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Help - Rate of Turn Question

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Old 14th Jun 2005, 05:10
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Help - Rate of Turn Question

Help

Is there a formula to work out a rate of turn eg degrees per sec
for a given airspeed and angle of bank.

60knots and 20 degrees angle of bank.

Cheers

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Old 14th Jun 2005, 06:27
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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.
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Old 14th Jun 2005, 07:46
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Thanks BEagle

Would you be so kind as to walk me through the formula please as it been a long time since I've used pi and tan

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Old 14th Jun 2005, 09:39
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Pi - press the "Pi" button on your calculator.
Tan - press the "Tan" button on your calculator.

Sorry!!!

FFF
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Old 14th Jun 2005, 13:00
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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.
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Old 14th Jun 2005, 16:00
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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)
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