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MetalGear
1st Mar 2009, 10:30
Hi, I'm doing the ATPL distance learning course and am having a bit of trouble with Point of Equal Time. Any help would be greatly appreciated. So here's the problem:

Formula:
Time to destination = D-X/O
Time to return = X/H
X/H = (D-X)/O
X = DH/(O + H)

Question:
A – B 1240 nm
TAS 340 KNOTS
Wind Component +20 knots outbound

D = 1240
H = 320
O = 360

X = 1240x320/(360+320)
X = 396800/680 = 583.5

So far I'm correct according to the answer in the book, however when I try to find the time it goes a bit awry.

So from the formula given above:
Time to return = X/H
583.5/320=1.82 = 1hr 42min

And to double check it's the same time to continue on to B
D-X = 1240-583.5 = 656.5

Again from the formula given above:
Time to destination = D-X/O
656.5/360 = 1.82 = 1hr 42min

Now, the answer given in the book is 1hr 37min. Which would be correct for the Time to return if we changed the formula to read X/O

In that case the time to continue to B cannot be 1hr 37min because 656.5 divided by either 360 or 320 does not give 1hr 37min. So in this case it is not a PET.

It all looks a bit confusing, I know, but I've done 7 of these questions all with similar results. It's probably something so obvious I'll kick myself. But that's the problem with distance learning, sometimes you've got to figure things out for yourself. Anyway, thanks for reading and again any help would be greatly appreciated.

r44flyer
1st Mar 2009, 12:10
Unless I've missed something, they've used the O groundspeed in the initial formula by mistake. Unless they mean start at B and continue to B by turning round!?

The error would give 1240x360/680 = 656.5, leaving 583.5 to run outbound at 360kts, which is 1h37m.

By the way, 1.82 in your calcs is not 1h42, it's 1h49.