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I Rotating bullet vs non rotating bullet

  1. Apr 2, 2017 #1
    I'm writing a program to simulate bullet impact on various materials.
    I need to know, will the angular momentum of a spinning bullet give it more impact force?

    I would assume that if so it would also mean that spinning bullets overcome more air friction if I am supposed
    to add the angular momentum to the linear forward momentum when calculating total forward momentum.

    I have to calculate two things:
    1. Penetration depth of the impact material
    2. How much air friction will slow down the bullet.
    Both will require assumptions about weather or not angular momentum adds to linear momentum when calculating total forward momentum felt by the air or the impact material.
     
  2. jcsd
  3. Apr 2, 2017 #2

    phyzguy

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    No, the angular momentum of the bullet does not add to the linear momentum. The reason bullets are given a spin is to keep them from tumbling. If they tumble the air resistance is significantly increased.
     
  4. Apr 2, 2017 #3
    So the spin adds ZERO EXTRA PENETRATING POWER?
     
  5. Apr 2, 2017 #4
    Is it because the torque on the bullet acts perpendicular to its linear motion?
     
  6. Apr 2, 2017 #5

    Dale

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    No, it is because linear and angular momentum just don't add. They have different units.
     
  7. Apr 2, 2017 #6

    phyzguy

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    To answer this, you'd have to tell me how you plan to calculate the penetration depth. If the bullet is tumbling, the penetration depth will be different from what it will be if it is not tumbling, because the cross-sectional area of the bullet will be different. However, the difference is not due to adding the angular momentum to the linear momentum. Can you outline how you plan to calculate the penetration depth?
     
  8. Apr 2, 2017 #7
    I'm going to use resources below, Impact Depth by Isaac Newton and material strength. I will also simulate gravity so the ballistics of the bullet may not always strike the simulated wall like material head on, it might hit on a deflection angle.

    I will calculate if the force of the bullet is enough to shatter the target, if not it will just deform it. Regardless it will make the same depth based on Isaac Newton's theory. The only difference is the 3D picture will look more like the surface was broken(shattered) if material strength was exceeded vs merely deformed.

    Bullet might still be in the hole if there was only deformation, with shattering there is a chance the bullet penetrated the material completely and exited depending on thickness.

    https://en.wikipedia.org/wiki/Impact_depth
    https://en.wikipedia.org/wiki/Ultimate_tensile_strength
    https://en.wikipedia.org/wiki/Specific_strength
    https://en.wikipedia.org/wiki/Strength_of_materials
     
    Last edited: Apr 2, 2017
  9. Apr 2, 2017 #8

    phyzguy

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    OK, Newton's approximation is a reasonable method, although I suspect it will only be accurate to a factor of two or so. So you see that if the bullet impacts "head-on", as in the Wikipedia picture, the penetration depth will be different than if it impacts "sideways" because it is tumbling, since the cross-sectional area A is different in the two cases. But again, this difference is not due to adding the angular momentum to the linear momentum.
     
  10. Apr 2, 2017 #9
    What do you mean by factor of 2?

    I can calculate the sideways force component of the ballistic trajectory so only the relevant force will be used to calculate the depth, the other component of the impact velocity can be used to calculate deflection path if any energy is left after impact.
     
  11. Apr 2, 2017 #10
    Since there is a certain amount of energy contained in the rotation and the bullet will come to an entire (non-rotational) stop, that energy will be imparted to the body as well. Whether that energy results in further penetration is however a different question. It may just as well end up shredding more tissue on the way.

    But, as discussed above, it's an academic question for the most part. Other than civil war era rifles that shoot lead pellets, you won't find a gun that doesn't impart rotation to the bullet through the boring of the rifle.
     
  12. Apr 2, 2017 #11

    phyzguy

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    I just mean that the assumption that momentum is only transferred directly ahead of the bullet and there is no interaction between the cylinder directly ahead of the bullet and the material to the sides of this cylinder sounds like a gross approximation to me. For example, this method says that the penetration depth is independent of the speed of the bullet. Does this sound reasonable to you? Do measured results agree with this?
     
  13. Apr 2, 2017 #12
    I dont like this Newton theory. Doesnt seem realistic enough for a commercial real time simulator.

    What if I used the Drag Formula and input drag coefficient of the bullet and input the material density (target) as a fluid density?
    I could then deduce how far the bullet can penetrate by distance travelled before velocity becomes zero and work out the heat produced by friction force produced in total and use the heat as a variable to determine if the material melted or not during the penetration.

    Only thing ill have trouble with left is the deflection of the bullet if it hits the target on an angle. Any ideas how I can solve this bit?
     
  14. Apr 2, 2017 #13

    phyzguy

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    I think these things are really complicated, and you are best building some sort of semi-empirical model. Here's a page where they try to measure and model the penetration depth of bullets.
     
  15. Apr 3, 2017 #14
    Everybody knows that it much easier to pull out a nail if you pull it and rotate simultaneously than if you only pull it
     
  16. Apr 3, 2017 #15
    Ok I have a strategy , please critique:

    I will use drag coeficient formula substituting material density for air pressure.

    Also if the force of the bullet is below the tensile strength of the material the bullet bounces off it according to deflection angles.

    Im guessing my "substituting" needs a little more work than just that?
     
  17. Apr 3, 2017 #16

    A.T.

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    I guess because it's easier to brake the initial static friction using mechanical advantage (tool grip radius vs. nail radius). But how is that relevant to a bullet?
     
  18. Apr 3, 2017 #17
    not only this
    directly.The angular momentum expense helps the bullet move inside а bulletproof vest for example. The mechanism is the same: the properties of the Coulomb friction
     
  19. Apr 3, 2017 #18

    jbriggs444

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    As I would word this, the force of kinetic (or static, for that matter) friction has a fixed [maximum] magnitude. If you change the direction of that friction, e.g. by imparting a spin then the component of friction that is aligned with the bullet's trajectory is reduced.

    This is similar in principle to doing doughnuts on a car on icy roadways -- if you rev the engine and run the tires at a high rate of speed, the forward traction of the tires does not increase, but their resistance to slow lateral motion becomes almost non-existent.

    In pulling a nail, e.g. with a pair of pliers, you will find yourself twisting the nail first one way, then the other, but always pulling. The motion of nail against the wood will be mostly in the twisting direction because that's the direction where the most force is being exerted. But that little bit of pulling tension means that the net motion has a shallow spiral component -- easing a tiny bit out of the hole. Worry the nail back and forth long enough and it'll eventually come free.

    If the force of friction were directly proportional to velocity rather than roughly constant then no advantage could be obtained in this fashion. The force of friction in any one direction would be independent of motion in any other direction.

    If the force of friction were more than directly proportional to velocity (e.g. quadratic rather than linear) then a disadvantage could apply. A spinning bullet could experience more resistance to motion than a non-spinning bullet.
     
  20. Apr 3, 2017 #19

    phyzguy

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    Note that nobody in this thread has said that the spin of the bullet doesn't affect the penetration depth. It seems reasonable to assume that it might, although I'd like to see some measurements. However, any impact of the spin on the penetration depth is not caused by adding the angular momentum of the bullet to its linear momentum, as was the original question.
     
  21. Apr 3, 2017 #20

    A.T.

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    Makes sense, thanks.
     
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