Calculating vertical time and distance in a stall turn (US Hammerhead)

I've been teaching aerobatics for years but last weekend (after a session of
stall turns) a student said he was surprised at how long we were vertical.
Afterwards it occurred to me that it would be fun/interesting to look at the
sums, later I realised that I didn't know how to DO all the sums!!

If the propellor thrust and form drag are initially ignored it's easy
(forgive the elderly units):
induced drag = 0 (in an ideal stall turn the CG track is vertical so the AOA
needs to be zero, therefore the induced drag should be close to zero)
initial velocity u = 120kts (200 fps)
final velocity v = 40kts (67 fps)
acceleration f =-32 fps/s

v**2 = u**2 + 2fs
so distance s = 555ft.
v=u+ft
so time in the vertical t = 4secs.

Now to bring in form drag and thrust.....
If we assume the deceleration is linear the form drag could probably be
based on the average
speed. And if we assume the engine is producing 200BHP with a propellor
efficiency of 75% we should be able to work out something useful.

Thinks - but Thrust is proportional to torque!..... elderly brain gives up.

How can I factor-in the propellor thrust and estimate the form drag to give
a realistic estimate of time and distance?

In practise I think it would be possible to estimate the net acceleration
due to thrust and form drag by flying a ballistic push (zero AOA) from some
angle above the horizon to the same angle below (to eliminate the effect of
gravity), starting at 60kts and then timing the acceleration to whatever
speed is reached. I'll try this next weekend and then see how the final
estimate matches the reality of a stall turn.

....but I'd still like to know how to calculate the approx thrust....??

Dave S