Can some one tell me the formula for finding potential velocity for a bullet combination. I wanted to find the potential velocity of a 250 gr. bullet for the 9.3x57. You have to use a velocity of a bullet you have and then you can calculate the other bullets potential. Lets say the 286 gr. is going 2000 fps.

Littleloadingblock,

Calculate the energy of your 286 grainer at 2000 f/s

Velocity Squared X Bullet Weight
-----------------------------------
64 X 7000

2000 X 2000 X 286 / 64 /7000 = 2550 ft llbs

To calculate velocity when you have the energy

Energy / Bullet Weight X 64 X 7000.

Take the square root of the above result.

2550 /250 X 64 X 7000 = 4,569,600

Square root of 4,569,600 = 2137 ft /sec

Mike

You buy a chrony, then you load a couple of rounds, then you fire those rounds through the chrony and then you record the results.

"Potential velocity"...geesh...

Thanks that is a lot of help. I was wondering how to do it, more than what that certain gr. bullet will do in my gun. The 64 / 7000 that comes from a proportion. Mike do you know where those numbers come from. I am sure that gentleman, I am not sure his name, in the 1900's from England came up with it during his work for the Royal Navy. I am sure that I am wrong on some of this, I just think it is a wonderful thing to ponder. Math just blows my mind. LLB

Those formulae do not give you the information you are seeking.

You stated that you have a 286 grain bullet traveling at 2000 FPS.

You want to know how fast a 250 grain bullet would be traveling.

Now...

I am assuming that you are using the same case, powder, primer and firearm?

And I will assume that you are loading that 250 grain bullet deep enough into the case to ensure that the free space in the case is exactly the same as the free space with the 286 grain bullet.

So tell, me, how do those formulae give you your answer?

To calculate the energy of a bullet is very simple when you know the weight and the velocity.

However, to "reverse engineer" to find the velocity of a lighter or heavier bullet, using the energy figure derived from the original bullet, using the same data, cannot be done.

You can easily find out what velocity is needed to deliver the same amount of energy.

But "potential velocity"?

Naw...no such animal.

My Hum-V will do 85 MPH, how fast would my Volkswagon go with the same engine?

210 MPH, give or take...


Steve, I certainly understand what you're saying, BUT Mike hasn't lost his mind on this one. Any given cartridge has the ability to generate approximately the same muzzle energy figures regardless of bullet if operating at the same pressure level, or so it seems if you puruse the ballistics tables. Yes it will vary a few hundred foot pounds and it is not prescise and the concept ain't necessarily healthy if you extrapolate load data from this BUT, it's there and that's that. If you reverse the mathmatical process it will give you a ballpark figure, probably within a few hundred FPS or less. In the original post, LLB said he was looking for POTENTIAL, not load data.

It's mental exercise, that's all.

Littleloadingblock

The 7000 is for the number of grains in a pound. The 64 (should be 64.32) refers to twice the acceleration due to graivity.

Of course if equal energy is to be given to different bullet weights then that assumes equal pressure and an equally suitable powder for each bullet weight.

In practical terms the bullet energy obtainable drops of a bit when very light bullets for calibre are used such as the 90 grain 270 or 110 grain 30 calibre etc.

Mike

Thanks Mike, I see how it works now. Thanks for your time. LLB