Submitted by Eastern-Ability8411 t3_11eu6f6 in askscience
Had a discussione with a friend the other night. Assume we fire two bullets. Same mass, same velocity upon exiting the barrels. One barrel is rifled, the other is not. Also assume the non spinning bullet magically maintains itself as stable as the spinning one, and both retain perfect shape. Is their velocity identical after traveling in air for a while? Or does the spinning somehow influence the fluid dynamics of the spinning one?
On point answer, thank you l!
>and a "negative pressure" behind the projectile.
And quite a lot. For an artillery shell this can be as high as 40% of the drag. Modern artillery shells have pyrotechnic charges at the back to reduce this. They aren't pushing the shell, they just fill up the empty void with gases, reducing the drag by a lot. That type of ammunition called base-bleed can have up to 30% longer range.
E.g. a German PzH2000 self-propelled howitzer firing a base-bleed shell in Ukraine:
https://pbs.twimg.com/media/FoVlKRDWAAACjsY?format=jpg&name=large
Very interesting, I wasn't aware this effect was so substantial! Our capacity for making weapons to kill each other is truly remarkable.
A lot of long-range shooting enthusiasts use projectiles known as "boat tails" for the same reason.
Instead of having a flat back, they taper down a bit.
By making the surface area of the rear smaller, the size of the negative pressure zone is also made smaller, reducing the effect and raising the ballistic coefficient.
[removed]
“Filling the empty void with gases” IS pushing the shell. It might not have a net positive acceleration, but it’s like pushing a car that’s out of gas - you’re still pushing even though it’s not accelerating.
I think their point was that the purpose wasn't propulsion, but a reduction in drag. If you look at the momentum imparted by the gas released it will be much smaller than the decrease in momentum caused by drag if the gas weren't there.
>IS pushing the shell
By a negliglible amount. There are actual range-boosted shells with rocket motors which gain significant thrust from the shells actually accelerated in flight.
Base-bleed is not that and for simplicities sake these shells aren't said to be propelled by it, they just have less drag.
Vacuum doesn't exert a force. The gases released do.
You can look at this from a force diagram. Having a vacuum behind the shell just means there isn't gas pushing on the shell. Adding a gas does push on the shell.
If you’re about to argue that a vacuum which is surrounded by matter doesn’t exert force, you are gonna make my day.
[removed]
Out of curiosity, in your force diagram, what happens to the matter around a vacuum? What forces are created by the pressure differential? If you found a way to decrease the pressure differential, what would happen to those forces? Which one would exert a greater net gain in speed, the equalization of the pressure differential or the ejection of mass to fill said void in this particular scenario?
>what happens to the matter around a vacuum?
Matter in a vacuum has no forces exerted on it.
>What forces are created by the pressure differential?
Pressure is just Force / Area.
Pressure ( in this example ) is the result of the summation of forces of gases bouncing off of a surface. There is very high pressure at the front of the shell, and very low pressure at the rear of a shell when fired.
>If you found a way to decrease the pressure differential, what would happen to those forces?
You can reduce drag if you change the shape of the shell. If you have more pressure at the rear of the shell the net force on the shell is reduced. You can do this with a boat-tail shell.
>Which one would exert a greater net gain in speed, the equalization of the pressure differential or the ejection of mass to fill said void in this particular scenario?
The gas sent out of the rear of the shell does not act like a rocket. In a bleed gas shell the gas is going nearly the same speed as the shell, and is trapped there for multiple interactions. The trapping of this allows the gas to exert a pressure onto the shell. If your force diagram model has this gas as part of the shell you have a new drag coeffienct, if you model the shell alone then you have this gas exerting a force ( or a pressure over an area ) on to the shell.
I didn’t say they are propelled by it, I said they are pushed by it. It exerts a force, which reduces its acceleration due to drag. This is called pushing.
[removed]
[removed]
Just to clarify, is that additional drag in the direction of travel? Or is it the same amount in the direction of travel plus a small lateral amount in the direction of spin?
Excellent point. The latter. If you divide the motion of the bullet into a linear component and a rotational component, the additional friction drag only acts against the rotation. So the spinning bullet will slow its rotation over time, but the two bullets should have almost exactly the same speed.
[removed]
[removed]
There is another effect that can result in additional spin drag in the form of viscous drag.
Preamble: both bullets fly with the exact same speed and perfect orientation, so zero angle of attack.
As you elaborated, the viscous friction in the spinning case is higher. This can be additionally contributed by an earlier boundary transition from laminar to turbulent.
In the spinning case, the flow spirals around the bullet, effectively traveling a longer distance in the same amount of time. The relative velocity between surface and air is higher. This increases the Reynolds number, (rho * v * l_ref / eta; rho, l_ref & eta are constant between the cases) possibly leading to earlier laminar-turbulent transition. Also, the flow distance along the spiral is bigger, so there's more distance for the transition to happen. If it were to happen at the same run length of the boundary layer, due to the spiral the transition would effectively happen earlier along the bullet spin axis. And since turbulent boundary layers exert more drag than laminar ones, this can increase drag for the spinning bullet.
[removed]
[removed]
>For a spinning projectile, this total shear stress value is larger than if it was not spinning, so by this reasoning a spinning bullet should experience more drag than a stable, non rotating bullet.
But the direction of this increase in drag should be orthogonal to the translational motion, no? If I'm right, this drag would contribute to slowing the rotation, but not to the arc of the projectile.
[removed]
Hi! This is my field. As phrased, my intuition is a spinning bullet has slightly more drag IF both projectiles had identical body dynamics. IRL though, the non spinning bullet is unstable and would immediately tumble due to tipoffs and would experience much higher drag. Fun fact, there are smooth bore guns out there for fin stabilized projectiles.
Apologies if already addressed, but does the weight of the projectile matter? Or does speed negate it?
Weight does not affect drag force, but it does have effects on the bullet’s flight path. Only the bullet geometry and the medium it flies through should affect the drag I believe.
No though you will have the same force acting on a heavier bullet, so the drag will affect the bullet less. That is why heavier bullets have higher ballistic coefficients.
Sounds counterintuitive but makes perfect sense, thank you. It's why downhill skiers are faster when they weigh more.
[removed]
[removed]
So, here’s a twist on the question that makes me wonder, offered more as food for thought than a rebuttal of any of the above analysis - I agree that the rotation effectively increases speed of airflow which should increase drag.
The thought was whether the spin also increased a boundary layer of air over the bullet to decrease drag. In this regard, I’m thinking of comparing two round objects to ignore the impact tumbling would have. Say, like a golf ball or a baseball. A knuckleball is slower than a fastball, for example, though I appreciate that’s largely due to the throwing mechanics of each - you can’t throw a knuckleball as hard as a fastball just due to the motion. Similarly, a spinning, dimpled golf ball travels farther than a non-spinning, non-dimpled ball - I understand the dimples create the boundary layer and help to give it lift by creating higher pressure on the underside of the ball (which has air traveling past it faster than the top part, due to the reverse spin).
Bringing back to the original question, comparing two smooth bullets it does seem that neither of these factors would be at play (and of course, the smoother the bullets the less of an impact spin would have on drag), but interesting to think about.
If a bullet could maintain straight line flight, with the bullet axis coincident with the flight axis, then the effect should theoretically be zero.
However, a variety of factors can cause the two axes to NOT be coincident. In this case, imagine you're driving down the road with the window down and you put your hand out. If you angle your hand relative to the flow of air past it, you can easily feel the increased drag on your hand.
The whole point of spinning the bullet (gyroscopic stabilization) is to keep the two axes coincident through the expected/desired flight path of the bullet.
I disagree. The vector sum of the air speed against the bullet's body is greater when the bullet spins, resulting in a (slightly) greater Reynolds number, which will (slightly) increase drag in the direction of travel.
If it implies tumbling I could see a higher drag from the smooth bore vs rifled. The spin imparts stability and keeps the bullet in a pointing direction of flight while tumbling will expose more surface area if the round turns pointing sideways to flight
[removed]
[removed]
[removed]
[removed]
[removed]
[removed]
[removed]
[removed]
Just from an energy point since i havent seen it addressed, assuming they were fired with the same amount of force, the one not spinning would be traveling faster since some of the kinetic energy got spread out to make the bullet spin. At least in the magic scenario
[removed]
[removed]
[removed]
In addition, it is pertinent to note that the rifling marks present on both bullets are identical, yet one of the bullets does not rotate when discharged. Provided the assumption that the velocity of the bullets at the point of discharge is equivalent and that both are aerodynamically stable, it follows that the rotational force generated by the rifling in the first bullet consumes a portion of its kinetic energy. Consequently, there is an increase in the drag experienced by the rifled bullet as compared to its non-rifled counterpart. Moreover, it can be inferred that the absence of rotation on the second bullet results in a more laminar flow, causing less distortion of drag and leading to lower drag values
[removed]
[removed]
[removed]
The_Illist_Physicist t1_jagtuez wrote
Seeing as though nobody is really answering your question I'll take a swing at things. As a disclaimer, while I am a student of physics, my specialty is optics and not fluid dynamics.
From the Wikipedia page on drag coefficients, it looks like the total drag coefficient (C_d) is defined as the sum of pressure drag (C_p) and friction drag (C_f). These can be thought of respectively as:
The component of drag that comes from the traditional resistive force we think of due to the fluid exerting a pressure on the face of the projectile, and a "negative pressure" behind the projectile. If you look at the expression to compute this, there's a dot product between the free stream flow unit vector and the negative surface element normal unit vector (inside a surface integral). What this means is for a cube in a non-turbulent fluid, only the front and back faces of the cube contribute to this C_p term. By thinking about the symmetry, a spinning bullet should have no effect here.
The component of drag that comes from a shear stress due to fluid friction along the sides of the projectile. Again in this surface integral there's a dot product, this time between the shear stress unit vector and and the free stream flow unit vector. This is then scaled by multiplying it by the total shear stress at the differential surface element. For a spinning projectile, this total shear stress value is larger than if it was not spinning, so by this reasoning a spinning bullet should experience more drag than a stable, non rotating bullet.
Now I suppose this is all for a very idealized scenario. However a quick search for literature on friction drag simulations turned up this paper https://www.hindawi.com/journals/ijae/2020/6043721/ which seems to suggest that the spinning effect does indeed have some small contribution to the total drag, although it gets becomes less important at higher speeds (below Mach 1).
TL;DR: Yes it appears a bullet does incur a little bit of extra drag from it spinning, however it's necessary to stabilize the bullet and is fairly negligible when the projectile speed is high.