Dude you are mixing things up, you wanted the results of the research paper in a graph.
I told you that the paper gives everything but you have to calculate yourself. it isn’t hard.
So let us check the median number of rounds needed for a kill on a P-47 and a B-25. I will only do it for the 20mmHeI M97, since you only need to swap the p_k (one hit kill probability) to that of another round. We assume the plane is fully loaded with fuel.
The formula is given in page 42 where it explicitly says that the formula does not take compounding damage into account. I will write down the formula in the limited capacity as it is possible in this forum
It give a probability for a kill as follows.
P_n = 1 - Product of (1-p_k) n times.
It’s an old way to write
P_n= 1- (1-p_k)^n
Where p_k is the one hit kill probability of the selected round.
We can proof this formula is correct since
p_k is 5.3% according to the table on page 41.
For 10 shots it gives us
1-(1-.053)^10 = 1-(.947)^10 = 1-.58 = .42
That checks out since it corresponds with the table on page 49. Which should be obvious due to the formula given at page 42, but since i cannot trust you to validate anything or even do simple math i am going to do it myself.
So as another validation. We do the same thing for the 30mm mk 108 HEI.
1-(1-.161)^10 = 1-(.839)^10 = 1-.173=.827
And that checks out as well.
So since we now know that the old connotations have been translated correctly. Lets do the actual work.
So since number of shots needed to kill a plane is obvioulsy not a symmetric distribution. We need the median to get the “average” number of shots to kill the bomber.
Pennstate university has a great explanation why we need to do this: 2.2.4.1 - Skewness & Central Tendency | STAT 200
Yes you have to figure this out yourself. But i cannot teach you a semester of basic math on an online forum. So just read it.
The median is easy since it is the value n for which:
1-(1-p_k)^n=.5
For the 20mm hei on the B-25 we get:
1-(1-.053)^n=.5
Solved for n we get n=12.7
So on a fully fuelled b-25 the “average” number of shots you need for a kill is 12.7.
And thats without cumulative damage!!!
So in reality it was actually lower.
But i hope this answers your Post
Now how would you test this in war thunder.
Well, one way would be to shoot b-25s with a single round of M97. Again each time shoot a new b-25. Then count how many b-25s you have attacked until the first one died from a single round. Not that down.
Repeat this until you have a fairly decent sample size. And then take the median of the noted down numbers. Of course you have to reset the counter each time a b-25 dies.
That would be the dumb way to do it.
The smart way is to just use the single hit death chance, since we do not look at cumulative damage,it doesn’t make any difference anyway.
So just shoot a b-25 once and then replace the b-25 and shoot a single shot in to the new b-25.
Do it about 200 times, as this is a good sample size (there is a way to calculate the needed sample size but at max you never need more than around 220 so 2000 will be fine).
So do it 200 times and it should be around 11 one hits. From 500yards rear above of course, since that is what the researchers did.
I did test it a few years ago we stopped at 50 tests as we didn’t get a single one hit kill from 500yrds rear above.
Of course that is anecdotal evidence. We do not have laboratory enviroments in WT where we can actually make double blinded tests.
No it wasn’t from the graphs, o just expected you to be able to read a research paper and do basic math on your own. Don’t worry i will never expect that from you again.
But now that i have shown you how it is done, i hope you can do the rest for the other rounds yourself.