Landing a Plane in a Crosswind: Relative Velocity and Vector Addition Explained (IB Physics)
To fly straight down a runway in a crosswind, a pilot does not point the nose along the runway. The nose has to point into the wind, and the amount can be worked out with a right-angled triangle. This is relative velocity, and it is a clean example of why velocities add as vectors.
The big idea: velocities add as vectors
An aircraft moves through the air, and the air itself moves over the ground. So the velocity of the aircraft over the ground is the velocity through the air plus the velocity of the air: v(ground) = v(air) + v(wind). Draw the two vectors head to tail and their sum is the third side of the triangle, the velocity you really have over the ground.
The problem
An aircraft has an airspeed of 200 km/h. A steady 50 km/h wind blows across the runway, at right angles to it. The pilot wants the aircraft to move straight along the runway. At what angle must the nose point, and how fast does the aircraft move along the runway?
The solution
- •For no sideways drift, the sideways part of the aircraft’s velocity through the air must cancel the wind: V sin θ = W.
- •sin θ = 50 ÷ 200 = 0.25, so θ = sin⁻¹(0.25) = 14.5°. The nose points 14.5° into the wind. Pilots call this the crab angle.
- •The 200 km/h airspeed is the hypotenuse, so the speed along the runway is √(200² − 50²) = 193.6 km/h. It is a little less than the airspeed because part of it is used to fight the wind.
What happens if you ignore the wind
Point the nose straight down the runway and the aircraft keeps its 200 km/h along the runway but also drifts sideways at 50 km/h. Over a 4 km approach, which takes 4 ÷ 200 = 0.02 hours or 72 s, the drift is 50 × 72 ÷ 3600 = 1.0 km. That is far off the runway.
Same method, a different river
A swimmer can swim at 1.5 m/s in still water and wants to cross a river straight across. The river flows at 0.60 m/s. The swimmer’s own speed is the hypotenuse again: sin θ = 0.60 ÷ 1.5 = 0.40, so θ = 23.6° upstream, and the speed across is √(1.5² − 0.60²) = 1.4 m/s. Aircraft in wind and swimmers in rivers are the same problem.
How to choose sin, cos or tan
Always draw the triangle first and ask which side you know. In the crosswind problem the airspeed (the longest side) is known, so the angle comes from sin θ = W ÷ V. In the umbrella problem (see the related post) the rain speed is a shorter side, so the angle comes from tan θ = W ÷ v. Same method, different triangle.
Common mistakes
- •Pointing the nose at the destination in a crosswind.
- •Adding the speeds instead of using Pythagoras for vectors at right angles.
- •Using the airspeed as one leg of the triangle when it is the hypotenuse.
Frequently asked questions
The wind adds a sideways velocity to the aircraft. Pointing the nose into the wind gives the aircraft an equal and opposite sideways part of its own velocity, so the sum points straight along the runway.
sin θ = wind speed ÷ airspeed, when the wind is at right angles to the desired track. For 50 km/h and 200 km/h, θ = 14.5°.
No. Airspeed is the speed through the air. Ground speed is the speed over the ground, which is the vector sum of the airspeed and the wind, so it is generally different.
The velocity of one object as seen from another: v(A relative to B) = v(A) − v(B). The ground velocity of an aircraft is its velocity relative to the ground.