Thanks for the replies. They started me thinking and, having thought back many years, I managed to put together a couple of formula...
Assumptions:
- Constant power: I know the power delivered (CS prop aside) will increase with speed.
- No friction / air resistance: In real life this will reduce acceleration as speed increases.
- No other factors: e.g. As weight comes onto wings reducing wheel drag vs increased aerodynamic drag
s = distance
t = time
P = power
m = mass
Energy at an instant = 1/2.m.v^2
Energy used from rest = P.t = 1/2.m.v^2
Thus:
v = sqrt(2.P.t/m)
From rest: distance = Integral, between 0 and n, of sqrt(2.P.t/m) with respect to time which is
(4/3.P.t^1.5)/m (time to the power of 1.5).
A graph of these 2 (normalised to 100% on each axis - 100% speed = lift off, 100% distance = end of take off run available) looks like:
Combining into a Speed over distance graph gives:
Anywhere you fall below the curve you are below your target speed to achieve lift off.
I've added 3 dots:
50% speed after 30% distance
71% speed after 50% distance
85% speed after 40% distance
I believe the wrong assumption for 71/50 suggestion is that it assumes constant acceleration - whereas, in reality, we have constant power.
Anyway - that's my latest theory.
OC619