To non-pilots, landing an aircraft in a crosswind looks all but impossible. When the wind is perpendicular to the direction of motion of the plane, the plane has to aim in one direction—its wheels not lined up with the runway—so it moves in another. To pull it off, the pilot must quickly change the orientation of the plane once it touches the runway. It’s *hard*. Whenever possible, a pilot would prefer to land flying into the wind and not perpendicular to it.

You don’t have to be a pilot to get a feel for this kind of landing, though. Crosswind landings follow the same concepts as a classic physics problem that goes something like this:

You have a boat that can travel with a speed of 4 m/s relative to the water. This boat will be used to cross a river that is 50 meters wide and has water moving at a constant speed of 2 m/s. What angle should you point the boat such that it travels across the river to a point directly on the opposite bank? What is the fastest way to cross the river?

I will answer the above question, but first I will go over the physics of relative velocity. Let me start with a simple case. Suppose you have a train car moving along with a constant speed of 1 m/s. Inside the car a person throws a ball so that it has a constant horizontal speed of 3 m/s. What would that look like if you were inside the car? Yes, this is a simple question. If you are inside the car and you throw a ball at 3 m/s, it will look like it is traveling 3 m/s.

Now imagine that you are standing on the ground outside the moving train car. As you look into the car at the ball, how fast does the ball appear to be moving? Ah ha! You can’t actually answer this question because I didn’t say which way the ball was thrown. If the ball is thrown in the same direction as the motion of the car, then it will appear to be moving with a speed of 4 m/s (1 m/s + 3 m/s). However, if the ball is thrown in the opposite direction of the car it would appear to be going at a speed of 2 m/s.

In general, we define velocities relative to some coordinate system—this coordinate system could be moving with the train or it could be on the ground. Heck, the coordinate system could even be on a different train car moving at a different speed. Really any coordinate system that moves with a constant velocity will work. But once I have two difference reference frames (like the car and the ground), then I can write the following vector equation relating velocities in different frames.

I actually wrote the equation twice (in case you couldn’t tell). In the first version, I explicitly included the velocities in terms of the object and the reference frame. So v_ball-ground_ is the velocity of the ball with respect to the ground and v_car-ground_ is the velocity of the car with respect to the ground. The second equation is written in the way you would normally see it with “b” representing the ball and “c” representing the ground. But here is the key—these are vector quantities that have to be added as vectors.

Just for fun, here is a Python model in which I can show the motion of a ball both as viewed from inside the car and from outside the car. First, this is the motion as viewed from the car. Just click the “play” button at the lower left to get the thing started (if you want to look at the code, click the “pencil”).

Here it is looking from the ground for *the exact same situation*.

Notice in the view from the car it appears as though the ball just goes straight up and then back down. However, when viewed from the ground you get something different. But your viewpoint doesn’t matter. Either way, the ball lands back on the car in the same spot.

But what about the case of crossing the river? How do you get straight across? How do you get across the quickest? Before going over the exact solution, I made a Python model so you can play with the different crossing angles. Below you see a river (yes, I made the river to the best of my artistic ability). The arrow is the boat and it is pointing in the travel direction with respect to the water (so this is how it would look as viewed from above). You can click and drag the direction of the arrow to set the launch angle of the boat. When you let go, it runs and shows you the motion of the boat with respect to the ground (not the water). If you want to run it again, click the “play” button. Once the boat gets across the river, the program will print out the time to cross and how far the boat traveled in the direction of the river.

Play around with the river crossing model and see what you can figure out.

Please tell me you tried at least a couple of different angles. Here is a hint: The fastest time you can cross the river is in 12.96 seconds. If you didn’t get that time, you can keep trying to get a faster time.

Now for the full solution. I will start by writing the two things I know—the velocity vector for the water with respect to the ground and the magnitude of the velocity of the boat with respect to the water. Actually, if I assume the boat is pointed at some angle, θ then I can also write this as a vector. Note that I am representing vectors as three components in the x, y, and z directions with the angle brackets. Of course there are many ways to represent a vector—use the format that makes you happy.

Just to be clear, the x-component of the water’s velocity with respect to the ground is negative since I have the water flowing to the left. Of course to solve the two problems about the river crossing, I need the velocity of the boat with respect to the ground. I can find that by adding the two vectors above together.

If the boat is to travel to a point directly on the opposite side of the river, then its x-velocity must be zero (with respect to the ground). When looking at a vector equation (like the one above), it is possible to just look at one component of the vectors. By just considering the x-components of the velocities and letting the x-velocity of the boat with respect to the ground to be zero, I get the following:

Try going back to the Python model above and see if this angle does indeed make the boat go straight across the river. Yes, I know it’s not trivial to get the arrow right at 60 degrees, but you can at least get close.

But what about the fastest crossing time? This will happen when the y-velocity of the boat with respect to the ground is the highest. There is no y-velocity of the water’s velocity, so it’s all just due to the boat. Look at that expression for the y-velocity of the boat and notice that it depends on the sine of θ. When is the sin(θ) the greatest? When θ is equal to 90 degrees. So just aim the boat straight across the river and it will get there in the least amount of time—but it doesn’t travel straight across since there is still the x-motion due to the water. Go ahead and try it with the model and see if you can get the lowest time.

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