Post by klmhq on Jul 5, 2005 7:18:09 GMT -5
I don't believe them.
Don't get me wrong... I'm not saying that they are full of it... yet.
They make some 'interesting' assumptions and stick a lot of stuff in that doesn't seem to apply.
Quoted below is my e-mail to them asking them to explain their formula and such. I may even get an answer... though I doubt it since they sell something. Also, I tend not to ask in the most politic way. Something in my personalilty says, get correct info first, worry about ruffled feathers last.
Anyway, mathematically and physicsally (), I believe that their work is BS, but I'll wait for a response.
Kev
Dear Sirs,
A friend of mine pointed me to your site and asked me to look at how you calculate recoil... we've had quite the discussion on it lately.
If I may ask a couple of questions about your formula...
***FROM www.loadammo.com/Topics/August01.htm
*** (1) I = (Wb*Vb + Wc*Vc)/225400
***Where:
***I = recoil impulse, lb.-sec.
***Wb = bullet weight, grains
***Vb = muzzle velocity of bullet, f.p.s.
***Wc = weight of powder charge, grains
***Vc = velocity of powder gases, f.p.s. (4,000 in these calcs.)
***225400 = unit correction factor (7000 gr./lb.*32.2 f.p.s.2)
***The free recoil velocity of the gun is easily found from the recoil mpulse and weight of the gun:
*** I = Wg/32.2*Vg or rearranging terms:
***(2) Vg = 32.2*I/Wg
***Where:
***Vg = free-recoil velocity of the gun, f.p.s.
***32.2 = acceleration of gravity, f.p.s.2
***I = recoil impulse, lb.-sec.
***Wg = weight of gun, lb.
***The recoil energy of the gun can then be calculated from the kinetic energy equation:
***E = ½*Mg*(Vg)2 or simplifying:
***(3) E = Wg*(Vg)2/64.4
***Where:
***E = free-recoil energy, ft.-lbs.
***Mg = mass of the gun (Wg/32.2)
***Wg = weight of gun, lb.
***64.4 = acceleration of gravity*2, f.p.s.2
***Vg = free-recoil velocity of the gun, f.p.s
1) Impulse in measured in (metric) kilogram meters per second. Here you are using lbs/sec. There is no component for distance as is required. How do you get the impulse then?
2) Why do you have a correction factor and where did it come from? Why is the correction factor 7000 grains/pounds (which doesn't seeem to make sense anyway) * the acceleration due to gravity? Gravity would play litte or no part in a normal one dimesional Newtonian mechanics problems. I realize that recoil isn't purely one dimensional, but your use of 1-d momentum implies that you think it is too.
3) How do derive the 2nd formula? Why is gravity included?
4) Equation 3, where does the 64.4 come from and why? You say that it is accel due to gravity*2, why?
Please tell me how you derived these equations.
Thank you.
Don't get me wrong... I'm not saying that they are full of it... yet.
They make some 'interesting' assumptions and stick a lot of stuff in that doesn't seem to apply.
Quoted below is my e-mail to them asking them to explain their formula and such. I may even get an answer... though I doubt it since they sell something. Also, I tend not to ask in the most politic way. Something in my personalilty says, get correct info first, worry about ruffled feathers last.
Anyway, mathematically and physicsally (), I believe that their work is BS, but I'll wait for a response.
Kev
Dear Sirs,
A friend of mine pointed me to your site and asked me to look at how you calculate recoil... we've had quite the discussion on it lately.
If I may ask a couple of questions about your formula...
***FROM www.loadammo.com/Topics/August01.htm
*** (1) I = (Wb*Vb + Wc*Vc)/225400
***Where:
***I = recoil impulse, lb.-sec.
***Wb = bullet weight, grains
***Vb = muzzle velocity of bullet, f.p.s.
***Wc = weight of powder charge, grains
***Vc = velocity of powder gases, f.p.s. (4,000 in these calcs.)
***225400 = unit correction factor (7000 gr./lb.*32.2 f.p.s.2)
***The free recoil velocity of the gun is easily found from the recoil mpulse and weight of the gun:
*** I = Wg/32.2*Vg or rearranging terms:
***(2) Vg = 32.2*I/Wg
***Where:
***Vg = free-recoil velocity of the gun, f.p.s.
***32.2 = acceleration of gravity, f.p.s.2
***I = recoil impulse, lb.-sec.
***Wg = weight of gun, lb.
***The recoil energy of the gun can then be calculated from the kinetic energy equation:
***E = ½*Mg*(Vg)2 or simplifying:
***(3) E = Wg*(Vg)2/64.4
***Where:
***E = free-recoil energy, ft.-lbs.
***Mg = mass of the gun (Wg/32.2)
***Wg = weight of gun, lb.
***64.4 = acceleration of gravity*2, f.p.s.2
***Vg = free-recoil velocity of the gun, f.p.s
1) Impulse in measured in (metric) kilogram meters per second. Here you are using lbs/sec. There is no component for distance as is required. How do you get the impulse then?
2) Why do you have a correction factor and where did it come from? Why is the correction factor 7000 grains/pounds (which doesn't seeem to make sense anyway) * the acceleration due to gravity? Gravity would play litte or no part in a normal one dimesional Newtonian mechanics problems. I realize that recoil isn't purely one dimensional, but your use of 1-d momentum implies that you think it is too.
3) How do derive the 2nd formula? Why is gravity included?
4) Equation 3, where does the 64.4 come from and why? You say that it is accel due to gravity*2, why?
Please tell me how you derived these equations.
Thank you.