Hi,
"The energy comparison shows us that each cannon hit is almost five times as powerful as a 12.7 mm hit, so cannon still do more damage."
If the .50cal hit unprotected not selfsealing fueltank, at the end it cause the same result like the 20mm, same count for the pilot.
Since the pilot and the fueltank cover a pretty big part of a plane, the question is how much different is the effect?
"You don't merely lack the language, you lack the basic concepts described by the terms that don't even exist in everyday language. A teacher might be inclined to explain the concepts you are lacking more thoroughly in plain language here, but I am no teacher, and I don't consider your "know-it-all, don't-need-math" attitude a good basis for teaching anyway."
This is what you wrote before:"........you don't have the vocabulary to talk about the different aspects of random experiments so that you don't understand my explanations."
Strangewise i dont say i know all, its rather you who dont read exact what i wrote, and without to waste a though that my therms could offer a correct describtion with for you not common words, you go on and refer to absolut not relevant things.
After you took notice that i talk about the gun hitprobability, you dont say, 'ah ok, now i understand', you dont come down from your high horse. You immediately switch and claim 'hitprobability is nothing, killprobability is all'.
"The energy comparison shows us that each cannon hit is almost five times as powerful as a 12.7 mm hit, so cannon still do more damage."
How you know that the cannon do more damage, if there is a not to smal possibility that the .50cal hit a fueltank or the pilot? In this case there dont will be a real different of the result.
"A good method of estimating the Pk of a round is to examine the total energy it carries because this energy will be converted into destructive power as soon as it strikes the target:"
I dont think this is a good method of estimating the Pk, cause you dont take the target into account.
As i wrote before, i think the killprobability change with the target, not only with the damagepower.
A not protected target, without selfsealing tanks offer a much higher killprobability for a .50cal than a on a plane with selfsealing tanks and good amor.
While i guess the cannon round provide a much more constant killprobability, cause it mainly aim for the structure of the plane and it isnt a big different if a 20mm hit a selfsealing tank or a not selfsealing tank, same count for the amor.
On a unprotected target specialy smal MG´s have a much higher kill probability than on a same sized, protected target.
If i hit the head of the pilot, it dont matter if i do this with a .30cal, .50cal or 20mm. The killprobability is pretty the same in this rare but without plating possible example. With plating to cover the pilot, the killprobability in this case will will be much higher for the cannons.
Btw. Somewhere i did read the Luftwaffe did estimate around 4 x Mk108 hits are needed to take down a B17 and around 20 x MG151/20mm to get the bombwer down.
catrige power MK108 = 58 * 3 = 174
catrige power MG151/20 = 12 * 20 = 240
I also did read that on smal fighters 1 * MK108 was enough, while around 4 * MG151/20 rounds did the job.
1 * 58 = 58
4 * 12 = 48
What a suprise, on smal(less tough) targets the effectiveness of the smaler gun increase in relation to the big gun.
The problem isnt to calculate something, its to know what we need to consider while the calculation.
Greetings, Knegel
