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Badmachine
8th Mar 2010, 00:00
In general, what is the affect if any, upon an aircraft turn radius for a given angle of bank (AoB) during ascending and descending turns, compared to level turns using the same AoB? Will a turn radius distance still remain constant during ascending/descending turns as long as the AoB and speed remains constant?

galaxy flyer
8th Mar 2010, 01:24
The turn radius in a climb or descent, viewed from the ground, will be smaller than a level turn. Think of an Immelmann or Split-S, the plane over the ground has a radius of zero, all the turn is in the vertical. How to measure it--beats me?

GF

Badmachine
8th Mar 2010, 06:02
Thanks.

It seems that if a rate of ascent/descent and AoB were unchanging, that a turn radius would also be unchanging. Perhaps the effect of changing rates of ascents/descents on a turn radius for a common AoB would be the same as a change of speed.

Wingswinger
8th Mar 2010, 07:42
How to measure it--beats me?

Here's a stab at it:

In a level turn the flight path angle is 90 degrees in the horizontal plane and 0 degrees in the vertical plane. The sine of 90 is 1 so the actual turn radius is 1 x the radius for the given bank angle and TAS derived from the turning nomogram.

If the flight path is 10 degrees below or above the horizontal then the turn radius will be the horizontal turn radius described above x sine 80 (90-10) = 0.98481.

If the flight path is 45 degrees below or above the horizontal, multiply the horizontal radius x sine 45 = 0.7071.

At 90 degrees below or above (vertical) the horizontal turn radius would be 0 (sine 0 = 0)

Every fighter pilot knows this instinctively but doesn't actually bother to work it out in the heat of combat.

All theoretical and of little practical use to any but fighter pilots since at extreme angles the challenge would be to keep the TAS constant while converting potential energy to kinetic energy and vice versa and thus eliminate dynamic changes to the horizontal turn radius.

Will that do?

WS (ex-fighter pilot)

waren9
8th Mar 2010, 09:12
Nicely done WS.

And theres your answer.

With a 10deg vertical slope you still use over 98% of the horizontal distance of a level turn.

Given that 3deg is relatively steep (for a commercial airliner), you can basically call it the same as a level turn

clunckdriver
8th Mar 2010, 12:05
Please, not ascent, its climb, there is an obvious reason for this!

Badmachine
8th Mar 2010, 12:12
Thanks WS.

For determining a climbing/descending turn radius for a given AoB, could one then simply utilize the basic level flight turn radius formula (R = v^2/11.26TanAoB), then muliply the level turn radius answer by the product of the angle of flight above/below the horizontal x sine, for the turn radius for the climbing/descending turn?