Hi Glider,
>Even with the poor firepower of the Ki43, Hurricanes could be shot down merely by one hole in the radiator.
Here is an interesting article on Ki-43 armament:
Nakajima Type 1 Model 1 Army Fighter (Ki 43-I) Armament
With regard to the "one hole in the radiator", the question is not so much one of the number of holes, but rather one of the number of rounds fired to achieve that single hole.
As we know, the 7.7 mm machine gun - technically entirely sufficient to punch holes in radiators - was recognized as badly inadequate as an air-to-air combat weapon in WW2.
This means that as a reality check, we can conservatively assume that - for example - four 20 mm cannon were better than 12 7.7 mm machine guns. (This is the "exchange rate" when the Typhoon received increased armament.)
The total probability of a kill is:
Pk = n * Ph * (Ac * Pkc + Au * Pku)
Pk = total probability of a kill
n = number of rounds fired
Ph = hit probability
Ac = percentage of critically vulnerable target area as projected in the plane perpendicularly to the line for fire
Au = percentage of uncritically vulnerable target area in the same plane
Pkc = average kill probability for a single round hitting a critical area
Pku = average kill probability for a single round hitting an uncritical area
If we consider a burst of a single second for each of the Typhoon's battery variants, this comes down to:
Pk7.7 = 240 * Ph * (Ac * Pkc7.7 + Au * Pku7.7)
versus
Pk20 = 40 * Ph * (Ac * Pkc20 + Au * Pku20)
Assuming hit probability to be equal for both weapons, it follows that
Pk7.7 / Pk20 = 6 * (Ac * Pkc7.7 + Au * Pku7.7) / (Ac * Pk 20 + Au * Pku20)
Due to the cannon armament being historically considered superior, we also know that
Pk7.7 / Pk20 < 1
Now we can make assumptions on the target area presented by the radiators in relation to the overall target area. Let's say for example that
Ac = 5%, Au = 95%.
By definition, 7.7 mm hits against uncritical areas have
Pku = 0
(Note: This ignores cumulative damage. This makes our estimate even more conservative in favour the capability of 7.7 mm guns against critically vulnerable areas.)
So we get:
6 * (0.05 * Pkc7.7) / (0.05 * Pkc20 + 0.95 * Pku20) < 1
Now there is a German statement regarding fighter vulnerability claiming that it took an average of six random 20 mm hits to bring down a fighter. That seems conservative since I've never heard of a single-engined fighter coming back with more than eight 20 mm hits, but we mean to be conservative for now.
Accordingly, Pku20 can be considered roughly 1/6. Additionally, a hit by a 20 mm projectile to a vulnerable area apt to be critically damaged by a 7.7 mm round is likely to be close to 100% critical, so let's say
Pkc20 = 90%
So we get:
6 * (0.05 * Pkc7.7) / (0.05 * 0.9 + 0.95 * 1/6) < 1
or
0.3 * Pkc7.7 / (0.045 + 0.0158) < 1
or
4.93 * Pkc7.7 < 1
Thus we arrive at
Pkc7.7 < 20.3%
In short, historical experience (conservatively analyzed) shows that in reality, an average of at least five hits by 7.7 mm machine gun round were necessary to destroy a part like the radiator that "could be pierced by a single machine gun bullet".
Now it's possible to make different quantitative assumptions than those I made above and arrive at different figures, but the real lesson here is that one should not confuse the optimum effect on the target with the normal effect on that target.
Of course the notion of the "single-bullet kill" is an attractive thought model, but as Priller noted in his "JG 26" history, during the Battle of Britain some Messerschmitts came back to France safely with as many as 80 bullet holes. Others might well have gone down to single bullets - but the effectiveness of air-to-air armament has to be measured by the average results, not by the extremes.
Regards,
Henning (HoHun)