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 Computing Groundspeed From Crosswinds
 Wardialer From United States of America, joined Sep 2001, 1201 posts, RR: 0Posted Sun Sep 22 2002 20:29:19 UTC (13 years 8 months 1 week 3 days 21 hours ago) and read 6266 times:

 OK, I'll try my best to explain this question. So bare with me. OK, suppose an aircraft is flying at FL350 with a TAS of 450 knots and with a MAG Heading of 180 and the wind is from 270@30 knots (Crosswind). Now how could I compute the Groundspeed with a non-aviation calculator? Is there a formula for this? Now if we say that the aircraft is flying at that heading (180) and the wind is from 360@30 knots then of course that would come out to be 480 knots GS. TAS being 450 + Tailwind 360@30 = 480kts GS Is there any way of figuring out the same situation for the Crosswinds? I dont have a aviation calculator so is it possible to do a formula with a standard calculator?
 Inbound From Trinidad and Tobago, joined Sep 2001, 856 posts, RR: 2 Reply 1, posted Sun Sep 22 2002 21:51:36 UTC (13 years 8 months 1 week 3 days 19 hours ago) and read 6253 times:

 umm there's an international diagram for this. But I have no idea on how to get it to you :-( It's like a graph, and you enter with the difference in MAG heading, and you can read across your Headwind and Crosswind Component. it applies to most aircraft. hopefully someone else will figure out what I'm talking about and post it up for you. sorry.
 Maintain own separation with terrain!
 PPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0 Reply 2, posted Sun Sep 22 2002 22:10:18 UTC (13 years 8 months 1 week 3 days 19 hours ago) and read 6251 times:

 Never have used a formula the E-6B is the easiest way to calculate it. Its only like \$20.
 At worst, you screw up and die.
 Wietse From Netherlands, joined Oct 2001, 3809 posts, RR: 53 Reply 3, posted Sun Sep 22 2002 22:36:26 UTC (13 years 8 months 1 week 3 days 19 hours ago) and read 6245 times:

 The art of vectors.... I could calculate it for you and give you a basic formula, something with sinus, cosinus and/or tangus. But I just dont feel like it now. It is pretty easy and straight forward, you probably had had it at high school or something. Wietse
 Wietse de Graaf
 Timz From United States of America, joined Sep 1999, 7227 posts, RR: 7 Reply 4, posted Sun Sep 22 2002 23:20:56 UTC (13 years 8 months 1 week 3 days 18 hours ago) and read 6231 times:

 You chose a simple example, so no trig needed, just Pythagoras. The square root of (450 squared plus 30 squared), i.e. the square root of 203400, is the answer-- 451 knots. If the wind isn't 90 degrees to the heading then we need the law of cosines, which goes like this: Let V equal airspeed Let W equal windspeed Let Z equal the wind angle where zero degrees means a direct headwind (not tailwind) Then groundspeed is the square root of V squared plus W squared minus (2 times V times W times the cosine of Z). The cosine of 90 degrees is zero, so this is identical to Pythagoras if the wind is at 90 degrees.
 Timz From United States of America, joined Sep 1999, 7227 posts, RR: 7 Reply 5, posted Sun Sep 22 2002 23:38:09 UTC (13 years 8 months 1 week 3 days 18 hours ago) and read 6224 times:

 Let's make an insignificant revision: call Z the wind angle where zero degrees means a direct tailwind, not headwind. That requires changing the minus (in the law of cosines) to a plus. No other changes.
 Ralgha From United States of America, joined Nov 1999, 1614 posts, RR: 5 Reply 6, posted Mon Sep 23 2002 00:15:18 UTC (13 years 8 months 1 week 3 days 17 hours ago) and read 6212 times:

 Sorry Timz, you're wrong. Right idea, but you got the vectors wrong. ground speed = sqrt( 450^2 - 30^2 ) A direct crosswind to your ground track will always slow you down. You have to crab into the wind, directing some of your thrust off of your track. Since the crosswind contributes nothing to your ground track, the loss of thrust along your track reduces your speed on that track.
 09 F9 11 02 9D 74 E3 5B D8 41 56 C5 63 56 88 C0
 Timz From United States of America, joined Sep 1999, 7227 posts, RR: 7 Reply 7, posted Mon Sep 23 2002 02:06:52 UTC (13 years 8 months 1 week 3 days 15 hours ago) and read 6204 times:

 Ralgha: try your formula with a 450-knot crosswind. If you consider the result correct then apparently we're talking about two different things. As you'll see when you look again at the original post, he assumed a heading of 180 degrees, not a ground track of 180.
 B747skipper From , joined Dec 1969, posts, RR: Reply 8, posted Mon Sep 23 2002 03:32:52 UTC (13 years 8 months 1 week 3 days 14 hours ago) and read 6193 times:

 Dear Wardialer - Look in your email for wind triangle drawing instructions... Enjoy...   (s) Skipper
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