wardialer
Posts: 1163
Joined: Fri Sep 14, 2001 1:08 pm

### Computing Groundspeed From Crosswinds

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
Posts: 614
Joined: Sat Sep 15, 2001 7:59 am

### RE: Computing Groundspeed From Crosswinds

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
Posts: 2398
Joined: Sun Sep 30, 2001 5:39 am

### RE: Computing Groundspeed From Crosswinds

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
Posts: 3630
Joined: Thu Oct 18, 2001 12:49 am

### RE: Computing Groundspeed From Crosswinds

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
Posts: 6224
Joined: Fri Sep 17, 1999 7:43 am

### RE: Computing Groundspeed From Crosswinds

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
Posts: 6224
Joined: Fri Sep 17, 1999 7:43 am

### On Second Thought

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
Posts: 1589
Joined: Tue Nov 09, 1999 6:20 pm

### RE: Computing Groundspeed From Crosswinds

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
Posts: 6224
Joined: Fri Sep 17, 1999 7:43 am

### No, I'm Right (I Think)

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.

Guest

### RE: Computing Groundspeed From Crosswinds

Dear Wardialer -
Look in your email for wind triangle drawing instructions...
Enjoy...

(s) Skipper

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