revolutions per minute debate

[goodgrouper]

Ok. I have several friends whom I have been talking with that have disagreed about this topic yesterday. I will ask the main questions here in a way as to not "steer" the answer so I can't be accused of wording the question in a way that leans the answer in my way of thinking. I will not say now what I believe to be the answer.
Ok, here it is:

If a bullet leaves a barrel that has a 1-8" twist, is it still spinning 1 time in every 8" all the way through it trajectory say out to 1000 yards? In other words, if we viewed the bullet at the 786th yard and followed it in slow motion throughout that yard, would it still be revolving 4.5 times in that yard? (lets assume the bullet has no flaws in the outer surface of the jacket, and there are no voids inside the bullet)



If you were to shoot a 55 grain bullet out of an 8 twist barrel at 4000 fps, and then shoot a 80 grain bullet out of the same barrel at 3000 fps, they have different rpm's, but do they both still spin 1 time every 8 inches?

 

[Guest]

A bullet's rpm slowes in flight but not as quick as its forward motion. The forces slowing the forward motion are much greater than the forces slowing the spin.
8 twist is 8 twist no mater what fps. rpm increases as fps increase.

 

[Ballistic64]

Yes,but as Hary says,the rotation will slow somewhat but nothing compared to the velocity loss.When compared to time and distance traveled,the bullet is spinning hundreds of thousands rpm.

 

[4ked Horn]

Yes GG the "rate of twist" as we use it in shooting is fixed by the rifling. 1:8 is 1:8 and that is it. The RPM however is governed by the velocity of the bullet leaving a barrel. the faster the bullet the faster the rpm OR if the muzzle velocity is constant RPMs wil increase with a faster twist rate.

In other words, If your gun has a 1:36 twist (assuming the revolutions of the bullet dont slow down during the few second flight) the bullet wil make 100 revolutions in 100 yards regardless of how fast the bullet is.

There is no "snap" factor to rifling. The bullet does NOT spin faster than the rifling just after it leaves the muzzle. I know of many people that think the bullet speeds up after leaving the barrel for a few yards as well and this is not the case either.

 

[jb1000br]

Yea, what they said, it only loses RPM's because the velocity is dropping off.

why does this matter anyway?

JB /ubbthreads/images/graemlins/confused.gif

 

[royinidaho]

GG, Go here: http://www.nennstiel-ruprecht.de/bullfly/fomo.htm

and go down to "Spin Dampening Moment".

BTW: CTR = Choose The Ruger (#1 that is.) Or maybe its Remington. Or maybe its Come-on Truckers Roll.

 

[LB]

I agree. Have heard, many times, that the RPM degrades much slower than velocity, and it makes sense, to me. However, other than the necessary (paper) computations, I really wonder if it can be ACTUALLY MEASURED, ....say, 500 yards downrange? Whatever, I accept the theory.

Good hunting. LB

 

[Mysticplayer]

LB, child's play. Set up a high speed camera in an appropriate booth or tunnel at 500yds. Shoot a bullet and watch the film.

Jerry

 

[Buffalobob]

Revolutions per minutes

We will neglect frictional forces here and assume an ideal world

bullet spins 1 revolution per 8 inches no matter what speed or time. This is revs per distance.

Shooting 786 yards is the same as shooting 36 inchesper yard times 786 yards = 28296 inches. How many revs occurred in that distance is 28296 inches divided by 1 rev per 8 inches = 3537 revs

Now then, how much time did it take the bullet to get there.

First gun @ 4000 feet per second

786 yards times 3 feet per yard = 2358 feet

Time of travel (ToT)= 2358feet divided by 4000feet/sec = 0.5895 seconds or 0.009825 minutes

Second gun @ 3000 feet/sec
ToT = 2358 feet divided by 3000feet/sec = 0.786 seconds or 0.0131 minutes


Now then, we have both the total number of revolutions for the distance and the time for the distance, so we can now get revolutions per time.

Gun one @ 4000fps

Rev/minute = 3537 Revs divided by 0.009825 minutes =360,000rpm

Gun 2 @ 3000fps

RPM = 3537 revs divided by 0.0131 minutes = 270,000 rpm

And by looking at ratios of the velocities we can conclude that a bullet coming out of this twist barrel at 2000 fps is 180,000 rpm and at 1000fps the rpm will be 90,000rpm

That was ideal world with no friction and the RPM would be the same at any distance and is really independent of distance. Varies only with the rate of twist of the barrel and velocity of the bullet

Now if you wish to include frictional forces from air on the spin and frictional drag forces due to different BCs you enter the world or calculus, differential equations and numerical analysis. The final answer which bullet has higher RPM does not change, just the magnitude of the answer.

The last time I had to do this type of calculation was 1967, I hope I got it right this time.

 

[LB]

Yeah, I suppose you are right, Jerry. However, I think I will just watch you do it. The necessary lighting, timing and high speed camera is not my concept of childs play, several thousand dollars in equipment to prove something that one writer (above) observed: "why does this matter anyway?"

Good hunting. LB

edit: thanks, BB. I have a simple spread sheet that tells me everything I need to know, converting velocity to RPM.

 

[Brown Dog]

At last! Ballistics Nerditry, haven't had any for a while! /ubbthreads/images/graemlins/grin.gif

Most of the comments above are correct:

At the muzzle, both rounds will be doing 1 turn per 8 inches.

Downrange the story will change.

Because the rate of decay of both rounds' horizontal (or forward) velocities is massive in comparison to the rate of decay of their rotational velocities both will start to do more than one turn per 8 inches.

For illustration lets ignore rotational velocity decay; the 4000fps bullet in your eg is doing 360000rpm (scratchpad sums that I haven't double checked -but my figures agree with buffalobob's!) at the muzzle. When it slows to 2000fps it's still doing 360000rpm, so now it's doing 1 turn in 4 inches.

Obviously, in the real world, rotational velocity does decay. But far more slowly than horizontal. The effect will be there, but far harder to calculate.

Assuming that the BCs of the two bullets only vary as a function of their weight rather than shape (ie both same spitzer BT design but one is heavier (ie the 80grainer isn't a flat point versus a 50 grain spitzer!)) The 50 grainer will lose its horizontal/forward velocity faster than the 80 grainer, thus it will start to do more turns per 8 inches than the 80 grainer.

 

[4ked Horn]

[ QUOTE ]
When it slows to 2000fps it's still doing 360000rpm, so now it's doing 1 turn in 4 inches.

[/ QUOTE ]

This, I feel is an important distinction to make. The bullets gyroscopic stability is the RPMs it is turning. The stability is not the rotations per distance while in flight. If we spin a bullet on a table like a top at 360,000 rpm it will remain (somewhat) stable even though it is not going anywhere. The rotation per distance is ONLY significant inside the barrel when coupled with the bullets launch velocity. In other words it is the 1:8 twist at muzzle velocity that brings it up to the correct RPM. At this point the distance traveled on each rotation is (almost) meaningless in regard to its stability in flight.

 

[LB]

Yeah, there is something I can agree with. Good point! LB

 

[Richard338]

4ked, that is not right. The rotational angular velocity (spin) required for a particular stability factor is dependant on the drag forces being experienced. Obviously this then depends on the translational velocity. As others have stated, spin decays more slowly than translational velocity, increasing stability downrange. This increases the relative spin rate from 1:8, but not the absolute spin rate. This may still not be enough to remain stable during the transonic transition. It is not either stable or not regardless of velocity (such as you suggest on a benchtop), or else a loss of stability through the transsonic region wouldn't occur.