Help - Rate of Turn Question
Joined: May 1999
Aviation Qualifications: ATP+Mil
Posts: 27,394
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From: Quite near 'An aerodrome somewhere in England'
Time taken to turn through n degrees
= (n x 2 pi x TAS)/(360 x g x tan AOB)
Careful with the units; if you use miles per minute, g = 19.065 nautical miles/min/min
At 60 KTAS and 20 deg AOB, 360 deg will be achieved in 54.3 sec; this equates to 6.63 deg per sec.
= (n x 2 pi x TAS)/(360 x g x tan AOB)
Careful with the units; if you use miles per minute, g = 19.065 nautical miles/min/min
At 60 KTAS and 20 deg AOB, 360 deg will be achieved in 54.3 sec; this equates to 6.63 deg per sec.
Sorry!!!Joined: Jul 1999
Posts: 3,931
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From: Warrington, UK
Well, Pi is 22/7. Tan gets a bit complicated. Draw yourself a right angled triangle, with one of the other angles 20 degs(or whatever AOB). Measure the length of the two shortest sides. Divide the shortest side by the other and that will give you the Tan of the angle. The bigger the triangle the more accurate will be the final answer. A triangle with sides of 42 and 114mm gives a Tan of 0.36842 as opposed to 0.36397 for a calculator.
Joined: May 1999
Aviation Qualifications: ATP+Mil
Posts: 27,394
Likes: 856
From: Quite near 'An aerodrome somewhere in England'
Radius of turn R = TAS squared, divided by (g x tan AoB)
Circumference of circle = 2 x pi x R
If TAS = V, time to cover 360 deg is thus 2 x pi x R / V
i.e. 2 x pi x (V squared / (g x tan AoB)) / V
or for n degress, n/360 of this time:
i.e. (n x 2 x pi x V) / (360 x g x tan AoB)
Circumference of circle = 2 x pi x R
If TAS = V, time to cover 360 deg is thus 2 x pi x R / V
i.e. 2 x pi x (V squared / (g x tan AoB)) / V
or for n degress, n/360 of this time:
i.e. (n x 2 x pi x V) / (360 x g x tan AoB)
