Engine Survivability

[HoHun]

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)

[mad_max]

Ho Hun posted:

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.

We all know of this one. 21 holes from them, but it's just one.


P-47 Thunderbolt: Aviation Darwinism - The Cradle of Aviation Museum - The Cradle of Aviation Museum

[HoHun]

Hi Madmax,

>We all know of this one. 21 holes from them, but it's just one.

Wow, thanks a lot Of course I knew of this incident, but I hadn't been aware that he collected a total of 21 cannon hits!

As a reality check for my above reality check, the likelihood of surviving 21 cannon hits under the above assumptions is:

Psurvival = (1 - (0.05 * 0.9 + 0.95 * 1/6)) ^ 21 = 0.84%

These are not odds anyone would like to face! However, with the great number of fighters exposed to cannon fire in WW2, I think we should have heard of similar cases if roughly one in hundred fighters exposed to such a barrage would have survived like Johnson's. So I'd say this case by its exceptional nature confirms that my above assumptions were conservative.

Unless of course other cases were not as widely published as Johnson's ... I'm not sure how likely that is, though.

Regards,

Henning (HoHun)

[mad_max]

Of the few that returned I've read of seemed most were under 5-6 hits. Let's see if I can find them on the old HD in the corner.

[slaterat]

Here's one from Bloody Shambles vol 2 p 320

RAF pilot Sandeman Allen

" The zeros came in underneath and we had a simple target for the first few vital moments, after which we were so heavily out numbered that we got into serious trouble. I was credited with 2 zeros destroyed ,one probable and one damaged, but crawled back to the aerodrome with 28 cannon shell and 43 bullet holes in the machine..."

The hurricane airframe was amongst the toughest single engine fighters of the war.

Slaterat

[HoHun]

Hi Slaterat,

>crawled back to the aerodrome with 28 cannon shell and 43 bullet holes in the machine...

Amazing! But can we be sure that the opponents really were Zeros in this case? If they were Ki-43s, the "cannon shells" would be explosive 12.7 mm rounds with much inferior destrcutive power, of course.

(Just asking because in "Hurrcane over Sumatra", the author points out that the RAF forces were completely unaware of the existence of the Ki-43 at the time and invariably identified the type as Zero - or "Navy Nought", as they put it

Regards,

Henning (HoHun)

[slaterat]

Accordind to Shores and Cull, the opponants were the 22nd air flotilla , so it was definately zeros that the Hurricanes were engaging. Also at this time , March of 42, I don't think that the Ki43-IIs were deployed yet.

I"ve got a few of Terrence Kellys books, Hurricane vs Zero [Battle for Palambang] , Hurricane and Spitfire Pilots at War and Nine Lives of a Fighter Pilot. I've got Hurricane Over the Jungle on order. All very good books, well researched and quite insightful . Good companions to the Bloody Shambles series .

Slaterat

[HoHun]

Hi Slaterat,

>Accordind to Shores and Cull, the opponants were the 22nd air flotilla , so it was definately zeros that the Hurricanes were engaging.

OK, so it sounds like we can be sure it were in fact 20 mm cannon shell hits in this case.

The Japanese 20 mm HE shells were not the most powerful around, but 28 hits certainly is a great record ... big points for Hurricane toughness here.

Here is a comparison of the total energy of some 20 mm shells, based on the data on Tony's site:

MG 151/20:      20x82 HE(MX):        161.1 kJ
VYa-23:         23x152B HE:          155.6 kJ
MG 151/20:      20x82 HE(M):         142.5 kJ
MG-FF/M:        20x80RB HE(M):       135.6 kJ
VYa-23:         23x152B API:         111   kJ
Hispano II:     20x110 HE (Mk II):   106.2 kJ
Hispano V:      20x110 HE (Mk V):    102.9 kJ
20mm Type 99-2  20x110RB HE:          96.5 kJ
Ho-1 / Ho-2:    20x125 HE:            92.4 kJ
20mm Type 99-2  20x101RB HE:          78.9 kJ
20mm Ho-5       20x94 HE:             74   kJ
20mm Type 99-1: 20x72RB HE:           65.9 kJ
20mm ShVAK      20x99R HEI:           61.5 kJ
MG 151/20:      20x82 API:            50.6 kJ
MG 151/20:      20x82 HET:            50.4 kJ
MG-FF/M:        20x80RB API:          40.3 kJ
MG-FF/M:        20x80RB HEIT with I:  40.2 kJ
20mm ShVAK      20x99R API:           37.7 kJ
20mm Ho-5       20x94 AP:             27.4 kJ
12.7 mm M2:     12.7x99 API (M8):     21.3 kJ
12.7 mm Ho-103  12.7x81SR HE:         13.8 kJ
7.7 mm Vickers  7.7x56R I:             5.5 kJ

(I have included a 23 mm cannon, the 12.7 mm Browning, the 12.7 mm Ho-102 and the 7.7 mm Vickers for comparison purposes.)

The 20mm Type 99-1 with the 20x72RB HE offers "just" 65.9 kJ of total (kinetic+chemical) energy, compared to 106.2 kJ for the Hispano Mk II HE shell or 135.6 kJ for the MG FF mine shell.

Regards,

Henning (HoHun)

[kool kitty89]

Accordind to Shores and Cull, the opponants were the 22nd air flotilla , so it was definately zeros that the Hurricanes were engaging. Also at this time , March of 42, I don't think that the Ki43-IIs were deployed yet.

The Ki-43-Ib was armmed with 1x 7.7mm and 1x 13mm MG, and the Ki-43-Ic was armed with 2x 13mm MG's. (the same as the Ki-43-II)