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Computing Groundspeed From Crosswinds  
User currently offlineWardialer From United States of America, joined Sep 2001, 1192 posts, RR: 0
Posted (12 years 7 months 12 hours ago) and read 4788 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?

8 replies: All unread, jump to last
User currently offlineInbound From Trinidad and Tobago, joined Sep 2001, 852 posts, RR: 2
Reply 1, posted (12 years 7 months 10 hours ago) and read 4775 times:

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.

Maintain own separation with terrain!
User currently offlinePPGMD From United States of America, joined Sep 2001, 2453 posts, RR: 0
Reply 2, posted (12 years 7 months 10 hours ago) and read 4773 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.
User currently offlineWietse From Netherlands, joined Oct 2001, 3809 posts, RR: 55
Reply 3, posted (12 years 7 months 10 hours ago) and read 4767 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 de Graaf
User currently offlineTimz From United States of America, joined Sep 1999, 6965 posts, RR: 7
Reply 4, posted (12 years 7 months 9 hours ago) and read 4753 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.

User currently offlineTimz From United States of America, joined Sep 1999, 6965 posts, RR: 7
Reply 5, posted (12 years 7 months 9 hours ago) and read 4746 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.

User currently offlineRalgha From United States of America, joined Nov 1999, 1614 posts, RR: 5
Reply 6, posted (12 years 7 months 8 hours ago) and read 4734 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.

 Big thumbs up

09 F9 11 02 9D 74 E3 5B D8 41 56 C5 63 56 88 C0
User currently offlineTimz From United States of America, joined Sep 1999, 6965 posts, RR: 7
Reply 7, posted (12 years 7 months 6 hours ago) and read 4726 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.

User currently offlineB747skipper From , joined Dec 1969, posts, RR:
Reply 8, posted (12 years 7 months 5 hours ago) and read 4715 times:

Dear Wardialer -
Look in your email for wind triangle drawing instructions...
 Wink/being sarcastic

(s) Skipper

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