.50 cal machine guns vs 20 mm autocannons on US aircraft

[davparlr]

Gun power was the Chemical energy of the projectile x the rof

I recognised that your calculation showed more hits by the 0.5 something I do not disagree with. All I did in posting 81 was multiply the He component of each shell by the number of hits that you calculated. This resulted in a significant advantage to the 20mm, something you haven't actually disagreed with

I cannot disagree with what you say. However, Mr. Williams takes chemical energy into account for his "damage" number, and I am sure the Navy did the same. This is what I based my assertions on.

 

[Vincenzo]

Mr Williams use a strange method per chemical energy, so 2 rounds with same HE filler but one with double of kinetic energy this get also the double of chemical energy

 

[Glider]

Mr Williams use a strange method per chemical energy, so 2 rounds with same HE filler but one with double of kinetic energy this get also the double of chemical energy

Can I ask which examples you are looking at?

 

[Glider]

I cannot disagree with what you say. However, Mr. Williams takes chemical energy into account for his "damage" number, and I am sure the Navy did the same. This is what I based my assertions on.

SO you agree that 0ne 20mm Hispano II is worth three 0.5 M2, which means that 4 x 20mm equals 12 x 0.5.

It therefore follows that the F86 with 6 x M3 (with approx 50% greater ROF) is equal to approx 9 x M2 0.5 which is still less than the 4 x 20mm Hispano II which we agree equalled 12 x 0.5 M2.

However the Hispano V used in the Panther, had a ROF 25% greater than the Hispano II which means that 4 x Hispano V is the equivalent of 15 x 0.5 M2 (12 x 125%) which is still a hell of a lot more than than 6 x 0.5 M3.

At the end of the day, however you slice it, the F86 was heavily outgunned by the Panther. The difference may change due to the way you want to approach the problem, but the Panther always has a clear advantage.

 

[Vincenzo]

none examples him explanation of calculation method

 

[davparlr]

SO you agree that 0ne 20mm Hispano II is worth three 0.5 M2, which means that 4 x 20mm equals 12 x 0.5.

No. Lets assume the damage of one 20 mm is equivalent to four 50s ala Williams (three estimated for the Navy), so two 20mm rounds are equivalent to eight 50cals rounds. If we have a burst time of two seconds at a 5% strike probability per projectile, the four 20s, at 50 r/sec, will land two rounds with a 96% probability in two seconds. The twelve M2 guns at 160 r/sec will land an equivalent 8 rounds in only 1.7 sec with a 96% probability. The number of M2 machine guns required to land eight rounds in two seconds, is about ten (10.2 actually). So, in an actual engagement with these conditions one 20 mm cannon is worth two and a half M2s. If strike probability goes up to 10% per projectile (very close target in WW2), one 20mm cannon is equivalent to just two M2 machine guns, where four 50 cal rounds are equivalent to one 20 mm round.

It therefore follows that the F86 with 6 x M3 (with approx 50% greater ROF) is equal to approx 9 x M2 0.5 which is still less than the 4 x 20mm Hispano II which we agree equalled 12 x 0.5 M2.

Again, no. You have to calculate the probabilities of round strikes verses time fired to determine equivalence, as I have done previously.

However the Hispano V used in the Panther, had a ROF 25% greater than the Hispano II

I am not sure of this, wikipedia shows a difference of 700 r/m for the II to 750 r/m for the V, not 25%.

which means that 4 x Hispano V is the equivalent of 15 x 0.5 M2 (12 x 125%) which is still a hell of a lot more than than 6 x 0.5 M3.

Not in real situations like I just described.

At the end of the day, however you slice it, the F86 was heavily outgunned by the Panther. The difference may change due to the way you want to approach the problem, but the Panther always has a clear advantage.

I would really like to look at it from the pilot's stand point when holding the trigger down for two or three seconds while the pipper is on the target, which is all the control the pilot has, and depending on how many hits there will be and what is the damage of those hits.

The real unknown in all of this discussion is the probability of kill given a hit, an highly complex calculation. It seems to be estimated at between three or four to one in favor of the 20mm.

Given all these assumptions of damage and equivalence, it appears to me that the F-86 and F9F are reasonably equal in probability of kill verses trigger time.

 

[Vincenzo]

i think that navy equivalent 3 50 for 1 20 is for guns, for round is much higher

taking data from Williams page
the kinetic energy, at muzzle, of a round of Hispano II is ~2,8 times that a .50 Browning, there are no reason for think that 20mm round loss speed fastest of .50", and generally heaviest round loss speedl slowly (a 20mm round weight 3 times that of a .50")
the chemical content is ~12,1 times

 

[tomo pauk]

I do not see that USN (1 x 20 = 3 x HMG) and Williams (1 x 20 = 4 x HMG) are in contrast.
USN regarded it's comparison for a weapon set-up as a whole (strike probability included), while Williams analyzed just a part of the weapon system ( the terminal effect). So I'd reckon that in 'static' firing one second burst of the Mk2 Hispano is equal to one-second burst from 4 M2 BMGs, while in the 'dynamic' firing the odds are thinned to 1=3, still in cannon's favor.

The Hispano Mk2 was able to make 600-700 rpm, the Mk5 700-800?
If we take 750 rps for Mk5, we need 3000 rounds for 4 cannons for one minute of 'trigger time' (7 kg/sec*), and 7200-7500 rounds for 6 HMGs (M3 from Sabre; 5,16-5,375 kg/sec). No great saving weight- or volume vise if we choose HMGs, while great deal of firepower is discarded. And we never talked about all aerial targets (not just the fighters), that a cannon-armed fighter is way better choice to deal with, along with a capability to hit enemy fighters.

*then we add explosive effect, and cannon is a true winner

 

[davparlr]

i think that navy equivalent 3 50 for 1 20 is for guns, for round is much higher

taking data from Williams page
the kinetic energy, at muzzle, of a round of Hispano II is ~2,8 times that a .50 Browning, there are no reason for think that 20mm round loss speed fastest of .50", and generally heaviest round loss speedl slowly (a 20mm round weight 3 times that of a .50")
the chemical content is ~12,1 times

I don't think the aerodynamics is as good on for the 20mm, its taper is shorter and it has a blunt nose. The 50 cal is much cleaner. This could have a big impact at the speeds these projectiles are traveling when they meet the ambient air. At 25k ft, 600 mph aircraft speed, the projectiles are leaving the aircraft at about 3.7 mach relative to the air, about 1.3 mach below hypersonic. Drag difference could be significant. I don't know but others have said the 20 mm slows down faster.

 

[davparlr]

I do not see that USN (1 x 20 = 3 x HMG) and Williams (1 x 20 = 4 x HMG) are in contrast.
USN regarded it's comparison for a weapon set-up as a whole (strike probability included)

Since the Navy does not record what is test definition consisted of, this is hypothetical. Doing an analysis by gun makes no sense since probability analysis shows that projectile strike quantity per time fired is a variable based on rof, accuracy, and time of engagement. Yes the 20 mm gun may be equal to three 50s, but at what variables, ROF is a known, but at what accuracy, what firing time? Changing accuracy, or time of fire, will/may invalidate the conclusion, and accuracy and fire time is a situational variable, how close, how good, etc. The solution is only good at the specified variables. The proper method would be to test individual projectiles to determine velocity deltas and target damage at varying ranges. With this empirical data, mathematical analysis can be performed to determine performance over a range, time, etc. and that will vary. Now maybe the Navy did this, but then they needed to specify at exactly what is the accuracy/range and firing time. If you read my response to Gilder, just comparing gun effectivity gives a much different answer to when you calculate in hit probability. When hit probability is factored in using 4 to 1 projectile effectivity, varying accuracy give a range of gun to gun ratio from 2.5 M2 to each 20mm cannon to 2 M2 to each 20mm cannon. Quite a bit of difference, especially from the original one to three Gilder used.

, while Williams analyzed just a part of the weapon system ( the terminal effect). So I'd reckon that in 'static' firing one second burst of the Mk2 Hispano is equal to one-second burst from 4 M2 BMGs, while in the 'dynamic' firing the odds are thinned to 1=3, still in cannon's favor.

Actually, I think he did it by projectile damage using both chemical and kinetic energy.

The Hispano Mk2 was able to make 600-700 rpm, the Mk5 700-800?
If we take 750 rps for Mk5, we need 3000 rounds for 4 cannons for one minute of 'trigger time' (7 kg/sec*), and 7200-7500 rounds for 6 HMGs (M3 from Sabre; 5,16-5,375 kg/sec). No great saving weight- or volume vise if we choose HMGs, while great deal of firepower is discarded. And we never talked about all aerial targets (not just the fighters), that a cannon-armed fighter is way better choice to deal with, along with a capability to hit enemy fighters.

*then we add explosive effect, and cannon is a true winner

I believe explosive effect is included in the data I used in calculations.

Tomo, you can probably make this comment if a plane has a 75mm firing 60 r/m. Just looking at throw weight vs. rof does nothing. If that 75mm hits, it will blast the plane to smithereens. Of course it must hit. Without calculating in the probability of hits vs. impact damage, one cannot estimate the impact of various armament configurations.

 

[Glider]

I will prepare a more detailed reply tomorrow as its late here but in brief.

I certainly agree that the 6 x 0.5 will land more hits than 4 x 20mm Hispano V. However when you factor in the HE content of those shells that hit, be it 1 to 4 or 1 to 2.5, the damage will be greater for the 20mm hits causing more damage as it has 13 times the HE per shell than the 0.5.

I also agree with the statement in the fighter conference was a vague general statement which is open to some guesswork.

Re the rate of fire of the Hispano II I have seen a copy of the manual issued to the RAF and the ROF was given as 570 to 620 per min so 600 rpm is a good working number. I don't know where Wiki got their information but if I find it I will add a link.

 

[Vincenzo]

somewhere there are data on loss speed of .50" browning and some germans round and in the pratical distance of fight and if i rememebr right there was no much difference from .50" browning and the 13 and 20 (mg 151) germans

 

[fastmongrel]

As most combat is taking place at relatively short ranges does the loss of velocity of the 20mm make much difference. Most stuff I have read seems to say that firing at much more than 300 meters was just a waste of ammo.

 

[Vincenzo]

As most combat is taking place at relatively short ranges does the loss of velocity of the 20mm make much difference. Most stuff I have read seems to say that firing at much more than 300 meters was just a waste of ammo.

the point is that the difference of loss velocity of the 20mm is not different to .50" in significant measure, some posted data time ago, unlucky i don0t remember if was in this forum or in other

 

[davparlr]

There may be some velocity differences, but I don't think my argument identified any, only the possibility.

 

[renrich]

The external ballistics of a projectile determine how fast the projectile is traveling when it reaches the target and how soon and how flat the trajectory is.
External ballistics are determined by three factors: Ballistic coeficient, sectional density and form factor and these factors are interrelated.
I have ballistic tables for rifle and pistol calibers but do not have them for 50 BMGs or 20 MMs so will quote from memory of earlier posts I made and from some info I found online.
Simply put a projectile which is long and slim will have a better ballistic coefficient (BC) and better sectional density (SD) than a short fat projectile.
A high BC means the projectile will maintain it's veloecity better and a high SD means the projectile will have more penetration of the target.
The form factor will heavily impact BC and thus the penetrative powers.
Form factor (FF) is determined by the shape of the projectile which means, obviously, a sharp pointed projectile has a better FF than a blunt nosed projectile. The shape of the projectile at the rear can be important also but the nose shape is the most important,
Therefore a long skinny sharp nosed projectile is going to have the best external ballistics, (if it can be stabilised but that is another story.)
The following is from a table found online. I cannot vouch for the exactness of the data but I do know that it is generally correct as to the characteristics of the two projectiles.

20 MM Hispano HE
Muzzle Velocity- 860 M/S
Velocity @ 100Ms- 804 M/S
Velocity @ 500 Ms-606 M/S

50 BMG AP
MV- 863 M/S
100M-841M/S
500M-767 M/S

As you can see the two projectiles start out at about the same speed but the 50 BMG because of better external ballistics retains it's velocity better and will therefore arrive at the target faster, will have a flatter trajectory and will penetrate better when it hits.

Another factor which plays a role in the comparison is the fuse of the 20MM. If the fuse does not work there is no explosion. If the fuse works prematurely the explosion takes place on the surface without penetration. If the fuse works too slowly, for example if the projectile hits a thin part of the wing, the explosion takes place outside in the air.

 

[tomo pauk]

Hi, David,

Since the Navy does not record what is test definition consisted of, this is hypothetical. Doing an analysis by gun makes no sense since probability analysis shows that projectile strike quantity per time fired is a variable based on rof, accuracy, and time of engagement.

We can shelve the USN 'statement' of one cannon being equal to 3 HMGs into an 'urban myth' folder, and I'll be just fine with that. The fact is that USN went for all-cannon fighters pretty soon after WW2, so that equation may hold the truth.

Yes the 20 mm gun may be equal to three 50s, but at what variables, ROF is a known, but at what accuracy, what firing time? Changing accuracy, or time of fire, will/may invalidate the conclusion, and accuracy and fire time is a situational variable, how close, how good, etc. The solution is only good at the specified variables. The proper method would be to test individual projectiles to determine velocity deltas and target damage at varying ranges. With this empirical data, mathematical analysis can be performed to determine performance over a range, time, etc. and that will vary.

A well made statement, no doubt. The issue is indeed that we know some variables, but don't know other ones that, together with previoulsy known variables can point us into the right answer.

Now maybe the Navy did this, but then they needed to specify at exactly what is the accuracy/range and firing time. If you read my response to Gilder, just comparing gun effectivity gives a much different answer to when you calculate in hit probability. When hit probability is factored in using 4 to 1 projectile effectivity, varying accuracy give a range of gun to gun ratio from 2.5 M2 to each 20mm cannon to 2 M2 to each 20mm cannon. Quite a bit of difference, especially from the original one to three Gilder used.

Think we need to factor in that a KE projectile (=.50 cal bullet) shares same issues with any other KE projectile - it looses a target efficiency at much greater scale when traveling in high speed at some distance, compared with HE shells.

Actually, I think he did it by projectile damage using both chemical and kinetic energy.

Agreed.

I believe explosive effect is included in the data I used in calculations.

If you could remind me where to read that?

Tomo, you can probably make this comment if a plane has a 75mm firing 60 r/m. Just looking at throw weight vs. rof does nothing. If that 75mm hits, it will blast the plane to smithereens. Of course it must hit. Without calculating in the probability of hits vs. impact damage, one cannot estimate the impact of various armament configurations.

I won't go to the extreme claiming that a really huge gun is the best choice for fighter
The cannons between 20-30 mm were favored by most of the air-forces as early as WW2, with USN joining in in late 1940s. Such weapons were able to provide a 'regular' fighter with armament to tackle both other fighters (one need good RoF along the weight of fire), twin-engined and multi-engined planes (where weight of fire/target effect is the preferable property of a weapon).
IMO the USAF admitted the .50 cals are lacking the punch by installing rocket trays for Mighty Mouse rockets on their Sabres (=Sabredog):

F-86D Sabre Dog by Aleksandar Sekularac (Revell 1/48)

 

[Vincenzo]

Renrich good table we can know what is the site it came?
probably hispano round has badest balistic of that of mg 151
the table take in the count the 12,7 AP not the API so w/o any chemical energy

 

[renrich]

Vincenzo, I googled 20MM ballistics and found the site but do not remember the name. It also had the German guns and the Oerlikon. The german cannon did not equal the exterior ballistics of the 50 BMG but were slightly better than the Hispano. I used the Hispano because the discussion is about US fighters and that would have been the cannon used.

Long ago and far away I used to do a lot of hand loading for rifles and handguns. I really got into it and began to have a little understanding of ballistics and realised that many of my ideas were false.

Here is an example. I owned a 3006 and a 270 Win. The 3006 bullet is .308 in diameter and the 270 is .277. Now everyone knows that the 270 is not as good for large game as the 3006 because the most popular load for the 270 is a 130 grain bullet and the most popular load for the 3006 is the 150 grain bullet. Bigger bullet, better for a big animal. Right? Not necessarily!
One can go into all sorts of argument about MV and shocking power but here is what I found out. Take a 308 pointed nose bullet in 150 grains and a 277 pointed nose bullet of 150 grains and both have a MV of say 2800 FPS. They will have the same energy at the muzzle but as soon as they leave the muzzle the 277 bullet starts outrunning the 308 bullet. Why? Because the 277 bullet, in order to weigh the same as the 308 has to be longer and therefore it has a better SD and a better BC. So at 300 yards, for instance, the 277 bullet is going faster, is flatter shooting, has more energy and greater penetration. That is one reason I wish that the US Army had adopted the 277 bullet instead of the 3006 back during the Depression when that was under consideration.

Looking at cannon versus MG rounds, the 20 mm is quite a bit bigger in diameter than the 50 bmg. The 20 MM could be made to have a better BC than the 50 BMG but it would have to get a lot longer and thus there is a weight penalty which is bad for the airplane and also slows down the projectile. The MV can be raised but that means a heavier gun and that is bad and the recoil gets worse which is also bad for the airplane. The fuse in the nose may cause the projectile to be more blunt also.

 

[Vincenzo]

i found this Weapons and Ammo
their source web site is off (almost is off for me, this days i got many errors with many sites)