Aircraft gun article

[HoHun]

Hi BlauK,

How about going back and reading the entire thread before you ask questions that have been answered before?

Regards,

Henning (HoHun)

[BlauK]

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

[HoHun]

Hi BlauK,

>I eyed it pretty carelessly

That much was pretty obvious.

Regards,

Henning (HoHun)

[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"  

Just FYI... there is no one and only formula for "hit probability". It always depends on the given conditions at each time.

[Karnak]

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.

[BlauK]

Karnak, point taken

While the hit probability with 6x0.50 cals is higher, delivered damage probability of 4 hispanos is likely higher.

[HoHun]

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)

[straffo]

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

[Knegel]

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.

[Knegel]

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

[BlauK]

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)

[Karnak]

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.

[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 and discussion forum

[HoHun]

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)

[gripen]

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"  

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