I saw some equation in a log recently and I believe it went something as follows:

$v_1$ : flying North, $\ \ v_2$ : flying West, $\ \ v_3$ : flying South.

$$ \text{TAS} = \frac 1 2 \sqrt{v_1^2+v_2^2+v_3^2+\frac{v_2^2}{v_1^2+v_3^2}} = 199.7 $$

Which seems a little high for an RV-4 with O-320, even at 8,000ft and max power.

I may be mixing up the subscripts on the $\frac{v_2^2}{v_1^2+v_3^2}$ part, but those are the only subscripts that make sense by symmetry.

How does this airspeed calculation work?

I can imagine flying in a square and then taking the average $\frac 1 4 \left( v_N + v_W + v_S + v_E \right)$ (GPS-calculated groundspeeds) and I see the problem with that is that while the headwind and tailwind components cancel, the crosswind will not. How does taking the euclidean distance between velocities fix this and what is the final term $\frac{v_2^2}{v_1^2+v_3^2}$?

I just found a paper on this that seems to correspond to the TAS measurement technique this test pilot was using: I believe it is David Rogers Horseshoe Heading technique.

But I don't see how the equation presented above relates to any on that paper.

So what is the equation above?

For context, the reason we would not want to use the normal equation based on CAS, is because when flight testing a new aircraft we do not know for sure what the IAS→CAS conversion is, so it is necessary to determine it (or verify it, if building a kit). One way of doing that is to determine TAS independently (here using GPS and DR's HHT) and then calculate CAS from the TAS so that you can create your IAS→CAS correction table.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Farhan
    Commented May 18, 2022 at 2:20

1 Answer 1


It seems what they could have done here is to make a right triangle with an angle and a side, using cosine angle to determine TAS$^1$.

What they did was fly a 3 sided "horse shoe" pattern to determine wind component speed from drift, then applied this information to the headings into, across, and away from the northerly component. The westerly component was across, with, and across for the 3 legs.

From there they applied the formula a$^2$ + b$^2$ = c$^2$ to determine the sides of each triangle. That part is fairly straightforward.

Interestingly, the Air Force Test Pilot School suggested simply flying 2 legs perpendicular to the wind, while the National Test Pilot School flew 2 legs into and away from the wind. The Horseshoe Heading Technique offers the advantage of verifying wind direction by flying one leg perpendicular to the other two (begging the question: would not a box pattern be even better?)

The RV-4 looks rather sporty at 1500 lbs and easily makes 200 mph with the 180 horsepower O-360 engine at 75% power. The O-320 is around 150 horsepower. At max power it may be able to make 200.

The mathematics beyond building the triangles and determining the sides may be a bit more understandable by deriving wind components first, TAS component from each triangle next, and finally averaging all TAS results.

$^1$ even if not perfectly into or perpendicular to the wind, opposite TAS runs can still be averaged from cosine angle of course deflection × distance traveled (the hypotenuse). The "horseshoe" does offer 3 sets of data, instead of 2. The more runs you do, the more accurate the final averaged TAS will be. As the aircraft moves with the airmass, it's TAS relative to the airmass is not affected by wind direction, so ground track and speed corrections can be made.

  • $\begingroup$ My understanding is that you could calculate TAS from CAS derived from the IAS you observe, but the derivation for your particular aircraft may differ slightly from the factory plans, so using the HHT with GPS to calculate TAS would give an entirely independent measure of TAS that you could use to verify your calibration equation. $\endgroup$
    – Zaz
    Commented May 16, 2022 at 19:14
  • $\begingroup$ I see that by moving the 1/2 inside, we are taking an average inside the square root, but I am confused about what that last term in that average represents. $\endgroup$
    – Zaz
    Commented May 16, 2022 at 19:15

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