Why still no dive acceleration difference?

[FC99]

Originally Posted by mayshine

come on!!!

in this extreme case it is not 2,5m/s advantage

its 50 % speed acceleration advantage

se FW = a1, La = a2
by the time la dive to from 300 to 500km/h , FW will get the speed of 600km/h

enought to do a verticle hammer action.


u know I can feel a lot of people are just defending theirself instead of
looking for truth

I should be more precise and use m/s^2, I meant acceleration in my post and I still call it just 2,5m/s^2.

Even with this numbers distance between the planes will be less than 200m. Now plug in real numbers for FW and La and do the calculation again. When you include real numbers for mass, drag and thrust difference between La and FW will be very small.

[BadAim]

LOL! You are so funny Mayshine. You tell us we aren't listening, and we don't know what we're talking about, but you have not listened to a bloody word anyone has said here, then you have the unmitigated gall to say that you would be happy to be proven wrong. That's Bull.

You are engaging in mental masturbation, pure and simple.

Have fun.

[Arrow]

Originally Posted by FC99

So in this extreme case plane have 2,5m/s advantage in acceleration. What will be the difference in distance after 12-13 seconds?

And in the end try with different values for drag for each plane like it is in most real life cases.

yep, and it is only valid for 90 degree dive. You will have to multiply gravity force (mxg) with sine of the dive angle (assuming zero degrees of AoA) in a very simplified case. In the end, the final effect is marginal as tests show and as it is correctly modeled in Il-2 that has the basic physics 100% right.