Aircraft Maneuverability (Final? Adding PR Spitfire)

[el cid again]

However, applying the same algorithm to the P-38 that you apply to the B-25 is a little bit ridiculous on the surface if you even take more than a casual look at the aircraft. One is obviously not going to have the drag coefficient that the other has, or it would not have had 150+ of the top speed. It may have something to do with the coding of what type of AC it was as well. I would hope that Matrix would have figured out that applying the same code to a twin engine fighter that they would to a twin engine bomber would be a little absurd, sort of like being a little bit pregnant.

In defense of Matrix - they DID EXACTLY what you wanted. I am the bad guy who said "this is wrong." THEY treat twin engine fighters and night fighters AS IF they had ONE engine! No divide by factor at all. Only bombers and transports got treated the other way. I think you are perfectly - well 99% - wrong. The reason is right in your remarks - the algorithm DOES count that extra speed. It ALSO counts the higher ROC.
In fact, I DOUBLED the impact of that ROC to give maneuverability less of a pure speed element.

I would have said "no algorithm this simple can be any good" - and did privately to Joe. But I looked at it and I was wrong. The algorithm is remarkably good - for any plane - WWI to now - nevermind P-38s and Nells. ROC really is proportional to overall maneuverability, and speed is, well proportional to speed. The combination matters, and that is why the need to say "multi engine is not the same as single engine" works - this particular multi engine has so much speed and maneuverability it outclasses EVERY other multi engine and it can play with the single engine types. Just as in real life - it should not try to dog fight - but it has firepower and durability - and statistically that means it works very well.
Add to that range - and you have a very P-38 like package.

[el cid again]

BTW - only data I have so far on climb rates are
P-38F - time to 20K feet - 5.94 min.
P-38J - rate of climb (SL) 4000 fpm
P-38J - rate of climb (critical altitude) 2900 fpm @ 23,400'
time to critical altitude - 6.19 min.
Service Ceiling - 40K feet

IF we can get it, what we need is INITIAL ROC.

By definition, ROC at service ceiling is 100 feet (or 30 meters) per minute.
That is the DEFINITION of service ceiling - the altitude at which that is your ROC. So the place planes are most different is at the other end - and that is also what matters if you are trying to intercept. So that is the value we are after.

[RevRick]

Well, if it took the P-38F roughly 30 seconds less to climb to 20,000 feet than it took the P-38J to climb to 23,400 feet with an initial (Sea Level) climb rate of 4,000 fpm, I'd think that the P-38F rate of climb would have to be very similar to achieve what is basically a very similar result - probably on the order of 3,700 - 3,800 fpm initial rate.

[witpqs]

ORIGINAL: el cid again

P-38J - rate of climb (SL) 4000 fpm <----{{

IF we can get it, what we need is INITIAL ROC.

Is ROC from sea level the same as Initial ROC, or does Initial ROC mean 'from a runway' (by definition)?

[el cid again]

Well, if it took the P-38F roughly 30 seconds less to climb to 20,000 feet than it took the P-38J to climb to 23,400 feet with an initial (Sea Level) climb rate of 4,000 fpm, I'd think that the P-38F rate of climb would have to be very similar to achieve what is basically a very similar result - probably on the order of 3,700 - 3,800 fpm initial rate.

OK - I will do an engineering estimate and get a value from your 30 seconds and 20,000 feet and 4,000 fpm - and see if that changes anything. It might help our algorithm if I used a lower value before.

[el cid again]

Is ROC from sea level the same as Initial ROC, or does Initial ROC mean 'from a runway' (by definition)?

It should mean "from a runway at sea level" since it could be different, say, at Denver or some other place not near sea level.