Tuesday, March 15, 2005

More Hold Timing

I figured out my hold timing puzzle in the shower. I had made it far more complicated than I had to.

See, while I need calculus to determine the travel of the aircraft at an arbitrary time t, I don't need it at all to determine the total travel during a 180 degree turn. I am considering only the distance the aircraft moves in the direction parallel to the inbound radial; the distance along the curved track and the distance the airplane moves perpendicular to the inbound radial are not relevant for this calculation. Performing a 180 degree turn in still air, the aircraft covers zero distance in the direction parallel to the inbound radial, so I can consider that the airplane is standing still for one minute, under the influence of the wind w. Speed times time equals distance. During the one minute turn, the aircraft travels a distance in nautical miles of w/60. (The 60 is the result of multiplying by 60 and dividing by 3600, because t is in seconds and w is in knots).

With this amazing escape from the need for definite integrals, I'll consider the case where we have a headwind on the outbound leg:

If there is a headwind on the outbound leg, the aircraft finishes its turn before passing abeam the fix, so the pilot travels that distance w/60, in order to draw abeam the fix, and then starts the timer. Thus the time taken to make up that distance is irrelevant. The pilot then flies outbound for one minute, covering a ground distance of (v - w) / 60 nautical miles, where v is the true airspeed and w is the wind speed. At the other end, of the hold, the turn to inbound ends up closer to the fix than it started, so the inbound track is w/60 shorter than the outbound track. Therefore the length of the inbound straightaway on the first turn is (v - 2w) / 60, and the time required to fly it will be speed divided by distance: 60 * (w + v) / (v - 2w).

Now the case where there is a tailwind on the outbound leg:

If there is a tailwind on the outbound track, the aircraft passes abeam the fix w/60 nautical miles before wings level, so the outbound distance (v + w) / 60 is increased by w/60. The outer turn drifts the aircraft a further w/60 nautical miles from the fix, so the inbound distance is (3w + v) / 60, and the time recorded for the first inbound leg will be 60 * (v - w) / (3w + v).

Initially I thought I could consider only one wind direction, without loss of generality, because the wind velocity could be positive or negative, but that doesn't work because the procedure changes.

Honestly, while I'm in a hold I think normal thoughts like "ack, the timer didn't start," "What was my EFC time again?" and "damn, I really have to pee." I only think about this stuff on the ground.

Next time I'm on the ground, I'm going to crunch some numbers through Excel and find out which is the best procedure to determine how long to fly outbound.


Aviatrix said...

If someone tells me the HTML command for "don't break this across two lines" I'll make that more readable.

flyboy said...

I'm not sure what you're trying to calculate...
KLM teaches this:

Outbound track= 1min plus or minus TWC
(kts=sec.) TWC you visualise on your DG.. if wind is 30 knots from a 45 degree angle (tail) you'll see on your DG that the TWC is about 2/3 say 20kts (cos45*30=21)
60-21= 39 seconds timing
et voila

(but you already knew this I think.. I'm just a student pilot with a big mouth) ;)