Why still no dive acceleration difference?

[BlackBerry]

Entry speed 300mph = 441fps
Angle of Dive 45 degrees

We need a ballpark of our equilibrium speed. A quick method is to use the relationship of Parasitic drag. It is the major drag component at high speed. More detailed analysis will give better results but this is accurate within 10%. 10% is acceptable for climb/dive performance.

441(SQRT 1000/500) = 624fps

At 624 fps, the acceleration about the CG will be zero.

Dive:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs + 6364lbs = 6864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 6864lb/279.5lb-s^2/ft
a = 24.6 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 624^2)/(2 * 24.6ft/s^2) = -3961ft (negative is a vector direction)

3961ft * sin 45 = 2801 ft of altitude lost!!

Again, constant acceleration formulars are wrong in calculating airplane's dive.

Do you mean when the airplane reach 624fps(424mph), the acceleration is zero?


OMG, why don't you open il2 4.11m and choose any piston aircraft, spawn at 7500m altitude and dive at 45 degree with full throttle?

I promise your aircraft will lose wings sooner or later. You could NEVER reach the equilibium point where the speed stops increase.

Crumpp, plz run il2 4.11m and try some tests, it's very easy for you to find the truth.