Information about loss of velocity is pretty easy to come by but I was wondering about loss of bullet spin. If a 55 grain bullet from a 20” barrel with a 1:7” twist is spinning 350,000 RPM at the muzzle, how well is that maintained down range? Ballistic Coefficient is a value to describe how fast a bullet looses velocity. Is there a similar value for spin?
Loss of Bullet Spin
- Thread starter DMW1116
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There is a formula for spin and how it acts upon a bullet.
Google "Bullet RPM for Accuracy."
Be aware though your definition of Ballistic Coefficient is incorrect. BC is defined as "the ability of a bullet to travel through air." Velocity is an effect bullets have because of drag and other environmental factors.
Google "Bullet RPM for Accuracy."
Be aware though your definition of Ballistic Coefficient is incorrect. BC is defined as "the ability of a bullet to travel through air." Velocity is an effect bullets have because of drag and other environmental factors.
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Dave DeLaurant
Member
Not the answer you asked for, but with regard to keeping bullets stabilized a significant event occurs when they transition from supersonic to subsonic. The team at CETME in Spain initially worked at creating an intermediate cartridge that would preserve rotational stability by remaining supersonic out to 1K. They used a very long, lightweight bullet with a very high BC:
tominboise
Member
You are talking about the angular acceleration (actually negative acceleration or deceleration) of a body around the center of its rotation. A change in acceleration requires a net external torque. In this case, the torque would be from the drag of the air on bullet as it spins (note - not the drag from the air hitting the front of the bullet, but rather the drag from the air 90 deg to the line of flight, dragging on the sides of the bullet as it spins). These formula's are published but the difficultly in this case is figuring out the net force on the bullet.
^ Excellent. Hard to extract those angular (spin) drag parameters since they aren't that important in terms of exterior ballistics. (They are in terms of terminal ballistics.) Keep in mind that if they were available, there would be a difference between a smooth unrifled bullet spin versus with the additional angular drag from the rifling grooves.
Time to remark about bullets spinning on ice. They go for quite awhile before stopping. Could be enlightening.
Search term: <bullets spinning on ice>.
Terry, 230RN
Time to remark about bullets spinning on ice. They go for quite awhile before stopping. Could be enlightening.
Search term: <bullets spinning on ice>.
Terry, 230RN
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mcb
Member
To be technically correct the BC is a scaling factor for a selected drag model, usually G1 or G7 though there are others. At each time step the ballistic calculator looks up the coefficient of drag for the selected drag model (Gx) that is a function of velocity and then scales the coefficient of drag by dividing the coefficient of drag by the particular bullet's Ballistic Coefficient (bullet mass and frontal area are hidden in their too). This scaled coefficient of drag is used in the ballistic models integrator (to calculate the acceleration, velocity, and position) and is updated with each time step of the simulation as the bullet moves down range.
As for spin the lost of spin in nearly all firearm projectiles happens at a much slower rate than the lost of velocity and is a factor that is hard to calculate and in most cases something that can be ignored for most exterior ballistic calculations.
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Thats kinda what I was trying to figure out. I have a pistol load that shoots very well at 10 yards but falls apart at 25 yards. One explanation offered was the bullet was becoming unstable, but I doubt that is the case. It's a standard load and velocity for a standard bullet shape and weight. The information I didn't have was how much the bullet spin decreases as it goes down range. Evidently it's not much influence, especially over handgun distances.
Dave DeLaurant
Member
Examine the holes in the 10 yard target. Usually degrading stability is apparent due to smears found at one edge of the hole. If that's the case, you should expect bullet tumble at slightly longer range.
Also, are you sure you've eliminated all other variables that might lead to an accuracy issue, such as loose sights or barrel lockup?
No, I'm slowly eliminating them as I go. The most likely issue is my eyesight, but I have other loads that shoot very well at both distances from the same pistol. This seems pretty unlikely, but I didn't know enough about it to rule it out. It's a revolver so barrel lockup shouldn't be an issue. Sights might be loose but for handgun distances I figure I'd see that at 10 yards too, as well as with some other loads I tested last time I shot it. These are 38's in a 357 revolver so it might not like the shorter case length. At least that's what I'm leaning towards. This was the first time I've heard spin changing enough during flight to be an issue, so I was looking for more and more reliable information.
Varminterror
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Folks say this kind of thing often, especially said in this manner, but we really have to focus on the reality of our application: plainly, if a bullet is shooting well at 10yrds but not doing well at 25yrds, either the issue is inherent instability, or something else entirely.
Folks want to picture a top on a table, spinning in place, until eventually the spin rate decays sufficiently to no longer stabilize the top, but at the scales we’re considering, that’s NOT what’s happening for a bullet between 10 and 25yrds. If this case IS a spin stability issue, then the bullet is not ever being stabilized and the difference between 10 and 25 is simply the amount of time the unstable bullet has had to fling itself out of path. Yaw induced instability doesn’t immediately slide bullets around a target, but rather takes time to plane and gyro, and further promote greater yaw. In a manner of speaking, an unstable bullet will “spin its wheels and eventually gain traction,” meaning it will fly into a relatively small group at close range because the initial inertia propelling it on course at first has not yet been overcome by the destabilizing yaw influence - but once that nose starts grabbing sufficient air to deflect the mass of the bullet, it’ll move dramatically off of course.
Naturally also, hole shape in the paper is a dead give-away.
So yes, the answer - or understanding that an answer exists - to the question of whether a formula exists for spin rate decay, is yes. But that isn’t the appropriate question to ask why your groups open so much at 25yrds - as spin rate decay and sudden destabilization is not the answer to that problem.
Jim Watson
Member
Well, the short answer is to use those other loads.
Not necessarily.
Gil Hebard saw that 40% of the bullet holes from a S&W M52 wadcutter automatic showed signs of tipping. Just a slight wipe mark at one edge of the bullet hole, not wild key holing. In 2" groups at 50 yards.
There was a top ranked PPC shooter who said he could intentionally tip a bullet by flipping the barrel as he fired. I think he was just seeing the same effect and mistakenly thinking he was causing it.
lysanderxiii
Member
The loss of rotational velocity is almost negligible.
In fact, stable bullets get better stability as they go down range (until they pass through the transonic region if supersonic), because the velocity drops so much faster than the spin, and the overturning moment is a function of the velocity and the righting moment is a function of the spin.
Unstable bullets will behave as Varminterror described.
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