The Gurney Equation

Discussion in 'Aviation' started by Garyt, Apr 12, 2014.

  1. Garyt

    Garyt Member

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    If anyone is familiar with the Gurney Equation for frag effects of shells, please let me know. I seem to be doing something wrong trying to work out some calculations.

    If I get it to work properly, I'm expecting a 30mm MK108 shell to deliver about 4-5 times the energy of a German aircraft 20mm, like the MK151 20mm
     
  2. davebender

    davebender Well-Known Member

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    Why don't you simply compare amount of HE filler?
    18.6 grams. MG151/20 mine shell.
    22 grams. 2cm Flak38 HEI.
    72 to 85 grams. WWII era German 3cm Mk101, Mk103 and Mk108 mine shells.
     
  3. Garyt

    Garyt Member

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    It's not quite that simple. The Gurney Equation calculates the velocity of the fragments of the casing and converts to kinetic energy in kilojoules. In the case of what the call "brittle metal" casing (I'm not sure what they mean by brittle), about 80% of the velocity is calculated, as some of the energy is spent breaking the case. The explosive also creates energy of 4612 kilojoules per kilogram of TNT.

    This is different than the energy from more or less solid shot like a .30 cal or .50 cal, which are still in kilojoules but are based on their velocity when they strike the target and the mass of the bullet.

    Problem with doing it merely by the HE load - it does not take into account the casing. so the German mine shell would get to high of a rating. For a comparative example, a 1000 pound GP bomb would get the damage rating of 4-5 18.1" HE shells from the Yamato.

    But the problems I see with the equation - Fragmentation will spend a good part of it's energy not striking the plane, or another way of saying this is that there would be overpenetration, such as when a battleship round strike a destroyer and passes clean through.

    Some applications I have seen will take the square or cube root of the joules produced - I think more often the cube, I believe there is a formula that shows something to the effect of the radius of the explosion being equal to the cube root of energy expended.
     
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  4. davebender

    davebender Well-Known Member

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    Is that an appropriate way to compute damage for a small shell such as 20mm which is likely to explode after penetrating aircraft skin? Blast inside an aircraft wing will rip a hole from the inside out. More HE = bigger hole ripped in aircraft skin plus possible damage to internal components such as fuel tanks. Most fragments will be pieces ripped off the target.
     
  5. Garyt

    Garyt Member

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    That is indeed an accurate way to model damage. Think about it, fragmentation is very similar to solid shot from say a .303 or .50. These fragments damage the internal components such as fuel tanks as well. The Gurney equation factors these fragments and their velocity based both upon the explosive charge and the weight of the casing. And more explosives result in higher velocity fragmentation. I might add, the general velocity of these fragments is a fair amount than the shell's initial muzzle velocity.

    But you indeed also need to calculate the energy in joules released by the explosion.

    The real question IMO is how to assign these values. Solid shot will do it's full damage other than if overpenetration occurs. The amount of energy left in a shell once a round fully penetrates it's target is useless. This is why hollow point does more damage than armor piercing ammo when striking a human. The armor piercing goes right thru with energy left over- the hollow point will usually not fully penetrate. So the modest .303 round will be more efficient in imparting damage to a plane than a .50 - but the .50 has many times the energy to impart, so it still does a fair amount more damage.

    Now with a fragment vs. a solid round - I think the fragment should only get a portion of it's KE applied. The fragments will go off in various directions. With their higher velocity, more will be lost to overpenetration.

    The KE from the explosion should not get it's full value either, but more than fragments. Blast energy seems to be more confined, it seems to want to fill the container it is in prior to expanding outside the container - not so with fragments.

    Now even with all this, a .303 bullet striking a pilot in the skull has almost as good of chance of disabling a plane as a .50 tht does the same thing, so some method of accounting for this helps, again perhaps using the square root of energy imparted.

    Just thought I'd also mention - a larger or heavier armored plane will contain energy from rounds, fragments or an explosion better than a smaller lighter plane - which is one reason why heavier rounds are more effective against heavier planes.
     
  6. davebender

    davebender Well-Known Member

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    Don't think so. 50cal creates a larger hole and shock from impact makes the hole larger still. Airframes don't have all that much empty space so there's a significant chance the round will impart it's energy on fuel tank, landing gear, weapons, ammunition etc. Of course explosive shells are better still as even a 20mm projectile is likely to blow a foot size hole in aircraft skin plus blast damage and fragments ripped from target aircraft.
     
  7. Garyt

    Garyt Member

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    #7 Garyt, Apr 14, 2014
    Last edited: Apr 14, 2014
    I think you misunderstood what I was saying. I said the .30 round was more efficient in turning KE into damage, I did not say it did more damage.

    a .50 round may have 10x the KE of a .30 round, but only transmits 7x the damage Thereby the .50 round transmits more damage, but the .30 round does it more efficiently.

    The "fragments" are not parts of the plane that are ripped off - the fragments are the pieces of shell casing that go flying when the plane is struck.
     
  8. Garyt

    Garyt Member

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    Another important part of the equation is calculating waht the dud rates were on shells. And a dud as well as a partial burst (forgot the term for this) would both be helpful.

    Anyone have any resource in regards to the percentages of dud rounds are various WW2 cannon?
     
  9. pbehn

    pbehn Well-Known Member

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    I know nothing about ballistics but in radiography the calculations for allowable doses for a radioactive particle in the body are completely different from a radioactive source outside the body. In an explosion within an aircraft the structure must absorb the explosion, an explosion outside is attenuated by distance, while an explosion on the surface is in between (I would assume).
     
  10. Garyt

    Garyt Member

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    Unless you are looking at rounds from cannon roughly 3" or more (75mm), they are going to be contact fused, which means they will not be exploding outside the aircraft, that is unless the shell penetrates through the entire craft, which should be very unusual.

    From what I know there was not an autocannon capable of taking a proximity fuse in WW2, due to the shell being too small. I also don't think timed fuses were used on 40mm or smaller autocannon.

    So for aircraft cannon or small caliber anti aircraft, whether it explodes inside or outside the craft is irrelevant, the only question is IF they explode, i.e. duds.

    Now when we get to the larger AA such as 88mm, or dual purpose Naval weapons like the US 5"/38, this is a completely different bird of course where distance indeed plays a role. And the destructive effects radius of merely the shell KE are much smaller than the destructive radius of the fragmentation.
     
  11. pbehn

    pbehn Well-Known Member

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    Like I said I know a nothing, but with the speed of a shell, if it explodes on "contact" surely the explosion by the time it progressed would be inside not on the surface. I think I should shut up now cos as stated I know nothing about ballistics
     
  12. Garyt

    Garyt Member

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    Correct. I don't know the specifics on the delays of fusing for 20mm-30mm autocannon off hand, but it would be a short enough of a delay to ensure it blows up inside the plane. Of course, not centered where it would be maxed on damage, but somewhere inside the plane. That is at least if the fuse functions properly. I think mid-late war German fuses may not functions as well as early war fuses, due to slave labor.

    I also think one of the reason you will not see small caliber cannon like the 20mm given a time delay fuse as opposed to contact - the time fuse is going to be very inaccurate, at least in terms of exploding in a very close proximity to the target, less accurate say than a proximity fuse. And unlike a 5" round, the 20mm is not going to have much of a burst radius making it rather ineffective unless contact fuse.

    A proximity fuse would get you a lot more closer on target when it explodes - but the smallest one could get a proximity fuse in WW2 was to fit a 75mm round. My understanding is the US had a 76mm autocannon that could take proximity fused shells - but it was not ready to go til after the war ended.

    Your fine. You bring up some good questions.
     
  13. pbehn

    pbehn Well-Known Member

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    proximity fuses were used quite extensively during ww2 but at the moment I cant see the calibre involved
     
  14. Garyt

    Garyt Member

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    Not sure what you mean here. That the caliber did not have a bearing on whether or not proximity fuses were used?

    The proximity fuse was miniaturized radar, and the problem is making it small enough to fit in a shell. The Allies had this, the Germans did not have the fuse til late in the war, I know they were planning on using in anti-air rockets, not sure if they ever used it in their own AA artillery. I think this would have made the B-17 missions far more dangerous. But, the smallest shell the Allies could fit a working model in during the war was about 75mm. This was probably just as revolutionary of a development for ground to ground artillery resulting in airbursts, though this rarely gets the attention that the anti-air proximity fuses get.
     
  15. Koopernic

    Koopernic Active Member

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    #15 Koopernic, Apr 15, 2014
    Last edited: Apr 15, 2014
    The HE MK 108 shell, and I believe the MK 151/20, were developed to be "mine shells" in other words they were to rely on explosive effects as much as possible rather than penetration and shrapnel.

    To do this they developed processes of deep drawing (sort of stamping out of sheet metal) the shell. There would be big advantages not only in production as well as saving of metal. The latter MK 108 shell is sometimes stated as having a hydrostatic fuze designed to trigger in a fuel tank, so it must have had some penetrative capability. Another advantage of mine shells is that when they self destruct (which was after about 5.5 seconds for a German ground based C28 20mm) you don't rain down as many splinters on your own population or troops.

    Technically 'shrapnel' is a specific type of fragment. The naval folks always refer to them as 'splinters' even if they are big enough to tear through Armour on a battleship.
     
  16. Garyt

    Garyt Member

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    I'm familiar with these shells. They were indeed designed more to do damage by explosion, but they would have fragmentation damage as well. And their fragments would be propelled at a higher velocity than a standard 20 or 30mm round, thereby making them potentially more damaging. Any shell does damage by both fragmentation and blast Kinetic energy (4612 Kilojoules per Kilogream of TNT). I would think that a HE round would still do a small portion of their damage the way ball ammo does - after all it imparts some of it's energy penetrating the plane.
     
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