Paul Lakowski Posted January 13, 2004 Share Posted January 13, 2004 Heres a question that has vexed me for ages! What is the velocity down range [iE 500m, 1000m & 2000m] for the following weapons? German 50mm L60 & 50L42 Russian 76mm L41 gun Russian 85mm gun Russian 100mm Gun All firing AP/APC ammo? 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted January 13, 2004 Share Posted January 13, 2004 I suppose you could relate penetration to 1/2*mV^2 and find V. http://www.waffenhq.de/panzer/kwk42.html Data like this would help. [ January 13, 2004, 09:50 PM: Message edited by: Mr. Tittles ] 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted January 13, 2004 Share Posted January 13, 2004 That reminds me: Does any of the local math gurus have the formula to compute how wind resistence slows down a passively moving object? I tried to get to the formula but there was an integration step I couldn't cross with my current algebra rating. I tried googleing but to no avail. (I know that the formula will be crap about the speed of sound, but one neglecting that would be better than nothing) 0 Quote Link to comment Share on other sites More sharing options...
flamingknives Posted January 13, 2004 Share Posted January 13, 2004 That sort of thing is best done using an interative formula where you feed the answer back into the formula for each time step. 0 Quote Link to comment Share on other sites More sharing options...
Simon Wagstaff Posted January 13, 2004 Share Posted January 13, 2004 Formula for air resistance is 1/2DpAv^2 | D is the drag coeff. for the shape of interest (value < 1, something you find experimentally or look up; p is the density of air; A is the cross sectional area of the shape; v is the velocity squared. This provides a value for the force in your F=ma=m*dv/dt. Substitution yields dt=-2mdv/[DpAv^2]. 2m/[DpA] is constant, so you have to integrate dt=dv/v^2; which yields -1/v. Combining the contant with the integral solution will give the time rate of change of velocity due to drag. Itegrating again will yield time rate of change of position. It involves a ln[v]. Hope that helps. Oh, and this only works for ideal gases. 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted January 14, 2004 Share Posted January 14, 2004 Thanks Simon. Unfortunately that is exactly where I was I know the wind resistence and know what integrations to make to get the speed over way out of it. Unfortunately that integration turned out to be bejond my rusty algebra. Note that I want speed over way, not time. flamingknives, just computing it out in steps is not sportive. Sane, but not sportive 0 Quote Link to comment Share on other sites More sharing options...
flamingknives Posted January 14, 2004 Share Posted January 14, 2004 Technically, being as it's an indefinate integral, 1/(v^2) dv should be -1/v + c. Substituting a known time and velocity (e.g. at the muzzle), you then get the value of c. 0 Quote Link to comment Share on other sites More sharing options...
rexford Posted January 15, 2004 Share Posted January 15, 2004 Originally posted by Paul Lakowski: Heres a question that has vexed me for ages! What is the velocity down range [iE 500m, 1000m & 2000m] for the following weapons? German 50mm L60 & 50L42 Russian 76mm L41 gun Russian 85mm gun Russian 100mm Gun All firing AP/APC ammo? German 50mm AP and APC and Russian AP rounds follow a relationship where penetration is proportional to velocity raised to 1.43 power, or velocity is proportional to penetration raised to 0.7 power. This comes from the DeMarre equation. Find the 0m penetration for each of the AP and APC rounds, divide the penetration at range by the 0m figure, raise the result to the 0.7 power and multiply by the muzzle velocity. That will be the velocity at range. Russian flat nose APBC rounds follow a more complicated equation that can be found in our book. 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted January 15, 2004 Share Posted January 15, 2004 Originally posted by flamingknives: Technically, being as it's an indefinate integral, 1/(v^2) dv should be -1/v + c. Substituting a known time and velocity (e.g. at the muzzle), you then get the value of c. No, it's a lot more complicated. Just for starters, because the wind resistence force is already quadratic to the speed, the power drawn from the kinetic energy of the projectile is cubic. That instantly turns the base formular for wind resistence into a small fragment mess like a 3" mortar shell on caffeine. 0 Quote Link to comment Share on other sites More sharing options...
rexford Posted January 15, 2004 Share Posted January 15, 2004 Velocity loss per second is equal to: air density x velocity squared x diameter squared x form factor x drag coefficient divided by projectile weight A form factor is needed because actual projectiles are rarely the same as the rounds used to generate the drag coefficient vs velocity curves. Major changes encountered include nose length relative to diameter, secant ogives instead of tangent ogives and whether ammo has a boat tail (standard Type 1 uncapped AP round used for curves has a flat base while many AP rounds used during WW II has a boat tail which reduced air resistance). A really rough estimate for velocity at range can be made using small increments of time (like 0.1 seconds or less) and going through iterations. A. Initial velocity at muzzle - velocity loss per second at 0m times 0.1 second = first estimate for 0.1 seconds. B. Since initial estimate used muzzle velocity and velocity at end of 0.1 seconds is lower, must use average velocity within interval (basedmuzzle velo plus first estimate at 0.1 sec.) C. Use average velocity for first 0.1 seconds from step B. to recalculate velocity loss within interval. D. Keep going until changes are acceptably small There is a computer program that automatically does all of this, generated by William Jurens and included with his article TERMINAL BALLISTICS FOR THE MICRO-COMPUTER which was published in Warships International. My copy is buried in boxes in a storage shed or I would offer to share some copies. 0 Quote Link to comment Share on other sites More sharing options...
flamingknives Posted January 15, 2004 Share Posted January 15, 2004 Originally posted by redwolf: </font><blockquote>quote:</font><hr />Originally posted by flamingknives: Technically, being as it's an indefinate integral, 1/(v^2) dv should be -1/v + c. Substituting a known time and velocity (e.g. at the muzzle), you then get the value of c. No, it's a lot more complicated. Just for starters, because the wind resistence force is already quadratic to the speed, the power drawn from the kinetic energy of the projectile is cubic. That instantly turns the base formular for wind resistence into a small fragment mess like a 3" mortar shell on caffeine. </font> 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted January 15, 2004 Share Posted January 15, 2004 Sorry, guys, I didn't want to get snippy, I had to eat chicken wings before I posted, and I hate them. No really. I should dig out the formular stages I had so far and post them. The fact that an iterative method was published by a magazine is reassuring, though, maybe I'm not the only one who can't get past that integration. Thanks, guys. 0 Quote Link to comment Share on other sites More sharing options...
Simon Wagstaff Posted January 15, 2004 Share Posted January 15, 2004 If you do not want a formula with respect to time, then set your F=ma | a= [v*dv]/[dx]. 0 Quote Link to comment Share on other sites More sharing options...
rexford Posted January 16, 2004 Share Posted January 16, 2004 Originally posted by redwolf: Sorry, guys, I didn't want to get snippy, I had to eat chicken wings before I posted, and I hate them. No really. I should dig out the formular stages I had so far and post them. The fact that an iterative method was published by a magazine is reassuring, though, maybe I'm not the only one who can't get past that integration. Thanks, guys. In the next few days I'll post an example of how the iterative method works. Oue book provides a simpler way to estimate velocity, where the equation is: velocity = muzzle velocity x 2.718282 raised to the following power ("meters range" x 0.7 x "K") The above equation works for rounds where the penetration is proportional to velocity raised to 1.43 power, which the equation does work for Russian flat nose windscreen capped APBC ammo. The thing about Russian APBC is that is follows an odd curve for penetration-vs-velocity against homogeneous armor, but lines up well with the penetration = velocity raised to 1.43 power against face-hardened armor, and the face-hardened penetration curves are what we used for APBC "K" figures. We analyzed the "K" factors for a wide spectrum of WW II rounds based on firing tables and ballistic documents, penetration curves and analysis using a trajectory computer program (BASIC language) that uses the iterative method. The results match the firing tables reasonably well. Lorrin [ January 16, 2004, 07:47 AM: Message edited by: rexford ] 0 Quote Link to comment Share on other sites More sharing options...
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