Bullet identification please.

It is approx 1/4 inch thick by the measurment you see here below.

Click to expand...

in that case it's approx a .257 caliber bullet.

The diameter of the bullet is going to tell us the most about it. Can you get a more accurate measurment with a pair of calipers?

As above, either a .257 or one of the many 6mms.

Something around a .25 cal. Lots of choices. You'd need an exact diameter and weight to try to narrow it down. You'd also need forensics to get down to model and manufacturer.

Same as above - -

More information needed.

How close are you to your State Police crime lab? They should be able to help. The question is, CAN they help out with an unofficial request? Some agencies will, some won't. I don't know anyone at your SP HQ.

Good luck
Johnny

I imported your pic into AutoCAD. I used the scale of the tape measure to find that 1" = 23.1313". The bullet diameter measured 7.7771" inches in the picture. 7.7771/23.1313 = .3362". It appears you have .338 remington core-lokt on your hands.



Continuing...
I also calculated the rifling pitch at 6 deg:
using the twist rate formula: 6 deg = tan^-1 x (pi x .338 / "Twist Rate"). Solving for "Twist Rate", you get a twist rate of 1:10.103, or 1:10 if you round. The twist rate for most .338 winchester magnums is 1:10.


I'd say you got a .338 Winchester Magnum on your hands

"Educated" Guess?

Okay, looks like an interesting game, so let's play...

On my monitor, your photo shows the projectile deformed & on a tape measure. Looks like it should straighten out to about 1 1/16 to 1 1/8 inch in length, as it appears to be an spitzer profiled soft point. Thats 1.0625" to 1.125", or a median length of 1.09375". This displays on my screen as approx. 4 1/2 inches. The diameter shows to be about 1 1/2 inches on my screen, which proportionally equates to about one third (4.50 divided by 1.50) of the length.

We assume approximations only, as you haven't measured the actual diameter or weighed the residual mass, it stands to reason that one third of your length will result in the diameter. Taking the middle ground, we guessed 1.09375 as the length, divide by 3, and end up with a diameter of 0.3645.

Applying this information to common bore diameters, we round-out to 0.366" (9.3mm) or so. Your bullet could actually be considerably different, as the photo image can be out of proportion due to angle, etc.

All of this is nothing more than an entertaining exercise without more precise information. If you care to get a more accurate measure of diameter and weight, you could at least establish a list of possible commercially loaded cartridges this bullet may have come from.

Now that I've come up with all this theory, I find the very reasonable & logical/scientific post by Michaelmcgo, above. I yield to his superior skill and resources, and I most heartily agree. Here in the United States, the popularity of the .338 is far more likely.

Nicely done michaelmcgo, but it looks to me like you've measured from one side of the bullet to the edge of the shadow/tape measure on the other side; would you be willing to re-run your calculations again?

Measure its diameter with a micrometer just aft of the cannelure.

.257 is possible. I think it may be even larger, i.e. 264. Put me down as a definite I don't know.

Well, I can tell you that it's not a 40mm grenade round.... so that narrows the search somewhat.

Hummmm,

Care to clue us in as to why you're asking?

Now, I know that tapes come in many different widths to be sure but two common widths are 1/2"w and 3/4"w. I'm guessing this might only be a 1/2"w tape? If so, the bullet isn't 1/4"dia as you claim as it looks to take up roughly 2/3rds the width of the tape. (.33)

Another thing, is that you threw out a preliminary guess of it possibly being from a .22-250? There is a big enough difference between a 22cal bullet and what this looks that you'd think it would be a bit obvious. Do you have any cartridges or bullets available you could have compared it to first?

Or is this really just a test?