BULLETS AND RPM

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joeb33050 posted this 27 February 2019

 

BULLETS AND RPM

 

Bullets in flight are spun by the rifling in the gun barrel. The rate of this bullet spinning is commonly measured in revolutions per minute, RPM.

 

MV = bullet muzzle velocity in feet per second.

 

T = twist, the number of inches for one bullet revolution in the barrel.

 

* = multiplied by.

 

RPM = (720 * MV)/T

 

MV = (RPM * T)/ 720

 

T = (720 * MV) / RPM

 

To increase RPM, either MV must be increased, the T number must be reduced = made faster, or both.

 

To decrease RPM, either MV must be decreased, the T number must be increased = made slower, or both.

 

Any RPM defines an array of T and MV.

 

Using the Greenhill formula we can solve for bullet length, and using a survey ratio, solve for bullet weight.

 

This table shows, for 120000 and 140000 RPM, an array of MV, T, bullet length “, and bullet weight grains; each of which solve for the RPM.

 

For example, .308 bullets at 120000 RPM, at 2000 fps, T = 12”, bullet length = 1.19”, and bullet weight = 207 grains.

 

This is one of the 16 solutions shown; there are essentially an infinite number of solutions.

 

These solutions may be divided into “feasible” and “unfeasible”, with the borders subjective.

 

For example: A .308” bullet, 2.37” long, weighing 415 grains, in a 6” twist barrel, at 1000 fps, spins at 120000 RPM. I would consider this solution to be in the unfeasible set.

 

This table contains a lot of feasible solutions; maybe the most for a 20000 RPM span. If so, increasing or decreasing RPM would diminish the number of feasible solutions, and probably accuracy.

 

Feasibility varies with RPM. The 120000 to 140000 span has a lot of feasible solutions; but see the .457” bullet diameter table.

 

There are not many feasible solutions on this table. A 502 grain bullet at 4000 fps? A 6.09” long bullet, 2344 grains, at 1000 fps? Perhaps there are none.

 

Bullet RPM, MV and Twist are mathematically related, if we know MV and Twist, we know RPM. RPM is a representation of MV and Twist. Feasible solutions are those that have been shown to work, to give at least reasonable results.

 

 

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M3 Mitch posted this 27 February 2019

Yet 45-70 bullets stabilize fine at a much lower RPM, unless I am doing the math wrong in my head.  Even the big old 500 grain bullets stabilize fine from an original Trapdoor Springfield, I forget what twist these have, but velocity is in the 1200-1300 range at best, and IIRC the twist is like 1 in 38, but am not sure.  But I have never seen any evidence that these bullets are not flying point-on out to the longest ranges I have tried it at.

I do not think I would be flagging myself as a sissy if I noted that a 502 grain bullet at 4000 FPS would kick a bit more than I want to put up with on a regular basis.

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Paul Pollard posted this 28 February 2019

Joe,

MV = (RPM * T) 

Should be:   MV = (RPM * T)/720

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joeb33050 posted this 28 February 2019

Thanks Paul, fixed.

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joeb33050 posted this 01 March 2019

And the .225 array:

:

 

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