lets start a blog with formulas, this one is for MODEL 48 who has a math mind but others will gain from them

I start with the simplest one:

muzzle energy

grain X velocity X velocity / .450240 = foot pound

please add another formula to this one

warren



and remember "it's 30% the gun and 70% the shooter"

meters per second
feet per second / 3.281 = meters per second


joules
fpe X 1.356 = joules

but we don't have enough bandwidth on this forum. (here come the nightmares)

those of you who took econometrics in B school know what i am talking about.

Looks to me like you might have an errant '.' - the divisor is 450240.

To compute ballistic coefficient:

bc = (d2 - d1)/(8000* ln(v1/v2)) where d1 and d2 are two different distances, and v1 v2 the velocities at those distances - ln is natural log i.e. log to base 'e'.

Terminal energy (energy at a point downrange):

te = me/exp(yds/4000*bc)

"Game Theory" is may be known to some from the movie "A Beautiful Mind".

Thats it, at least for now. I've got to get out the text books for formulas that I don't use anymore....

Good idea for a thread!

Model48

I have the proof, but there isn't room to write it in this margin.


=or (B2,NOT (2b))

Dave @vabch

"TKO". Developed to see what it would take to knock out a bull elephant (important if you happen to be hunting one, unless you want to become toe jam.) Applies to air guns, as well. Good to put the .177 vs .22 debate into perspective.

Taylor's formula is:

This formula is simple. You take the weight of the bullet (in grains), times the speed (in feet per second), times the diameter (in inches - like .451) and after multliplying the results, you divide by 7000 to give you a reasonable number to work with, somewhere between 1 and 100. Perhaps he used 7000 because that's how many grains makes a pound, but in any event it gives a workable number.

TKOF = weight speed diameter / 7000

warren





and remember "it's 30% the gun and 70% the shooter"

Here is a question from another user that seemed to not get answered, maybe you can help, My calculator is broken.

Shooting a block up an incline - work energy
A block of mass m is placed in a smooth-bored spring gun at the bottom of the incline so that it compresses the spring by an amount x_c. The spring has spring constant k. The incline makes an angle theta with the horizontal and the coefficient of kinetic friction between the block and the incline is mu. The block is released, exits the muzzle of the gun, and slides up an incline a total distance L.

Find L, the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside the gun. Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance x_c while inside of the gun). Use g for the magnitude of acceleration due to gravity.
Express the distance L in terms of mu, theta, g, x_c, m, and k.


My energy equation is as follows:

0.5k(x_c)^2 = 0.5mv^2 + (mu)mg((x_c) + L)cos(theta)
0.5k(x_c)^2 = (mu)mg((x_c) + L)cos(theta)
[0.5k(x_c)^2]/[(mu)mgcos(theta)] = x_c + L
L = [0.5k(x_c)^2]/[(mu)mgcos(theta)] - x_c

I know you have to add X_c to L to find the total distance the block is moved by the spring, and I really don't understand why this is wrong. Anyone know?

dave