wuzak
Captain
Isn't penetration partly a function of momentum, not velocity? Or is it energy?
What use is a round that penetrates if it doesn't hit something vital internally? Like the pilot.
.50 cal machine guns vs 20 mm autocannons on US aircraft
- Thread starter DogFather
- Start date
Ad: This forum contains affiliate links to products on Amazon and eBay. More information in Terms and rules
Isn't penetration partly a function of momentum, not velocity? Or is it energy?
What use is a round that penetrates if it doesn't hit something vital internally? Like the pilot.
wuzak
Captain
From that website:
12.7mm Browning MG (.50 calibre)
Code:
AP
Dimensions 12.7x99mm
Ammo Mass 43.3g
Explosive Mass 0g
Muzzle Velocity 863m/s
100m Velocity 841m/s
500m Velocity 767m/s
100m Penetration 15mm
500m Penetration 13mm20mm Hispano-Suiza Mk II Cannon
Code:
AP HE
Dimensions 20x110mm 20x110mm
Ammo Mass 130g 130g
Explosive Mass 0g ~6g
Muzzle Velocity 834m/s 860m/s
100m Velocity 777m/s 804m/s
500m Velocity 575m/s 606m/s
100m Penetration 10mm 11mm
500m Penetration 7mm 7mmAnd for comparison
20mm Mauser MG 151/20 Cannon
Code:
AP HE HE
Tracer Mine
Dimensions 20x82mm 20x82mm 20x82mm
Ammo Mass 134g 134g 104g
Explosive Mass 0g ~6g ~18g
Muzzle Velocity 836m/s 701m/s 792m/s
100m Velocity 792m/s 623m/s 711m/s
500m Velocity 633m/s 377m/s 439m/s
100m Penetration 20mm 10mm 11mm
500m Penetration 14mm 5mm 5mm
davparlr
Senior Master Sergeant
I don't think you can put this into an urban myth. The comment is documented officially and the presenter was the ordinance chief of the Navy. He didn't present all of the data necessary to validate his comment, but it doesn't mean he didn't have it, it just remains a question.
I agree with this completely.
In my earlier post I assumed the Navy data included explosive effect and Williams site certainly claimed to.
When I got renrich velocity numbers, I did some quick analysis of Navy data and Williams estimate. It looks like the Navy mistook momentum for being energy. Williams calculations seemed to fairly good estimates.
I think you will find that all those cannons have a high rate of fire making such a selection a moot point.
I don't know about this. The F-89A was equipped with six 20 mm cannons yet the F-89D removed the twenties for rockets, so I don't think the adding of rockets on the F-86 implies the fifties were a problem, rather they thought the rockets were the real anti-bomber weapons.
davparlr
Senior Master Sergeant
These are significant drop offs for the 20 mm. My calculation of energy in Joules shows that kinetic energy for the 20 mm at muzzle velocity is 3.2 times the kinetic energy of the 50 cal at muzzle velocity. However, at 500 meters kinetic energy of the 20 mm is only 2.01 times the 50 cal. The Navy indicated this number was 2.5. I think they were calculating momentum, which is the value I came up with when I calculated momentum.
Putting these numbers into Williams calculations I got the following ratios of damage at various ranges for .50 cal vs. 20mm.
Muzzle 4.46 to 1
500 ft 4.3 to 1
1000 ft 4.0 to 1 (est)
1500 ft 3.5 to 1
This falls into the area of what I had been using, although this is still only a certain thumbnail using momentum and not energy. This is still an educated guess.
I think, by making some simple assumptions we can answer the question of power vs rate of fire. Comparing two cases we can see the impacts. Case one, heavy projectile 100% kill with two hits (can't use one, the probabilities go to one), slow rate of fire 50 rounds/sec. Case two, gun has one third probability of kill (need six hits for 100% kill), and three times rate of fire of case one 150 rounds/sec. There are no energies and velocities deltas for all ranges. First I'll fix firing time and vary accuracy. This will be somewhat redundant to analysis to what I have already posted, I am sorry for that but I wanted to zero out all the undefined variables and do, basically a one to one comparison.
Test case 1. 2% accuracy, two second firing time, the probability of the case one gun hitting 2 round out of 100 is 64%. At 64% kill probability it will take case two gun 1.04 seconds, getting 6 six hits. Advantage case one.
Test case 2. 5% accuracy, two second firing time, the probability of case one gun landing 2 round out of 100 is 96%. At a 96% kill probability, it will take case two gun .71 seconds, getting 6 hits. Advantage case two.
Test case 3. 10% accuracy, two second firing time, the probability of case one gun landing 2 round out of 100 is 99.97% (I have to go to these many digits to detect difference. At this accuracy, there is effectively a 100% kill probability). At 99.97 probability of kill, it will take case two .58 seconds, getting 6 hits, a big advantage over case two.
Now I will fix accuracy and vary firing time. I am choosing 5% accuracy.
Test case 1. 5% accuracy, one second firing time. Case one gun will hit the target with 2 rounds out of 50 with a 72% probability. At that probability of kill, Case two will hit the target with 6 rounds in .94 sec compared to 1 sec, a small advantage with case one.
Test case 2. 5% accuracy, two second firing time. Case one gun will hit the target with 2 rounds out of 100 with a 96% probability. At that probability of kill, Case two will hit the target with 6 rounds in 1.5 sec compared to 2 sec, a larger advantage with case two.
Test case 3. 5% accuracy, three second firing time. Case one gun will hit the target with 2 rounds out of 100 with a 99.6% probability. At that probability of kill, Case two will hit the target with 6 rounds in 1.9 sec compared to 3 sec, almost half the time of case one.
This analysis indicates that, mathematically, except for poor accuracy (long ranges?) and short time of firing, and all power/velocities/weight being equal, the lower power, higher firing gun has definite advantages over the slower firing high power gun. Advantage of the faster firing gun improves with accuracy and with firing time. If you were making a decision on these two cases, all things being equal except for rate of fire and projectile power, you would choose the fast firing gun. It allows a kill at less firing time over a wider operating environment, an important capability
The real world is not this clean, and unknowns and uncertainties are rampant. But, using all of this analysis, it still appears to me that the F-86, over varying accuracies and firing, is roughly equivalent in probability of kill versus firing time as the F9F Panther, against another fighter. It certainly is not vastly inferior.
Last edited:
Glider
Captain
I admit this could be me having a blank moment but how do you factor the HE content into your summary?
I can cetainly see that the USN could have been talking about momentum instead of energy. However momentum doesn't include the explosive content of the shell or its better penetration, that is a significant additional factor.
Can I ask how you included this nto the section
Putting these numbers into Williams calculations I got the following ratios of damage at various ranges for .50 cal vs. 20mm.
Muzzle 4.46 to 1
500 ft 4.3 to 1
1000 ft 4.0 to 1 (est)
1500 ft 3.5 to 1
I think we agree that one 20mm shell has a massive advantage in Explosive content, The momentum will obviously change with range but the explosive content is constant. I just dont see how this has been included in the calculation
Other points
Official penetration figures for the 20mm and 0.5 are attached and the manual for the 20mm Hispano II showing the ROF on page 5
I can cetainly see that the USN could have been talking about momentum instead of energy. However momentum doesn't include the explosive content of the shell or its better penetration, that is a significant additional factor.
Can I ask how you included this nto the section
Putting these numbers into Williams calculations I got the following ratios of damage at various ranges for .50 cal vs. 20mm.
Muzzle 4.46 to 1
500 ft 4.3 to 1
1000 ft 4.0 to 1 (est)
1500 ft 3.5 to 1
I think we agree that one 20mm shell has a massive advantage in Explosive content, The momentum will obviously change with range but the explosive content is constant. I just dont see how this has been included in the calculation
Other points
Official penetration figures for the 20mm and 0.5 are attached and the manual for the 20mm Hispano II showing the ROF on page 5
Attachments
davparlr
Senior Master Sergeant
The Anthony Williams site you referenced me to had a formula for calculating projectile "damage" based on velocity, projectile weight and explosive weight. When I put in the velocity data from Renrich for ranges the ratios shown was the results. 1000 ft. velocity was not available so I made a guess. This seems to be close to the Navy assertions.
davparlr
Senior Master Sergeant
Good question. Probability of hits is one of those values that are hard to get to in a combat situation. Some pilots, real good shooters, may have a probability of hit at 50% or more for a single shot. Certainly those willing to fill the screen with the enemy would have a very high probability of hit. Other variables include, range, vibration of the aircraft, usefulness of gunsight, angular velocity of the target, target profile, etc., etc. I picked 2% as a reasonable low value and 10% pretty high in that at 3 second burst at 10%, probability of massive damage, for the two weapons I looked at is near 100%.
Edgar Brooks
Senior Airman
This is a subject that I've, largely, kept clear of, but I fear that you're going to founder under a welter of statistics, which will be mostly irrelevant.
I don't know about the advice given to Americans, but RAF pilots were advised never to open fire at ranges greater than 300 yards (read metres, if you prefer, or 900') so talk of what would happen at 1500' or even 2000' should be immaterial.
The pilot needed to fly precisely, too, so that, in a deflection shot, he kept the same "angle off," ahead of his target, with his turn bank needle centred to avoid sideslip, and no use of the rudder, which induced skidding.
RAF pilots were advised that, up to a range of 250 yards, they could ignore any "gravity drop."
I don't know about the advice given to Americans, but RAF pilots were advised never to open fire at ranges greater than 300 yards (read metres, if you prefer, or 900') so talk of what would happen at 1500' or even 2000' should be immaterial.
The pilot needed to fly precisely, too, so that, in a deflection shot, he kept the same "angle off," ahead of his target, with his turn bank needle centred to avoid sideslip, and no use of the rudder, which induced skidding.
RAF pilots were advised that, up to a range of 250 yards, they could ignore any "gravity drop."
renrich
Chief Master Sergeant
How do you keep the ball centered if you are turning without use of the rudder? The USN taught their pilots in a full deflection run to open fire at about 1000 feet which is close to the 300 yard or meter business.
Edgar Brooks
Senior Airman
You use ailerons and elevators; it's the standard manouevre. Never use rudder alone.
Last edited:
Shortround6
Major General
I would take another look at some of those figures and try applying a little logic. There seem to be a number of errors. Some of the figures don't match other sources for weights and velocities either.
A few basic things about ballistics. Ballistic co-efficient is made up of two things, the form (shape) of the projectile and it's sectional density or weight per unit of frontal area. Putting the two together the .50 does have an advantage but it is not so pronounced as just comparing the shapes would show. The difference between the two rounds shows little practical difference until the range gets to 500-600yd or beyond. As the range gets closer to 1000yds the difference becomes more significant and beyond the .50 shows an ever increasing advantage in velocity retention, time of flight and trajectory. However very little air to air combat took place over 500-600yds.
A few other points are the actual "accuracy" of the guns or the gun/installation. These guns DO NOT fire with laser like precision. They fire into group sizes, zones or areas. As a for instance the British have targeting diagrams showing the area covered by 1/3 minute grouping and 1 minute grouping of the .303 Brownings in the Spitfire. 1 minute is about 5 feet at 100yds and 20 feet at 400yds. There are other diagrams showing the convergence pattern superimposed on a He 111 of the 75% zone and the 100%zone(75% of bullets fired will fall in the circle or 100% of bullets fired will fall in the circle) at 4 ranges with a 350 yd convergence.
Simple math can also show that wing guns may not have been as bad as some people think from a convergent stand point. If guns 12 feet apart are set to converge at 300yds the center of the bullet streams will be 8 feet apart at 100 yds and only 4 feet apart at 200yds, cross at 300yds and be 4 feet apart again at 400yds and 8 feet apart at 500yds. Unless the target in question has a fuselage of 4 feet or less across and is directly end on to the shooter the wing guns will hit any where from about 200 to 400yds from the firing plane. If the target plane is at an off angle to the firer the "hit zone" just gets bigger.
Something else to consider in the .50 vs 20mm debate is that most of the velocity figures are taken at sea level. At 20,000 ft or bit higher where the air is 1/2 as dense both projectiles are going to slow down at roughly 1/2 the rate they do at sea level and so the difference in speed of the projectiles at 500yds ( or any other range) at 20,000ft is going to about 1/2 the difference that it is at sea level. The higher you go the less difference and lower you go the more difference until you hit the sea level figures.
wuzak
Captain
I agree. It didn't seem right to me that the 0.50 would have more penetration than the HS 20mm at all ranges. The energy and momentum advantage was still in the HS's favour, even if you use those numbers.
drgondog
Major
Edgar - you can't carve a turn in a recip airplane without feeding rudder. Pure aileron turn will cause the wing with greater leift (high) wing to yaw in that direction.
The reason a tuft of yarn is palced on training gliders is to illustrate that point to the pilot so that he learns to feed rudder in a co-ordinated way to make the turn without pulling the nose 'off'.
Secondly, many pilots would automatically feed more rudder to get an instananeous yaw (favorable) to increase to temporary lead if the were shooting behind the enemy in the turn.
renrich
Chief Master Sergeant
If you are turning you are already using ailerons and elevator. Because of adverse yaw, in order to keep the ball centered, you have to "step on the ball" use the rudder. I recently got to do a few turns and try to maintain course and altitude in a Stearman. It was a constant dance on the rudders. I don't think a Stearman would make a good gunnery platform. Makes me wonder how the WW1 aviators hit anything.
Edgar Brooks
Senior Airman
Having never flown (and never likely to) a fighter, I can only go by the advice of W/C Donaldson, "Sailor" Malan, and "Zura" Zurakowski, who, in a manual, recommended never to use rudder alone to correct aim.
davparlr
Senior Master Sergeant
I agree. Certainly the armor penetration appears off. I have tried to find ballistics tables for velocities of the .50 and 20 mm, but have been unsuccessful to my satisfaction. If the velocities are near correct, then so are my calculations, if they are not, then neither are my calculations.
Max drag differential should occur at max velocity, but you are right in that time is so short for combat distances. I could not find good velocity data and could not even find something to correlate.
Good point. However, the aircraft velocity must also be added the aerodynamic calculations. At 600 mph, the velocity of the fired projectile into the ambient air is approximately 1.3 times the velocity of a static gun. In the drag formula, density is linear term and velocity is a squared term. At the 20k ft. example, drag verses sea level is half less due to density, however drag due to increased velocity in 1.7 more, so, delta velocity the two projectiles over time at 20k vs. SL is a bit less, but not much. As combat degenerates to sea level, density difference goes to zero, but velocity forces remain at 1.7 times sea level, so at low altitude, delta velocity differences of the two projectiles, as compared to static firing, are magnified by as much as 1.7 times in favor of the .50 cal.
renrich
Chief Master Sergeant
Edgar, I think you misunderstand what we are saying and we misunderstand what you are saying. The three control axies (sp?) are roll, pitch and yaw. The ailerons control roll, the elevators pitch and the rudder yaw. Anyone who has ever flown an airplane knows that trying to turn with rudder alone is not efficient. The plane will turn a little but also skid. If you have ever skiied, you know that carving a turn is efficient and can control your speed. Trying to turn on skis while sideslipping is a recipe for a fall. An airplane is similar. By combining the use of ailerons to bank or roll the airplane with the elevator to pitch the nose up, you have a level turn but because of adverse yaw (aileron drag) like Bill mentioned you must use the rudder to counteract that yaw (opposite to the way you want to turn) and then you have a coordinated turn. I am not sure why but jets don't seem to have the adverse yaw problem as much as prop planes. I once did a few turns in an L39 (jet) and the pilot told me to just keep my feet on the floor. Of course they were gentle turns.
