How to calculate the real Ground Speed from True Air Speed?

In most of the formulas I've found online GS = TAS + Vw, i.e. true air speed plus wind.

However, on the simulator, GS changes drastically if I dive or climb which is obvious because I'm covering 0 ground distance if I dive vertically.

What is therefore a real GS formula from TAS? It has to take into account the wind (Vw) but also the "3D angle of the aircraft" (for the lack of better expression).

• If you have a non-null GS when flying vertically, then your simulator has a problem. Likely duplicate of Why is there a difference between GPS Speed and Indicator speed?
– mins
Sep 8, 2017 at 8:47
• Not necessarily: With pitch of +/- 90 degrees, you might still have a lift generating AoA, depending of angle of incidence etc. Even considering a zero-lift vertical dive, horizontal wind component could still cause positive GS. Sep 8, 2017 at 8:51
• ...(this of course depends on how one defines "flying vertically", if you mean "with indicated attitude of +/- 90 degrees", then my comment applies) Sep 8, 2017 at 9:07
• @Waked: "Vertically" as the OP described very accurately: "I'm covering 0 ground distance"
– mins
Sep 8, 2017 at 11:06
• @mins and in that case you are correct, by definition. I was allowing for the possibility that the OP incorrectly deduced that "nose pointing straight up/down" automatically results in "covering 0 ground distance". Sep 8, 2017 at 11:19

First calculate horizontal component of airspeed, then add the wind:

$$v_{GS} = cos(\theta) * v_{TAS} + v_{wind}$$ with $\theta$ being the angle between the horizon and the path of the aircraft in the vertical plane.

Or, if you are unfamiliar with trigonometry (using Pythagora's theorem):

$$v_{GS} = \sqrt{v_{TAS}^2-v_{verticalSpeed}^2} + v_{wind}$$

Both formulas assume the same units being used for all speeds ($v_{TAS}$, $v_{verticalSpeed}$, $v_{wind}$), and only take horizontal wind into consideration. $v_{wind}$ is only considering the headwind/tailwind component.

• Pitch angle being pure pitch value, regardless of roll and yaw?
– Ska
Sep 8, 2017 at 11:21
• Well, more correctly, the angle between the horizon and the actual flight path. This example is in straight flight, ie not turning. If you'd like to involve turning flight (roll/yaw) you'd also have to decide along what you want to calculate ground speed. Along the turn radius or along an arbitrary vector? Sep 8, 2017 at 11:27
– Ska
Sep 8, 2017 at 12:59
• Actually, although acceleration is directed towards the center of the turn, at any given moment, the velocity of the aircraft will always be tangential to the turn (given coordinated flight). This means that the formula still applies. If you wanted to, you could parameterize head/tailwind ($v_{wind}$) as a function of time, $t$ to calculate $v_{GS}$ at any given $t$. This would involve finding the rate of turn (function of gravity constant, bank angle and $v_{TAS}$). What is it that you need the formula for? Sep 8, 2017 at 16:10
• @Weaked I need it for getting the ground speed to calculate waypoint arrivals for missions. I can read IAS, but rest I have to calc. Now, how to get "angle between the horizon and the path of the aircraft in the vertical plane"? With "rate of turn " as you mentioned?
– Ska
Sep 10, 2017 at 19:22

A real GS formula from TAS takes into account two velocity triangles: one with the vertical velocity, and one with the wind velocity.

1. Vertical velocity. Here's the velocity triangle. With no wind, we get:

$$cos(\Phi) = \frac {GS}{TAS} \tag{1}$$ $$sin(\Phi) = \frac{V_C}{TAS} \tag {2}$$

And we know from math lessons that $sin^2(\Phi)$ + $cos^2(\Phi)$ = 1, so:

$$\frac {GS^2}{TAS^2} + \frac{V_C^2}{TAS^2} = 1 => GS^2 + V_C^2 = TAS^2 =>$$ $$GS = \sqrt{TAS^2 - V_C^2} \tag{3}$$.

1. Wind speed. The equation in the OP just adds wind speed to the TAS, and this is only valid if the wind direction is the same as the flight direction. This is usually not the case, and we will need to consider another velocity triangle, this time from the viewpoint of looking down on the plane:

In this example, $\Phi$ = 70-30 = 40°. The cosine of the wind speed we can add directly to the ground speed, the sine component will need to be added in a Pythagoras way.

$${V_{TOT}}^2 = (V + V_W \cdot cos (\Phi))^2 + (V_W \cdot sin (\Phi))^2$$

=> $${V_{TOT}}^2 = V^2 + 2 \cdot V \cdot V_W \cdot cos(\Phi)+ {V_W}^2 \cdot cos^2(\Phi) + {V_W}^2 \cdot sin^2(\Phi)$$ and again since $sin^2(\Phi)$ + $cos^2(\Phi)$ = 1 $${V_{TOT}}^2 = V^2 + {V_W}^2 + 2 \cdot V \cdot V_W \cdot cos(\Phi) \tag{4}$$

1. Combine equations (3) and (4)

$$GS = \sqrt{TAS^2 - {V_C}^2 + {V_W}^2 + 2 \cdot \sqrt{TAS^2 - {V_C}^2}\cdot V_W \cdot cos(\Phi)} \tag{5}$$

• Is Vertical velocity coming directly from instruments?
– Ska
Sep 11, 2017 at 4:46
• There is an instrument that directly indicates the vertical speed, but it has a time delay. Since vertical speed is measured as altitude compared to altitude some time ago, the time delay is inherent since we're measuring a time derivative. Sep 11, 2017 at 5:24