How to Solve the Wind Triangle

[RingLaserGyroSandwich]

If you have a CX-2 flight computer (or equivalent), you likely plug in the true course, true air speed, wind direction, and wind speed, without giving it much thought. The computer spits out the required true heading in order for the plane to follow the proper true course, and also tells you what ground speed to expect. How does the computer do this? An E-6B does some cool mechanical rotation of vectors which I don't use and won't analyze, but I'll show the math behind the direct calculation method. If you don't like how I explain it but are interested there are other explanations floating around the internet. I'll start with a diagram to illustrate the problem.

The length of each arrow is the "magnitude" of the velocity it represents.

The red arrow represents the magnitude and direction of the wind, which is known (based on forecasts). The black arrow represents the magnitude and direction of the actual course of the plane, assuming proper wind corrections are made. We know the direction (based on drawing a line on a chart connecting our departure airport and our destination airport) but do not know the magnitude, since we don't yet know what ground speed to expect. The blue arrow represents the magnitude and direction of the plane's velocity relative to the air. The direction of the arrow is the heading of the plane after correcting for wind, and the magnitude is the true air speed which is known based on performance calculations. In short, the plane's velocity relative to the air, plus the air's velocity relative to the ground, gives the plane's velocity relative to the ground.

We know two sides of this triangle (it is not a right triangle so we can't use sin/cos/tan or the Pythagorean theorem!), TAS and W. We want to know the third side, GS (ground speed), but need to determine an angle before we can calculate it. Since we know what direction the wind is blowing, and we know the desired course, we can easily determine the angle between W and GS, which I labeled A.

In order to solve for GS, the length of the black arrow, we can use a math rule that most people forget about and never use again. The law of cosines is the more advanced version of the Pythagorean theorem and works even for non-right triangles. a^2 = b^2 + c^2 - 2 b c cos(A). In the formula, a, b, and c, are sides of the triangle, and A is the angle opposite side a. In the case of the wind triangle, A is opposite TAS, so TAS serves the role of "a". We'll say W serves the role of b (although it could be c too... they are interchangeable). Substituting:

TAS^2 = W^2 + GS^2 - 2 W GS cos(A)

That is an equation where we know the value of everything except for GS. Let's create a real example and try it out:

You want to fly direct from Ring City Airport to Laser Gyro Airport, which requires a course of 010 degrees (we'll keep directions true vice magnetic for simplicity). Your V-Tail Skychicken will achieve a TAS of 100 kts. Wind is from 315 at 20 kts. What is your required true heading and resulting ground speed to fly this course?

Just visualizing the problem, the plane will need to aim left of the destination in order to compensate for the quartering headwind (this matches the drawing, fortunately). I need to compute angle A before we can use the formula above.

By drawing the wind and the TC, then filling in all the angles (remember the angles inside triangles add up to 180 degrees), you can see that angle A is 45 + 80 = 125 degrees. Now we can solve the Law of Cosines:

100^2 = 20^2 + GS^2 - 2 20 GS cos(125 degrees)

This is a quadratic equation, which can be solved using the quadratic formula. See, all that crap from high school is useful when you decide to solve the wind triangle for fun. Rearranging the equation to make it simplest to solve:

GS^2 - [40 cos(125 degrees)] GS + (20^2 - 100^2) = 0

Further simplifying:

GS^2 + 22.943 GS - 9600 = 0

The quadratic formula says for f(x) = a x^2 + b x + c, the solution is x = -b/(2a) +/- sqrt(b^2-4ac)/(2a). In this case, a is 1 making the equation easier to use. Plugging in:

GS = -11.4715 +/- (1/2) sqrt(526.3812 + 38400) = -11.4715 +/- 98.6488. We can reject the negative solution, so GS = 87.177 kts. If you plug this problem into a CX-2 flight computer, it says your GS will be 87.2 knots. Pretty darn close.

It also says to expect a true heading of 0.6 degrees. How can we determine that without using the flight computer? Look back at the first figure. Now that we know GS, it is relatively easy to compute the angle between GS and TAS. That angle will be how far left of the destination we should aim the plane, aka, the wind correction angle (WCA). We will use the Law of Cosines again, except this time the angle will be B, the angle opposite W (remember, W was considered side b in the Law of Cosines before). The TAS and GS will be "a" and "c" respectively.

b^2 = a^2 + c^2 - 2 a c cos (B)

Notice how you can put b on the left side so long as the angle in the cosine is B. That's the neat thing about the Law of Cosines. Substituting:

W^2 = TAS^2 + GS^2 - 2 TAS GS cos(WCA)
2 TAS GS cos(WCA) = TAS^2 + GS^2 - W^2
cos(WCA) = (TAS^2 + GS^2 - W^2)/(2 TAS GS)

We need to substitute in the values for the right hand side of that equation, then take the inverse cosine (aka arc cosine) to find the WCA. By my calculations, the right hand side of that equation is 17,199.829/17,435.4 = 0.986. The arc cosine of 0.986 is 9.599 degrees. Correcting left from the course of 010 by a WCA of 9.599 degrees gives a true heading of 0.4 degrees. Oops, I was off by a two tenths of a degree. Actually, that's pretty close! The CX-2 is actually a little more accurate than I was because it uses somewhat more sophisticated mathematical methods to perform the calculations, but the concept is exactly the same in either case.

Once you turn the procedure above into an algorithm, a simple flight computer can calculate the ground speed and WCA in a short fraction of a second. Or, you can play around with the mechanical E-6B if you enjoy it and like battery independence. Or... you can just let ForeFlight do it all for you.

 

[kath]

Personally, I prefer to split both the vectors into x- and y-components using sin/cos, then perform vector addition (or subtraction) using the components by just adding (or subtracting) the components up, and then convert the resultant back into a magnitude and direction using arctan(y/x) and Pythagoras.

But that's just me.

(Actually, my favorite is skyvector.)

 

[Dave Theisen]

My head hurts.

 

[texasclouds]

My head hurts.

Ya, do you even magenta line bro?

 

[Landing Fees]

Like the algebra I never used after high school, I was never taught wind triangles and I still don't use them.

 

[Doc Holliday]

Let's take something simple and make it super complex.............

 

[EdFred]

Or use a transparent circle and a sliding rectangle.

 

[MuseChaser]

..
In order to solve for GS, the length of the black arrow, we can use a math rule that most people forget about and never use again. The law of cosines is the more advanced version of the Pythagorean theorem and works even for non-right triangles. a^2 = b^2 + c^2 - 2 b c cos(A). In the formula, a, b, and c, are sides of the triangle, and A is the angle opposite side a. In the case of the wind triangle, A is opposite TAS, so TAS serves the role of "a". We'll say W serves the role of b (although it could be c too... they are interchangeable). Substituting:

TAS^2 = W^2 + GS^2 - 2 W GS cos(A)

That is an equation where we know the value of everything except for GS. Let's create a real example and try it out:

You want to fly direct from Ring City Airport to Laser Gyro Airport, which requires a course of 010 degrees (we'll keep directions true vice magnetic for simplicity). Your V-Tail Skychicken will achieve a TAS of 100 kts. Wind is from 315 at 20 kts. What is your required true heading and resulting ground speed to fly this course?

Just visualizing the problem, the plane will need to aim left of the destination in order to compensate for the quartering headwind (this matches the drawing, fortunately). I need to compute angle A before we can use the formula above.

By drawing the wind and the TC, then filling in all the angles (remember the angles inside triangles add up to 180 degrees), you can see that angle A is 45 + 80 = 125 degrees. Now we can solve the Law of Cosines:

100^2 = 20^2 + GS^2 - 2 20 GS cos(125 degrees)

This is a quadratic equation, which can be solved using the quadratic formula. See, all that crap from high school is useful when you decide to solve the wind triangle for fun. Rearranging the equation to make it simplest to solve:

GS^2 - [40 cos(125 degrees)] GS + (20^2 - 100^2) = 0

Further simplifying:

GS^2 + 22.943 GS - 9600 = 0

The quadratic formula says for f(x) = a x^2 + b x + c, the solution is x = -b/(2a) +/- sqrt(b^2-4ac)/(2a). In this case, a is 1 making the equation easier to use. Plugging in:

GS = -11.4715 +/- (1/2) sqrt(526.3812 + 38400) = -11.4715 +/- 98.6488. We can reject the negative solution, so GS = 87.177 kts. If you plug this problem into a CX-2 flight computer, it says your GS will be 87.2 knots. Pretty darn close.

It also says to expect a true heading of 0.6 degrees. How can we determine that without using the flight computer? Look back at the first figure. Now that we know GS, it is relatively easy to compute the angle between GS and TAS. That angle will be how far left of the destination we should aim the plane, aka, the wind correction angle (WCA). We will use the Law of Cosines again, except this time the angle will be B, the angle opposite W (remember, W was considered side b in the Law of Cosines before). The TAS and GS will be "a" and "c" respectively.

b^2 = a^2 + c^2 - 2 a c cos (B)

Notice how you can put b on the left side so long as the angle in the cosine is B. That's the neat thing about the Law of Cosines. Substituting:

W^2 = TAS^2 + GS^2 - 2 TAS GS cos(WCA)
2 TAS GS cos(WCA) = TAS^2 + GS^2 - W^2
cos(WCA) = (TAS^2 + GS^2 - W^2)/(2 TAS GS)

We need to substitute in the values for the right hand side of that equation, then take the inverse cosine (aka arc cosine) to find the WCA. By my calculations, the right hand side of that equation is 17,199.829/17,435.4 = 0.986. The arc cosine of 0.986 is 9.599 degrees. Correcting left from the course of 010 by a WCA of 9.599 degrees gives a true heading of 0.4 degrees. Oops, I was off by a two tenths of a degree. Actually, that's pretty close! The CX-2 is actually a little more accurate than I was because it uses somewhat more sophisticated mathematical methods to perform the calculations, but the concept is exactly the same in either case....

Personally, I prefer to split both the vectors into x- and y-components using sin/cos, then perform vector addition (or subtraction) using the components by just adding (or subtracting) the components up, and then convert the resultant back into a magnitude and direction using arctan(y/x) and Pythagoras....

Both excellent solutions for quick calculations during flight. Thank you!

 

[RingLaserGyroSandwich]

Both excellent solutions for quick calculations during flight. Thank you!

Hah. Well, I too don't use the Law of Cosines while I'm flying. In addition, I do not try to understand what my calculator is doing when I'm flying... I just treat everything like useful black boxes. This thread is the type of stuff you can do on the ground if you don't want to just accept what you are given.

Presumably, the CX-2 was not designed by a pilot who was flying at that particular moment.

Or use a transparent circle and a sliding rectangle.

This is fine if you want to do some personal cross country calculations, but won't help you make a digital flight computer or an app. If you are satisfied with using the E-6B and/or using digital tools created by others, that is up to you.

 

[MuseChaser]

Just having fun with you and Kath. I, too, have the need to know the "why" and "how" logistics behind almost everything, and appreciated and followed your original post. Thanks!

 

[dtuuri]

Um, no. Waay too complicated. Many pilots never took a physics class. You'll lose them on the first "magnitude". Heck, you lost me and I did take physics.

Try this way instead: "What you want to find out is what heading to fly, so that at the end of an hour you'll be right on course. The easy way is to plot a starting point from where the wind would drift a balloon from the departure airport in one hour then, from there, draw the still-air distance that will intercept the course to your destination at exactly one hour at your cruising true airspeed by swinging an arc. That'll be the true heading and the intercept point will measure the ground speed from the real departure point."

 

[RingLaserGyroSandwich]

Um, no. Waay too complicated. Many pilots never took a physics class. You'll lose them on the first "magnitude". Heck, you lost me and I did take physics.

Try this way instead: "What you want to find out is what heading to fly, so that at the end of an hour you'll be right on course. The easy way is to plot a starting point from where the wind would drift a balloon from the departure airport in one hour then, from there, draw the still-air distance that will intercept the course to your destination at exactly one hour at your cruising true airspeed by swinging an arc. That'll be the true heading and the intercept point will measure the ground speed from the real departure point."

Well, this isn't exactly something that should be taught in ground school. Perhaps I should put a disclaimer in the OP that this not for everyone. It's for people who want to go a step further and see how you can solve the wind triangle mathematically, as people do when creating flight aids other than the mechanical E-6B. You do raise a good question, if I was teaching ground school to new pilots, how would I teach this topic? An approach like yours would make more sense, but I haven't developed for myself exactly what I would do. I might have to some day, but not now.

 

[dtuuri]

Well, this isn't exactly something that should be taught in ground school. Perhaps I should put a disclaimer in the OP that this not for everyone. It's for people who want to go a step further and see how you can solve the wind triangle mathematically, as people do when creating flight aids other than the mechanical E-6B.

I see. Maybe there ought to be a better forum than "Pilot Training"?

 

[RingLaserGyroSandwich]

I see. Maybe there ought to be a better forum than "Pilot Training"?

I'm not aware of any forums that are more appropriate. I selected this forum because, for people who are interested, they may learn something about flying and how it all works. I don't think this forum has to be limited to discussing the process of earning certificates and ratings. Of course, when there's a more specific forum (example: how do I properly communicate this issue with ATC?) the thread can go there.

 

[dtuuri]

I'm not aware of any forums that are more appropriate. I selected this forum because, for people who are interested, they may learn something about flying and how it all works. I don't think this forum has to be limited to discussing the process of earning certificates and ratings. Of course, when there's a more specific forum (example: how do I properly communicate this issue with ATC?) the thread can go there.

Just meant maybe there ought to BE a better forum, not that you should have posted in another one. This place is lousy with nerds, they'd go for it.

 

[Brad Smith]

All is good and fine until the wind changes direction and velocity.

 

[sarangan]

Um, no. Waay too complicated. Many pilots never took a physics class. You'll lose them on the first "magnitude". Heck, you lost me and I did take physics.

Try this way instead: "What you want to find out is what heading to fly, so that at the end of an hour you'll be right on course. The easy way is to plot a starting point from where the wind would drift a balloon from the departure airport in one hour then, from there, draw the still-air distance that will intercept the course to your destination at exactly one hour at your cruising true airspeed by swinging an arc. That'll be the true heading and the intercept point will measure the ground speed from the real departure point."

It is ironic that this topic was discussed recently. Some insisted that doing it with paper and E6B demonstrates a better understanding than using an electronic flight planner. Well, here is why that argument is BS. If you can use an E6B but can't follow the math, then you are doing it just as blindly as someone plugging it into a website.

 

[Landing Fees]

All is good and fine until the wind changes direction and velocity.

Then you just take the triangle you drew and twist it a bit. Duh.

 

[Capt. Geoffrey Thorpe]

if I was teaching ground school to new pilots, how would I teach this topic?

First, remember that the math behind weight and balance is over the heads of many pilots...

 

[James331]

Not reading all that lol

Maybe don’t read into things so much, I’m a ATP/CFI and that makes my head hurt

 

[MuseChaser]

..., I’m a ATP/CFI ...

THAT alone makes my head hurt, even just grammatically...

Just kiddin' around...couldn't pass up the opening..

 

[PaulS]

Holy smokes. Tldr. 30 degrees. .5 for x wind. 45 .7 for x wind,. 60. .9 for x wind. Estimate in between those numbers or use worst case. Wind is never as forecast or even reported, so it's always an estimation.

 

[EdFred]

Holy smokes. Tldr. 30 degrees. .5 for x wind. 45 .7 for x wind,. 60. .9 for x wind. Estimate in between those numbers or use worst case. Wind is never as forecast or even reported, so it's always an estimation.

thats good for crosswind, but for ded reckoning you may want a little more info.

 

[dtuuri]

Holy smokes. Tldr. 30 degrees. .5 for x wind. 45 .7 for x wind,. 60. .9 for x wind.

Just plug these into the autopilot do you?

 

[Landing Fees]

Here is my wind triangle.

 

[woodchucker]

Just for a fun project I created a flight planner in excel. Everything on the paper nav log is calculated via excel.

 

[Matthew]

Enough of your wizardry, I look out the window.

 

[Kelvin]

It IS entirely possible to overthink something.

 

[RingLaserGyroSandwich]

A few people have accused me of overthinking. If this derivation is what I'm doing before a flight, then perhaps you are quite right. However, without people thinking about this topic the way I am (or similarly), there is no CX-2, there is no Foreflight, etc.

I'm an engineer/physics type, so of course I am not going to approach questions the same way as someone limited almost entirely to practical experience and what is written in FAA publications for student pilots. At the end of the day, a pilot does not need to solve the wind triangle manually using formulas. That doesn't mean it's bad to know how, or that there aren't people who do need to know how.

 

[Capt. Geoffrey Thorpe]

Personally, I prefer to split both the vectors into x- and y-components using sin/cos, then perform vector addition (or subtraction) using the components by just adding (or subtracting) the components up, and then convert the resultant back into a magnitude and direction using arctan(y/x) and Pythagoras.

That. Do it the easy way.

 

[RingLaserGyroSandwich]

Personally, I prefer to split both the vectors into x- and y-components using sin/cos, then perform vector addition (or subtraction) using the components by just adding (or subtracting) the components up, and then convert the resultant back into a magnitude and direction using arctan(y/x) and Pythagoras.

But that's just me.

(Actually, my favorite is skyvector.)

After @Capt. Geoffrey Thorpe seconded your suggestion, I decided to add it to the OP for comparison. I'm not sure I'm understanding the method you use. If you break the TAS and W into components, then you are taking the sin and cos of the orientation of the TAS, which is not known. If you add up the x and y components, and set them equal to the x and y components of the GS respectively, then you are introducing a second variable, the magnitude of the GS, which results in a system of two equations which are not that easy to solve. Exactly how do you do this? If the process is easier than mine, I will be happy to say as much.

 

[Capt. Geoffrey Thorpe]

Hmmm. Need to think about it more. You may be right.

 

[PaulS]

Just plug these into the autopilot do you?

Naw, mainly worry about landing x wind, but end of the day these docs are estimations and planned wind never equals actual winds.

 

[EdFred]

What sort of wind triangle do you draw up to demonstrate the angle of your arm patting yourself on the back?

 

[RingLaserGyroSandwich]

What sort of wind triangle do you draw up to demonstrate the angle of your arm patting yourself on the back?

It depends on whether you want to use the Law of Cosines or Law of Sines, but however you draw it, a particular finger unavoidably protrudes.

 

[EdFred]

It depends on whether you want to use the Law of Cosines or Law of Sines, but however you draw it, a particular finger unavoidably protrudes.

That usually happens when its pointed out about how awesome someone thinks they are in reference to themselves.

 

[sarangan]

If you have a CX-2 flight computer (or equivalent), you likely plug in the true course, true air speed, wind direction, and wind speed, without giving it much thought. The computer spits out the required true heading in order for the plane to follow the proper true course, and also tells you what ground speed to expect. How does the computer do this? An E-6B does some cool mechanical rotation of vectors which I don't use and won't analyze, but I'll show the math behind the direct calculation method. If you don't like how I explain it but are interested there are other explanations floating around the internet. I'll start with a diagram to illustrate the problem.

The length of each arrow is the "magnitude" of the velocity it represents.

The red arrow represents the magnitude and direction of the wind, which is known (based on forecasts). The black arrow represents the magnitude and direction of the actual course of the plane, assuming proper wind corrections are made. We know the direction (based on drawing a line on a chart connecting our departure airport and our destination airport) but do not know the magnitude, since we don't yet know what ground speed to expect. The blue arrow represents the magnitude and direction of the plane's velocity relative to the air. The direction of the arrow is the heading of the plane after correcting for wind, and the magnitude is the true air speed which is known based on performance calculations. In short, the plane's velocity relative to the air, plus the air's velocity relative to the ground, gives the plane's velocity relative to the ground.

We know two sides of this triangle (it is not a right triangle so we can't use sin/cos/tan or the Pythagorean theorem!), TAS and W. We want to know the third side, GS (ground speed), but need to determine an angle before we can calculate it. Since we know what direction the wind is blowing, and we know the desired course, we can easily determine the angle between W and GS, which I labeled A.

In order to solve for GS, the length of the black arrow, we can use a math rule that most people forget about and never use again. The law of cosines is the more advanced version of the Pythagorean theorem and works even for non-right triangles. a^2 = b^2 + c^2 - 2 b c cos(A). In the formula, a, b, and c, are sides of the triangle, and A is the angle opposite side a. In the case of the wind triangle, A is opposite TAS, so TAS serves the role of "a". We'll say W serves the role of b (although it could be c too... they are interchangeable). Substituting:

TAS^2 = W^2 + GS^2 - 2 W GS cos(A)

That is an equation where we know the value of everything except for GS. Let's create a real example and try it out:

You want to fly direct from Ring City Airport to Laser Gyro Airport, which requires a course of 010 degrees (we'll keep directions true vice magnetic for simplicity). Your V-Tail Skychicken will achieve a TAS of 100 kts. Wind is from 315 at 20 kts. What is your required true heading and resulting ground speed to fly this course?

Just visualizing the problem, the plane will need to aim left of the destination in order to compensate for the quartering headwind (this matches the drawing, fortunately). I need to compute angle A before we can use the formula above.

By drawing the wind and the TC, then filling in all the angles (remember the angles inside triangles add up to 180 degrees), you can see that angle A is 45 + 80 = 125 degrees. Now we can solve the Law of Cosines:

100^2 = 20^2 + GS^2 - 2 20 GS cos(125 degrees)

This is a quadratic equation, which can be solved using the quadratic formula. See, all that crap from high school is useful when you decide to solve the wind triangle for fun. Rearranging the equation to make it simplest to solve:

GS^2 - [40 cos(125 degrees)] GS + (20^2 - 100^2) = 0

Further simplifying:

GS^2 + 22.943 GS - 9600 = 0

The quadratic formula says for f(x) = a x^2 + b x + c, the solution is x = -b/(2a) +/- sqrt(b^2-4ac)/(2a). In this case, a is 1 making the equation easier to use. Plugging in:

GS = -11.4715 +/- (1/2) sqrt(526.3812 + 38400) = -11.4715 +/- 98.6488. We can reject the negative solution, so GS = 87.177 kts. If you plug this problem into a CX-2 flight computer, it says your GS will be 87.2 knots. Pretty darn close.

It also says to expect a true heading of 0.6 degrees. How can we determine that without using the flight computer? Look back at the first figure. Now that we know GS, it is relatively easy to compute the angle between GS and TAS. That angle will be how far left of the destination we should aim the plane, aka, the wind correction angle (WCA). We will use the Law of Cosines again, except this time the angle will be B, the angle opposite W (remember, W was considered side b in the Law of Cosines before). The TAS and GS will be "a" and "c" respectively.

b^2 = a^2 + c^2 - 2 a c cos (B)

Notice how you can put b on the left side so long as the angle in the cosine is B. That's the neat thing about the Law of Cosines. Substituting:

W^2 = TAS^2 + GS^2 - 2 TAS GS cos(WCA)
2 TAS GS cos(WCA) = TAS^2 + GS^2 - W^2
cos(WCA) = (TAS^2 + GS^2 - W^2)/(2 TAS GS)

We need to substitute in the values for the right hand side of that equation, then take the inverse cosine (aka arc cosine) to find the WCA. By my calculations, the right hand side of that equation is 17,199.829/17,435.4 = 0.986. The arc cosine of 0.986 is 9.599 degrees. Correcting left from the course of 010 by a WCA of 9.599 degrees gives a true heading of 0.4 degrees. Oops, I was off by a two tenths of a degree. Actually, that's pretty close! The CX-2 is actually a little more accurate than I was because it uses somewhat more sophisticated mathematical methods to perform the calculations, but the concept is exactly the same in either case.

Once you turn the procedure above into an algorithm, a simple flight computer can calculate the ground speed and WCA in a short fraction of a second. Or, you can play around with the mechanical E-6B if you enjoy it and like battery independence. Or... you can just let ForeFlight do it all for you.

Here is another way to do it.

 

[RingLaserGyroSandwich]

That usually happens when its pointed out about how awesome someone thinks they are in reference to themselves.

Since I'm much less of a veteran here than you are, I figured I'd check with the mods and see where they stand on the acceptability of our exchange (of which we've both participated). There was a determination that nothing was out of line, so I consider that I can proceed with what I wanted to ask you instead of keeping quiet. What is it you take issue with? Can you be specific? I'll even withdraw my joke finger.

Your fist post seemed to suggest that you thought the OP was rather pointless given how you can use an E-6B (in addition to other tools). You weren't the only person to have that opinion, and I don't take it personally. Then you came back later to suggest I was patting myself on the back. Was that due to something I said in a subsequent post? You followed up with me thinking I'm awesome. Once again, was that due to something I said in a post after the initial post, or is this all revisiting the first post?

 

[EdFred]

I've got 5 of them on speed dial. Let me make a few calls.

My patting yourself on the back comment was more in reference to you follow up posts of "I'm too smart to be bothered with the simplistic approach the FAA and you simpletons follow in regards to WCA."

You're a bit preachy and self congratulatory and don't realize how you're coming across. Lots of engineering types are that way.

 

[RingLaserGyroSandwich]

I've got 5 of them on speed dial. Let me make a few calls.

My patting yourself on the back comment was more the follow up posts of "I'm too smart to be bothered with the simplistic approach the FAA and you simpletons follow in regards to WCA."

Okay now we are getting somewhere. I believe you are referring to, "I'm an engineer/physics type, so of course I am not going to approach questions the same way as someone limited almost entirely to practical experience and what is written in FAA publications for student pilots." That is different than "I'm too smart to be bothered with the simplistic approach." I DO use the simplistic approach, and I have no problem with anyone else using the simplistic approach either. I usually just plug stuff into the CX-2 or Foreflight.

As for the second part of your statement, "you simpletons [follow a simplistic approach]," I consider there to be a big difference between people who don't want to or are not immediately able to solve the wind triangle, and simpletons. Even if we limit our discussion to just the capability, you aren't a simpleton for not studying math or not being good at math. I like math and obviously encourage others to use math when they can, but lacking a math background doesn't make you simple... it makes you less likely to do things like try to understand exactly how, mathematically, the calculator is figuring out your wind correction angle and expected ground speed.

Perhaps I wasn't clear enough that I was using the word "limited" to refer to background associated with understanding the wind triangle, specifically... not anything farther reaching than that. Also I should have made it clear I was not referring specifically to the folks saying I was overthinking the problem.