Apparent Wind is the wind, both direction and strength, that is experienced by an observer who may or not be moving. In the most common example, the observer is a sailboat, ice boat, or airplane that is moving with respect to any "stationary" spot on earth. The true wind is the wind that would be felt by an observer standing still, or a craft not moving with respect to the earth. By 'not moving', we mean that the observers GPS coordinates are not changing.

Suppose that the true wind measures zero, and that the observer is a car which is driving along a road at 30 mph. Then, the observer would experience a 30 mph breeze coming from exactly the direction of motion. That breeze is called "apparent wind" since there is no true wind in our thought experiment.

Suppose we are sailing downwind in a boat and that the wind and current are causing the boat to go in exactly the same direction and at exactly the same speed as the true wind. In this case, the observer on the boat would experience no wind at all, and the "apparent wind" would therefore be zero.

At the bottom of this page is an applet that allows you to set the boat direction and speed as well as the "true wind" direction and speed and it calculates and shows us the "apparent wind" direction and speed. This is a very simple vector problem that is enormously complicated by two facts!

Mathematics assumes counterclockwise rotation for trigonometry computations, while compasses use clockwise rotational increase.

Mathematics assumes the zero angle to be the right side horizontal axis, while compasses assume the zero angle to be vertical.

Except for these two issues, all of the maths are the same. In fact, the problem is much easier now that we routinely can measure the observers speed over land using GPS. So let's do a maths example.

Exercise: Suppose the "true wind" is coming from $225^{\circ}$ at $6$ knots and that a sailboat is heading $340^{\circ}$ at 3.5 knots. What is the apparent wind direction and speed on the boat. Note this is the wind that will be measured by an anemometer or wind vane fixed to the boat.

Answer: The maths will work here, its just that some things are not obvious. First, true wind direction and speed have to be composed into a vector. We will call it $\mathbf{w}$ and $$\mathbf{w}=6\cdot\left(\begin{array}{c} \cos(225^{\circ}\\ \sin(225^{\circ}\end{array}\right)$$ That is, $$\mathbf{w}\approx \left(\begin{array}{c} -4.243 \\ -4.243\end{array}\right)$$
The two values are identical because $225^{\circ}$ is exactly halfway between south and west. For most wind headings, they will be different.

Now we have to do the same thing with the boat heading. Call it $\mathbf{bv}$
$$\mathbf{bv}=6\cdot\left(\begin{array}{c} \cos(340^{\circ}\\ \sin(340^{\circ}\end{array}\right)$$ That is, $$\mathbf{bv}\approx \left(\begin{array}{c} 3.289 \\ -1.197\end{array}\right)$$

Now the trivial part is adding these two vectors together to obtain the apparent wind vector.
$$\left(\begin{array}{c} -4.243 \\ -4.243\end{array}\right) + \left(\begin{array}{c} 3.289 \\ -1.197\end{array}\right) = \left(\begin{array}{c} -0.954 \\ -5.440\end{array}\right)$$
Seems rather useless doesn't it? What we need to do is decompose the vector back into an angle and a magnitude (speed). There is a maths function for that called "atan2". On most calculators it is input as atan2(y,x), but be careful because on some calculators, it is atan2(x,y)!
$$atan2(v_y,v_x)=-99.944^{\circ}$$
Well, that nice, but a) we want degrees between 0 and 360 and b) what about speed?
OK. To get proper degrees we just need to add $-99.944^{\circ} + 360^{\circ}=260.056^{\circ}$, and as for speed, it is the vector magnitude.
$$\left| \begin{array}{c} -0.954 \\ -5.440\end{array} \right| = \sqrt{v_x^2+v_y^2}=5.523 \text{ knots.}$$
A quick plot will also show that the maths work.