pof help needed
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Join Date: May 2002
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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
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
Join Date: Mar 2002
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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
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
Last edited by '%MAC'; 10th Jul 2002 at 19:40.