Identifying which bullet hole on the target was made by which round

[Jeffvn]

I use permanent magic markers. I put the individual colors on the tip almost all the way out to the ogive; the most colors I've ever needed to try was 2 per projectile.

+1 Pushing / closing in on the holes from behind gives a larger colorized area to see and discern the color.

I shoot the paper when its mounted on a piece of cardboard; sometimes its easier to tell the color of an individual hole (e.g. is it a red or a red/blue combo) by looking at the cardboard rather than the paper.

Jeffvn

 

[Cassidy]

It's interesting that a 2-colored bullet, with the huge spin it has enroute (what, 60-70K rpm?), can leave marks from both colors on the target hole as it passes through the paper. I am not doubting, just think this is kind of amazing. Maybe this is why more than 2 colors might not be discernable because the spin mixes the colors on the paper.

 

[sailhertoo]

A Q-Tip with alcohol will make the colors brighter to see on the paper.

 

[mr45man]

The spin would be negligible, when you think about it.
Say a 8 twist barrel, the bullet is making one revolution in 8 "
A 1" long bullet passing through the target would be making 1/8 th
of a revolution
John H

 

[GSRswapandslow]

The spin would be negligible, when you think about it.
Say a 8 twist barrel, the bullet is making one revolution in 8 "
A 1" long bullet passing through the target would be making 1/8 th
of a revolution
John H

um.....are you being serious???

 

[Killshot]

The spin would be negligible, when you think about it.
Say a 8 twist barrel, the bullet is making one revolution in 8 "
A 1" long bullet passing through the target would be making 1/8 th
of a revolution
John H

Someone failed math.

 

[RDavies]

Mr45man probably went to the same school as I did. His maths seems fairly close to me.

On a slightly different tangent, I was wondering if anyone uses moly in in 600yd BR purely to make the bullet holes more visible on the white targets?

 

[mr45man]

Maybe i did not word it right.
A 1" long bullet passing through the target would be making 1/8 th
of a revolution as it passes through the target
So what is wrong with that assumption?

 

[Cassidy]

Mr4 - your assumptions look OK to me. I'm wondering if there's a second way to calculate this to see if they agree. But your reasoning seems OK to me.

 

[GSRswapandslow]

sigh

so say you're shooting a bullet 2850fps....and that's in a 1:8 barrel....your bullet will leave the barrel spinning at a rate of 256,500 rpm......................................little bit faster than you were calculating...

 

[BobGibson]

I dip each bullet in camo pattern and with a large expensive microscope match up the patterns on the target.

 

[Toby Bradshaw]

The spin would be negligible, when you think about it.
Say a 8 twist barrel, the bullet is making one revolution in 8 "
A 1" long bullet passing through the target would be making 1/8 th
of a revolution
John H

Someone failed math.

Apparently. But it wasn't John H.

 

[Toby Bradshaw]

Mr4 - your assumptions look OK to me. I'm wondering if there's a second way to calculate this to see if they agree.

Sure. Assume that the bullet's rotational speed (in rpm) remains constant after leaving the muzzle. That's very close to true for typical small arms ranges.

For a 1" bullet at the muzzle of an 8-twist barrel, of course the bullet will make 1/8 revolution as it passes through the (negligibly thin) target paper placed at the muzzle. As the bullet's velocity slows going downrange, the number of revolutions made going through the target increases by the ratio of the muzzle velocity to the retained velocity. For example, at 500 yards the bullet's velocity might be down to 2000 ft/sec (from a 3000 ft/sec muzzle velocity), and the number of revolutions made in 1" of bullet travel would be (1/8)*(3000/2000)=3/16.

Close enough to John H's estimate for all practical purposes, I'd say.

 

[GSRswapandslow]

lol....you guys aren't math majors

 

[Toby Bradshaw]

lol....you guys aren't math majors

Feel free to show us the error of our ways. Start by calculating the time it takes the bullet to travel 1" (the bullet's length), and then multiply that time by the rotational speed of the bullet to find out how many turns it makes going through the paper.

And find a good recipe for crow.

 

[Toby Bradshaw]

Well, it looks like GSRswapandslow and Killshot ran out of fingers and toes to do their "calculations," so here's how it's done:

1) Calculate the rotational speed of the bullet. rpm = [muzzle velocity (ft/sec) * 60 sec/min * 12 in/ft] / twist (in)

2) Assume that the rotational speed of the bullet remains constant during its (brief) flight.

3) Call the length of the bullet "L" (inches) and the current velocity of the bullet "v" (ft/sec).

4) The time (in seconds) it takes the bullet to travel 1 bullet length: S = L (in) / [v (ft/sec) * 12 (in/ft)]

5) The number of turns that the bullet makes when traveling 1 bullet length is simply (S * rpm)/60 (sec/min)

Here's a numerical example, for the algebraically challenged.

muzzle velocity = 3000 ft/sec
downrange retained velocity @ 500yd = 2000 ft/sec
twist = 1 turn in 8 inches
bullet length = 1 inch

So, at the muzzle the bullet is spinning at 3000 * 60 * 12/8 = 270000 rpm. It takes the bullet 1 / [3000 * 12] sec to go 1 inch, which is S = 0.000028 sec. The number of turns made by the bullet as it travels through the target is 0.000028 * 270000 / 60 = 0.126. Of course anyone who understands barrel twist could have skipped this calculation. [The only reason that the answer doesn't exactly equal 1/8 (0.125) is rounding error in the S calculation.]

Downrange at 500 yards, the bullet has slowed to 2000 ft/sec, but the rotational speed is still 270000 rpm. It now takes S = 0.000042 sec for the bullet to travel 1 inch, hence the bullet makes 0.189 turns (and would have been exactly 0.1875 -- 3/16 -- except for rounding error in the calculation of S).

Didn't they have an 8th grade where you guys went to school? ;D

 

[bighorn06]

I hate to get in this, but how do you measure rotational rpm at 500 yds or any other distance. It has to slow down some time. mike

 

[Toby Bradshaw]

Calculating the bullet rotational speed at the muzzle is easy. Imagine a piece of target paper placed against the muzzle. As the nose of the bullet leaves the muzzle and penetrates the target, the base of the bullet is still engaged with the rifling and is forced to rotate at the twist rate. So, a 1" long bullet in a 1-8 twist barrel will advance 1/8 turn as it goes through the target at the muzzle, regardless of the velocity of the bullet. The rotational speed of the bullet at the muzzle simply depends upon the muzzle velocity and twist rate.

Downrange, spin decay can be approximated by the Kolbe formula:

N = Nm exp(-0.035 t / d)

Where N is the downrange spin rate, Nm is the spin rate at the muzzle, t is the elapsed flight time in seconds, and d is the projectile diameter in inches. For a 6mm bullet with a BC of 0.440 and a muzzle velocity of 3000 ft/sec, time of flight to 500 yards is about 0.6 sec. Using the formula above, bullet spin rate will decay less than 1% over those 500 yards. So, for all practical purposes in small arms, bullet spin can be assumed to be constant, determined only by muzzle velocity and twist rate. If you think about it, the slow rate of spin decay is how we can get away with using twist rates that marginally stabilize the bullet, because that's what produces the best accuracy.

 

[Cassidy]

Thanks, Toby! All your effort and the information are really helpful!

 

[Link]

Well gentlemen I have a degree also I have a Doctorant. I am a D.O.P.E. I have a Doctorant Of Personal Experience. If you can't dazzle em with brilliance, baffle em with BS.

My vote goes to Mr4.

;D