Calculating Drift On Approach

[tgrayson]

On approach you note you are holding 10' drift and that your G/S 180kts, what is the X/W?
(A/C still doing 3nm/min, 3 X 10 = 30, ANS X/W 30kts)

See what your E6B says on this:

You're landing 36. You have a wind of 033@79 knots, your airspeed is 250 knots, what's your GS and and WCA?

(According to an online E6B, your WCA is 10 degrees and GS is 180 knots.)

 

[fish314]

Wind is 150/20kt what is the crosswind on RWY10? = 14kts

Lot's of ways to solve this one, but the exact answer is:

wind velocity*cos (angle), where "angle" is the difference between the runway and the wind

150-100=50

20*sin(50)=12.855

There are several rules of thumb for finding sines. Check out the thread entitled "2 questions" in the Tech Talk forum from a few days ago. The 20% rule that you mentioned is one of the rules of thumb. That's where your 14 knots of cross came from, and is a decent wag at it.
(14-12.855)/12.855= 8.9% error or so.

See what your E6B says on this:

You're landing 36. You have a wind of 033@79 knots, your airspeed is 250 knots, what's your GS and and WCA?

(According to an online E6B, your WCA is 10 degrees and GS is 180 knots.)

He was showing you the comparison between that answer and the one that you came up with in your post here:

On approach you note you are holding 10' drift and that your G/S 180kts, what is the X/W?
(A/C still doing 3nm/min, 3 X 10 = 30, ANS X/W 30kts)

Basically, you both arrived at the same ground speed and the same wind correction angle, but you have vastly different cross-winds (about 43 knots vs. the 30 knots that you came up with). That was the point of his post above where he said that the question wasn't solvable using the Ground speed. Now if you change it to TRUE airspeed, I think you'll make some headway.

Tgrayson's example (worked in reverse) becomes a 10 degree wind correction at 250 true and what's the crosswind? 10*4.3333 NM/minute= 43.333 knots (pretty close to the real answer of 43.026)

You're example (using 180 TRUE instead of 180 GROUND) 10*3 NM/minute=30 knots (31.74 knots is the exact crosswind to give a 10 degree wind correction.)

Hey taylor, how'd I do?

 

[subpilot]

I still don't see how you can have a tailwind with the wind 50 degrees off the nose.

My bad, I was thinking RWY1 (as in 010), not RWY 10 (as in 100). Damn, I need to get back to aviation.