Rotating bullet vs non rotating bullet

In summary: What do you mean by factor of 2?I can calculate the sideways force component of the ballistic trajectory so only the force applied at a certain point on the trajectory will be used to calculate the penetration depth.
  • #1
darkdave3000
242
5
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
phyzguy said:
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.
So the spin adds ZERO EXTRA PENETRATING POWER?
 
  • #4
phyzguy said:
No, the angular momentum of the bullet does not add to the linear momentum.

Is it because the torque on the bullet acts perpendicular to its linear motion?
 
  • #5
Eclair_de_XII said:
Is it because the torque on the bullet acts perpendicular to its linear motion?
No, it is because linear and angular momentum just don't add. They have different units.
 
  • #6
darkdave3000 said:
So the spin adds ZERO EXTRA PENETRATING POWER?

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?
 
  • #7
phyzguy said:
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?

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:
  • #8
darkdave3000 said:
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 if material strength was exceeded vs merely deformed.

https://en.wikipedia.org/wiki/Impact_depth

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.
 
  • #9
phyzguy said:
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.

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.
 
  • #10
darkdave3000 said:
So the spin adds ZERO EXTRA PENETRATING POWER?

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.
 
  • #11
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?
 
  • #12
I don't 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 traveled 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?
 
  • #13
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.
 
  • #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
 
  • #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?
 
  • #16
zwierz said:
Everybody knows that it much easier to pull out a nail if you pull it and rotate simultaneously than if you only pull it
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?
 
  • #17
A.T. said:
I guess because it's easier to brake the initial static
not only this
A.T. said:
But how is that relevant to a bullet?
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
 
  • Like
Likes A.T.
  • #18
zwierz said:
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
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 anyone 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.
 
  • Like
Likes A.T.
  • #19
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.
 
  • #20
jbriggs444 said:
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.
Makes sense, thanks.
 
  • #21
Im glad I got the community talking about this topic and I'll continue to monitor replies, I will now go and post another thread regarding a real time formula to calculate penetration of a bullet. Thanks for contributing everyone, I'll continue to monitor this thread as I post a new one. The topic of spin of the bullet contributing to a deeper hole is quite interesting , I am sure the spin will cause the bullet to some how act like a drill bit especially if it has a helix shape as a result of the rifling.
 
  • #22
The effect with rotating bullet can be illustrated by the following analogy.
8cbc71749684.png
Let a horizontal conveyor belt moves with velocity ##v##. There is a fixed rail above the conveyor belt. A matchbox of mass ##m## lies on the belt and rests at the rail. The friction coefficient between the belt and the matchbox is ##\gamma##. The rail is smooth. Then somebody pushed the matchbox along the rail with initial velocity ##u##. Which distance does the matchbox pass till it stops?
This distance is bigger than ##u^2/(2\gamma g)##
 
Last edited:
  • #23
darkdave3000 said:
The topic of spin of the bullet contributing to a deeper hole is quite interesting , I am sure the spin will cause the bullet to some how act like a drill bit especially if it has a helix shape as a result of the rifling.

I have my doubts. A bullet does not spin very quickly relative to its forward velocity. For each bullet-length it travels, it may spin perhaps a quarter turn or so. Just look at the video posted in your other thread, located here: https://www.physicsforums.com/threa...culate-bullet-hole-depth.910145/#post-5732931
 
  • Like
Likes jim mcnamara and jbriggs444
  • #24
Drakkith said:
I have my doubts. A bullet does not spin very quickly relative to its forward velocity. For each bullet-length it travels, it may spin perhaps a quarter turn or so. Just look at the video posted in your other thread, located here: https://www.physicsforums.com/threa...culate-bullet-hole-depth.910145/#post-5732931
And even if it spun much faster, most of the resistance comes probably from normal forces at the tip, from pushing the material out of the way, rather than from friction. So reducing the friction component along the path won't have that much effect.
 
  • Like
Likes jbriggs444
  • #25
jbriggs444 said:
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 anyone direction would be independent of motion in any other direction.
this is the reason why I speak about the motion of the bullet through bulletproof vest; in this case it is reasonable to assume friction to be obeyed the Coulomb law
 

Related to Rotating bullet vs non rotating bullet

1. What is the difference between a rotating bullet and a non-rotating bullet?

A rotating bullet has spiral grooves on its surface, which causes it to spin as it travels through the barrel of a gun. This rotation stabilizes the bullet's flight and improves accuracy. A non-rotating bullet does not have these grooves and relies solely on its shape and velocity for stability.

2. How does the rotation of a bullet affect its trajectory?

The rotation of a bullet helps to counteract the effects of air resistance and other external forces, allowing it to maintain a more consistent and predictable trajectory. This is especially important for long-range shots where even small variations in trajectory can greatly impact accuracy.

3. Does a rotating bullet have a higher velocity than a non-rotating bullet?

No, the velocity of a bullet is determined by the force of the gunpowder explosion and the design of the bullet. However, the rotation of a bullet can help it to retain its velocity and energy over longer distances, making it more effective at longer ranges.

4. Are there any disadvantages to using a rotating bullet?

The main disadvantage of a rotating bullet is that it requires a gun with rifling (the spiral grooves in the barrel) in order to function properly. This limits the types of guns that can use rotating bullets and may make them more expensive to produce.

5. Can a non-rotating bullet be just as accurate as a rotating bullet?

In theory, yes, a non-rotating bullet can be just as accurate as a rotating bullet if it is designed and fired correctly. However, the added stability and consistency provided by rotation can greatly improve accuracy, especially at longer ranges. In general, rotating bullets are considered to be more accurate than non-rotating bullets.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
236
Replies
19
Views
4K
  • Classical Physics
Replies
10
Views
1K
Replies
23
Views
1K
Replies
9
Views
1K
Replies
17
Views
2K
Replies
22
Views
3K
Replies
3
Views
1K
Back
Top