Wind triangle
Course is where you want to go. Heading is where you have to point to get there. The wind is the reason those differ and the groundspeed is what it costs you — and for a glider that cost is best expressed not in minutes but in feet.
The track you want to make good over the ground.
True, not indicated. At altitude they differ — convert here.
Heading to steer
Crab
Groundspeed
kt
for the leg
- Crosswind component
- kt
- kt
- Wind relative to course
Height the leg costs
Against in still air — % . Your glide ratio over the ground is :1.
sin (WCA) = W · sin α ÷ V
GS = V · cos (WCA) − W · cos α
- WCA
- wind correction angle — how far to crab
- GS
- groundspeed
- V
- true airspeed
- W
- wind speed
- α
- angle between the wind and your course
The first line is the one a flight computer hides: the crab depends ONLY on the crosswind component and your airspeed. Headwind does not steer you, it only slows you down — which is why a direct headwind needs no correction at all however strong it gets.
The case with no answer
If the crosswind component exceeds your airspeed, no heading holds the course. You cannot out-crab it: point as far into it as you like and the wind still carries you downwind of the track. The equation says so honestly — sin (WCA) would have to exceed one, and there is no such angle.
For a powered aircraft this is a curiosity. For a glider it is a Tuesday: thirty-five knots of wind and a weak day flown at fifty is well inside the range where a cross-track leg simply stops being available. The page tells you the airspeed you would need rather than returning nothing, because that is the next question.
Why height, not minutes
A powered aircraft converts a headwind into extra time and extra fuel. A glider converts it into extra height, and the conversion is worse than people expect, because height goes as the reciprocal of groundspeed rather than with it.
Twenty knots on the nose at sixty knots true leaves forty over the ground. That is a third of your speed gone — but your glide ratio over the ground is also down by a third, to two thirds of the still-air figure, so the height the leg costs goes up by half again. A 750-foot leg becomes 1,125.
The asymmetry runs the other way too, and is why a downwind final glide feels so generous: the same twenty knots behind you gives eighty over the ground, a third faster, but only saves a quarter of the height. Wind helps you less than it hurts you, at every speed.
What this page does not model is that your speed-to-fly should change with the wind as well. Into a headwind the right answer is to fly faster than still-air MacCready, which recovers some of the loss. The polar page does that properly, and the final glide page shows how sensitive the arrival is to getting it wrong.
What this assumes, and what it can't know
One wind, one leg, no gradient. Real wind changes with height and with where you are over the ground, and a glider crosses a lot of both. This solves a single triangle in a uniform wind, which is the right first answer and not the last one.
True airspeed, and the same units throughout. Feeding it indicated airspeed at altitude understates your speed and overstates the wind's effect. Wind and airspeed must be in the same unit; the page cannot tell knots from km/h.
The glide ratio is the one you will achieve. Not the one on the placard. Bugs, water, the air you are crossing and how accurately you fly all take their cut, and a sensible working figure is well below the book number.
No arrival margin. The height figure is the height to arrive at zero, over the ground, in the air as modelled. It contains no circuit, no safety margin and no allowance for the sink you will meet.
Not yet checked by anyone but me. If you instruct, or one of these assumptions is wrong, I would genuinely rather hear it than not — tell me and I will credit you here. This is ground school, not a flight computer, and not an authority on your aircraft. Fly the numbers in your own flight manual and the instruments in front of you.