Equation for groundspeed while crabbing

jrh

jrh

Well-Known Member
I got into an interesting discussion with another CFI last night. We were talking about how a strong crosswind slows you down on a cross country almost as much as a headwind.

That would be because the aircraft has to crab into the wind in order to track a course. If you think of the aircraft's true airspeed as a vector quantity, crabbing would redirect the vector to the side. The forward vector (the ground track) would then be shortened.

This is the same concept as changing the vertical component of lift in a turn. As crab angle increases, the "forward component of airspeed" decreases.

This friend I was talking to had heard before that even if a crosswind is a quartering tailwind, if it is not more than 20 degrees behind the wingtips, it will slow the aircraft's groundspeed. I wasn't sure about that, but I can see the logic (because the aircraft must crab into the wind, regardless of a quartering headwind or quartering tailwind). If the "tailwind component" that increases the aircraft's groundspeed does not offset the loss of forward speed caused by crabbing into the "crosswind component", the net effect would be a loss in groundspeed.

So my question is, does anybody have a deep understanding of this relationship? Are there any math equations that calculate the effects of different situations? I know the variables are true airspeed, desired course direction, wind direction, and wind speed. Otherwise, I don't know how to put it into solid mathematical terms. I think it would be cool to be able to show my students how a tailwind can slow them down!
 
My Jeppesen CR-2 computer has the ability to account for a crosswind and crab angle in groundspeed calculations, but I'll have to dig it out.

I would agree that a wind at a small angle would present more of a crosswind component than a tail/headwind component. Using a crosswind component chart will give you an idea of the angle at which it becomes significant.
 
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I got into an interesting discussion with another CFI last night. We were talking about how a strong crosswind slows you down on a cross country almost as much as a headwind.

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As PSCRAIG says, the Jeppesen CR computers will compensate your true airspeed for crab angle and it is not as significant as you might think. Since I am bored and happened to have my CR-3 laying next to the computer, I tried it out. It has been awhile since I have done a VFR flight plan, but this is what I came up with:

If you have a TAS of 100kts and are flying North with a 50 kt direct cross wind from the West, you will require a 30 degree crab angle, so will by flying a heading of 330. This will result in an effective true airspeed of 87kts.

If we are trying to fly a course of 25 degrees, and the wind is still 50kts from 270 (45 kt x-wind), we will need a 27 degree crab angle. This will result in a heading of 358 and an effective TAS of 89kts. However, we now have a 21kt tail wind, so our ground speed is back to 100 kts.

In a typical trainer that cruises at 125 kts, this is how crab angle effects TAS:
10 deg - 2 kts
15 deg - 4 kts
20 deg - 8 kts
25 deg - 12 kts
30 deg - 17 kts

If you have a 250 kts TAS, then a 30 deg crab will cost you 33 kts.
 
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If you have a TAS of 100kts and are flying North with a 50 kt direct cross wind from the West, you will require a 30 degree crab angle, so will by flying a heading of 330. This will result in an effective true airspeed of 87kts.

[/ QUOTE ]I'm having a little trouble with the concept that the speed of the air mass over the ground in some way affects the speed of an object located in the air mass relative to the air mass.

Can someone explain it to me?
 
Think about it like swimming in a river with a strong current. If you try to swim upstream, you go slow. Downstream, fast. And if you want to make it straight across to the next bank, you need to angle upstream a little as you swim or you'll get swept away.
 
Methinks someones is confuseding GS and TAS. (I think that's Mark's point, anyway.)

grin.gif
 
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This will result in an effective true airspeed of 87kts.

[/ QUOTE ]Effective true airspeed? This is a new term to me. I think you mean ground speed. This does not bode well for the rest of your calculations:

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If we are trying to fly a course of 25 degrees,...However, we now have a 21kt tail wind, so our ground speed is back to 100 kts.

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I don't know what your method is, though I can guess, but your result is wrong. The ground speed in this example is 110.3Kt

Since the original post has a subject of "equation for groundspeed..." I'll indulge in a little trigonometry. no magic, no opinions, just the same simple math used by electronic flight computers

The Law of Sines:
sin a / A = sin b / B = sin c / C,
where a, b, and c are the sides of the triangle and A, B, and C are the angles opposite to their respective sides.

In the wind triangle of the example a = 100kt and A = 65° (90-25; the wind vector is 90° and the ground track is 25°)
Wind velocity is b = 50 kt. The crab angle is B = arcsin(b * sin A /a) = 26.95°.
That allows us to find C, the angle opposite the ground track side C = 180 - 26.95 - 65 = 88.05°
so
c = a * sin C / sin A = 100kt * 0.999 / .906 = 110.3kt

The required heading is the course made good (ground track) minus B:
25° - 26.95° = -1.25° or 25° - 26.95° + 360° = 358.05°

These are the same answers you E6B, electronic or mechanical wil give.

If you really care, and can't get the triangle to look right, I'll try to post a graphic of it. But it shouldn't be too hard if you remember that the the ground track is the vector sum of the heading-speed vector and the wind vector.
 
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Think about it like swimming in a river with a strong current. If you try to swim upstream, you go slow. Downstream, fast. And if you want to make it straight across to the next bank, you need to angle upstream a little as you swim or you'll get swept away.

[/ QUOTE ]I know the swimming analogy. But that's all relative to the =shore= not the water. The speed of my body relative to the =water= isn't changing based on how quickly the stream is going.

Suppose I don't move at all. Just float. Isn't my speed relative to the =water= zero, no matter how fast or slow the river's flowing?

Is it, as Steve suggests, just that folks are confusing TAS (speed relative to the =air=) with GS (speed relative to the =ground=) or is there some new concept I never heard of before?

To me, TAS is IAS corrected for pressure, and I frankly never noticed my airspeed indicator change much enroute because I was crabbing.
 
I didn't make this up! Honest.

You are all correct that your actual TAS is unaffected by the wind. The term 'Effective TAS' is used on the wind side of the Jeppesen Circular computers to account for ground speed changes caused by large crab angles, which are not accounted for using the traditional method of figuring ground speed by taking your TAS and either adding a tailwind or subtracting a headwind.

Even though the term has caused much confusion here, it does make things easy to understand when using the Jeppesen CR-3. After you find the headwind and crosswind components, you get your required crab angle. If your crab angle is more than 10 degrees you can then modify your original TAS (giving 'Effective TAS') and you add or subtract your tailwind/headwind from this new 'effective TAS' number.

Sorry for the confusion.

If you think about this it makes sense that a large crab will effect your groundspeed. The normal E6B does not really account for this (At least I don't think it does, I have never really used one. I'm sure one of you will let me know.). If you crab into a direct crosswind by a large amount, you actually end up giving yourself a slight headwind. If you take this to an extreme example, with our 100 kt trainer trying to fly in a 100 kt direct cross wind, we would end up with a 90 degree crab into the wind and our ground speed would be zero.
 
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I don't know what your method is, though I can guess, but your result is wrong. The ground speed in this example is 110.3Kt

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You are correct, I added 89 and 21 and some how ended up with 100! It should have looked like this:

"If we are trying to fly a course of 25 degrees, and the wind is still 50kts from 270 (45 kt x-wind), we will need a 27 degree crab angle. This will result in a heading of 358 and an effective TAS of 89kts. However, we now have a 21kt tail wind, so our ground speed is now 110 kts."

What I was trying to do was give an example where you could have a tailwind, but still not get an increase in groundspeed due to the large crab angle.

If we use our 100 kt trainer, with the 50 kt wind from 270, flying a course of 015, I think it will work (hopefully no math mistakes!). We will still require about 30 deg of crab, for a heading of 345 deg. This will give us an 'efffective TAS' (I could say 'effective ground speed in still air') of 87 kts and a 13 kt tail wind component. So, our groundspeed should be back to 100 kts.

In my previous post, our course of 025 deg cut the required crab angle and increased the tail wind by too much, so it didn't quite work out as I intended.
 
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If you think about this it makes sense that a large crab will effect your groundspeed.

[/ QUOTE ]Of course it does.

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The normal E6B does not really account for this

[/ QUOTE ] Of course it does,

I haven't use the CR-3, but from your description, all it is really doing is adding an extra calculation of components and giving it a name Jepp made up for the purpose. Just a different take on the standard wind triangle.

In your example of the 100 kt trainer, with the 50 kt wind from 270, flying a course of 015, the E6B gives the same answer.
 
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