B24 ceiling vs. B17 ceiling

33k in the air · Jan 6, 2023

GregP said:
When loss rates for some month go up, so do losses, exactly in proportion to sorties.

Loss rates could vary considerably by individual mission. The 6 March 1944 mission to Berlin, for example, lost 9.45% of the bombers despatched (69 out of 730). The 8 March 1944 raid on Berlin had a bomber loss rate of 5.94% (37 out of 623). The 22 March 1944 mission to the Berlin area lost only 1.74% of the bombers sent up (12 out of 688). The general rule of thumb was 5%: above that was not sustainable, and below that was preferable. But even a seemingly low 5% loss rate adds up quickly: a force of 100 bombers would be down to 63 aircraft after just 9 missions at a 5% loss rate per mission.

GregP · Jan 6, 2023

Actual losses varied considerably by individual mission, too, right along with loss rate per mission. A LOT of the loss rate depended on the target, the route to the target, how many and which German fighters were operational while simultaneously having pilots, fuel, and ammunition that day.

I get 64 to 66 aircraft remaining after 9 missions at a 5% loss rate, but ±1 or 2 isn't much of a quibble. Basically depends on whether or not you round off the calculated loss downward or upward to the nearest whole number since 0.3 fighters isn't going to either fly, get shot down, or be scrapped.

pbehn · Jan 6, 2023

The point of fighter defence is to make losses unsustainable for the enemy. The enemy chooses missions that keep loss rates to a level they can sustain, or they change tactics.

JDCAVE · Jan 6, 2023

GregP said:
Actual losses varied considerably by individual mission, too, right along with loss rate per mission. A LOT of the loss rate depended on the target, the route to the target, how many and which German fighters were operational while simultaneously having pilots, fuel, and ammunition that day.

I get 64 to 66 aircraft remaining after 9 missions at a 5% loss rate, but ±1 or 2 isn't much of a quibble. Basically depends on whether or not you round off the calculated loss downward or upward to the nearest whole number since 0.3 fighters isn't going to either fly, get shot down, or be scrapped.

Its a simple survival function:

where p s is probability of surviving s operations, s is operations and L is rate of loss.
for 30 operations and 5% loss rate: e to the power of (30*-0.05) = 22.3%

Jim

GregP · Jan 7, 2023

pbehn said:
The point of fighter defence is to make losses unsustainable for the enemy. The enemy chooses missions that keep loss rates to a level they can sustain, or they change tactics.

The enemy choose missions that damage the opponent. If loss rates get too high, they change tactics or stop the mission before losing the entire force.

They don't choose missions based on loss-rate; they monitor the loss rate to see how much pain they can stand before not flying those missions on a regular basis.

By way of example, any mission to Ploesti was going to have heavy losses, but that mission needed to be flown on occasion.

pbehn · Jan 7, 2023

GregP said:
The enemy choose missions that damage the opponent. If loss rates get too high, they change tactics or stop the mission before losing the entire force.

They don;'t choose missions based on loss-rate; they monitor the loss rate to see how much pain they can stand before not flying those missions on a regular basis.

By way of example, any mission to Ploesti was going to have heavy losses, but that mission needed to be flown on occasion.

Exactly, the losses suffered decide how often you undertake such a mission. Also the rate expressed as a percentage can be misleading, the Luftwaffe were shooting down as many aircraft as they could, so if you send 300 or 900 bombers the losses dont increase by a factor of three. Losing 60 aircraft from 300 it is an unsustainable 20% but 60 lost from 1,200 would be a "sustainable" 5%.

GregP · Jan 7, 2023

JDCAVE said:
Its a simple survival function:
View attachment 701408
where p s is probability of surviving s operations, s is sorties and L is rate of loss.
for 30 operations and 5% loss rate: e to the power of (30*-0.05) = 22.3%

Jim

The Rand Corporation had a completely different formula, but that one works if you are interested in survival of the aircraft as opposed to aircrew survival.

So, if we have 100 sorties and lose 5 aircraft, your formula works out to 67.38% probability of survival when, in fact, you lose 5 of 100 aircraft, leaving 95 intact. Something very wrong there.

However, if we assume an "operation" is actually a mission and 1) we lose 5% per mission and 2) we start with 100 airplanes, then we have 22 left after 30 missions, which follows your formula. So, S is not sorties; it is really number of missions, and L is the average loss rate PER MISSION. So, if we start with 100 airplanes, then 0.223 * 100 = 22.3 airplanes remaining, which is 22 airplanes. That works.

The really tricky part is the loss rate, which depended largely on the target, the route, the state of German fighter availability (fighters losses), the effectiveness of German flak crews (AAA losses), and overall reliability of the airplanes (Other losses). They all combined to give you a loss rate. The entire reason for calculating this is to predict losses.

But in the Statistical Digest of World War Two, we HAVE the losses in Europe. So, we don't really need to predict them at all since they are listed. Predicting them was great if you were in the war at the time. But, the war is over and we KNOW the losses and the sorties.

We flew 332,904 sorties (don't know how many missions) and lost a total of 5,548 heavy bombers. That's an overall loss rate of 1.67% for the ETO war. It is also the average loss rate per mission in the ETO for the war.

If we start with 1,000 bombers and fly thirty missions with a loss rate of 1.67%, we get a probability of survival of 60.59%. I calculate 603 bombers remaining after 30 missions (with rounding to whole numbers of losses). That's very close to 60.59% (actually 60.3% with rounding). But ... again ... we KNOW the losses by month, so why predict them? I mean, it works, after a fashion, but we HAVE the loss numbers, so we don't have to predict them.

GregP · Jan 7, 2023

pbehn said:
Exactly, the losses suffered decide how often you undertake such a mission. Also the rate expressed as a percentage can be misleading, the Luftwaffe were shooting down as many aircraft as they could, so if you send 300 or 900 bombers the losses dont increase by a factor of three. Losing 60 aircraft from 300 it is an unsustainable 20% but 60 lost from 1,200 would be a "sustainable" 5%.

5% losses is ALWAYS better than 20 % losses. I think we get that.

But I seriously doubt that any combination of fighters and flak that shot down 60 of 300 aircraft would ever shoot down only 60 of 1,200 aircraft. They already showed they are better than that when they got 60 of 300.

But, hey, maybe.

JDCAVE · Jan 7, 2023

GregP said:
The Rand Corporation had a completely different formula, but that one works if you are interested in survival of the aircraft as opposed to aircrew survival.

So, if we have 100 sorties and lose 5 aircraft, your formula works out to 67.38% probability of survival when, in fact, you lose 5 of 100 aircraft, leaving 95 intact. Something very wrong there.

However, if we assume an "operation" is actually a mission and 1)we lose 5% per mission and 2) we start with 100 airplanes, then we have 22 left after 30 missions, which follows your formula. So, S is not sorties; it is really number of missions, and L is the average loss rate PER MISSION. So, if we start with 100 airplanes, then 0.223 * 100 = 22.3 airplanes remaining, which is 22 airplanes. That works.

The really tricky part is the loss rate, which depended largely on the target, the route, the state of German fighter availability (fighters losses), the effectiveness of German flak crews (AAA losses), and overall reliability of the airplanes (Other losses). They all combined to give you a loss rate. The entire reason for calculating this is to predict losses.

But in the Statistical Digest of World War Two, we HAVE the losses in Europe. So, we don't really need to predict them at all since they are listed. Predicting them was great if you were in the war at the time. But, the war is over and we KNOW the losses and the sorties.

We flew 332,904 sorties (don't know how many missions) and lost a total of 5,548 heavy bombers. That's an overall loss rate of 1.67% for the ETO war. It is also the average loss rate per mission in the ETO for the war.

If we start with 1,000 bombers and fly thirty missions with a loss rate of 1.67%, we get a probability of survival of 60.59%. I calculate 603 bombers remaining after 30 missions (with rounding to whole numbers of losses). That's very close to 60.59% (actually 60.3% with rounding). But ... again ... we KNOW the losses by month, so why predict them? I mean, it works, after a fashion, but we HAVE the loss numbers, so we don't have to predict them.

So what is the formula? I don't see the RAND corporation formula described.

The loss model I have posted is not new. It is a simple exponential decay function.

The RAF used the term "operation" rather than mission. Dad always corrected me, when I was young and used "mission". He never used the term "Mission".

So, S is not sorties; it is really number of missions, and L is the average loss rate PER MISSION.

You are correct. Sloppy use of the term. I will correct original post.

Jim

GregP · Jan 7, 2023

I found the Rand paper on my first search. The formula they used operated on different data. They knew sorties, fighters effectiveness per attack, number of attacks before running out of ammo, and the like. It was very neat, but the data required to use the formula were quite extensive. If you are in the war and flying combat missions, the formula you posted was most likely used, or one close to it. I'll see if I can't find the Rand pdf again. It came right up when I Googled my original search, but now it doesn't. Can't be too tough to find unless it mysteriously got erased right when I was looking for it. Doubt that.

After the war, you can go back and predict loss rates based on declining losses and actually check out your loss model but, during the war, you can't really do that. Predicting losses is what the brass wants to see during the conflict to plan missions and plan replacement aircraft and crews.

Cheers.

pbehn · Jan 7, 2023

GregP said:
5% losses is ALWAYS better than 20 % losses. I think we get that.

But I seriously doubt that any combination of fighters and flak that shot down 60 of 300 aircraft would ever shoot down only 60 of 1,200 aircraft. They already showed they are better than that when they got 60 of 300.

But, hey, maybe.

It is the same number of losses. On the famous Mosquito raid to shut down Goerings radio transmission 6 aircraft in two flights made the attacks, one aircraft was lost, so it can be expressed as 17% of the whole mission or 33% of one part of it. However you express it the loss rate in percentage is unsustainable, but it was one plane and crew. Using single engined aircraft, a raid of 300 planes has more than enough targets to use all ammunition trying to shoot them down, if you have to refuel and re arm you cant re-join the fight, so unless the LW were holding forces in reserve the numbers shot down dont automatically increase. Since statistically it took a certain number of hits of each calibre to bring an aircraft down, having more aircraft would increase the number damaged and decrease the number shot down.

GregP · Jan 7, 2023

Let's say we disagree on that one and let it go since we know the numbers for WWII.

If the Flak is an 88 (or likely anything above 30 mm), one hit is all it takes to bring down even a larger aircraft, assuming a more or less center-mass hit. Only smaller-calibre stuff needs multiple hits; machine guns and maybe up to 20 mm cannons.

But, you might be right. While I love to shoot, I'm more into airplanes than the effects of flak. Fortunately, when I'm in a Cherokee, nobody is shooting at me. Good thing since it's SLOW and would make a great target.

Cheers.

pbehn · Jan 7, 2023

GregP said:
Let's say we disagree on that one and let it go since we know the numbers for WWII.

If the Flak is an 88 (or likely anything above 30 mm), one hit is all it takes to bring down even a larger aircraft, assuming a more or less center-mass hit. Only smaller-calibre stuff needs multiple hits; machine guns and maybe up to 20 mm cannons.

But, you might be right. While I love to shoot, I'm more into airplanes than the effects of flak. Fortunately, when I'm ina Cherokee, nobody is shooting at me. Good thing since it's SLOW and would make a great target.

Cheers.

No argument with a direct hit from an 88mm that explodes on contact, most damage from flak was from shells that exploded in the air which then becomes a function of accuracy, tightness of the formation and statistics.

Timppa · Jan 8, 2023

JDCAVE said:
Its a simple survival function:
View attachment 701408
where p s is probability of surviving s operations, s is operations and L is rate of loss.
for 30 operations and 5% loss rate: e to the power of (30*-0.05) = 22.3%

Jim

(1-0.05)^30 =0.215 (21.5%) is the correct answer.
Using binomial and Poisson distributions give nearly the same results though.

GregP · Jan 8, 2023

Timppa said:
(1-0.05)^30 =0.215 (21.5%) is the correct answer.
Using binomial and Poisson distributions give nearly the same results though.

I get 22.313% which agrees with JDCAVE, and the formula is P = e^(30*(-.05)) = e^(-1.5) = .22315. The 30 is missions, which JDCAVE calls operations, and the .05 is the aggregate average loss rate per mission.

I could be wrong, but the results agree with the calculations I have used for 40+ years for this type of survival calculation, though with different definition of the variables. It isn't the probability of survival of a crew member, it is the probability of the aircraft getting shot down or disabled so it goes down and is considered lost.

JDCAVE · Jan 8, 2023

Timppa said:
(1-0.05)^30 =0.215 (21.5%) is the correct answer.
Using binomial and Poisson distributions give nearly the same results though.

Hmmm! Interesting Timppa. Thank you for that post. I stand corrected! The exponential decay model I used is an approximation (in my ignorance). Timppa's model exactly matches the discretized results at all steps through 30 Operations (missions if you like). The exponential decay model results in slightly higher survival rates.

"Personally, I am always ready to learn, although I do not always like being taught."-Winston Churchill.

Jim

GregP · Jan 8, 2023

If you calculate out the two formulas, they diverge a very small bit, but Timppa's equation is exact. That is, it is exact if you can have a partial loss, like 0.6512478 losses. Unfortunately, we can't have partial losses. We lose an airplane or we don't.

If you round off, do you round the losses downward, to the nearest whole number, or add 0.49 to the loss calculation and round to the nearest whole number? Do you round down or round up? Depending on how you calculate and round it, you can get anywhere from 214 (pessimistic) aircraft left out of 1,000 on the 30th mission all the way up to 230 (optimistic).

Most statisticians would come up with 214 remaining after 30 missions with a 5% average loss rate and would likely use a cumulative survival formula to do it. No sense coming up short on resources, unless you are recruiting pilots.

Spreadsheet attached.

JDCAVE · Jan 9, 2023

Greg: These are models, not an accounting of what will actually happen to get to that result. So while having partial aircraft is hard to wrap one's mind around, not rounding is the correct approach. "adding 0.49" biases the outcome and is not an appropriate addition to the equation.

So the model is to provide an estimate of the p survival of aircraft or aircrew or whatever, given whatever assumption you wish to impose. i.e. what is the probability of survival a "tour of 30 operations" given a specified rate of loss at which point the crew is screened and move to a non-operational unit. It does not necessarily project how you get there, during the course of a tour, only what the expectation would be of the end result.

That's my interpretation.

Jim

GregP · Jan 9, 2023

This discussion would be a model for whether or not an aircraft is lost on operations to any cause, not for aircrew survival in the event of an aircraft loss.

In a B-17, we have a crew of 10 people, not all of which are always lost when the aircraft is lost. In a B-24, we have a crew of 11. The B-24 has a separate nose gunner while the B-17 has a nose gunner/bombardier. Sometimes a crewman is lost even when the aircraft gets home relatively undamaged. Sometimes, all survive when the aircraft is lost. Loss of a crewman is a completely different discussion.

The probability of survival for any particular position was a well-debated subject during the war.

JDCAVE · Jan 9, 2023

GregP said:
This discussion would be a model for whether or not an aircraft is lost on operations to any cause, not for aircrew survival in the event of an aircraft loss.

In a B-17, we have a crew of 10 people, not all of which are always lost when the aircraft is lost. In a B-24, we have a crew of 11. The B-24 has a separate nose gunner while the B-17 has a nose gunner/bombardier. Sometimes a crewman is lost even when the aircraft gets home relatively undamaged. Sometimes, all survive when the aircraft is lost. Loss of a crewman is a completely different discussion.

The probability of survival for any particular position was a well-debated subject during the war.

Sure it is! A probability of survival, is a probability of survival, whether it is fatalities, wounded, missing in action, accidents, and so on. The starting point is different. I agree that a bomber loss is not the same as fatalities, however, the model (whatever model you choose) is the same. The parameterization is different, that's all.

Jim

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