fmardani
14th Jan 2002, 12:51
An aircraft has a crosswind limitation of 14kt for both T/O and landing. The commander of this aircraft has the choise of three runways. Surface wind = 25018kt.
The choises are runways 22, 29, 33
The correct answer is runways 22 and 29.
What I'd like to know is how and why these two runways are the correct answers.
Thank you
Hersham Boy
14th Jan 2002, 13:29
Have you checked it on the flight computer?
haven't got mine to hand, but it looks right to me - use the square grid at the bottom of the wind-side of the wheel. You should find that the 18kt wind vector from 250deg will exceed 14kt either to the left or to the right when you turn the card as far as 330deg (ie. runway 33) - ie. out of crosswind limit.
Don D Cake
14th Jan 2002, 15:59
To work it out exactly....
The crosswind component is the sine of the angle between the runway heading and the wind heading, multiplied by the wind speed so....
For runway 22, the angle is 30deg, the sine of 30 is 0.5 so the crosswind component is 9kts.
For 29, sine 40 = 0.64, so crosswind = 11.6kts
For 33, sine 80 = 0.98, so crosswind = 17.7kts
I suspect the windspeed being 18kts is no coincidence and they may have wanted you to use the "rule of sixths" ie
To obtain crosswind component, for 10deg difference multiply windspeed by 1/6, 20deg by 2/6, 30deg by 3/6, 40deg by 4/6, 50deg by 5/6 and 60deg and over by 6/6.
So we get
runway 22, angle = 30deg, 18 * 3/6 = 9kts
runway 29, angle = 40deg, 18 * 4/6 = 12kts
runway 33, angle = 80deg, 18 * 6/6 = 18kts.
Either way, runway 33 is outside your limits.