A ship travels westward at 6.5 m/s and wind is blowing from the south-west at 3.5 m/s. What wind speed and wind direction will an anemometer on the ship record relative to the ship’s course?

What the ship actually measures

A wind device on a moving ship does not measure the true (ground) wind. It measures the air’s velocity relative to the ship, called the apparent wind: $$\vec v_{\text{app}} = \vec v_{\text{air}} - \vec v_{\text{ship}}.$$

Set axes and write each velocity as components

Take $+x$ east and $+y$ north.

Ship (west at 6.5 m/s): $$\vec v_{\text{ship}} = (-6.5,\,0)\,\text{m/s}.$$

Wind “from the south-west” at 3.5 m/s: “From SW” means it blows toward NE, i. e. $45^\circ$ north of east. $$\vec v_{\text{wind}} = \left(\frac{3.5}{\sqrt2},\,\frac{3.5}{\sqrt2}\right) \approx (2.475,\,2.475)\,\text{m/s}.$$

Compute the apparent wind vector

$$\vec v_{\text{app}} = (2.475,\,2.475) - (-6.5,\,0) = (8.975,\,2.475)\,\text{m/s}.$$ So, relative to the ship, the air flows mostly eastward and slightly northward.

Apparent wind speed

$$|\vec v_{\text{app}}| = \sqrt{(8.975)^2 + (2.475)^2} \approx \sqrt{86.676} \approx 9.31\,\text{m/s}.$$

Apparent wind direction relative to the ship’s course

Direction of the airflow relative to the ship is $$\theta = \tan^{-1}\left(\frac{2.475}{8.975}\right) \approx 15.4^\circ,$$ which is $15.4^\circ$ north of east (direction the air is moving).

But wind direction is usually given as the direction it is coming from, which is opposite to the airflow. So it is coming from

• $15.4^\circ$ south of west, relative to the ship’s westward heading,
• i. e. about $15^\circ$ off the port bow (port side is south when heading west).

Quick reasonableness check

Since the ship is moving into the wind’s westward component (ship is going west, apparent wind tends to come from the front), the measured wind speed should be larger than $3.5\,\text{m/s}$, and $9.31\,\text{m/s}$ makes sense.