Hi Knegel,
>the lethality dont have to do anything with hitprobability!!
>It have to do with kill probability!
We agree.
>If your bullet with 5 times as high lethality simply dont hit, cause the much smaler hitprobability, its worthless, at least less worth than the smaler gun.
The hit probability Ph is the same. It has nothing to do with rate of fire.
>Specialy if the smaler gun is enough to strike the kangaroo down.
To be exact: If the lethality of the greater gun is so high that it achieves a Pk of 100% and still has some lethality reserve against the kangaroo. That is the overkill phenomenon I wrote about, and it certainly does not apply to the 20 mm cannon.
>Me neighter,t hats why i wrote:"Better bring 3 emenys damaged downward, than 1 exploded"
Well, I misunderstood that as normal WW2 air force use was "kill" to describe a target that was probably shot down, "probable" a target that might have been shot down, and "damaged" a target that was not shot down. Anyway, then your point is overkill again, and the 20 mm cannon does not overkill.
>Often the decission was based on the order.
How about some clear historical evidence?
>>"What's the desired event? "Plane is shot down"? Nothing to gain from rate of fire if it comes at the cost of lethality of the individual rounds. Nothing to gain from an increase in dispersion either."
>If a .50cal is good enough to bring a target down with some lonly hits, cause missing selfsealing tanks or cause the bullet was strong enough to get through the plating( according to all i did read, this was the case vs Japanese planes and smal fighters), there is nothing to gain from lethality at the cost of rate of fire.
Not anything that could happen will happen. You could bring down a Zero with a single 12.7 mm hit, but even the Zero is not all unprotected fuel tank. Even the chance for hitting the fuel tank, never mind doing lethal damage to hit, might be as low as 15% even if you manage to hit the Zero. A 20 mm mine shell can do serious damage on 90% of the airframe, so it will end up with a higher Pk in the end, making cannon superior aginst Zeros as well. (Interesting side note: The un-armoured, un-protected, machine-gun-armed Curtiss CW-21 light-weight fighter fought some battles against cannon-armed Zeros in the Pacific, suffering much heavier losses than they could afflict to the Zeros.)
Anway, it's all rolled up in this formula:
Pdestruction = Pkill * Nf * Ph
>Only if you consider the big dispersion even while good aiming, and so the high quote of not hitting bullets, you can see in what way the hitprobability get increased by the rof.
Hit probability does not increase through rate of fire. Basic stochastics.
>Its simply more likely that a bullet hit, if more rounds fly toward the target.
That's the completely different event "at least one bullet hits the target". This is not connected to the probability of destruction.
>Its like shooting with rifle to a bird and a shotgun. The hitquote of the rifle may be similar and the destructive power is much more big, but the hitprobability of the shotgun is much more big on a moving target.
Actually, it's not. The trick about the shotgun is that it shoot a vast number of small projectiles, most of which miss. Basic stochastics.
What you are considering is the event "at least one of the shotgun pellets hits", which does indeed have a higher probability, but makes the example different from the MG vs. cannon example.
The probability for each shotgun pellet to hit is just the same as the probabilty for each rifle bullet to hit. It's just that the shotgun fires hundreds of projectiles in a fraction of second, but most of them miss just the same as hundreds of rifle projectiles would miss.
The shotgun analogy typically is completely misunderstood by simulator pilots.
>As higher the rof, as longer the practical effective range was(the theoretical range of a .50cal got assumed by the USAF with 900yard).
Best keep that out of the discussion since "effective range" is another can of worms. Just to illustrate this: What is the effective range of one 12.7 mm machine gun? And if rate of fire increases effective range, what is the effective range of eight 12.7 mm guns? And of course lethality enters the picture, too, since "effective" implies the capability not just to hit but to damage upon hitting. (If I remember correctly, when I asked Tony, he mentioned that there was no scientific definition of the term "effective range" anyway.)
>I think you mix up hitquote with hitprobability.
Read this first:
http://de.wikipedia.org/wiki/Erwartungswert(First paragraph could be enough.)
I regret that there seems to be no similar concise explanation in the English Wikipedia, it would be quite useful for our discussion here.
Regards,
Henning (HoHun)
