We can all agree, flying from point A to B, a headwind will increase fuel consumption, and a tail wind will reduce it. The "jet stream" inspires these types of thoughts.

Quartering headwind, crabbing into, still more fuel, right?

Quartering tailwind, probably less, but the aircraft still must oppose the cross wind component with some force which is ultimately derived from burning fuel.

Perhaps, from someone with airliner experience, at what point off my tail does wind stop assisting in fuel economy?

With the tail as 0 degrees, nose as 180, and beam as 90, my guess would be around 80 degrees, a cross wind with a bit of tailwind component.

Is there evidence from practical experience to support or refute this guess?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Farhan
    Commented Jun 3, 2022 at 12:50

1 Answer 1


To solve this problem, let's look at the vector diagram of TAS, GS and Wind Speed (WS):

Wind Correction Angle Diagram

(image source)

In this diagram, the angle between the desired course (ground track) and the wind is $\delta$ (where 0° is tailwind and 180° is headwind). The resulting wind correction angle $\theta$ can be calculated as follows:

$$ \theta = \arcsin \left( \frac{\text{WS}}{\text{TAS}} \sin\delta \right) $$

The resulting ground speed (length of vector labelled "Desired course") is then given by

$$ \text{GS} = \text{TAS} \cos\theta + \text{WS}\cos\delta $$

The following plot shows the wind correction angle and ground speed for an example of 400 KTAS and 50 kt wind speed at different angles:

WCA and GS plot

We are now looking for the angle $\delta$, where the ground speed is equal to the TAS. This will typically be close to 90°, but not exactly. In the example above, this angle is at ~86.4°. When solving the equation above with GS = TAS for $\delta$, the result is:

$$ \boxed{\delta = \pm \arccos \left( \frac{\text{WS}}{2 \text{TAS}} \right)} $$

This result depends on the ratio of WS/TAS, meaning there is no single answer to the question. We can however plot this angle, where the wind has no impact on ground speed, as a function of this ratio (only positive solution plotted):

Angle of no wind impact

When the wind angle is less than this value, we benefit from the tailwind. When the wind angle is higher than this value, we suffer from the cross- and headwind.

  • 2
    $\begingroup$ @GdD It only depends on the ratio, not the individual values. I highlighted that there is no single answer to the question. $\endgroup$
    – Bianfable
    Commented May 30, 2022 at 10:56
  • 4
    $\begingroup$ To be clear, I was emphasizing that crabbing does not increase aerodynamic drag. It will effect fuel consumption on a fixed distance leg due to increased or decreased distance (time) flown throughout the airmass as compared to over the ground. Any wind component aft of relative 90/270 line helps, anything forward hurts. The amount depends on direction and magnitude of the wind vector. Trig it out if you like… $\endgroup$ Commented May 30, 2022 at 14:46
  • 3
    $\begingroup$ @MichaelHall This answer calculates the breakevent point. The last plot shows the wind angle below which you would benefit from the wind as a function of wind speed over true airspeed. $\endgroup$
    – Bianfable
    Commented May 30, 2022 at 14:59
  • 1
    $\begingroup$ @Bianfable, I suppose it would be correct to say that in any aircraft equipped with GPS groundspeed, (most) the breakeven point would simply be displayed GS and computed TAS being equal, right? No need for trig while flying... $\endgroup$ Commented May 30, 2022 at 16:32
  • 1
    $\begingroup$ @MichaelHall Yes, that's correct (it doesn't have to be GPS ground speed, could be IRS based as well). $\endgroup$
    – Bianfable
    Commented May 30, 2022 at 16:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .