Author Topic: Aircraft gun article  (Read 7846 times)

Offline Knegel

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Aircraft gun article
« Reply #45 on: March 26, 2006, 12:58:44 AM »
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

Offline HoHun

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« Reply #46 on: March 26, 2006, 02:27:34 AM »
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)

Offline Knegel

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« Reply #47 on: March 26, 2006, 03:22:44 AM »
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
« Last Edit: March 26, 2006, 03:42:08 AM by Knegel »

Offline Knegel

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« Reply #48 on: March 26, 2006, 04:22:14 AM »
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

Offline HoHun

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« Reply #49 on: March 26, 2006, 04:24:22 AM »
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)

Offline HoHun

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« Reply #50 on: March 26, 2006, 04:35:40 AM »
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)

Offline HoHun

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« Reply #51 on: March 26, 2006, 04:44:04 AM »
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)

Offline Knegel

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« Reply #52 on: March 26, 2006, 08:37:44 AM »
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
« Last Edit: March 26, 2006, 08:52:56 AM by Knegel »

Offline HoHun

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« Reply #53 on: March 26, 2006, 12:18:43 PM »
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)

Offline Karnak

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« Reply #54 on: March 26, 2006, 02:03:44 PM »
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.
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Offline Knegel

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« Reply #55 on: March 27, 2006, 03:35:43 AM »
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
« Last Edit: March 27, 2006, 03:43:57 AM by Knegel »

Offline straffo

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« Reply #56 on: March 27, 2006, 04:25:39 AM »
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"

Offline HoHun

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« Reply #57 on: March 27, 2006, 06:41:25 AM »
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)

Offline Knegel

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« Reply #58 on: March 27, 2006, 07:17:15 AM »
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
« Last Edit: March 27, 2006, 07:20:27 AM by Knegel »

Offline straffo

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« Reply #59 on: March 27, 2006, 07:50:59 AM »
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 :
Quote

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 :)