Novairia wrote:Sorry for taking so long, had some stuff to do first, then I had pizza! Now, down to brass tacks.
The Holy Dominion of Inesea wrote:So, would A 96 Kilogram round traveling at approx. 1,600 Meters persecond Do ANY damage?
Well..
(96kgx0.5)x(1,600^2) = (48x2,560,000) = 122,880,000 J of kinetic impact energy at muzzle escape OR 122.88 MJ when LEAVING the gun.
Then by using the ballistic conversion chart with a ballistic coefficient avg of .172, at an assumed 45* angle of fire, at sea level, with no wind, at "
the 66%-69% hit range (47 kilometers) " the round has lost all but 0.06 of its initial muzzle energy.
Meaning it has 7.37 MJ left. Which then converts to 7370 Kilonewton per meter [kNm] of force.
Then we gather the data needed next. Such as shell radius for indentation (a^2/R=d), the shell cone angle (used for Tan), and SAE 4340's MoE rating (E).
Then by using the "Contact between a rigid conical indenter and an elastic half-space" equation:
SO: Fn= (((2/3.14159265)x205,000,000,000 Pa)x(0.0009996^2/tan(70)) = 47,462.69 Newton meters
Meaning that it takes 47.5 kNm of force to indent SAE 4340 steel 0.1cm with the Paris Gun's 238mm steel jacket round (using equal grade steel).
As such, to pierce the 26cm side armor, it would require aprox 12,350kN of force. And the 18cm top armor would require 8550kN of force [and that was only IF the incoming round also did not deform, which it will].
So no, at that range the 238's cant hurt the
ICMF Armageddon's armor.. Also remember tho that the 238mm shell wasnt ever designed to be AP, but rather it was a High Explosive delivery system to wreak buildings.
PS: You would not believe the stuff I had to look up in my books and the internet to get that all figured out.