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Kharan -> (9/28/2001 11:46:00 PM)

Sorry, Paul, I was ignorant that you were leading the v7 OOB development. Knowing it's you pretty much eliminates my concerns. And the "prior achievements" were not yours, although you did list them in a past thread (don't want to dig up that old dirt, probably should've just kept quiet).
quote:

I even lowered the vehicle FT values just for you
LOL, you just had to do that . *grumble grumble* thanks.




Charles2222 -> (9/29/2001 12:11:00 AM)

I will add that I seen, but can't find it now, a diagram which I believe fprado's site had. This diagram was from a Tiger manual to all appearances. Some of the print of the diagram was very tiny, particularly the ranges it was stating, but the target showed some of the fire coming in what appeared to be between 30 and 45 degrees from that longer range (probably 3000yds). While many of the fire demonstrated was at near flat trajectory, the longer ranges were considerably arched. It seems a lot of people would not believe what that Tiger diagram showed about the arch of their further shots. I'm of the thinking that this is because a lot of nations may had refused to train their tankers to fire at longer ranges, due to not having powerful velocity in the first place, and also because it was easier to train people to fire on flat trajectory only, plus of course the German superiority in optics had to aid in the idea of firing from great distances. What this means, is that if the shell were to hit in a 30 degree trajectory, that suddenly the tank with the 30 degree slope is in the worst slope possibility. Some monthes ago I mentioned that in such an instance a zero degree slope on a tank would be better, that position was countered by the statement that if the 30 degree shell had not penetrated, that it would bounce for the most part down instead of up (as with flat trajectory shots on sloped armor), but the more I think about it it shouldn't matter a whole lot (I could be wrong of course). Let's say the upper hull is hit froma 30 degree shell on a zero degree hull. On such a hull, it will either bounce into the ground, the shell would've been largely damaged on original impact, or it would hit a lower portion of the hull with less impact, where hopefully the target had just as much hull armor. OTOH, if a flat trajectory shot were to hit a hull slope of 30, we are told it will bounce up. While there is less things to hit up, than down, there is still the chance that the turret will receive the bounce or the gun itself. What I imagine, with a shell hitting at 30 degrees, is that there are some instances where a tanker in a Tiger, for example, was actually better off firing on well sloped tanks, such as the T34, from a further distance, where the slope was all but nullified, assuming the loss in some accuracy was worth the gain in penetration through a flat hit.




Paul Vebber -> (9/29/2001 1:42:00 AM)

Thanks Kharan! I had a feeling that was the case.
Pz Leo I will check Lorrins book (which has proceedures for calculating the angle of trajectory) but even at a 2000 yards I don't think its a great as 7 degrees. Certainly not 30 - which I'm sure was done for emphasis in the diagram. OK I found some of Lorrin's equations in an old email: velocity at range expressed as muzzle velocity x 2.71828^(0.7 x "k" x range in meters)
=606m/s (using -.000174=k) at 2000m Time of Flight to 2000m(is
flight time = range (meters)/(0.5 x (muzzle velocity + velocity at range)) 2000/(.5x(773+606)) = 2.9 sec tangent gun elevation = 0.5 x 9.81 x (flight time to aimed range)squared/aimed range =.5x9.81x2.9^2/2000 = .0206 = 1.18 degrees There are some tweaks to thiss, but unless I made a major error it is certainly not much more than 1.25 degrees barrel elevation to get to 2000m and the angle at the other end will be slighty more, but certainly not 7 degrees. I will double check this tonight. Note that the random +/- 10 degrees is not a uniform random variable, but "triangular" distribution from adding several "die rolls" together so most results will be clustered toward the mid region. Note that the table you post is for a single T/D ratio, the multipliers vary based on T/D, (though what you post seems to be close to the "rule of thumb" 1/cos(angle)^1.4 for unknown T/D. Using this rule the 88L71 with 230mm @0 is estimated to pen 190. Is teh 170 number an "experimental result" if so it is probably due in part to the skewed distribution I mentioned in assigning an actual penetration number for a given shot (ranging in effect from +10% to -15% again in a distribution that has small tails - this represents the shot - to - shot variability in muzzle velocity, physical characteristics of the projectile, etc. Note that a "vulnerable location hit simply reduces the effective armor after all calculations by 50%. As to ammo type, you won't be able to select it specifically, but will be able to assign "rules" to use certain types in certain situations (based on target type, range etc) You are Regimental Commander (or Battalion)in Combat Leader, not the tank commander I will shift this thread up to TO&E and Lorrin may chime in.




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