pof help needed
 

CAN ANYONE HELP ME
DOES ANYBODY KNOW HOW TO CALCULATE RADIUS OF TURN............I HAVE POF EXAMS NEXT WEEK AND NEED TO KNOW HOW TO CALCULATE THIS. A WRITTEN EXAMPLE WOULD MAKE THINGS ALLOT EASIER


THANKS

Always wanting to do other's homework, and never my own.

For a turn with zero side slip:

mV^2/R = L sin (bank)

You can draw this out, it's a simple centripetal problem,
F = ma, a = V^2/ R, the airplane force in that direction is supplied by the horizontal component of lift L sin (bank).

Great, so now solving for R we get

R = mV^2/ L sin (bank)

Problem, we don't know what the lift is.

But, looking at the turning airplane from the front we can visualize the lift vector and see that it has vertical and horizontal components. We figured the horizontal, let's use the vertical to solve for L (simple substitution).

mg = L cos (bank)

So, L = mg / cos (bank)

plugging this into the above horizontal equation we get:

R = mV^2/ mg sin (bank)/ cos (bank)

A bit gratuitous, simplifying we get

R = V^2/ g tan (bank)

(watch out, g must be in the same units as V)

R = V^2/ 68416 tan (bank)
Where V = kts

While I'm here:
Time to complete a 360 turn = .0055 V/ tan (bank)

Gforce = 1/ cos (bank)

Stall speed in turn = Stall speed normal X Gforce

My Bristol notes state:

Radius of turn = Tas/ Rate of turn x 60 pi

so, in a rate 1 turn, at 250 knots,

250/ 1 x 60 pi

= 1.32 nm

Apologies if this is a simplified version of the above, or I've missed the point completely...!!;)