Heading Problems

Heading problems are navigation problems that arise when pilots and boaters must correct for winds or currents. A simple example would be when someone wants to cross a river to get to the opposite side as shown in the diagram below.

If the person were to try to travel straight across, due North in this example, the river current would carry them west of where he or she wanted to be as shown below.

So to travel directly across, the person must actually head upriver shown in the following diagram.

Note that we assume that the person always travels at a constant velocity.

The question may ask you to find the heading (angle θ in the diagram), how long it takes to get to the destination, the resultant speed, or how fast and which way the current flows. All these questions involve the same starting point. You must the first draw the vector triangle that relates the three velocity vectors in the problem; the desired direction, the current, and the actual direction the person heads in. The vector triangle always has the same form shown in the diagram below.

The relationship between the three vectors is given by the vector equation,

Our vector equation is really two equations, one for the x direction and one for the y direction:

Now each vector is defined by two quantities, its magnitude and direction. Since we have three vectors, we have three magnitudes and three directions total. So to solve the problem, we must have four knowns and only two unknowns.

You must decide which quantities you are given and which are unknown. If the problem gives you the speed that a boat can travel at on the water, or a plane in the air, that is vessel speed V_vessel. If the problem gives you speed that a stationary observer sees that is the desired or ground speed, V_desired. Try not to mix the two up. Also you may have to calculate the ground speed from the distance to the destination and the time involved.

Example

A boater is at the south bank of a river. He wants to get to a point on the other side of the river which is 150 m east of his position. The river is 250 m wide. The river flows at 1.50 m/s due west. His boat can manage 3.70 m/s on the water. Which way must the boater head? How long will take the boater to get to the other side?

First we draw a quick sketch of the problem.

We quickly see that we know the direction that the person wants to head: θ = arctan(150/250) = 30.97° east of north. We are also given the vessel speed. Our vector diagram is therefore

The equations we know have are:

Equations such as these aren't easy to solve because you have sine and cosine involved. Typically in such cases you need to use the trig identity cos²φ + sin²φ = 1. First we isolate the terms involving φ to get

Substituting these into the trig identity yields

Squaring and rearranging into a quadratic equation yields

Solving the quadratic equation yields V_desired = 2.555 m/s or V_desired = -4.269 m/s. Since V_desired is a magnitude, it cannot be negative. So V_desired = 2.555 m/s is our answer. We find the angle φ from sinφ = 0.2318V_desired = 0.2318 × 2.555 which give φ = 36.32°.

So the boater must head a 36.3° north of east.

The boater has to travel a distance D = [2502 + 1502]½ = 291.5 m. Since D = V_desiredt, we find t = 291.5/2.555 = 114.1 seconds.

Questions? mike.coombes@kwantlen.ca