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

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.

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.

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

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.

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

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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.

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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.