Mathematical Solution to the Triangle of Velocities

The motion of an aircraft relative to the surface of the Earth is made up of two velocities. These are the aircraft moving relative to the air mass, and the air mass moving relative to the surface of the Earth. Adding these two vectors together gives us the aircrafts motion over the ground. Together, they form the “triangle of velocities”.

A wind that is blowing from the left will carry an aircraft onto a track that is to the right of the heading. In order to achieve a particular track from A to B the aircraft must be turned into-wind by an amount that corrects for the drift.

Each of the three vectors in the triangle of velocities has two properties – magnitude and direction. This means that there are a total of six components. These are the True Air Speed (TAS) and heading (HDG) of the aircraft, the speed and direction of the wind (W/V), and the Ground Speed (GS) and track (TR) of the path over the ground. This is shown in Figure 1 – The Triangle of Velocities.

Figure 1 - Triangle of Velocities

Figure 1 - Triangle of Velocities

Typical navigation problems involve finding two of these properties when given the other four. For example, most of the time we know what track and air speed we would like and we also have the forecast wind velocity. What heading do we need to steer to follow that track, and what ground speed will we achieve?

The usual method of solving this problem is with a “Flight Computer” such as the E-6B also known as a “whiz wheel”. The wind side of the flight computer consists of a circular rotating compass rose which is marked with an index at the top, and has a transparent screen in the middle to allow viewing of the slide plate underneath. The slide plate is marked with concentric speed arcs and radial drift lines. The computer allows you to physically visualise the triangle of velocities, and read off the answer you require. But what calculations is the flight computer performing? How do you solve the problem mathematically?


Calculating the Required Heading

In order to find the heading required, we need to make use of the sine rule. The sine rule states that for any triangle the ratio between a given side length and the sine of the corresponding angle is equal for each side of the triangle – Figure 2.

Figure 2 - The Sine Rule

Figure 2 - The Sine Rule

We simply substitute in the parameters from Figure 1, and rearrange to solve for our heading (HDG). Thus,

Heading Equation


Calculating the Ground Speed

The ground speed is simply the magnitude of our track vector. The easiest way to determine this value is to divide the triangle of velocities up into two right-angled triangles – Figure 3.

Figure 3 - Ground Speed

Figure 3 - Ground Speed

The length of the track vector is then just the sum of the x-component of our velocity through the air mass and the x-component of the wind velocity.

Ground Speed Equation


Flight Planning

These equations are quite cumbersome, and if working out the solutions by hand then by far the quickest solution is to use the whiz wheel. However, now that we understand the mathematical solutions it is possible to enter them into a spreadsheet and speed up the flight planning process considerably.

Posted in Geekorama, Rotorcraft.

Steven Hale leads the operation and development of the international Birmingham Solar Oscillations Network (BiSON), a global network of automated robotic solar telescope run by the University of Birmingham in the UK. His research interests are instrumentation and electronics, and high-resolution optical spectroscopy techniques. In his spare time he has many interests including photography and aviation, and has a private helicopter license rated on the Robinson R22 and R44 aircraft.

This is a private blog and in no way represents opinions or endorsements from the University of Birmingham.

37 Comments

  1. Pingback: A Windy Problem – Steven Hale Photography

  2. Thank you for this. I tried to work it out from first principles and being an engineer, started with cartesian co-ordinate geometry which leads to some tricky simultaneous equations. Then I realised that Euclidean geometry was the answer but got in a muddle again by trying to use the cosine formula. I hadn’t used the sine formula since school (> 40 years) and had forgotten it ! A happy rediscovery.

    One thing I would add to your excellent monograph and that is to remind your readers that they must convert wind angles from “from” to “to” for consistency with the Heading and Track vectors – then everything falls into place. Since you mention spreadhseets, it might also be worth warning that they work in radians (or at least the ubiquitous Excel does). A further picky point, you don’t define Wd and Ws although it’s pretty obvious to anyone who has thought about the problem.

    Best wishes,

    John Medhurst

  3. Pingback: Flight Planning « Programming Praxis

  4. this does not help me at aallla soo take it down and yall know that stuff don not help haha so stupid ughh

  5. I’m sorry this post did not help you Lisa. But tell me, why should I take it down just because it didn’t contain the information you needed? Other people have found it useful so I think I will leave it here. Thanks for your comments, and good luck!

  6. All very interesting! You take me back to my school days. But are all these calculations necessary? If your looking to look clever in the clubhouse, I would suggest that your looking to have your ego stroked!
    Let me ask the question… How long will it take you to calculate 5 legs that your trip may require?
    We conducted an experiment with the above scenario. Using my CRP-1 to calculate my drift and not forgetting variation and lastly my GS, all 5 legs took 6 minutes! I ask you, all who read this to try doing it the ego method. And we used a maths teacher, he laughed and told us “don’t be so f****** stupid.
    THIS METHOD IS OF NO VALUE!

  7. Thanks for your comments Gill.

    Method is of no value? How do you think people worked out how to make a CRP-1? And do you never want to know something, just for the sake of knowing and understanding?

    Working this out certainly had value when I wanted to put all the calculations onto a spreadsheet.

    But thanks anyway. Happy flying 🙂

  8. Your missing the point Steve. We all learnt this at school. My dear Steve, it seems your arrogance is indeed only preceded by your encyclopaedic opinion of yourself. Yeh,, keep it simple… stupid!! But no doubt you’ll try to justify your superior intellect.
    Must go… no flying today. xx

  9. Nope, not justifying anything. Just happy to help. If someone finds it useful, that’s great.

    Thanks for your comments.

  10. Steve;

    It is useful, as you said, to those who prefer to understand. I am familiar with the calculations but was looking for a shortcut – someone who put it together into a flight planning spreadsheet, or at least a cross check that I’ve not made a careless error in entering my own excel formulas.

    I, like most, are totally comfortable using an E6-B but prefer a less tedious means to getting a good knee board flight plan together. Your explanations are excellent and clearly stated, and with diagrams to boot.

    I don’t quite get the comments from those who attacked you, but applaud the restrained politeness in your responses. You’ve gone to the trouble to share information that is of interest to you, and I, unlike some others, really do appreciate it.

    Thank-you.

  11. Well I am glad to see that a similar named Hale is almost (just in case) Christian in his charity (unconditional love or peculiar kindness) in answers to folk that seem to desire to be obnoxious. Keep up the good work so that the rest of us Stevie Ray Hales in America are not scandalized by association…and it helps to have some kindness in this hateful world to offset that hate.

    I have never flown an aircraft but as a geologist, I am interested in aircraft to land photography in providing stereoscopic photos so that we can do some surface geology calculations of the terrain in areas of interest. I found this concept of stereoscopic photography of use when interpreting 1 Corinthians 13:12 depicting the information found in one eye (gnosis as knowledge in Greek) and that found in both eyes (giving EPIGNOSIS or knowledge upon knowledge in Greek) and how that relates to the gift of prophecy requiring TWO or more prophetic scriptures with a single focus in order to interpret prophecy as doctrine. (2 Peter 1:20 is Peter’s amen to Paul’s writings on this subject where he in this corrected from the Greek translation should be “this first knowing that every prophecy of scripture is NOT of its own interpretation” which is one of the corollaries to the Torah law of multiple witnesses which says that in the mouth of one witness shall no man be put to death) and what Jesus says in John “If I bear witness of myself my witness is not true). The concept of TWO photos come in as a clear example of seeing with two eyes or according to the Torah law of multiple witnesses which says that “in the mouth of two or three (or more) witnesses shall a matter be established.” The same thing Jesus says in Matthew 18 concerning matters dealing with church discipline. Sorry this seems way out of line, but I do get excited about these sort of things that give my little interest in photo geology an exalted position in the true manner of interpreting scripture. The Bible is actually an archive of testimony, out of which no doctrine can be established without two or more scriptures with a similar focus to confirm as the mind of God of what He desires us to know.

  12. Ahh ha, at last I found a flightplan to double check my “manual” calculations. Gives me everything I need, good job and it is very accurate.

  13. Just had to say what an excellent post, well done Steve.

    It’s a shame that you have had to endure abusive posts from the likes of Gill Rowan who as you say, has clearly missed the point by a country mile.

    In my experience attitudes like that are born from a previous bad experience at school and an innate phobia of Mathematics, it probably hindered him getting his PPL I suspect.

    If he’s still flying (I suspect not with an appalling attitude like that), then he’ll likely come into contact with an Examiner soon for a renewal who’ll put paid to his flying days. One can hope!

    You keep up the good work. Very informative and fun to read.

  14. Hi Steve,

    Thanks for your effort on your explanation of ‘Triangle of Velocities’.
    Mate I’m struggling little with it.
    Could you possibly introduce the figures for your ‘figure 2’? So I can check my workings.

    I thank you again for your time. I hope you don’t let negative buggers like Gill get to you.

    Cheers

  15. Hi Graeme,

    You just need to compare the triangle in figure one with the triangle in figure two, and substitute the values. They are both the same triangle, just rotated 90 degrees in the diagram. So,

    Angle A would be your track minus the wind direction.
    Angle B would be your heading minus your track.

    Length A would be your air speed.
    Length B would be the wind speed.

    Drop those values into the sine rule and you get the first equation immediately after figure two. Then you just need to re-arrange the equation to get the heading.

    Hope that helps?

    Steve.

  16. Thanks for the help in getting the formulas. Unlike some uncharitable folks on here, I see great value in this, and was wanting my spreadsheet to do it for me. According to one individual, 5 legs may take only 6 minutes, but what about 50 legs? Then I am saving some real time. What about wh, en things change shortly before takeoff? Or the weather keeps delaying your flight day after day and you have to replan multiple times? Much help here. I had already generated a spreadsheet to print out my kneeboard flight plan, I have modified your formulas to fit into my format, and greatly appreciate your help. One comment tho on one formula, it states: IF(cell > 359.5. cell-360) this would give a negative number if the original answer was 359.8. 359.8 is a valid number. I changed it to IF(cell > 360, in my spreadsheet, it allows the full range of 0 to 360 values. Thanks again!

  17. Many thanks for your comments Ken.

    Thanks also for the bug report! You’re quite right the condition should be if greater than 360. Not sure where 359.5 came from, I think I might have had a round() in there at some point. I’ll update the spreadsheet.

    Steve.

  18. Hi Steven,

    Many thanks for this. I’m not sure if I’ve missed something or done soemthing incorrectly but I found that the ground speed formula only worked for me if I changed the + to a minus. Plus gave me a higher GS with a headwind. When there’s a tailwind, the COS( TR – Wd) returns a negative value which then results in minus a minus i.e. plus, giving an increase in GS.

    Am I wrong ?

    Best regards,
    Ray

  19. Hi Ray,

    I think you’ve got the wind vector pointing the wrong way. Remember, a “northerly wind” comes FROM the north. So if the wind is listed as 0 degrees, you need to add 180 degrees to the wind direction to flip the vector and turn it into a headwind on a 0 degree track. In the ground speed calculation, you’re quite right that COS(TR – W_d + 180) is exactly the same as doing COS(TR + W_d).

    However, if that was the case then all your drift angles should be reversed and so your headings should be opposite as well. If that’s not the case and your headings are coming out correct, then I’m not sure what’s going wrong without seeing how you’ve set up the calculations. But I think it’s most likely a problem with how you’re defining the wind direction. You might like to take a look at my flight planner spreadsheet and see how the calculations compare with your own.

    Hope that helps!

    Steve.

  20. Many thanks, Steven.

    I haven’t done any flying for years, but I’m playing with JavaScript by modelling the two-body problem from scratch.

    I too had forgotten the sine rule, and it’s most useful. The moon’s straight line velocity combines with the velocity induced by us, which we know thanks to Isaac, and hence the resultant velocity.

  21. Wow, I can’t believe some of the negativity? What horrible people. It helped me and I appreciated the brain power so thanks!

  22. This is pretty straight forward. Thanks for posting. I have in into a problem with regards to parachuting. I calculate a high altitude release point (HARP) and I am looking for a region around that HARP where a jumper can make it to the drop zone (DZ). The chute has forward drive characteristics so I am using the formulas just like I would an aircraft to calculate drift. I get a circle for the release area, where I had imagined an elipse, due to the ground track of faster speeds downwind vs upwind. Have you ever looked at parachute release point characteristics?

  23. Hi Wes,

    I’ve never looked at parachute release point characteristics. But I should imagine that it is reasonable for the release area to be a circle rather than an ellipse. The drift error is additive rather than multiplicative. That is to say it’s an offset rather than a scale-factor.

    Hope that helps in some way!

    Steve.

  24. This is just what I was looking for.

    I’m surprised that those readers like Gill Rowan and Lisa Ray are reacting like that. They may not get that your post is just for fun or for implementing an automated flight computer. They may even be right in thinking that you feel proud of having worked it out and presented it neatly. But it’s bizzare to conclude that that is a bad thing.

  25. hi steven . i found this post so much valuable and useful the only thing i want to add is that both the formulas have sign problem in between they must have minus (-) instead of plus (+) in between the R.H.S

  26. Thanks for your comments lakhan.

    I disagree about the signs. Perhaps you’re forgetting that wind is defined as coming FROM the direction and not TO, and so you’re 180 degrees out?

    Steve.

  27. have been trying to solve this problem in Excel,(ok I know I’m a massochist) but i cant get over the 360-0 problem. Any ideas

  28. Could you explain how Drift Angle (Angle b) has been written as HDG – TR.

    Also how is Wind-Track Angle (Angle a) equal to Tr – Wind ?

    Cheers

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.