View Full Version : bank angle for a specific speed

5th Nov 2005, 13:18
Hi, could anyone tell me the equation for working the bank angle, if speed is given and for a rate 1 turn.

the question is:
An a/c is travelling at 120kts, what angle of bank would be required for a rate one turn?

a) 30 degrees
b) 12 degrees
c) 18 degrees
d) 35 degrees

the answer is 18 degrees, but i cant for the life of me work out how they got the answer.

thanks for your help, CJH

5th Nov 2005, 13:32
Rate 1 turn bank angle = (Tas in kts divided by 10) +7
=12 + 7
nearest answer c) 18

6th Nov 2005, 20:00
AoB = inv tan .0027463 x TAS in knots

In this case, AoB = inv tan 0.329556 = 18.24 deg

The approximation of (TAS/10) + 7 works pretty well and gives 19 deg

7th Nov 2005, 15:48
The approximation of (TAS/10) + 7 works pretty well and gives 19 deg
and is much easier to do in the head......;)

7th Nov 2005, 18:03
Beagle, though it obviously works, can you explain what exactly you are inverse tanning please?:confused:

kiwi chick
7th Nov 2005, 23:00
Haha! Well over here in New Zealand we do it the easy way:

Knock of the zero from the airspeed, and add seven, ie 120 becomes 12, add 7.

18 is the nearest answer (I would bow to anyone who can fly an 18 degree bank angle instead of 19 degrees, haha!!)


8th Nov 2005, 08:13
AoB = inv tan (0.0027463 x (TAS in knots))

i.e., at 120 KTAS:

AoB = inv tan (0.0027463 x 120)

so AoB = inv tan (0.329556)

thus AoB = 18.24 deg

Obviously the (10% TAS) + 7 method is the one to use in flight!

If you want to work it out for yourself, use the formula for radius of turn R = (TAS squared) / (g x tan AoB) and the radius R required to complete a full 360 deg turn in 2 minutes at the TAS in question. Be careful with the units!

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