What I tested seems correct then JW
ok.........
I found this not sure how applicable/correct it is and might explain what TD have done.
"
v= v(initial) + at works in a vacuum in which WWII bomber did not fly. v=obviously velocity. a= acceleration
D= 1/2 CpAv^2 Where C=Drag Coefficient, p= rho=air density, A=crossectional area v= obviously velocity.
However at some some value of v Drag will equal mg this point is called terminal velocity. m=mass, g=gravity
so, 1/2CpAv^2 = mg
v= (2mg/CpA)^(-1/2)
This is only a rough formula to calculate the terminal velocity of the bomb because aerodynamics get very complicated, but this should get you somewhat close.
All you need is the mass of the object, gravity which is 9.8 m/s^2, the drag coefficient,
air density at the point that you want to know presumably fairly close to sea level, and the cross sectional area."
Might be rubbish.............
Chart monkeys apply within
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