I have managed to clear up that confusion now :-)
Regards,
Henning (HoHun)
-
Hi,
"If a hunter fires his shotgun, the shot spreads out so much that maybe only 10% of them hit the rabbit, which drops dead. Hit probability is 10%, rabbit-kill probability is 100%."
This show exact that you mix up the hitquote with hitproabbility.
"That's exactly the confusion I pointed out in the beginning: The hit probability describes the likelihood of one shot fired at a target hitting that target. That's a basic random experiment."
We are talking of the hitprobability differents between a gun armament of 6 x .50cal and 4 x Hispano II, not about the hitprobability of a single shot!!
"As long as we talk about a very large number of rounds fired, we can treat expectancy value and probability as equivalent.
If a guy has fired 110 rounds in one fight, of which 11 hit, that doesn't mean that we can say with any kind of reliability that the hit probability for that weapon under those circumstances is 10%. However, if at the end of the war, we're looking at 1.1 million rounds fired and 110000 hits, we can say with great certainty that the hit probability was 10%."
Thats true, but is that of any relevance here?
Do you have a exact hitqote of a 6 x .50cal and 4 x hispano II armament?
I dont know any reliable hitquote of WWII at all, most are only rough estimations. The example of Marsaille and a average luftwaffe pilot show that the hitprobability vary much. If we dont take this big differents into account, the expectancy value only can estimate the need of rounds all over, but it cant give a hint how to change things to a better way.
And for sure this dont help while estimating which of the two armaments show a better hitprobability.
The confusion is that you talk about the hitprobability of a single bullet, while we wanna find out which armament show a better hitprobability.
Greetings, Knegel
-
btw:
"If a hunter fires his shotgun, the shot spreads out so much that maybe only 10% of them hit the rabbit, which drops dead. Hit probability is 10%, rabbit-kill probability is 100%."
"If a hunter fires a rifle, he will miss the rabbit 90% of the time, but that still means he has a 10% hit probability. (And if he hits, the rabbit drops dead.) This translates into a rabbit-kill probability of 10%."
In both of this examples you estimate a hitquote, but what if this estimations are wrong??
Actually i doubt that a average hunter have a 10% hitquote with a rifle while shooting to a fast running rabbit!! My experience is that the hitquote goes rather to 0,5%. For an average hunter its realy more luck to hit with an rifle.
But here again it depends to the skill of the hunter and other circumstances, like how fast is the rabit, how often he turn, how far away is it etc.
If the rabbit dont move the rifle will have a much higher bullte/ pellet hitquote, while the hitquote of the gun is the same for both.
But here again this is only the hitquote after both made the shot and did hit.
The hitprobability of the shotgun will remain higher, cause we can expect that while shots in the future the rifle armned hunter will make a smal mistake, or he get disturbed somehow so his hit will fail, while the shotgun have amuch higher probability to hit, even the aim isnt exact.
Greetings, Knegel
-
Hi Knegel,
>The confusion is that you talk about the hitprobability of a single bullet, while we wanna find out which armament show a better hitprobability.
And therein lies your fallacy. Hits are nothing, destruction is everything.
Here is the mechanism that ties everything together:
Pdestruction = Pkill * Nf * Ph
You are suggesting that it's beneficial to increase Nf at the expense of Pkill. Well, quite obviously, it isn't. Pkill is just as important a factor as Nf.
Here is the gun comparison again with number of rounds per second and relative total energy content of an average projectile added:
2x Hispano II - 193 rpg - 195 kg - 125% firepower - 20 rps - 486%
4x 0.60" MG 151 copy - 281 rpg - 373 kg - 103% firepower - 48 rps - 168%
6x ,50 Browning M2 - 313 rpg - 381 kg - 100% firepower - 78 rps - 100%
12x Browning ,303 - 782 rpg - 402 kg - 62% firepower - 240 rps - 20%
This demonstrates quite nicely that total number of rounds does not automatically increase firepower because in practice, an increase in number of rounds fired can only be achieved at the expense of other factors.
>In both of this examples you estimate a hitquote, but what if this estimations are wrong??
Then you can still use the example to understand the basic mechanisms of multiple-projectile gunnery.
Regards,
Henning (HoHun)
-
Hi,
only if we have estimated a hitprobability of the armament, we can start to estimate a killprobability!!
We DONT have Nh cause we dont know the hitprobability!!!
You always estimate the same hitprobability for all guns/armaments, but thats wrong!!!
My estimation is that 87 rps provide a more big probability to hit the target than 40 rounds, as more as the hitprobability of a single round decrease.
Of course this hitprobability advantage get evened out by the higher power of the 20mm, but same we can say on close distance(high hitprobability of a single round), where the 6 x .50cal are more than enough to bring a unprotected plane down. It simply dont need a cannon to kill a unprotected plane without selfsealing tanks.
Thats why i estimate the kill probability of 4 x hispano II and 6 x .50cal while shooting onto unprotected planes as pretty similar, probably even 8 x .30cal(160 rps) would provide a similar killprobability(many japanese pilots simply crashed cause one hole in the tank, and the tank´s are the biggest critical targets in a plane).
Selfsealing tanks and plating already decrease the killprobability of the .30cal´s, cause the damagepower simply wasnt strong enough, so although the hitprobability was very high, the killprobability was bad, but as the results while BoB show, the kill probability still wasnt that bad vs fighters, even with selfsealing tanks and plating, only bombers was a real probelm.
While shooting to fighters and unprotected planes in general(main targets of US fighters) at the end the time to shoot make the different, and 30sec for .50cal in a F4U are simply more worth than 12sec of 20mm´s in the same plane, if the killprobability is roundabout the same.
Iam pretty sure, if the IJAAF or the Luftwaffe would have had something similar to the B17 (also with same high numbers) the US planes would have carried 20mm´s in a much higher number.
Greetings, Knegel
-
btw, now when is reread all with the knowledge that your hitprobability is the hitprobability of a single round, i understand what you are talking about.
Nh = ROF * Ph(bullet)
Ph(bullet) vary much with the circunstances, quality of gunsight and/or pilotskill.
Ph(gun) increase with decreasing Ph(bullet) and increasing ROF.
Ph(gun) is always 100% (ROF dont matter) if Ph(bullet) is 100%.
If the assumtion is right that the hitquote of a averange pilot was around 2%(maybe 5% or 10%?), the rof will increase the Ph(gun), same like Ph(gun) get increased by using a shotgun under same circumstances.
Although the balistic of the shotgun pallets(bullets) are much worse to the rifle bullet, Ph(gun) increase much while shooting on targets which cause a horrible Ph(bullet) .
Same i assume while shooting with a higher rof!
The balistic actually isnt that a big influence to the Ph(bullet), at least not while common shooting ranges of 50-250m. Every pilot had to know the needed lead for his gun anyway and the gunsight was adjusted to this gun. Ph(bullet) got imfluenced much much more by the dispersion and piltoskill. Of course while shooting on long range (groundtargets) a higher middle bullet velocity turn to be a big aspect. Thats why the Mk108 was good while aircombat but bad while groundattacks.
Greetings, Knegel
-
Hi Knegel,
>You always estimate the same hitprobability for all guns/armaments, but thats wrong!!!
Did you ever learn about stochastics at school or at university? I'm just asking because I have explained things in a way that should suffice to get someone who knows a bit about stochastics to at least understand the terms I am using, and it's my impression that you (still) don't.
>My estimation is that 87 rps provide a more big probability to hit the target than 40 rounds, as more as the hitprobability of a single round decrease.
Put that into a formula. That will make your definitions clear and allow a rational analysis.
(All of your previous posts are confused to the point of nonsense. When I posted my formula in my first reply to you, I had hoped you'd reply with one, too, so we'd be able to sort it out quickly.)
Regards,
Henning (HoHun)
-
Hi Knegel,
>btw, now when is reread all with the knowledge that your hitprobability is the hitprobability of a single round, i understand what you are talking about.
OK, then we have made one important step towards agreement! :-)
>Nh = ROF * Ph(bullet)
It should actually be:
Nh = Nf * Ph(bullet)
The shotgun fires hundreds of "bullets" at one pull of the trigger, which is the "hidden trick" in its apparently higher hit probability. Fire the same number of rifle projectiles, and you'll get the same number of hits. (But it will take all afternoon :-)
Regards,
Henning (HoHun)
-
Hi Knegel,
>While shooting to fighters and unprotected planes in general(main targets of US fighters) at the end the time to shoot make the different, and 30sec for .50cal in a F4U are simply more worth than 12sec of 20mm´s in the same plane, if the killprobability is roundabout the same.
Well, big "if".
But with regard to machine guns, the assumption that they provide a better ammunition duration at the same battery weight is quite misleading.
Compare two batteries of roughly equal firepower and weight:
2x Hispano II - 570 rpg - 380 kg - 125% firepower - firepower per weight: 125%
6x ,50 Browning M2 - 313 rpg - 381 kg - 100% firepower - firepower per weight: 100%
The Hispano have superior firepower and 57 s of ammunition, while the 12.7 mm machine gun have just 24 s duration.
Regards,
Henning (HoHun)
-
Hi,
"Put that into a formula. That will make your definitions clear and allow a rational analysis."
Sorry, iam not able to put a estimation into a formula. Sorry for my bad math, but the fact that a shotgun provide a higher hitprobability on smal fast moving targets than a rifle should be enough.
Logic dont always need math if comparisons are good enough to explain the principle. Math dont help much if we have a range of hitprobabilitys between 2% and 90%.
"(All of your previous posts are confused to the point of nonsense. When I posted my formula in my first reply to you, I had hoped you'd reply with one, too, so we'd be able to sort it out quickly.)"
From my point of view your posts are confused. While i obvious try to compare the hitprobability of two different armaments, you start to talk about the hitprobability of a single bullet. You start to pop out formulas which include values we dont have, where is the sence in this?
This sounds to me you did learn something without to know to use it.
I rarely did hear about Stochastik, but according to the defination its 'the art of estimation'. I dont had such calculations in school, but this dont mean iam not able to think in a logical way.
Why you dont show a formula of the hitprobability??
For now you only did show a formula of the hitquote, which is equel to the expected hitprobability, based on hitquotes after many tests!!
The formula of the hitprobability, while shooting from a plane to a plane, would need to consider the influence of the turbolences, a special pilotskill, the way of convergence adjusting, the speed of the own plane in relation to the target, even the airthickness may influence this(thin air = higher middle bullet velocity) and so on.
With other words, the formula for a hitprobability is pretty complex.
Your formulas only help if we already know the hitprobability, but we dont!
So all your nice formulas regarding the kill probability only look nice.
">Nh = ROF * Ph(bullet)
It should actually be:
Nh = Nf * Ph(bullet)"
Nh = ROF * Ph <--- This formula YOU did post!!!
"Compare two batteries of roughly equal firepower and weight:
2x Hispano II - 570 rpg - 380 kg - 125% firepower - firepower per weight: 125%
6x ,50 Browning M2 - 313 rpg - 381 kg - 100% firepower - firepower per weight: 100%
The Hispano have superior firepower and 57 s of ammunition, while the 12.7 mm machine gun have just 24 s duration."
Why i should compare this???
If we wanna see why the USA did use .50cal instead of 20mm´s we need to compare the armaments like they was.
F4U1A: 6 × 12,7 mm Browning MG53-2 fixed forward-firing in the wings, 400 rounds each (4 inboard guns), or 375 rounds each (2 outboard guns)
F4U-1C: 4 × 20 mm Hispano M2 cannon, 120 rounds each,
P51D : 2 × 12,7 mm Browning MG53 fixed forward-firing in the wing, 350 rounds each (inboard pair)
2 × 12,7 mm Browning MG53 fixed forward-firing in the wing, 280 rounds each (outboard pair)
HurricaneIIC: 4 × 20 mm Hispano Mk I or Mk II fixed forward-firing in wing leading edges, 91 rounds each.
(i hope the source i found isnt that far off).
Resulting we have longer time to shoot for the MG armned planes, combined with a higher hitprobability and satisfying damagepower result in a more effective armarment.
There must be a reason why the Spitfires seldom carried 4 x 20mm or why they did keep MG´s at all, while the british ground attackers got only 20mm´s.
If the 20mm would have provided a satisfying hitprobability, they would have kicked the the MG´s and would have used the free space for 20mm amo.
Also, if the Hispano 20mm was so much better, why the P38 only carried 1??
But maybe you be right and the allied ing´s simply was stupid and also their thoughts what is the best armement based on experiences simply was wrong.
Greetings, Knegel
-
Hi Knegel,
>I rarely did hear about Stochastik, but according to the defination its 'the art of estimation'. I dont had such calculations in school, but this dont mean iam not able to think in a logical way.
The thing is, you have fallen into a logical trap and don't recognize it. My logic, even the bits you consider unlogical, is based on tried and tested methods developed by competent mathematicans over the last few centuries.
I seem to be unable to help you out of your trap, partly because you don't have the vocabulary to talk about the different aspects of random experiments so that you don't understand my explanations.
>So all your nice formulas regarding the kill probability only look nice.
All my nice formulae can be used quite nicely to estimate relative effectiveness of the guns because the unknown factor hit probability is eliminated as soon as we calculate an effectiveness ratio.
>Nh = ROF * Ph <--- This formula YOU did post!!!
Yes, sorry for that, I made a mistake in one my posts. It should have been
Nh = ROF * Tf * Ph = Nf * Ph
>If we wanna see why the USA did use .50cal instead of 20mm´s we need to compare the armaments like they was.
If we want to see if the USA made a good decision when they decided to use the 12.7 mm machine gun, we need to compare the armament they had to the armament they could have had if they had made a different decision.
>There must be a reason why the Spitfires seldom carried 4 x 20mm or why they did keep MG´s at all, while the british ground attackers got only 20mm´s.
Come on, the Spitfire simply carried 2 x 20 mm because of performance concerns. The so-called "ground attackers" Hurricane and Typhoon were indeed purpose-built fighters relegated to ground attack duties because of insufficient performance. In the Far East where no Spitfires were available at first, they occasionally pulled one pair of Hispanos from the Hurricane in a similar manner.
>Also, if the Hispano 20mm was so much better, why the P38 only carried 1??
Nonsense conclusion. If the 12.7 mm gun was so much better, why did the P-38 need to carry a 20 mm cannon?
>But maybe you be right and the allied ing´s simply was stupid and also their thoughts what is the best armement based on experiences simply was wrong.
Well, the Germans had a lot more combat experience and went to cannon against all types of targets as quickly as possible. The British also had two years of a headstart in combat experience and got rid of their machine guns as quickly as possible, too.
Regards,
Henning (HoHUn)
-
Knegel,
The F4U-1C's 20mm cannons had 230 to 232 rounds per gun. I have no idea where you got the 120 rounds per gun from for the F4U-1C. Spitfires had 120 rounds for their 20mm cannons. When fitted with .50s the Spitfire had 120 rounds for each 20mm cannon and 250 rounds for each .50 cal.
Use real numbers to compare, not made up ones.
-
Hi Karnak,
thanks for the correction, i dont had a other source.
HoHun,
"All my nice formulae can be used quite nicely to estimate relative effectiveness of the guns because the unknown factor hit probability is eliminated as soon as we calculate an effectiveness ratio"
Whats about this?
ROF(MG) = 13rps
ROF(Ca) = 10rps
Tf = 2sec
Ph = 2%
Nh(2*MG) = 2 x 13r/sec * 2sec + 2% = 1,04
Nh(2*Ca) = 2 x 10r/sec * 2sec + 2% = 0,8
With other words: in this example the 2 x MG´s got at least a hit, while the cannons tend to fail.
If we now compare the armaments:
Nh(2*MG) = 6 x 13r/sec * 2sec + 2% = 3,04
Nh(2*Ca) = 4 x 10r/sec * 2sec + 2% = 1,6
With other words a average pilot have a double high hitprobability with the .50cal´s armament, while attacking a strait flying big target!
The interesting point is: While attacking a smal or unprotected target, the 3 hits of the .50cal provide a high probability to hit a vital or overkill area, the 1-2 hits of the 20mm also may bring the plane down or not.
If the estimation is right that it need around 4 x 20mm hits to bring down a fighter, the 4 x 20mm armarment need a 5 sec burst.
In this time the 6 x .50cal armament already did hit 8 times, what also provide a good probability to hit a vital or overkill area(fuel tank, pilot, engine, radiator).
All damage power estimations need to depents to the target, same like on the bullet/gun.
The hitprobability get influenced by even more.
If we assume a hitprobability of 30%(maybe a ace on close distance).
4 x HispanoII provide 40 hits = Absolutly overkill(vs fighters) = not needed
6 x .50cal provide 78 hits = Deadly = same result like with the 20mm´s.
In the case of an unprotected plane without selfsealing tanks ALL used guns in WWII was relative easy able to bring down the plane with one bullet.
Therefor the big advantage of the 20mm´s regarding structural damages get minimized, in relation to a tough good protected plane with selfsealing tanks.
As result the hitprobability and killprobability are not linear with all targets and also not with the distance.
The importance of the hitprobability increase even more, if we consider that while a escort flight or while helping a wingi its more important to be able to 'ring the bell', than to knock the door off.
And we always need to keep in mind that MG armned planes tend to carry more amo(longer time to shoot), what enhance the hitprobability(armament + amo) even more.
"If we want to see if the USA made a good decision when they decided to use the 12.7 mm machine gun, we need to compare the armament they had to the armament they could have had if they had made a different decision."
They made the different decission and what i compare are the different armaments they did use for the same plane. As Karmak did point out, my source regarding the F4U-1C amoload was wrong, so the time to shoot increase to around 22 sec, what is still good below that of the .50cal´s.
"Come on, the Spitfire simply carried 2 x 20 mm because of performance concerns. The so-called "ground attackers" Hurricane and Typhoon were indeed purpose-built fighters relegated to ground attack duties because of insufficient performance. In the Far East where no Spitfires were available at first, they occasionally pulled one pair of Hispanos from the Hurricane in a similar manner"
Yes, but why they pulled two hispannos and used MG´s instead, if the Hispano did provide a higher kill probability? They could have used a higher amoload for the remaining hispanos.
Actually i think you be in a trap of pseudo logic. Math dont help much if we have to many unknown values. In our case math is only good to style a statistic. Show me a statistic and i show you how to style it to my need!!
In this case math get reduced to a same unexact speach like english and german, so why not staying in out speack, where we dont need to learn new therms?
If i talk to someone who dont know my laguage, but i know his language, i tend to use the language we both know. You seems to be aware of my missing knowledge regarding the currently common therms at school, nevertheless even dont try to think in 'normal' therms, you simply expect that i learn this new language, just while a discussion.
Or do you forgot the language of the 'normal' people?
Only cause someone dont know your language, it dont mean he dont know what he talk about!!
Anyway, probabilitys in a complex system cant get calculated exact without many datas based on former made experiences.
You try to calculate a killprobability, without to consider the hitprobability, without to consider the real used amoloads, without to consider the damagepower probability (we dont have a static damagepower, it depents much to the structural toughness, size and protection of vital areas of the target).
You do what many people do who did study, you take a statistic and count it as static fact, without to take the possibility into account that every statistic is more a approach than a fact. Every little mistake in such a statistic lead to bad mistakes while calculating with such values.
Thats why i avoid to use math in such cases, as long as i dont have exact datas for every single situation.
Statistics are only good as base for a good estimation, but to be able to use it you need to know the exact circumstances where the values comes from.
Tony Williams statistic is a very good work, but do you realy seems to think a Hispano II round always had damagepower of 201, while the .50cal always had a damagepower of 46.
You dont seems to consider that a .50cal was able to generate a overkill shot in the same way like a 20mm.
How likely this is depents to the "overkill area/target area relation". As more overkill area a plane have, as higher the probability to hit this area.
If there would be a plane with 100% overkill area for a MG, no cannons would be needed at all, cause every hit would lead to a failsure.
I assume that fighters had a much higher "overkill area/target area relation" than a Bomber. This relation is much different if we compare planes with and without selfsealing tanks.
(overkill area = area on a plane where one hit is able to cause damages to bring down the plane)
As higher the "overkill area/ target area relation" as more important get the hitprobability over the damagepower.
Of course the gun and the target determine which area is a overkill area.
Probably the overkill area will be more big for a 20mm than for a .50cal.
Additionanlly we need to consider 'vital areas', where some hit cause damages to bring the plane down.
For a 20mm probably on a fighter the whole plane is a vital area, while some bombers can absorb even a 20mm on big areas without to suffer much.
Thats the reason why 30mm´s was better vs Bombers, the gun alone did increase the vital area much.
"Well, the Germans had a lot more combat experience and went to cannon against all types of targets as quickly as possible. The British also had two years of a headstart in combat experience and got rid of their machine guns as quickly as possible, too."
Not all was happy with the MGFF and its poor rof and MuzzVel, even more wasnt happy with the single cannon in the nose, compared to the 2 x MGFF.
Maybe if the MG131 would have been available in 1937, the 109E would have carried four of this guns instead of cannons?
Maybe the Me109E would have been better able to protect the bombers with 4 x MG131 instead of the 2 x MGFF? We dont know this.
German fighters switched from smal MG´s to big MG´s in the nose, not to a 'only cannon setup' (only some planes, supposed to fight bombers, had only cannons).
Germany had to fight tough bombers with selfsealing tanks from the beginning of war!!
Afaik, the brits dont was unhappy with their 8 x .30cal vs Fighters, they was unhappy with this armament vs Bombers mainly.
Though later , with more plating the .30cal got to be a real problem. Thats probably why they did introduce the .50cal in 1944 to the Spits.
The brits had mainly the .30cal and Hitpano II, so of course they did switch to the 20mm, cause the .30 cal wasnt strong enough to penetrate the common amor in the needed way. The USA had the .30cal, .50cal and their 20mm.
They switched to the .50cal and the results in war show that they got pretty good results, even at times when the enemy still was strong in numbers and high in pilot skill(pacific mainly).
As i wrote before: Iam sure that they would have switched to 20mm´s if they would have had B17´s(or Wellingtons, Mosquitos, Lancasters, Ju88´s, IL-2´s) as oponent.
If you read how suprising fast US fighters, also with only 4 x .50cal, was able to destroy Japanese fighters and bombers, i realy dont see the need to carry 20mm´s for the price of around 30% time to shoot.
Greetings, Knegel
-
I'm trying to understand this discussion but there is a lot of terminology problem who need to be solved first.
Being a bit a math head (who said di*k head ???) I've never seen the word killprobability,hitprobability used previously :)
Ps : you should reason on a per round basis,as each round is unique
PS: it remind me one of the hardest part I had to understand the "Markov chain"
-
Hi Knegel,
>The interesting point is: While attacking a smal or unprotected target, the 3 hits of the .50cal provide a high probability to hit a vital or overkill area
The 12.7 mm hits have their Pk, the 20 mm cannon have a different Pk. Obviously, the 20 mm cannon's Pk is considerably higher.
A good method of estimating the Pk of a round is to examine the total energy it carries because this energy will be converted into destructive power as soon as it strikes the target:
Hispano II - 486%
0.60" MG 151 copy 168%
,50 Browning M2 - 100%
Browning ,303 - 20%
>If we now compare the armaments:
Hm, I fixed some typos here:
>Nh(6*MG) = 6 x 13r/sec * 2sec * 2% = 3,12
>Nh(2*Ca) = 2 x 10r/sec * 2sec * 2% = 1,6
>With other words a average pilot have a double high hitprobability with the .50cal´s armament, while attacking a strait flying big target!
The energy comparison shows us that each cannon hit is almost five times as powerful as a 12.7 mm hit, so cannon still do more damage.
>In this time the 6 x .50cal armament already did hit 8 times, what also provide a good probability to hit a vital or overkill area(fuel tank, pilot, engine, radiator).
You are implicitely assigning a Pk figure for the 12.7 mm machine guns here. You are also (sort of) implying that a large proportion of the target area consists of critical targets, and that the Pk for the 12.7 mm round against critical areas is high. You have to put numbers on these effects to compare them.
>Only cause someone dont know your language, it dont mean he dont know what he talk about!!
You don't merely lack the language, you lack the basic concepts described by the terms that don't even exist in everyday language. A teacher might be inclined to explain the concepts you are lacking more thoroughly in plain language here, but I am no teacher, and I don't consider your "know-it-all, don't-need-math" attitude a good basis for teaching anyway.
I'd recommend you read Tony's books on WW2 gunnery to clear up the remaining misunderstandings in your posts.
Regards,
Henning (HoHun)
-
Hi Stratos,
yep i agree, but read my 1st sentence to this theme:
"the .50 cal was the better weapon for the US need, cause, while the gun was good enough to destroy all japanese planes rather fast and to keep the german fighters away from the Bombers, it did provide a much better hitprobability than the relative slow firing Hispano."
I clearly speak of gun hitprobability, same like every bullet is unique, every gun is unique and we can estimate different hitprobability for every gun under different circumstances.
As the calculations with a possible bullet hitprobability show, the hitprobability of 6 x 50.cal is twice as high as the 4 x Hispano II.
The result in war show that the .50cal was powerfull enough to bring the japanese planes down with suprising few hits.
With other words, if HoHun would have read what i wrote, this discussion wouldnt have been needed!
How i shal know that he speak about the hitprobability of every single round, while i clearly wrote about the hitprobability of the gun(armament)?
I dont know a law that the word hitprobability refer always to a single round, although i agree that it can be usefull!
And of course the hitprobability per gun(armament) and also the hitprobability per amoload is one aspect, if we wanna find out why the the USA did choose the .50cal over the HispanoII.
Greetings, Knegel
-
Look like Hohun is reasoning in term of probability and Knegel in expected value (espérance).
But some post make me wonder what you really mean ,for example this :
Nh(2*MG) = 6 x 13r/sec * 2sec + 2% = 3,04
Nh(2*Ca) = 4 x 10r/sec * 2sec + 2% = 1,6
With other words a average pilot have a double high hitprobability with the .50cal´s armament, while attacking a strait flying big target!
%2 is the hit probability !
Mathematically this as no sense ,as you need to throw a percentile dice 156 or 80 times to obtain a real result.
ok , the laws of large numbers apply here and I guess you are thinking of a normal law distribution but ... well it's not a valid statement :)
-
Hi Knegel,
>As the calculations with a possible bullet hitprobability show, the hitprobability of 6 x 50.cal is twice as high as the 4 x Hispano II.
This is not the hit probability but the number of hits.
What you choose to ingore is Pk. Pk ist approximately proportional to the total energy of the round, and greatly favours cannon shells over bullets.
Instead of properly continuing the calculation to its logical conclusion, you break off here and assert the superiority of 12.7 mm guns with some handwaving arguments that don't hold water.
Here is the full calculation:
Pdestruction (6 MG) = Pk (MG) * Nh(6 MG) = 3.4% * 6 * 13 rps * 2 s * 2%
Pdestruction (4 Ca) = Pk(Ca) * Nh (4 Ca) = 16.7% * 4 * 10 rps * 2 s * 2%
=>
Pdestruction (6 MG) = 10.6%
Pdestruction (4 Ca) = 26.7%
This means that a 4-cannon-battery has a firepower of 250% of that of a 6-machine-gun-battery.
>How i shal know that he speak about the hitprobability of every single round, while i clearly wrote about the hitprobability of the gun(armament)?
Because I pointed that out in my first reply to you here in this thread:
"I know what you mean, but hit probability is really just the ratio of hits to total shots fired and has nothing to do with rate of fire."
I read what you wrote, and replied. Browse back and read it again. It's worth it as you still still haven't got over your misunderstanding.
Regards,
Henning (HoHun)
-
Hi,
"The energy comparison shows us that each cannon hit is almost five times as powerful as a 12.7 mm hit, so cannon still do more damage."
If the .50cal hit unprotected not selfsealing fueltank, at the end it cause the same result like the 20mm, same count for the pilot.
Since the pilot and the fueltank cover a pretty big part of a plane, the question is how much different is the effect?
"You don't merely lack the language, you lack the basic concepts described by the terms that don't even exist in everyday language. A teacher might be inclined to explain the concepts you are lacking more thoroughly in plain language here, but I am no teacher, and I don't consider your "know-it-all, don't-need-math" attitude a good basis for teaching anyway."
This is what you wrote before:"........you don't have the vocabulary to talk about the different aspects of random experiments so that you don't understand my explanations."
Strangewise i dont say i know all, its rather you who dont read exact what i wrote, and without to waste a though that my therms could offer a correct describtion with for you not common words, you go on and refer to absolut not relevant things.
After you took notice that i talk about the gun hitprobability, you dont say, 'ah ok, now i understand', you dont come down from your high horse. You immediately switch and claim 'hitprobability is nothing, killprobability is all'.
"The energy comparison shows us that each cannon hit is almost five times as powerful as a 12.7 mm hit, so cannon still do more damage."
How you know that the cannon do more damage, if there is a not to smal possibility that the .50cal hit a fueltank or the pilot? In this case there dont will be a real different of the result.
"A good method of estimating the Pk of a round is to examine the total energy it carries because this energy will be converted into destructive power as soon as it strikes the target:"
I dont think this is a good method of estimating the Pk, cause you dont take the target into account.
As i wrote before, i think the killprobability change with the target, not only with the damagepower.
A not protected target, without selfsealing tanks offer a much higher killprobability for a .50cal than a on a plane with selfsealing tanks and good amor.
While i guess the cannon round provide a much more constant killprobability, cause it mainly aim for the structure of the plane and it isnt a big different if a 20mm hit a selfsealing tank or a not selfsealing tank, same count for the amor.
On a unprotected target specialy smal MG´s have a much higher kill probability than on a same sized, protected target.
If i hit the head of the pilot, it dont matter if i do this with a .30cal, .50cal or 20mm. The killprobability is pretty the same in this rare but without plating possible example. With plating to cover the pilot, the killprobability in this case will will be much higher for the cannons.
Btw. Somewhere i did read the Luftwaffe did estimate around 4 x Mk108 hits are needed to take down a B17 and around 20 x MG151/20mm to get the bombwer down.
catrige power MK108 = 58 * 3 = 174
catrige power MG151/20 = 12 * 20 = 240
I also did read that on smal fighters 1 * MK108 was enough, while around 4 * MG151/20 rounds did the job.
1 * 58 = 58
4 * 12 = 48
What a suprise, on smal(less tough) targets the effectiveness of the smaler gun increase in relation to the big gun.
The problem isnt to calculate something, its to know what we need to consider while the calculation.
Greetings, Knegel
-
Originally posted by straffo
Look like Hohun is reasoning in term of probability and Knegel in expected value (espérance).
But some post make me wonder what you really mean ,for example this :
%2 is the hit probability !
Mathematically this as no sense ,as you need to throw a percentile dice 156 or 80 times to obtain a real result.
ok , the laws of large numbers apply here and I guess you are thinking of a normal law distribution but ... well it's not a valid statement :)
Hi,
yes, 2% is the hitprobability of the bullet, thats what the Luftwaffe did estimate when a average pilot attack 4mot bomber.
Greetings, Knegel
-
Hi,
"this is not the hit probability but the number of hits."
Rather the expected number of hits, but i would say it isnt wrong to call this the hitprobability of the gun or armament.
Or is Tony Williams also wrong if he once write "CARTRIDGE POWER" and next time "Gun Power"?
The word Power is not limitted to the smalest unit, same like i dont think hitprobability is limited to the smalest unit.
Gunpower is also 'only' catrige power * ROF!!
Therfor we can define "catrige hitprobability" and "gun hitprobability", i dont see any mistake in this!
""I know what you mean, but hit probability is really just the ratio of hits to total shots fired and has nothing to do with rate of fire.""
If you keep in mind that i was thinking of the the gun hitprobability, this sentence is misleading like mad!! Why you dont write: "In school the hitprobability is related to only one bullet/Pellet, not to a gun or shotgun shell."??
But actually on different shotgun pages they use the therm 'probability of hit' when they talk about a shotgun shell.
Anyway, this misunderstanding is solved now, now we can go on with the damage power thingi! :D
Greetings, Knegel
-
HoHun.. and all
You and your friend are equally good and accurate throwers.
You have a baseball in your hand and your friend has a handful of small stones.
You both are trying to hit a bird sitting in a tree not too far away with what you have in your hand.
Which one of you (people, not projectiles) have a higher probability to hit the bird (=hit probability) until it flies away?
Is it not true that the number of projectiles would increse the hit probability... not talking about what will happen to the bird if/when it is hit.
-
Hi Knegel,
>Why you dont write: "In school the hitprobability is related to only one bullet/Pellet, not to a gun or shotgun shell."??
Because it's the same in real life. Non scholam, vitam discimus.
>But actually on different shotgun pages they use the therm 'probability of hit' when they talk about a shotgun shell.
Sure, everything you read on the internet is correct.
Regards,
Henning (HoHun)
-
Hi Knegel,
>While i guess the cannon round provide a much more constant killprobability, cause it mainly aim for the structure of the plane and it isnt a big different if a 20mm hit a selfsealing tank or a not selfsealing tank, same count for the amor.
You have established a hit probibility for the entire aircraft, then you draw conclusions for the critical areas of the aircraft, thus breaking the logical chain in your argument. You are again neglecting the higher destructiveness of cannon shells which mean that they can easily cause destruction by attacking non-critical areas.
> Somewhere i did read the Luftwaffe did estimate ...
The Luftwaffe estimated that it was the chemical energy in the cannon shells that brought down aircraft. That is not entirely accurate since it neglects the kinetic energy, which admittedly in the case of cannon shells doesn't make that much of a difference.
>What a suprise, on smal(less tough) targets the effectiveness of the smaler gun increase in relation to the big gun.
The reports with the numbers of hits required define different desctruction probabilities, so you are most likely comparing apples and oranges. From what reports are your numbers?
Regards,
Henning (HoHun)
-
BlauK,
But that isn't what we are talking about. the .50 installations are not a shotgun. They produce a bullet stream just as the Hispano installation does, simply with about half again as many projectiles. If the stream misses, it misses and how many projectiles are in the stream are irrelevant. If it hits, it hits and likewise the number of projectiles in it is irrelevent.
When you realize that you realize that the number of times the .50 cal installation will score hits while the 20mm installation does not is vanishingly small and talking about the hit probability of a given round does not accurately describe the practical reality.
Given that, the 20mm weapons were clearly superior.
-
Hi BlauK,
>Is it not true that the number of projectiles would increse the hit probability... not talking about what will happen to the bird if/when it is hit.
The difference between the two throwers is that one of them is only allowed to throw one object, while the other is allowed to throw multiple objects. The hit probability is constant.
The question "How likely is it that n out of m thrown objects hit the bird" is a complex random experiment even though the hit probability of each object ist the same.
Applying the same term to both experiments is really means comparing an apple to a grocery store, and since hit probability is a well-defined term, it's plain wrong to tag it on the grocery store situation.
The latter leads to a dead end anyway since we have to consider what will happen to the bird when it is hit.
Regards,
Henning (HoHun)
-
Originally posted by Karnak
If the stream misses, it misses and how many projectiles are in the stream are irrelevant.
So you are assuming a situation where neither the shooter nor the target are not manouvering in any manner?
Why would you want to speculate on such? In such case you are no tspeculating on hits, but on destructive power.
-
Originally posted by BlauK
So you are assuming a situation where neither the shooter nor the target are not manouvering in any manner?
Why would you want to speculate on such? In such case you are no tspeculating on hits, but on destructive power.
No. I am saying that in the vast majority of events such manuvering will not produce different results for the Hispano or .50 cals. In most cases either they would both hit or both miss. In some rare cases the Hispanos would miss while the .50s hit, but not nearly often enough to remotely redress a gross disparity in firepower between the Hispano and .50.
-
Originally posted by HoHun
The difference between the two throwers is that one of them is only allowed to throw one object, while the other is allowed to throw multiple objects. The hit probability is constant.
So, are you saying that the probability of winning in lottery (="winprobability") is not affected by the amount of tickets bought? ... "One guy is only allowed to buy 1 ticket while the other can buy 10" ... "the winprobability (of these 2 guys) is constant" ?
I am not really interested in the probability of on esingle ticket.. if I have the option to choose between 1 or 10, which one do you think I should choose to achieve beter chances of a hit = win
The latter leads to a dead end anyway since we have to consider what will happen to the bird when it is hit.
So you are not actually wanting to talk abot hit probability, but instead about teh probability of killing the.. or destroying an aircraft. That is not hitprobability, that is destruction probability or something like that ;)
-
Originally posted by Karnak
No. I am saying that in the vast majority of events such manuvering will not produce different results for the Hispano or .50 cals. In most cases either they would both hit or both miss. In some rare cases the Hispanos would miss while the .50s hit, but not nearly often enough to remotely redress a gross disparity in firepower between the Hispano and .50.
Ok.. so the velocity is so high that the amount of rounds is not relevant anymore? Or the dispersion is equal.. or what ever.. then it would make sence.
I suppose this is only considering one weapon of each type.. and not considering e.g. 6 mgs against 1 cannon?
IMO, again, it would be pretty useless to consider only 1 weapon against 1 weapon if e.g. their weight is dramatically different etc.
NO kind of equations really matter until the question is clearly defined :)
-
Hi BlauK,
>So you are not actually wanting to talk abot hit probability, but instead about teh probability of killing the.. or destroying an aircraft. That is not hitprobability, that is destruction probability or something like that ;)
What you are talking about is not hit probability, but the probability that out of n shots fired, m hit.
Hit probability the the probability that if one shot is fired, one hits.
Regards,
Henning (HoHun)
-
Originally posted by HoHun
Hit probability the the probability that if one shot is fired, one hits.
Where or by whom is such defined?
You are simply talking about the hit probability of one bullet.. but then, what are the factors affecting the probability?
And what would you try to prove with that kind of probability? Something like the random behaviour of the bullet because of surface defects or such?
-
Hi BlauK,
How about going back and reading the entire thread before you ask questions that have been answered before?
Regards,
Henning (HoHun)
-
I eyed it pretty carelessly, but it seemed like you guys pretty much speculated (among the personal punches) on how one or the other interprets this and that word.
I would be grateful for a very compact summary or explanation on why in your opinion "hit probability" means on the probability of hitting with one bullet and nothing else.
But if not, maybe the whole discussion is not worth it :)
-
Hi BlauK,
>I eyed it pretty carelessly
That much was pretty obvious.
Regards,
Henning (HoHun)
-
:D
Well, just go back to the 1st page and check out what one of your "privileged contibutors" wrote about "hit probability, Knegel and RoF" :lol
Just FYI... there is no one and only formula for "hit probability". It always depends on the given conditions at each time.
-
BlauK,
I was comparing six .50s to four Hispanos. Six .50s compared to two Hispanos doesn't change it much, except that the two Hispanos are lighter and deliver at least as much firepower.
As to the muzzle velocities, they are are relatively the same at effective WWII gunnery ranges. The rate of fire for each package is sufficient to deny the target the ability to fly between the rounds in most cases, though moreso in the case of the four Hispano instalation than the two Hispano installation.
Keep in mind that the Browning .50 weighs a lot, so you cannot compare six of them to a single Hispano. Three Hispanos would, IIRC, weigh about the same as six Browning .50s.
-
Karnak, point taken :aok
While the hit probability with 6x0.50 cals is higher, delivered damage probability of 4 hispanos is likely higher.
-
Hi Blauk,
>Just FYI... there is no one and only formula for "hit probability". It always depends on the given conditions at each time.
Well, this formula does in fact cover all as long as the law of large numbers applies:
Nf = number of rounds fired
Nh = number of hits
Ph = hit probability
Ph = Nf / Nh
Regards,
Henning (HoHun)
-
Originally posted by BlauK
Where or by whom is such defined?
You are simply talking about the hit probability of one bullet.. but then, what are the factors affecting the probability?
And what would you try to prove with that kind of probability? Something like the random behaviour of the bullet because of surface defects or such?
it's defined by the probability theory .. in short mathematics :)
-
Originally posted by straffo
it's defined by the probability theory .. in short mathematics :)
And the probability theory dont allow to use the therm gun hitprobability??
Originally posted by HoHun
Well, this formula does in fact cover all as long as the law of large numbers applies:
Yes, nice formula, but if someone explecit talk about the gun or armament hitprobability, there is nothing worng.
If you dont think so, please ask Tone Whilliams to remove the word "Gun Power" from his table, he need to insert "number of rounds fired * catrige power" instead. He made more of such errors, but since its clear what he mean, noone would complain.
Noone would start to discuss that the typhoon dont had a "gun power" of 800, cause "gun power" stands for the power of only one gun, what is 200 for the hispanoII. Everybody see that he talk about the "armament power"
The therm of Nh = number of hits is simply wrong, if we calculate into the future, cause we dont know the number of hits, even if we know the hitprobability of one round.
This therm is only correct, if we calculate a hitquote out of already existing testdatas!
If we calculate into the future, its the "probable number of hits" or " Gun hitprobability".
It may be that the probability theory use the simplyfied therm cause while a calculation they work in the same way, but thats no reason to exclude the correct therms while a normal discussion.
Originally posted by HoHun
You have established a hit probibility for the entire aircraft, then you draw conclusions for the critical areas of the aircraft, thus breaking the logical chain in your argument. You are again neglecting the higher destructiveness of cannon shells which mean that they can easily cause destruction by attacking non-critical areas.
Looks like you avoid to read what i wrote!!!
Here again:
"As i wrote before, i think the killprobability change with the target, not only with the damagepower.
A not protected target, without selfsealing tanks offer a much higher killprobability for a .50cal than a on a plane with selfsealing tanks and good amor.
While i guess the cannon round provide a much more constant killprobability, cause it mainly aim for the structure of the plane and it isnt a big different if a 20mm hit a selfsealing tank or a not selfsealing tank, same count for the amor. "
I take the higher(more constant) damagepower ot the 20mm´s into account, but you dont take much different damage power(probably "resulting damage" would be more exact) of the MG´s on different targets into account.
If we have the hitprobability to the whole plane, its not that difficult to get the hitprobability for the critcal areas.
Its absolutly not logical to ignore the critical areas at all, specialy not on a plane without selfsealing tanks.
Therfor its absolut not logic to keep on calculating with a absolut constant damagepower for the whole target area!!
If you look to the rear of a A6M ot Ki43, how much % of the total area, do you think, get covered by the tank, pilot and other real critical point??
Iam dont know, but i guess this to 30%-50%.
Of course, if you look from above this may be smaler, but the normal attack´s was made from the rear and front.
If around 30%-50% is true(maybe someone know it better, a drawing maybe) the damagepower advantage of the 20mm get minimized much, cause the MG also cause critical damages on a not smal targetarea.
Greetings, Knegel
P.S.: Hi BK. :)
-
Hi,
Originally posted by Karnak
No. I am saying that in the vast majority of events such manuvering will not produce different results for the Hispano or .50 cals. In most cases either they would both hit or both miss. In some rare cases the Hispanos would miss while the .50s hit, but not nearly often enough to remotely redress a gross disparity in firepower between the Hispano and .50.
The .50cal armament tend to hit more often, on light amored planes the damagepower advantage of the 20mm get minimized.
My assumption is that a cannon shell is as more effective as bigger the not vital area is.
So we have the two extremes:
1. The rounds hit a area absolutly without important parts.
In this moment the 20mm round cause damages to the structure/aerodynamic, while the .50cal ´s cause around 5 times less damages, a .30cal will be even worse.
2. The round hit the head of the pilot.
All round cause a kill.
So we have on the one edge the 5 times higher theoretical damagepower of the 20mm, on the other edge we have the same result for both.
To calculate all impacts with 5:1 damagepower dont seems to be very realistic to me!!
Specialy if we talk about targets without selfsealing tanks and amor!!
Greetings, Knegel
-
Originally posted by straffo
it's defined by the probability theory .. in short mathematics :)
Certainly, but even mathematics has to be applied correctly and the formulas have to be built accordingly ;)
Hohun's approach is a statistical one, and he believes in one formula almost like in some religion ..."there is only one Hit probablitity" ;) Surely that is the bottom line of the theory, but what good is a theory if one cannot build on it and put it to practise.
If one relies solely in statistics when considering probabilities, what can he do in a completely new situation.. what if such gun has never been fired before and we simply do not have any data on previously fired shots or hits?
Hohun,
was your argument only about semiotics? What the term means or does not mean? (... in math theory, obviously)
-
Knegel,
I think you underestimate the difference between the 20mm hit and the .50 cal hit and that is where your disagreement stems from.
The USN, a miltary branch likely to face light aircraft, considered one Hispano to have as much value as three .50s. That wasn't a per hit basis, that was as a total package.
The 20mm will do far more than 5 times the .50 cal when hitting a non critical part of the aircraft and will be more likely to do critical damage when hitting through armor, which later Japanese aircraft certainly had.
You are exagerating the differences in ways that favor the .50 cals.
-
I am no statistician and don't want to get involved in complex mathematical arguments. However, it seems to me from what I can understand of some of the more abstruse posts that there it might be worth clarifying a couple of points in plain language, specifically, the difference between hit probability and kill probability.
The hit probability is simply the chance that the target will be hit - somewhere. It can be expressed as the hit probability per round, or per gun for a burst of fire of a given length, or for a burst of fire from a plane's entire armament.
The kill probability can be expressed as the chance that the target will be destroyed; either by a single round, or by a burst of fire as above. It incorporates the hit probability (because you're not going to kill something if you don't hit it) but multiplies that by the destructive power of each hit.
So to give a simple example, if one plane armed with many MGs has a hit probability for its entire armament (in a one-second burst, say) five times higher than that of another plane firing cannon, but that each cannon hit is five times as destructive as a bullet hit, then the kill probability of both armaments is equal.
Now let's turn to the argument that if an MG bullet kills the pilot, its kill probability is equal to that of a cannon shell (because you can't kill someone any deader). The validity of that depends on the probability that the MG bullet will in fact hit a vital point. This itself depends on two factors: the area of the vital point as a percentage of the area of the aircraft exposed to fire, and the probability of a bullet getting through the aircraft's structure and armour to hit that vital point.
I have a paper from RAF studies of the Bf 109F, in which they carried out some practice firings with different weapons which showed that in a rear attack the chance of a hit with a .50 AP bullet in the area of the pilot or vital controls being effective was 3%. That should not be too surprising in that passing through aircraft structures tended to destabilise the bullets, either deflecting them away from the vital point or tumbling them so they hit the armour side-on rather than point first.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi Knegel,
>Looks like you avoid to read what i wrote!!!
You are wrong. The first time you accused me, you were wrong as well, and I noted that you didn't apologize. If you keep that up, you'll find yourself on my Troll list rather sooner than later.
>The therm of Nh = number of hits is simply wrong, if we calculate into the future, cause we dont know the number of hits, even if we know the hitprobability of one round.
http://de.wikipedia.org/wiki/Gesetz_großer_Zahlen
I have mentioned that in about every second post, provided the Link to the German Wikipedia, and explained it in plain language here, and you still ignore it. No amount of attitude can make up for ignorance, so you better read up.
Here is an example for the information required to keep the logical chain intact. This information was missing from your previous posts:
>If around 30%-50% is true(maybe someone know it better, a drawing maybe) the damagepower advantage of the 20mm get minimized much, cause the MG also cause critical damages on a not smal targetarea.
Now that will enable us to to expand our formula to get more accurate results:
Pdestruction = ROF * Tf* Ph * (Pkcrit * Acrit/Atot + Pkstd * (1 - Acrit/Atot))
What still is missing for both weapons is:
Pkcrit - Probability of a kill when the projectile hits the critical area
Pkstd - Standard probability of a will when the projectile hits the non-critical area
Regards,
Henning (HoHun)
-
Originally posted by BlauK
Well, just go back to the 1st page and check out what one of your "privileged contibutors" wrote about "hit probability, Knegel and RoF" :lol
I don't know if you mean me but I have been not arguing against the benefits of the 20mm cannons over the larger number of the 12,7mm MGs. My point in this thread was just that the Americans had an obivious option to increase the ROF the M2 which would have been possible using the existsing technology (FN version of the M2) and minor modifications to existing airframes. Basicly 50% ROF increase has roughly same effect to the fire power as increasing the number of the guns by 50%; as an example the P-51B would have reached roughly same firepower as the P-51D with very little modifications (naturally that would have caused shorter total time of shooting using the original lenght of the ammunition belts, but firepower is usually calculated for one second).
As you can read Finnish, there is a pretty good book called "Lentoampumaoppi" published by "Puolustusvoimien Pääesikunta" (copies can be found from antikvarities or "Kasinhäntä" and also most aviation museum here have it in their collections). While the approach of the book is bit simplified, it contains some data from German sources on probabilities and accuracies of aerial shooting against the American heavy bombers. As an example it gives so called "r50" values which tells what is the average radius of the 50% dispersion caused by the skill of the shooter in the promilles of the shooting range against steady target (right behind of directly front):
Good shooter r50=0,008
Average shooter r50= 0,015
Bad shooter r50=0,025 (or more)
As an example a good shooter has an 1,6m radius for 50% dispersion at 200m range while a bad shooter has 5m (or more) radius for 50% dispersion at same range. Note that these are for ideal conditions, it is claimed in the book that at the pressure of the combat these values might be doubled according to some studies. Note also that these values are far larger than dispersion claimed for the guns itself or the caused by installation of the gun (except at very short range for wing mounted weapons) at practical shooting range.
Another thing claimed in the book is the German definition of the firepower and Combat value against the heavy bombers:
Ta = (n * V0) / [m * (Gw * Gl)]
Ta = Fire power
n = ROF
V0 = initial velocity
m = destruction factor
Gw = weight of the weapon itself
Gl = weight of the mounting
Tt = (n^2 * t) / (m * G)
Tt = Combat value
t = shooting time for total ammunition
G = total weight of the installation including ammunition.
As an example calculated firepower and Combat values for the MK 108 and and the MG 213 against heavy bombers (using same amount of HE in the projectile and same total shooting time and roughly same initial velocity) to show the advantages of the ROF:
MK 108 Ta = 9,1 Tt = 1,2
MG 213 Ta = 16 Tt = 2,9
Note that it is claimed in the book that against the fighters parameters would be different (favoring the ROF and somewhat lower calibres).
Originally posted by straffo
it's defined by the probability theory .. in short mathematics :)
That is OK if the probability theory can be utilized to a particular case. But in the case of HoHun's argumentation against so called "shotgun romanticism" there is no reason why the dispersion caused the shooter is somehow evenly distributed around the correct point (ie following the gaussian distribution). That means that the error is practically allways systematical.
Note that the dispersion of the gun itself is practically allways much smaller than the dispersion caused by the shooter so it's quite logical that some amount of dispersion in the gun and higher ROF will help the shooter to destroy the target due to (very probable) systematical error.
But again: I'm not arguing against the benefits of the cannon but merely pointing out the benefits of ROF and (some) dispersion and the connection with the shotgun case. Infact regarding the theory, I mostly agree with HoHun while Knegel seem to have some problems to understand the theory.
gripen
-
Originally posted by Tony Williams
I am no statistician and don't want to get involved in complex mathematical arguments. However, it seems to me from what I can understand of some of the more abstruse posts that there it might be worth clarifying a couple of points in plain language, specifically, the difference between hit probability and kill probability.
The hit probability is simply the chance that the target will be hit - somewhere. It can be expressed as the hit probability per round, or per gun for a burst of fire of a given length, or for a burst of fire from a plane's entire armament.
The kill probability can be expressed as the chance that the target will be destroyed; either by a single round, or by a burst of fire as above. It incorporates the hit probability (because you're not going to kill something if you don't hit it) but multiplies that by the destructive power of each hit.
So to give a simple example, if one plane armed with many MGs has a hit probability for its entire armament (in a one-second burst, say) five times higher than that of another plane firing cannon, but that each cannon hit is five times as destructive as a bullet hit, then the kill probability of both armaments is equal.
Now let's turn to the argument that if an MG bullet kills the pilot, its kill probability is equal to that of a cannon shell (because you can't kill someone any deader). The validity of that depends on the probability that the MG bullet will in fact hit a vital point. This itself depends on two factors: the area of the vital point as a percentage of the area of the aircraft exposed to fire, and the probability of a bullet getting through the aircraft's structure and armour to hit that vital point.
I have a paper from RAF studies of the Bf 109F, in which they carried out some practice firings with different weapons which showed that in a rear attack the chance of a hit with a .50 AP bullet in the area of the pilot or vital controls being effective was 3%. That should not be too surprising in that passing through aircraft structures tended to destabilise the bullets, either deflecting them away from the vital point or tumbling them so they hit the armour side-on rather than point first.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
Hi Tony,
i agree absolutly to all you wrote(well written!)!!
Do you also have a test similar to that of the Bf109F to a unprotected target without selfsealing tanks?? Include this test the tank area?
On what distance this test was made?
Wouldnt the probability to hit a critical area increase much if there isnt plating and selfsealing tanks?? Wouldnt the resulting higher killprobability, of this guns, increase much more than the killprobability of a 20mm, which have much less trouble with selfsealing tanks and plating anyway?
Hi HoHun,
if you did read what i wrote, why you dont saw that i did took the higher catrige power of the 20mm into account??
After Tony Williams statement regarding the term "hitprobability", there is not much sence in many of your replys. I dont say that the formulas are wrong, but at the end my initial statement of the higher hitprobability of the .50cal armament was not wrong.
Can you please point me to the point in the 'Gesetz der großen Zahlen' where they refer to Nh??
As far as i know, any value that i multiply with a probability (P ) result in a new probability, therfor the value would be P(Nh), what would be the same like P(gun) or P(armament).
We can use Nh( a exact known value) and divide it by the ROF*Tf to use the resultig hitquote as bullet hitprobability, to estimate comming results under similar circumstances. But we cant use a unexact value, a probability, multiply it with a exact value to get a exact value Nh(number of hits) as result. It remain a probability.
If the law of probability use a simplyfication, cause it dont matter for the calculation, it dont mean its a correct describtion!
Hi gripen,
maybe i dont understand the theory like it is written, or maybe i dont talk in the common therms so you have problems to follow my thoughts, but strangewise my conclusions are the same like the germans had(at least according to the book).
And looks like the Ing´s in the USA thought the same.
"Note that it is claimed in the book that against the fighters parameters would be different (favoring the ROF and somewhat lower calibres)."
If they was right to favoring the ROF and somewhat lower calibres, while shooting to fighters, this must be more important while shooting to unprotected fighters.
Nothing more and nothing less i told in my 1st post, till the end!
ROF and the target itself are important parts while determining the killprobability.
btw, i never told that the .50cal is more powerfull than the 20mm´s in general! I only disagree to the 5:1 Killprobability relation (same number of rounds hit) while shooting to smal targets in general and specialy unprotected targest!
Greetings, Knegel
-
Knegel,
It sounds to me like you are overfocused on the unprotected pre and early war Japanese designs as the primary American targets. While the Ki-43 and A6M never had really significant protection, later Japanese fighters such as the Ki-44, Ki-61, Ki-84, J2M and N1K did have protection. The same is true for their bombers and attack aircraft. The notorious G4M "One Shot Lighter" traded much range for protection in the G4M3 model. The Ki-67 and H8K2 also had significant levels of pretection.
And that ignores the large number of well protected German aircraft they faced completely out of the picture.
-
Hi HoHun,
>if you did read what i wrote, why you dont saw that i did took the higher catrige power of the 20mm into account??
I read your handwaving arguments and told you that you broke the logical chain. I pieced together the logical chain for you and posted the appropriate formula for your use. Go and use it.
>After Tony Williams statement regarding the term "hitprobability", there is not much sence in many of your replys.
I knew Tony would be misunderstood. Each probability is the probability for a defined event happening under defined circumstances. Tony quoted the requirement for stating the burst length, for example. "Hit probability" without any qualifier is just that, the probability for one hit in one shot.
>I dont say that the formulas are wrong, but at the end my initial statement of the higher hitprobability of the .50cal armament was not wrong.
You stated:
"the .50 cal was the better weapon for the US need, cause, while the gun was good enough to destroy all japanese planes rather fast and to keep the german fighters away from the Bombers, it did provide a much better hitprobability than the relative slow firing Hispano."
This sentence fulfills none of the requirements for stating a complex probability (which could safely be called "hit probability" in inaccurate slang if all defining elements were stated), but you compared one weapon against another weapon. The only valid interpretation of that sentence is that you suggest that for each bullet, the Browning has a greater hit probablity than the Hispano, which - as I was able to conclude from the context - probably was not what you meant.
In short, you were wrong.
>Can you please point me to the point in the 'Gesetz der großen Zahlen' where they refer to Nh??
I already told you that I am not your teacher, and your question is of a rethoric and thus insulting nature as it's obvious that the law of great numbers is so general that it doesn't need to refer to WW2 gunnery directly to apply to it.
One more offense like that, and I'll put you on my ignore list.
Kind regards,
Henning (HoHun)
-
Gripen,
I was pretty much agreeing with you an dsimply trying to direct HoHun to look at what you wrote... since he seems to believe you better that he believes others.
HoHun,
Knegel wrote about fast firing and slow firing guns in what you quoted. You simply cut away the RoF from the equation yourself when you believe it talks about both of those guns firing only a single bullet.
Keep the time as a factor in the equation and you will have the amount of bullets there as well. Without time there is no talk of fast or slow RoF.
"I already told you that I am not your teacher, and your question is of a rethoric and thus insulting nature as it's obvious that the law of great numbers is so general that it doesn't need to refer to WW2 gunnery directly to apply to it.
One more offense like that, and I'll put you on my ignore list. "
I was very tempted to comment on this... but, well.. it's not worth it :lol
-
Hi Blauk,
>Knegel wrote about fast firing and slow firing guns in what you quoted. You simply cut away the RoF from the equation yourself when you believe it talks about both of those guns firing only a single bullet.
As I stated above, I was aware what Knegel meant to say. Unfortunately, it was not what he had said - which is what I have quoted verbatim above - so I pointed out the confusion in terms.
>I was very tempted to comment on this... but, well.. it's not worth it :lol
My ignore list is a simple tool that helps me not to waste time on people who have nothing to contribute. Consider yourself added to the list.
Kind regards,
Henning (HoHun)
-
Hi HoHun,
it looks to me you mix up math with logic!!
We also can put logic into words, but of course this need the will and skill to read and understand the words. I have to admid that my skill in reading fomulas is not the best, but as BK´s example point out, your skill to read and understand words isnt the best eighter.
Maybe it wasnt as clear in the sentence you did quote, but the whole contex make it pretty clear:".........it did provide a much better hitprobability than the relative slow firing Hispano....... specialy if the smaler gun provide a much higher hitprobability."
I clearly talk about the different in ROF and the gun and its hitprobability. If i talk about a MG, why shal i talk about a single shot??? Yes sure, now you will come and say, no you dont told that both shoot a burst of fire of a given length, but realy, do i need to say that the sun dazzles??
I explained several times exact the same, but you always didnt agree!!
Example: "......if you agree that a shotgun provide a higher hitprobability, while shooting to a smal fast moving target, than a normal rifle, you also must agree that a MG, which shoot 12 rps have a higher hitprobability than a rifle.
And of course also 6 x 13rps(78 rps) provide a higher hitprobability, while shooting to this smal target, than 4 x 10rps(40 rps)."
Your answer:
">if you agree that a shotgun provide a higher hitprobability ...
The point is, I don't."
lol i just saw this in your last reply:
">Knegel wrote about fast firing and slow firing guns in what you quoted. You simply cut away the RoF from the equation yourself when you believe it talks about both of those guns firing only a single bullet."
BK dont believe i talk about the guns only shoot one round, thats the point you dont get. I didnt talk about the guns only shot one round, i was talking about the gun hitprobability!! You remeber?? Hitprobability(bullet) * ROF = Gun hitprobability. Since ROF implement rounds per sec, we have a clear determined timespan and number of rounds. Or will you tell me the Gunpower on Thonys page also is a wrong term?
Of course if you only relay on your formulas, without to see other possibilitys of defination, you will get problems to understand someone who dont know the exact terms you use.
And if you only refer to pages, where someone who dont know many of the displayed sighns cant find a correlation to your examples, noone will be able to understand you, if he dont understand you already before!!
As i told, i dont would expect from someone to learn a new language just in a discussion, while iam able to talk in the known language.
And if you refuse to point more exact to what you refer to, how shal i understand?
"One more offense like that, and I'll put you on my ignore list."
Is a question, where you dont be willing to answer and offence for you??
Or are you not able to answer?
From my point of view your way to discuss is pretty much a offence. Looks like you dont be able to jump over your shaddow and to agree that my 1st statement wasnt wrong, though not good formulated from your knowledge of terms.
I many times told that iam not used to your terms, but you simply didnt care and go on to tell me that i dont understand this, but now i would say its the other way around.
You know many formulas and how the theory work in some special situation, but it looks to me you dont be able to implement it into a different contex, into the real world.
I think this is enough, you also can set me onto your ignore list(you realy seems to think you be very important, if you threat with this), for now i dont got any new information out of your formulas. The only new information i got is that the 'possibility theory' seems to use some simplyfied terms.
Karnak,
as far as i know the Ki43 and A6M was the main japanese fighter even in 1945.
When newer fighters came in more big numbers, the numerical and skill advantage of the US fighters(pilots) already was much to high.
All statements i found regarding the Ki44 was like this: Lack of fuel and crew protection; very vulnerable. The main version of the Ki61 was armned with 4 x .50cal and had roughly the performence of the Bf109F2 and that in 1943-45, who realy need to fear this planes anyway in a F6F, F4U, P51D or P38J/L??
Most other japanese planes was also smal, without effective tailgunners and not nearly as good protected like many other allied and axis planes.
Many times i told that this thought specialy count for the in general light protected japanese planes, and that while a escortmission the time to shoot and hitprobability was a not a unimportant factor( to bring the interceptor off the target), but if you look to the result in war, you clearly can see that the 6 - 8 x .50cal armaments was pretty successfull, even vs the german fighters(even the P51B was successfull).
The brits did use only two hispanos in their Spits, this cannon hitprobability is even smaler than the 4 x 20mm of the F4U-1C, whould the P51 also have had only 2 x 20mm with 120 rounds(12 sec to shoot) + 2 x 50cal to keep the flight performence high?? Is there any hint that the Brits had a higher kill quote than the US boy´s?? Didnt the brits shot down well protected 109E´s and 110C´s without many problems with their poor 8 x .30cal??
The question remain: How many armament power is needed to bring a fighter down in a time and on a distance that the pilot would call satisfying and is a bit more satisfying worth to give up much of the time to shoot.
My assumtion is and was: If the pilot follow the instruction and go very close to shoot, it dont matter if he carry 6 x .50cal or 4 x 20mm while attacking a fighter, cause both armaments are able to cause deadly damages if the hitprobability is high. When the hitprobability go down, much of the damagepower advantage get lost by the more bad armament hitprobability.
To prove me wrong, you need to prove that the US armament was insufficient for their need.
Most of you realy sounds like the US pilots had real problems to shoot a enemy figher down, just like the brits while BoB vs the Bombers.
I realy start to wonder how the Ki43, Ki61 and Machi pilots was able to shoot something down at all.
Greetings, Knegel
-
Hi Knegel,
>>"One more offense like that, and I'll put you on my ignore list."
>Is a question, where you dont be willing to answer and offence for you??
Or are you not able to answer?
Oh, I would have tried to answer a genuine, polite question.
As somehow, your questions regularly miss this mark, I'll just add you to my ignore list now.
Kind regards,
Henning (HoHun)
-
Happy ignorance HoHun :lol Enjoy it!
eh.. so that privileged contributors is his ignore list???? Doesn't that term actually mean the opposite :rolleyes:
-
Originally posted by Knegel
Hi Tony,
i agree absolutly to all you wrote(well written!)!!
Do you also have a test similar to that of the Bf109F to a unprotected target without selfsealing tanks?? Include this test the tank area?
On what distance this test was made?
[/B]
Yes, the British assessed that a hit in the fuel tank area by a .50 stood a 10% chance of being effective if the tank was self-sealing, 40% if it wasn't.
Wouldnt the probability to hit a critical area increase much if there isnt plating and selfsealing tanks?? Wouldnt the resulting higher killprobability, of this guns, increase much more than the killprobability of a 20mm, which have much less trouble with selfsealing tanks and plating anyway?
The probability of a hit wouldn't change, but the probability of a kill would. Yes, it was the toughening-up of aircraft (not just in terms of armour and self-sealing tanks, but also in the strength of their structure) which drove the development of more powerful guns.
The RAF's 8x.303 armament was fine against unprotected aircraft, right at the start of the war, but they soon found it increasingly difficult to score kills because the aircraft toughened up in ways which were not always recorded, e.g. armour added to existing aircraft during the BoB.
As I've said, the USAAF's preference for the .50 was the right military decision at the time, because it did an adequate job and the targets it was firing at were generally not as tough as other air-forces faced. And that decision brought lots of production, logistical, maintenance and training benefits.
Incidentally, I should perhaps clarify a couple of points from my previous post, which should be obvious, but (for the record):
First, that a very important factor affecting the hit probability is of course the nature of the target: its size, the firing range, whether it is flying steadily or manoeuvring.
Second, that the kill probability will be affected by the toughness of the target.
My post assumed that these variables would be constant for comparison purposes. To try to allow for them would dramatically multiply the complexity of the calculations.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi Thony,
Originally posted by Tony Williams
Yes, the British assessed that a hit in the fuel tank area by a .50 stood a 10% chance of being effective if the tank was self-sealing, 40% if it wasn't.
The probability of a hit wouldn't change, but the probability of a kill would. Yes, it was the toughening-up of aircraft (not just in terms of armour and self-sealing tanks, but also in the strength of their structure) which drove the development of more powerful guns.
[/B]
Yes, regarding the hitprobability of the tanks, if there is no plating around, of course you be right, the hitprobability dont change.
In this case i was more thinking of the 3% probability to hit the pilot or vital controls or a Bf109F.
Without plating the probability to hit the pilot or vital controls should increase much.
My thought is that while calculating(i like the term estimating more) the killprobability of a special armament, we need to take the different hitprobability of this armament into account, same like the different probabilitys of kill, while hitting the tankarea and structural area and the different hitprobability and resulting probability of kill, while shooting to different good protected pilots or vital controls into account. (crappy sentence, i know :) )
My estimation is: Many smal guns, which provide a high armament RoF, can be same effective vs smal unprotected targets, like a armament of cannons, which provide a much smaler armament RoF.
As more though the plane and the tanks get and as better the vital areas get protected, as less the high ROF count, simply due to the extreme decreasing killprobability of the smal bullet, while the 20mm damagepower result in a more constant killprobability on a wider range of 'plane toughness'.
Originally posted by Tony Williams
The RAF's 8x.303 armament was fine against unprotected aircraft, right at the start of the war, but they soon found it increasingly difficult to score kills because the aircraft toughened up in ways which were not always recorded, e.g. armour added to existing aircraft during the BoB.
As I've said, the USAAF's preference for the .50 was the right military decision at the time, because it did an adequate job and the targets it was firing at were generally not as tough as other air-forces faced. And that decision brought lots of production, logistical, maintenance and training benefits.
Incidentally, I should perhaps clarify a couple of points from my previous post, which should be obvious, but (for the record):
First, that a very important factor affecting the hit probability is of course the nature of the target: its size, the firing range, whether it is flying steadily or manoeuvring.
Second, that the kill probability will be affected by the toughness of the target.
My post assumed that these variables would be constant for comparison purposes. To try to allow for them would dramatically multiply the complexity of the calculations.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/) [/B]
I absolutly understand why the variables are constant for comparison!
But i think you should point more to this not linear Killprobability of different damagepowers, while shooting to different sized, different tough and/or protected targets. Specialy the killprobability per amoload is not unimportant in this contex.
Otherwise many more people start to calculate incredible cannon killprobabilitys and think the US engeeners was absolutly stupid not to use cannons. Same like others could calculate that 12 x .30cal had a higher damagepower than a 1 x MG151/20 and so a higher killprobability vs a IL-2.
Greetings, Knegel
-
Any fighter armament fit was of course a compromise - within quite tight weight limits - between the volume of fire and the destructiveness of the projectiles.
At one extreme, you got up to 12 RCMGs pumping out a phenomenal volume of fire (over 14,000 rpm), which soon proved to be sub-optimal because the projectiles were insufficiently destructive. At the other extreme there were a few German fighters mounting a 50mm cannon, which was massively destructive but had great difficulty in hitting anything.
In practice, the calibre range which proved useful in WW2 was 12.7-30mm. Obviously, the 12.7mm armament was at its best against weak targets, whereas the 30mm were needed against the strongest targets. A decent (i.e. not low-velocity) 20mm cannon fit was, I believe, the best all-round compromise in WW2.
It is obvious that everyone except the USAAF agreed with this at the time, because all other air forces moved to 20mm (and even the USN wanted to, were it not for the reliability problems they experienced). In a way, the USAAF was lucky because the .50 proved well suited to their particular needs. If they had had tougher opponents to deal with, they would have been in a right mess.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi Tony,
>A decent (i.e. not low-velocity) 20mm cannon fit was, I believe, the best all-round compromise in WW2.
>It is obvious that everyone except the USAAF agreed with this at the time, because all other air forces moved to 20mm
A modern cannon was the best all-round compromise, and the state of the art advanced as the war progressed. At the end of the war, the Luftwaffe considered the 30 mm cannon the best compromise and the 20 mm a secondary weapon like the cowl guns had been secondary to the 20 mm cannon before.
>As I've said, the USAAF's preference for the .50 was the right military decision at the time, because it did an adequate job and the targets it was firing at were generally not as tough as other air-forces faced.
If you had asked any fighter pilot who had to go up against the manoeuvrable Japanese or fast-climbing German fighters in a F4F or P-40 whether it would have helped him to save 400 lbs by installation of a cannon battery of equal firepower, there can be no doubt what his answer would have been. (For a P-40E, a 400 lbs weight decrease would have resulted in a 10% climb rate increase, for example.)
You only rate the 12.7 mm machine gun from an armament point of view, and it might be adequate from that perspective. However, from an aviation point of view, not getting rid of those 400 lbs of ballast was a bad decision, and there can be no doubt that American aviators paid for that. In a way, every US pilot who just couldn't pull tightly enough to evade that lethal burst of fire was killed by his own machine guns! With 400 lbs less weight, his turn could have been just the decisive little bit tighter.
That's the reason why the US decision in favour of the 12.7 mm machine gun, if it was a decision at all, was clearly wrong.
Regards,
Henning (HoHun)
-
Thats assuming you replace 6 x 50s with only 2 x 20mm: as was the case with the F4U-1C and A-36, in all likelyhood, the US probably would have gone to 4 x 20mm as a battery in their fighters. The weight difference would have been negligable. The F4F, F6F, F4U, P-51, P-40, P-47 all more than likely would have carried 4 x 20mm with @ 200 rpg.
The 400 lbs of weight is speculation at best. Who knows what Grumman or Curtiss would have gone with, we will never know.
Also, there seems to be a tendancy for people to debate these issues as if they just "waved a magic wand" < and a different armament comes along like presto. There are many reasons why countries go to war with the guns they do. The USA was not happy with the early 20mm Hispano, so they went with the .50 caliber, they couldnt wait around, there was a war on.
Btw, its interesting to note that the axis did not go to an all 20mm force in most of their fighters, Fw190, Bf109, Ki-84, Ki-61, Ki-44, A6M (later), MC series, all employed either 12.7 or 13.1mm weapons. So they obviously thought they were good for something, and not just "useless weight". The Soviets used 12.7s on many fighters as well.
...All that being said, I think the 20mm is the better weapon, and had the USA been able to get the problems sorted out sooner, would have been preferable I think to the .50 cal, but as it was the .50 gave very good service, and was widely liked by most that used it, from what I read.
-
Hi Squire,
>Thats assuming you replace 6 x 50s with only 2 x 20mm: as was the case with the F4U-1C, in all likelyhood, the US probably would have gone to 4 x 20mm as a battery in their fighters.
If they had appreciated the weight problem in the first case, they might well have used the 2 x 20 mm option, especially with the early model fighters that were not so powerful.
>Also the F4F and P-40 would not have become uber fighters with the 400 lbs your talking about, and neither was ever employed as a "dogfighter" especially in the Pacific. They wouldnt be stall turning Zeros because they were armed with 20mm cannon. Thats grossly over stating it I think.
You are right, but I didn't claim that anyway :-) However, weight (or more accurately: mass) is a major factor for almost every aspect of performance, and there should be no doubt that the early US fighters were at a serious disadvantage there and would have greatly benefitted from a 400 lbs reduction.
And 400 lbs less with no disadvantage at all, that's something an airframe designer must work long and hard for. Just think of the P-40N or the FM-2 that actually had machine guns removed to save weight. For comparison: The gondola cannon of the Me 109 that reportedly turned the plane from a nimble interceptor into an unmanoeuvrable battleship weighed 474 lbs. Not that I consider the cliché justified, but 400 lbs more or less will affect the performance of a fighter.
>Also, there seems to be a tendancy for people to debate these issues as if they just "waved a magic wand" < and a different armament comes along like presto.
True enough. Engines and guns often require years of development time. However, in the case of the US Hispano copy, the British original had already reached a level of satisfactory reliability, so it seems not unreasonable to assume that a more determined effort would have resulted in a reliable US copy realtively quickly.
>Btw, its interesting to note that the axis did not go to an all 20mm force in most of their fighters, Fw190, Bf109, Ki-84, Ki-61, Ki-44, A6M (later), MC series, all employed either 12.7 or 13.1mm weapons. So they obviously thought they were good for something, and not just "useless weight".
Well, no simple answer to that. The Ki-44 was in fact an all-machine gun fighter, it appears no evidence of 20 mm armed versions could be found. (Of course, there was the caseless cannon, in a few examples.) In the A6M and the early Me 109, the MG had the advantage of a higher muzzle velocity and than the cannon and the advantage of a ballistally favourable centre-line position, too. In the late Me 109 types and in many of the Soviet fighters, only one cannon was carried so that the firepower increase was noticable. (The MG131 was a particularly light "heavy" machine gun.) In the case of the Fw 190, the machine guns seem not to have been worth it, especially considering that they not only increased weight but also drag. They were eliminated in some of the Sturmböcke, but I'd say they could have been eliminated from all 4-cannon Focke-Wulfs without anyone noticing. Of course, there is a school of thought that considers machine guns useful for sending a few tracers at the target to check aim, and then cut in with the cannon. This works well in simulation games, but I'm not sure it would work in real life, too.
>...All that being said, I think the 20mm is the better weapon, and had the USA been able to get the problems sorted out sooner, would have been preferable I think to the .50 cal, but as it was the .50 gave very good service, and was widely liked by most that used it, from what I read.
Well, I'm not sure contemporary literature has fully understood the disadvantages of the 12.7 mm machine gun. I have a dozen books here on my shelf that bash the Messerschmitt gondolas for their 500 lbs weight penalty, but somehow, there's not one book that bashes the US batteries for being 400 lbs overweight. I don't think the Messerschmitt gondolas were quite as bad as my books would have them, but I don't believe the extra 400 lbs of the US batteries were a non-issue either.
Regards,
Henning (HoHun)
-
Hi,
"A modern cannon was the best all-round compromise, and the state of the art advanced as the war progressed. At the end of the war, the Luftwaffe considered the 30 mm cannon the best compromise and the 20 mm a secondary weapon like the cowl guns had been secondary to the 20 mm cannon before."
Looks like he still cant see the different need while fighting B17´s, B24´s, IL-2´s, P47´s, La(GG)´s, Wellingtons, Mosquitos, Lancasters and other very tought planes, in oposide to the FW190A as most though oponent in big numbers, while the biggest number of enemys was Ki43´s, A6M´s, Me109´s and other rather light protected and/or relative smal targets.
Even the FW190´s did fear the .50cals of the B17´s pretty much.
And it looks like he realy think a nation, which did pop out CV´s, many other ships, thousands of Bombers, fighter, tanks etc, one after the next, wouldnt have been able to introduce a 20mm in big numbers, if there would have been a real need. They had the 20mm in the P38 very early, so they must have been able to produce it somehow, but no need = no motivation.
Btw, 2 x Hispano II(dont the US 20mm had a smaler ROF?) would have provided around a 4 times smaler armament-hitprobability than the 6 x .50cal, without to consider the shorter time to shoot, vs smal fighters this would have been a pretty bad handycap. 4 x 20mm would have been needed, but this would have increased the weight and reduced the ammo load, and wouldnt the long barrels also decrease the performence(afaik it did in the SpitVa to SpitVb, or was it the weight?)?
Anyway, at the end of the war the USA did introduce the new Gunsight, which did increase the killprobability much, specialy while a fightercombat. So the already good working armament got even more effective.
Greetings, Knegel
-
I actually think the real end to the .50 came with the Jet Age, where time on target issues began to make things more difficult, ala MiG-15 vs F-86, and the adoption of "cannon only" really came into being post WW2.
As far as the ammo issue goes, I also think that is a relevent partial defence of the 12.7mm, you did tend to get almost twice the ammunition as a 20mm in many fighters, but that only offsets it somewhat.
Still an interesting debate.
-
Hi,
Yep, i agree, in the jetage, when cannons with less relative weight, more reliability and ROF got produced, the MG´s got obsolet, specialy if we see that all nations (at least west allieds and east allieds) got tough bombers.
But its very interesting to see that the F-86 still did carry 6 x .50cal and was rather successfull vs the MG-15. VS fighters with probably easy to damage jet engines the .50cal still seemed to work pretty good.
Dont the north coreans had tough bombers?
Greetings, Knegel
-
Originally posted by Squire
Btw, its interesting to note that the axis did not go to an all 20mm force in most of their fighters, Fw190, Bf109, Ki-84, Ki-61, Ki-44, A6M (later), MC series, all employed either 12.7 or 13.1mm weapons. So they obviously thought they were good for something, and not just "useless weight". The Soviets used 12.7s on many fighters as well.
In the case of the German planes, the reason is simple: there was no room to fit cannon in the space for the synchronised cowling guns. The little 13 mm MG 131 was the biggest gun which could be fitted there, until the long-nosed Ta 152 and Do 335 came along, but they never saw service.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Originally posted by Knegel
But its very interesting to see that the F-86 still did carry 6 x .50cal and was rather successfull vs the MG-15. VS fighters with probably easy to damage jet engines the .50cal still seemed to work pretty good.
Actually the armament of the F-86 was considered unsatisfactory by the USAF - it took over 1,000 rounds fired to down a MiG-15 on average, and many MiGs got home despite being hit by up to fifty bullets. As a result, the USAF initiated Project GunVal, in which they fitted 20mm cannon to some F-86s and sent them out for operational testing in Korea. The USAF switched to 20mm for all new planes after that.
Fortunately for the USAF, the MiG-15 also had unsatisfactory armament for the opposite reason; its heavy, low-velocity cannon were designed to shoot down bombers and it was difficult for them to connect with with Sabres.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi Squire,
>I actually think the real end to the .50 came with the Jet Age, where time on target issues began to make things more difficult, ala MiG-15 vs F-86, and the adoption of "cannon only" really came into being post WW2.
Well, "the end" really came in the Korean war, but only because the USAF didn't understand the message in WW2. They were pretty disappointed by their 12.7 mm machine guns in Korea though they were using a version with considerably increased rate of fire - I'm sure you are aware of Project GUNVAL which was initiated to rush the 20 mm cannon into service quickly.
>As far as the ammo issue goes, I also think that is a relevent partial defence of the 12.7mm, you did tend to get almost twice the ammunition as a 20mm in many fighters, but that only offsets it somewhat.
I'm afraid it only looks like that as long as the weight penalty is accepted. If you look at batteries of equal weight, the heavy machine gun actually has the poorest endurance:
1x MK 108 - 634 rpg - 431 kg - 295% firepower - duration of fire: 63 s
1x MK 103 - 314 rpg - 430 kg - 240% firepower - duration of fire: 44 s
2x MG 151/20 - 811 rpg - 431 kg - 149% firepower - duration of fire: 67 s
2x Hispano V - 705 rpg - 431 kg - 145% firepower - duration of fire: 58 s
2x Hispano II - 673 rpg - 431 kg - 125% firepower - duration of fire: 67 s
2x MG-FF - 553 rpg - 430 kg - 92% firepower - duration of fire: 69 s
4x MG 151 - 361 rpg - 431 kg - 103% firepower - duration of fire: 30 s
8x MG 131 - 472 rpg - 431 kg - 99% firepower - duration of fire: 31 s
6x ,50 Browning M2 - 390 rpg - 431 kg - 100% firepower - duration of fire: 30 s
12x Browning ,303 - 863 rpg - 431 kg - 62% firepower - duration of fire: 43 s
In other words, if a 20 mm battery had only half the endurance of a machine gun battery, this was the result of a design choice that either preferred light weight over endurace or (more likely if we're talking about the 4 x 20 mm installations) firepower over endurance.
I don't know if you have seen the chart with the burst limits for the 12.7 mm machine gun that had to be observed to avoid overheating and "cook-offs". You could fire a 15 burst round, than pause 30 s, fire the next 15 burst round, and so on - and after 10 secondary bursts fired that way, the barrel would be so hot that you'd have to stop anyway (165 rounds in five minutes).
If the pauses were extended to 60 s, you could start with a 75 round burst, then fire 10 secondary bursts of 15 rounds each. (225 rounds in 11 minutes).
That doesn't look to me as if 390 rounds per gun as outlined above were likely to be expended. I believe Tony has posted once about the British Operations Research people tracking how many rounds of ammunition the bomber gunners were bringing back after an engagement with Luftwaffe night fighters - it would be interesting to know if similar evaluations were done for fighter guns!
Regards,
Henning (HoHun)
-
Originally posted by Tony Williams
Actually the armament of the F-86 was considered unsatisfactory by the USAF - it took over 1,000 rounds fired to down a MiG-15 on average, and many MiGs got home despite being hit by up to fifty bullets. As a result, the USAF initiated Project GunVal, in which they fitted 20mm cannon to some F-86s and sent them out for operational testing in Korea. The USAF switched to 20mm for all new planes after that.
Fortunately for the USAF, the MiG-15 also had unsatisfactory armament for the opposite reason; its heavy, low-velocity cannon were designed to shoot down bombers and it was difficult for them to connect with with Sabres.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
Hi,
afaik the two 23mm cannons was high velocity guns, but low in ROF. The 37mm was low in both aspects.
Since the cannons did suffer the same problem, i guess the needed 1000rounds was more caused by the poor hitprobability while fighting at highspeed than by missing damagepower.
I only wonder if the NATO didnt expected to meet more tough targets than a Mig-15. What they thought to do vs big and medium Bombers??
Greetings, Knegel
-
Originally posted by Knegel
afaik the two 23mm cannons was high velocity guns, but low in ROF. The 37mm was low in both aspects.
The 23x115 ammo fired by the MiG was a low-velocity round (690 m/s, the same as the N-37's 37x155), not at all the same as the much more powerful 23x152B fired by the WW2 VYa-23. Two 23mm guns were used in the MiG-15; initially they had the NS-23 (550-700 rpm), but the MiG-15bis had the NR-23 (850-950 rpm). The N-37 fired at 400 rpm.
Since the cannons did suffer the same problem, i guess the needed 1000rounds was more caused by the poor hitprobability while fighting at highspeed than by missing damagepower.
[/B]
That would have been the major cause, but note that I said that MiGs got back after receiving up to 50 hits. MiG pilots also reported seeing the .50 bullets deflected off their planes. This may sound odd, but the metal skinning of the jets was much tougher than WW2 prop planes, and the bullets would often strike at a very fine angle so could be deflected without penetrating. When you look at the back view of a MiG, you'll see what mean.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi Tony,
>Fortunately for the USAF, the MiG-15 also had unsatisfactory armament for the opposite reason; its heavy, low-velocity cannon were designed to shoot down bombers and it was difficult for them to connect with with Sabres.
Hm, maybe you have better sources, but I have a GDR booklet here that describes the performance of the Soviet guns in Korea as superior. It also mentions that the cannon were the result of a deliberate switch to medium velocity in order to maximize rate of fire while reducing weight. Now that might have been propaganda, but while the USAF completely got rid of what had been their primary air-to-air weapon, the Soviets continued to design medium-velocity guns for their fighters, like the NR-30 for the MiG-19 and the NR-23 for the MiG-21.
If for example instead of 3 x NR-30, they would have installed 2 x NR-30 and two smaller-caliber high-velocity guns in the MiG-19, we'd have an hint they were disappointed with the performance of their guns, but as it is, no paradigm change (as with the USAF) took place after Korea.
>MiG pilots also reported seeing the .50 bullets deflected off their planes. This may sound odd, but the metal skinning of the jets was much tougher than WW2 prop planes, and the bullets would often strike at a very fine angle so could be deflected without penetrating.
Actually, this effect had been observed in WW2, too.
Regards,
Henning (HoHun)
-
Originally posted by Knegel
And it looks like he realy think a nation, which did pop out CV´s, many other ships, thousands of Bombers, fighter, tanks etc, one after the next, wouldnt have been able to introduce a 20mm in big numbers, if there would have been a real need.
If you read the articles available on for example Emmanuel Gustin's and Tony's web pages or the Flying Guns: World War II book you'll find excellent expnalation how such a nation was not able to introduce the 20mm cannon in big numbers because technical problems. Another reason was of course the logistics.
Btw, 2 x Hispano II(dont the US 20mm had a smaler ROF?) would have provided around a 4 times smaler armament-hitprobability than the 6 x .50cal, without to consider the shorter time to shoot, vs smal fighters this would have been a pretty bad handycap.
Nope. With machineguns you'll need to put a long burst into the target, yet there are no guarantees of it going down. Machinegun bullets do minimal damage on their own, and there must be lots of them hitting the target to cause damage. And for example, especially if shot from rear angles most bullets tended to deflect from airplane's surface, even from fabric surface, or tumble after penetrating, causing the bullets to lose much of their power. One Allied study showed that only 3% of .50 bullets fired from rear damaged critical items, pilot or control cables inside fuselage. Rest just deflected or made holes to the airplane skin, nothing else.
With a cannon there are fewer bullets in the air, but when those hit they make immediate damage.
What point there is if you can hit the enemy plane, but can't kill it?
All air forces acknowledged the need for destructive power, instead mass of bullets, either during or after the WW2, and some only during the Korean war.
-
Originally posted by Knegel
Hi,
Yep, i agree, in the jetage, when cannons with less relative weight, more reliability and ROF got produced, the MG´s got obsolet, specialy if we see that all nations (at least west allieds and east allieds) got tough bombers.
But its very interesting to see that the F-86 still did carry 6 x .50cal and was rather successfull vs the MG-15. VS fighters with probably easy to damage jet engines the .50cal still seemed to work pretty good.
Dont the north coreans had tough bombers?
Greetings, Knegel
At jet dogfight altitudes, the API rounds from the Sabre's had severe problems getting ignition. You could chew off chunks of the MiG, but they'd just soak it up and keep flying. What let the Sabres down so many Migs was the radar gunsight. If I've got a radar gunsight, and you've got the Mk1 eyeball, in a dogfight with 1000mph closure rates, I've got an extreme advantage over you, cannons or not.
-
Originally posted by Knegel
Btw, 2 x Hispano II(dont the US 20mm had a smaler ROF?) would have provided around a 4 times smaler armament-hitprobability than the 6 x .50cal, without to consider the shorter time to shoot, vs smal fighters this would have been a pretty bad handycap. 4 x 20mm would have been needed, but this would have increased the weight and reduced the ammo load, and wouldnt the long barrels also decrease the performence(afaik it did in the SpitVa to SpitVb, or was it the weight?)?
Why do you keep repeating this debunked argument? It has been demostrated to be absolutely false, yet you keep repeating it? Do you actually read what others have said? You keep talking about RoF of the instalation like there are bullets flying out randomly in sphere, thus crediting RoF with a direct coralation to chance to hit.
[The gun instalations produce a bullet stream. This is true of both the .50s and the 20mm cannons. Both have a high enough RoF to stop the target from being able to fly between the shots with the possible exception of near 90 degree crossing shots, which in WWII resulted in less than 1% of shoot downs.
Therefore your main claim to the superiority of the .50 is that in less than 1% of cases it will hit and the 20mm will miss. TO hold that sub 1% as the deciding issue when all of the other strengths of the 20mm have been covered is ridiculous.
If a shot will hit with a six pack of .50s it will almost certainly hit with a two pack of Hispanos, and do more damage at the same time.
-
Originally posted by HoHun
Hm, maybe you have better sources, but I have a GDR booklet here that describes the performance of the Soviet guns in Korea as superior. It also mentions that the cannon were the result of a deliberate switch to medium velocity in order to maximize rate of fire while reducing weight.
Well yes, it didn't happen by accident ;) The issue is whether the decision was the right one. I don't think that a GDR source can be taken as unbiased.
There are really three separate issues here: the performance of the gun (power/weight ratio), the performance of the ammo, and the gun selection for the aircraft. Like most Russian guns, the NR-23 and N-37 were good. The NR-23 fired more quickly than the Hispano, for less weight, and the shells were much heavier. The N-37 was also quite a remarkable performer; compared with the US equivalent, the 37mm M10, the weight was the same, the ammo 1.5x more powerful and the RoF 2.5x faster.
But the choice in both calibres of a heavy shell/low velocity combination was better suited for attacking bombers rather than agile fighters.
I have no doubt that the MiG-15 would have been far more appropriately armed for dealing with fighters if it had four NR-23. Better still if the projectile weight had been dropped to boost the muzzle velocity.
Now that might have been propaganda, but while the USAF completely got rid of what had been their primary air-to-air weapon, the Soviets continued to design medium-velocity guns for their fighters, like the NR-30 for the MiG-19 and the NR-23 for the MiG-21.
The NR-30 was an excellent gun with a remarkable power/weight ratio (three of them weighed the same as two 30mm Aden revolver cannon, fired at least as quickly, and used much more powerful ammo), but it's worth noting that the muzzle velocity was usefully higher than the earlier cannon, even though the shell weight was still high.
I don't think that the MiG-21 ever carried the NR-23; that gun was obsolete by then. It started with the NR-30, went gunless for a period, then adopted the little GSh-23, which fired basically the same ammo as the NR-23 for little more weight, but at 3,000 rpm...
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi Tony,
>I don't think that a GDR source can be taken as unbiased.
Well, in this case it is matched the realities of Soviet weapon development, so it wasn't all wishful thinking.
When it praises the MiG-21 and MiG-23, and then there is the paradigm change to the MiG-29, that's the point where one should get suspicious :-)
>But the choice in both calibres of a heavy shell/low velocity combination was better suited for attacking bombers rather than agile fighters.
Actually, 690 m/s is not so far from the ca. 750 m/s of the MG 151/20, and that was an excellent anti-fighter weapon. High muzzle velocity is not as important in fighter-vs.-fighter combat as the ability to make the most out of time-limited shooting opportunities.
The way to kill an evading fighter is to hit him hard when he momentarily presents a good target, for example a large planform or a low deflection, or when he manoeuvres in only one plane for an instant. You can anticipate these opportunities for some fractions of a second (enough to make muzzle velocity secondary), but there is no way you can extend the duration of these moments, so you want a weapon with great knock-down power against fighters just the same as against bombers.
When talking about air-to-air gunnery, people seem to be fixated at maximum effective range, but in practice, the range of maximum effectiveness has much more relevance - and that is much shorter regardless of the muzzle velocity.
>it's worth noting that the muzzle velocity was usefully higher than the earlier cannon, even though the shell weight was still high.
Well, 780 m/s to 690 m/s according to my GDR booklet, useful but still in the same class. No paradigm change for sure.
>I don't think that the MiG-21 ever carried the NR-23; that gun was obsolete by then.
You are right, I thought of the twin-barrel cannon but erroneously kept using the NR-23 designation. Anyway, still in the same class muzzle-velocity wise.
Regards,
Henning (HoHun)
-
Originally posted by HoHun
The way to kill an evading fighter is to hit him hard when he momentarily presents a good target, for example a large planform or a low deflection, or when he manoeuvres in only one plane for an instant. You can anticipate these opportunities for some fractions of a second (enough to make muzzle velocity secondary), but there is no way you can extend the duration of these moments, so you want a weapon with great knock-down power against fighters just the same as against bombers.
Hm... In the case of the MiG-21 (NR-30 or GSh-23canon) the pilot have a luxury of the gyroscopic gun sight with automatic range measuring. That makes the high velocity of the projectiles less important than it was in the WWII fighters.
gripen
-
Originally posted by HoHun
Actually, 690 m/s is not so far from the ca. 750 m/s of the MG 151/20, and that was an excellent anti-fighter weapon. High muzzle velocity is not as important in fighter-vs.-fighter combat as the ability to make the most out of time-limited shooting opportunities.
The MG 151/20 was a good anti-fighter weapon in WW2, when aircraft speeds were much lower. As they increased and firing opportunities became briefer, so a reduced projectile flight time became more useful.
The RAF did some studies of gun armament in the 1950s (by which time they had adopted the 30mm Aden) and ran all sorts of simulations to determine what would be the best way of improving the kill probability of the armament: larger calibre? Higher rate of fire? Or higher velocity? They concluded that a higher velocity would be most useful - but of course, the Aden of the time was a low-velocity piece, and its calibre was already sufficient.
It's fair to say that there are currently two different views on muzzle velocity: the Russians, who still favour heavy shells at medium velocity (the current 30mm GSh-301 has an MV of 860 m/s) and the West, which prefers at least 1,000 m/s velocity.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Hi Tony,
>The MG 151/20 was a good anti-fighter weapon in WW2, when aircraft speeds were much lower. As they increased and firing opportunities became briefer, so a reduced projectile flight time became more useful.
It's not the absolute speed, but the speed differential that counts. WW2 aircraft, were slower, but they were much more manoeuvrable, too - and thus harder to hit. Firing opportunities were brief to begin with if we are talking about an evading target - fast jets up in the stratosphere dance at a far slower pace than Messerschmitts and Mustangs at 15000 ft.
We are not talking about flak guns after all where the gun is stationary while the jet flies by - these situations certainly became more difficult for the gunners. The pursueing aircraft has the same speed as its intended target, it's only rate of angle change that counts.
>They concluded that a higher velocity would be most useful - but of course, the Aden of the time was a low-velocity piece, and its calibre was already sufficient.
I agree, especially as the Aden was carried at least in pairs.
>It's fair to say that there are currently two different views on muzzle velocity: the Russians, who still favour heavy shells at medium velocity (the current 30mm GSh-301 has an MV of 860 m/s) and the West, which prefers at least 1,000 m/s velocity.
Hm, but I think the West ecept for the USA has employs primarily 27 mm guns, which fire heavy shells at a high muzzle velocity, so it's not clearly the one or the other view. The USA seems to have planned to replace the 20 mm with a 25 mm at least, which indicates their have abandoned their school of thinking, but did not manage to develop (or deploy) a suitable gun. The Russians seem to use an advanced fire control in their weapon which might help to achieve parity with the higher-velocity guns of other nations, so the nature of the different views is a bit different than in the past.
Regards,
Henning (HoHun)
-
Hi,
Originally posted by Tony Williams
The 23x115 ammo fired by the MiG was a low-velocity round (690 m/s, the same as the N-37's 37x155), not at all the same as the much more powerful 23x152B fired by the WW2 VYa-23. Two 23mm guns were used in the MiG-15; initially they had the NS-23 (550-700 rpm), but the MiG-15bis had the NR-23 (850-950 rpm). The N-37 fired at 400 rpm.
That would have been the major cause, but note that I said that MiGs got back after receiving up to 50 hits. MiG pilots also reported seeing the .50 bullets deflected off their planes. This may sound odd, but the metal skinning of the jets was much tougher than WW2 prop planes, and the bullets would often strike at a very fine angle so could be deflected without penetrating. When you look at the back view of a MiG, you'll see what mean.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
ah ok, i thought the weapon was similar to the WWII gun. And of course if the surface of the jets gave a much better protection itself, to a point where the .50cal simply can get through it, they turn to be obsolet!
Hi Grendel,
if there would have been a real need, they could have used the british Hispano II. Of course this would have needed sources and energy. Iam pretty sure they thought its not this big effort worth it.
"If you read the articles available on for example Emmanuel Gustin's and Tony's web pages or the Flying Guns: World War II book you'll find excellent expnalation how such a nation was not able to introduce the 20mm cannon in big numbers because technical problems. Another reason was of
What point there is if you can hit the enemy plane, but can't kill it?"
As i wrote many times before, where Tony William also agreed, vs the common US oponets in WWII, the .50cal catrige had a satisfying probability to damages the planes. As the result, the more bulltes provide a similar killprobability, like the much more powerfull but less numberus cannon rounds.
As i also wrote many times, vs more tough targets the advantage turn fast toward the cannons!
That the surface of the Korean planes was much tougher than the WWII planes was new to me, so of course i agree that the cannons was the more effective weapon.
Hi Karnak,
"Why do you keep repeating this debunked argument? It has been demostrated to be absolutely false, yet you keep repeating it? Do you actually read what others have said? You keep talking about RoF of the instalation like there are bullets flying out randomly in sphere, thus crediting RoF with a direct coralation to chance to hit."
With a hitprobability of maybe 5%(Luftwaffe say 2% while attacking straint flying Bomber by a average pilot), you dont think we can call this flying out randomly in sphere??
"[The gun instalations produce a bullet stream. This is true of both the .50s and the 20mm cannons. Both have a high enough RoF to stop the target from being able to fly between the shots with the possible exception of near 90 degree crossing shots, which in WWII resulted in less than 1% of shoot downs.
Therefore your main claim to the superiority of the .50 is that in less than 1% of cases it will hit and the 20mm will miss. TO hold that sub 1% as the deciding issue when all of the other strengths of the 20mm have been covered is ridiculous.
If a shot will hit with a six pack of .50s it will almost certainly hit with a two pack of Hispanos, and do more damage at the same time."
Maybe you should read what other write!!!
I never wrote the 20mm will fail in general, i told that the more hits of a higher ROF cause a similar result, as long as the smaler round have a good probability to hit a critical area.
If the planes plating get increased (or a much more tough surface) over a special point, what minimize the probability to get to a critical area, of course the killprobability decrease dramatically in relation to the 20mm´s.
But it looks like in WWII most oponets of the USA simply wasnt that tough. At least i never did read that the US pilots did cry for the Spit armament, at least not while fighting fighters(the main target of the US fighters).
Even the brits didnt introduce the 20mm in their P40´s and P51´s.
This is what Thony Williams wrote:
"As I've said, the USAAF's preference for the .50 was the right military decision at the time, because it did an adequate job and the targets it was firing at were generally not as tough as other air-forces faced. And that decision brought lots of production, logistical, maintenance and training benefits. "
and
"In a way, the USAAF was lucky because the .50 proved well suited to their particular needs. If they had had tougher opponents to deal with, they would have been in a right mess."
The germans had the bad luck to be in a situation that even the 20mm´s wasnt good enough.
The USAAF made the right decision in WWII, but failed to introduce the needed 20mm´s, vs the now much more tough planes in Korea.
After the infos Thony did offer, its a mystery for me that they didnt use at least the HispanoV (2 -3 of them) in the F86.
Edit:
Hi indy007,
"If I've got a radar gunsight, and you've got the Mk1 eyeball, in a dogfight with 1000mph closure rates, I've got an extreme advantage over you, cannons or not."
Good info!
Edit end:
Greetings, Knegel
-
Originally posted by Knegel
That the surface of the Korean planes was much tougher than the WWII planes was new to me, so of course i agree that the cannons was the more effective weapon.
After the infos Thony did offer, its a mystery for me that they didnt use at least the HispanoV (2 -3 of them) in the F86.
Greetings, Knegel
Purely conjecture on my part, but if I had an airforce of many ww2 vets, I wouldn't want to change the armament they've used since they've been in combat.
Also, you can put the Korean war's strategic goals up as a reason to keep the .50 onboard. UN forces enjoyed air supremecy. When the Chinese came pouring into the fight, The Meatgrinder was employed. Ultimately it was a contest between manpower (communist) and firepower (UN). The essence of The Meatgrinder was to concentrate all available artillery, tanks, infantry, and air power into wiping out large bodies of enemy troops. Since the enemy has little to no armor support, and supply convoys made out of handcarts, lorries, and people on foot... there's no need to attack them with 20mm cannons. Large numbers of .50s with API have a far larger potential for damage in this specific case.
Lastly, is the simple logistics of it. If you're using a heavy machine gun that's the most common heavy machine gun in all of your armed forces it greatly simplifies maintenance and logistics. Hypothetically, if we're fighter pilots waiting on our guns to be fixed, yours are Hispanos and mine are .50s... my parts & ammunition are going to be alot easier to come by in a war situation where supplies and parts are not guaranteed :)
-
Hi indy007,
as the edit in my last post show, i saw the gunsight thingi pretty late.
So your just stated arguments for the .50cal, in combination with the still equal or maybe better killprobability (US gunsight + 6 x .50cal vs less good gunsight + cannons) make sence to me, why they did keep the .50cals.
btw, your example with the not or bad protected groundtroops and the resulting higher killprobability of the "higher ROF armament", is a good example for what i try to explain the whole time.
Greetings, Knegel
-
Hi Indy,
>Ultimately it was a contest between manpower (communist) and firepower (UN).
I'm afraid that argument doesn't hold water as there was a division of roles in the USAF, with F-86 equipped squadrons in the North to screen the fighter bombers in the South from the MiGs. The fighter bombers actually had a number of distinct types, mostly F-80 and F-84 aircraft, and Navy and Marines operated fighter bombers as well.
The division in armament was not according to role, as the air-superiority F-86s with their disappointing armament had been given the same guns as the fighter bombers. The division in armament was along service boundaries: The USAF used the 12.7 mm machine gun for everything, and the Navy/Marines used the 20 mm cannon for everything.
The Navy had actually considered the 20 mm cannon superior for ground strafing in WW2 already, so I see no reason to assume any superiority of the 12.7 mm gun for this purpose.
>Large numbers of .50s with API have a far larger potential for damage in this specific case.
And even larger numbers of 0.30s should have a far larger potential again by the same logic, yet none of the fighters bombers in that conflict used them at all. The argument "more bullets is better" should be just as valid here.
>Lastly, is the simple logistics of it.
If you can run an air force, the few extra parts for guns are the least of your worries. Many of the parts for the batteries are aircraft-specific anyhow, and I'm pretty sure the Korean war era M3 had little in common with the M2 as used by the ground forces. Logistics is a bogus argument in my opinion.
Regards,
Henning (HoHun)
-
Originally posted by indy007
Since the enemy has little to no armor support, and supply convoys made out of handcarts, lorries, and people on foot... there's no need to attack them with 20mm cannons. Large numbers of .50s with API have a far larger potential for damage in this specific case.
Actually US tests done during the Korean war showed, that .50s were insufficient even against truck targets. 20 mm cannon fire on the other hand totally wrecked the trucks, and cannons could be effective against light armored vehicled and sometimes even against tanks.
So according to US studies, there definitely was need for cannon firepower.
-
Some thoughts:
"Large numbers of .50s with API have a far larger potential for damage in this specific case."
AFAIK, this holds some truth. When counting .50 and 20mm AP projectile mass and velocity it is apparent that .50 is nearly as good as 20mm because of greater velocity and smaller diameter if calculated as energy on diameter of impact (=concentration of mass=penetration).
The .50 IS a good gun in punching holes in a/c. In comparison to 20mm the projectile is, however, too light and streamlined so it easily deflects and tumbles upon impact, but then again tumbling is not necessarily bad. A tumbled projectile does not penetrate very well, but it is good in releasing its energy on softer structures such as aluminum wing spars etc. The tip of MinenG grenade is probably not very aerodynamic but as it is blunt it is less likely to glance off but likes to push through surfaces. AFAIK the shape of MinenG was altered only for 30mm caliber to allow better ballistics?
According to Finnish weapon training book from the '50s the WW2 experiences showed that a 50mm gun is just adequate to take down a large bomber on one shot. The guns were however cumbersome and had a difficult recoil, so rockets were considered a better choice. So there were some experiments with wing tip rocket pods on post war bomber interceptors. However, rockets with radar detonators were not very accurate so this lead logically to rockets with radar detonators and guidance systems and controls -ie. missiles.
-C+
-
Hi Charge,
>AFAIK, this holds some truth.
Hm, it was meant to describe air-to-ground gunnery (where, as Grendel pointed out, it was wrong), but you read it as applying to air-to-air gunnery.
>When counting .50 and 20mm AP projectile mass and velocity it is apparent that .50 is nearly as good as 20mm because of greater velocity and smaller diameter if calculated as energy on diameter of impact (=concentration of mass=penetration).
Tony might have the figures, but 12.7 mm was worse than 20 mm to the point that most planes' armour was considered to protect reliably againt 12.7 mm, but not against 20 mm.
>The .50 IS a good gun in punching holes in a/c.
Tiny little 12.7 mm holes that don't have much effect.
The largest areas of the aircraft consist of thin metal sheet a 7.7 mm bullet has no difficulties in punching through. The trick with mine shells is that they enter easily, and blow up within the structure, ripping apart the thin but structurally vital sheet metal. Insteal of a 12.7 mm diameter holes, you get a 300 mm by 300 mm hole. With such holes, the structually integrity of the target aircraft is compromised so that it tends to break apart in mid-air.
Mine shells were developed to kill a plane instantly instead of allowing it to retreat damaged, as that will often lead to the pilot escaping capture of even to the aircraft being landed and repaired.
Regards,
Henning (HoHun)
-
Hi,
"Tony might have the figures, but 12.7 mm was worse than 20 mm to the point that most planes' armour was considered to protect reliably againt 12.7 mm, but not against 20 mm."
I wonder who he think shot down the lot of german fighters while intercepting US planes??
Had every US plane a british escort with their super cannons, or did the germans forget to fit the armour??
I would say history show that most WWII planes' armour wasnt that reliable while protecting the plane againt 12.7 mm, and it looks like the much tougher planesurface + plating in Korea still wasnt what i would call reliable.
Most Tanks had a protection i would call reliable vs .50cal, and maybe the IL-2 too(actually i dont know if the API would go though its plating)!
"(by indy007) Large numbers of .50s with API have a far larger potential for damage in this specific case.
(by Charge)>AFAIK, this holds some truth.
(by HoHun) Hm, it was meant to describe air-to-ground gunnery (where, as Grendel pointed out, it was wrong), but you read it as applying to air-to-air gunnery."
For now Hohun didnt agreed that indy007s argument, what is exact what i try to explain since some days, is valid in Aircombat. Will he now tell us that it is valid in air-to-air combat?
"(by Charge)>The .50 IS a good gun in punching holes in a/c.
(by HoHun)Tiny little 12.7 mm holes that don't have much effect."
He should tell this the lot of german and japanese pilots who died by this "12,7mm´s without much effect".
I also would like to know how he explain the success of the the finnish pilots in their Brewsters and other more light amrned planes and the british Pilots in their Hurris and Spit1a´s, Glosters even vs german Bombers, but specialy vs german fighters. Why someone got shot down by a Ki43, MC202(early) at all?
He realy think the planes structure is the only vital point of a plane and it looks like he dont see that a MG round, which hit the plane somewhere in the rear fuselage or rear area of the wing, have a pretty long way through the plane, where it can cause much more than a tiny little 12,7mm hole.
Do someone know how many 20mm´s dont get through the surface??
From the rear, the planesurface show a pretty flat angle, how many rounds did explode outside the surface? I did read that some shells had a bad construction and did explode much to early(i think it was japanese round?).
If the 20mm HE hit the tailfuselage and explode short after that, would it still penetrate the plating?
Dot get me wrong again!!! I know a 20mm HE round is much more dangerus in most cases than the 12,7mm API!
Greetings, Knegel
-
You guys start to obscure the topic by your need to pick on each other.
Too bad. Interesting topic wasted.
-C+
-
Hi Charge,
>You guys start to obscure the topic by your need to pick on each other.
Well, I did at least try to adress specifically those points you raised in your post. If I have missed some of your points, please consider that a normal misunderstanding between us two specifically, probably caused by tunnel vision on my part ;-) Sorry for that!
Of course, as far as picking on me is concerned, that's what I have an ignore list for. I wish it weren't necessary ...
Regards,
Henning (HoHun)
-
Originally posted by Charge
According to Finnish weapon training book from the '50s the WW2 experiences showed that a 50mm gun is just adequate to take down a large bomber on one shot. The guns were however cumbersome and had a difficult recoil, so rockets were considered a better choice.
I don't know if you are talking about the book "Lentoampumaoppi"?
If so please take a look to the page 94, and check the values for say MG 213C/30, MK 214A/50 and RAM 55. First the destruction factor (tuhokerroin) against the heavy bombers as defined by the Germans (mainly the HE content of a projectile):
MG 213C/30 => 6,67
MK 214A/50 => 1,22
RAM 55 => 0,753
The rocket has the best destruction factor of these and the 50mm canon comes quite close while the MG 213 is far behind. Note that in this case the destruction factor does not tell much about the penetration capability of the projectile because it's estimated using mainly the HE content.
The fire power (Aseteho, Ta) values are following using the German definition:
MG 213C/30 => 16
MK 214A/50 => 3,7
RAM 55 => 115
Again the RAM 55 is the best with very large margin but now the MG 213 is better than the MK 214 mainly due to higher ROF and much lower weight
The Combat value or rating (Taisteluarvo, Tt) turns the table quite completely:
MG 213C/30 => 2,9
MK 214A/50 => 0,1
RAM 55 => 1,47
Now the MG 213 is the best with quite large margin...
Generally one can choose what ever yard stick depending what he/she wants to prove.
gripen
-
Originally posted by Charge
You guys start to obscure the topic by your need to pick on each other.
Too bad. Interesting topic wasted.
-C+
Hi,
there is no need to salamander on him, if someone salamander on him, its himself!
I simply pointed to his tunnel vision(his words), which seems to run seperated from historical facts and to a statement, which seems to agree to the "high ROF over bullet power advantage vs light amoured targets theory".
We can have many ideas what is true or not, we never will know it exact, but to negate historical facts seems to be a bit strange to me.
"Generally one can choose what ever yard stick depending what he/she wants to prove."
This is correct for more unknown datas, but will be someone able to prove that the .50cal was obsolet in WWII, or even underpowered vs the common targets?
If we look to the results in war, we dont need to find a calculation(clarification) that proof that the .50cal was obsolet, we need to find a calculation(clarification) why it wasnt, despite its obvious smaler damagepower per bullet.
In this contex, to insist on "the .50cal was much disadvanced", even vs smal less tough targets, only cause a very simplyfied calculation say so, seems to be a bit strange to me.
So, are there any proofs that the .50cal was insufficient vs WWII fighters??
Are there any proofs that the british Spitfire armament was better than the 6 x .50cal(vs common US targets)?(please no pilot statements, there are many where the pilot like the shotgun effect of the 6 x .50cals, but thats no proof)
For now i dont found any, i only found that different airforces was pretty successfull while fighter combat with big MG´s, specialy if they got used in large numbers.
Greetings, Knegel
-
Tony, what was the gun in MiG-21 , 23mm cross feed with twin barrels? Extremely short and light construction with really good rate of fire. Only drawback I know is that it wears the barrels down very fast.
I have installed and loaded that gun couple of times into MiG-21 and found out it was easy to handle and the system it operated (operating 2 barrels from same feed) was nicely done.
How good was it in combat?
Was it some GSh-23 or?...
Heh, imagine 2 of these , 1 in each wing with about 400 ammo each.. what a firepower!!!
-
The MiG-21F in FAF service had the NR-30 and the MiG-21Bis had the GSh-23L. Some people in FAF rate the GSh-23L in some degree better than the Vulcan in the F/A-18 because the GSh-23L reaches it's peak rate of fire instantly while the Vulcan needs some time to accelerate due to it's Gatling type operation.
Notable thing is that NR-30 was developed from the MG213 just like Aden.
gripen
-
Originally posted by gripen
Notable thing is that NR-30 was developed from the MG213 just like Aden.
Not so, the NR-30 was a recoil-operated gun (like the current GSh-301) which used a gas buffer. It had a single firing chamber, not like the multi-chamber revolver developed for the MG 213C.
Tony Williams: Military gun and ammunition website (http://www.quarry.nildram.co.uk) and discussion forum (http://forums.delphiforums.com/autogun/messages/)
-
Thanks, I stand corrected. Found a picture (http://home.snafu.de/veith/images/NR-30.jpg), no revolver drum.
gripen