Timing of turns less than rate 1
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Timing of turns less than rate 1
I currently fly the A320 but the question would apply to any aircraft at high altitude. When entering a holding pattern at altitude the angle of bank you get is considerably less than a rate one turn (because the TAS is so high) which leads to my problem - how do I know how long it is going to take to turn the full 360 degrees?
I often need to enter a pattern in order to leave the fix at a particular time and can adjust the timing of the outbound leg to suit this, does anyone have a rule of thumb though to calculate the timing of the turns??
Cheers!!
I often need to enter a pattern in order to leave the fix at a particular time and can adjust the timing of the outbound leg to suit this, does anyone have a rule of thumb though to calculate the timing of the turns??
Cheers!!
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Hello PSP1224
that's an interesting question. If you follow PANS-OPS down under your bank angle at high speeds would be 25 degrees. So the question would be: How long does a turn at 25 degrees take?
Since the time for a turn at a given bank angle is a linear function of speed the answer is quite simple (by my reckoning).
time [seconds] = TAS [knots] * 0.7
If you were flying with 300 kts TAS you would hence need 210 seconds for a three-sixty.
This is however just an approximation. Firstly, TAS is pretty useless in flight mechanics since what you really need is translatory speed against the earth's surface. Secondly the factor 0.7 is just an approximation.
that's an interesting question. If you follow PANS-OPS down under your bank angle at high speeds would be 25 degrees. So the question would be: How long does a turn at 25 degrees take?
Since the time for a turn at a given bank angle is a linear function of speed the answer is quite simple (by my reckoning).
time [seconds] = TAS [knots] * 0.7
If you were flying with 300 kts TAS you would hence need 210 seconds for a three-sixty.
This is however just an approximation. Firstly, TAS is pretty useless in flight mechanics since what you really need is translatory speed against the earth's surface. Secondly the factor 0.7 is just an approximation.
Last edited by hvogt; 5th Sep 2011 at 22:29.
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Give this a go.... It's easy enough for me to use and seems to work a treat!
2 mins for 180 degree turn for every 10000 feet.
180 degrees at 10000 feet.......... 2 mins
180 degrees at 20000 feet.......... 4 mins
180 degrees at 30000 feet.......... 6 mins
Give it a go
bbbbbbbbbbbzzzzzzzzzzzzbbbbbbbbbbbbbzzzzzzzzzzzzzzzzzzzzzzzz
2 mins for 180 degree turn for every 10000 feet.
180 degrees at 10000 feet.......... 2 mins
180 degrees at 20000 feet.......... 4 mins
180 degrees at 30000 feet.......... 6 mins
Give it a go
bbbbbbbbbbbzzzzzzzzzzzzbbbbbbbbbbbbbzzzzzzzzzzzzzzzzzzzzzzzz
