Relative velocity

Suppose that, on a windy day, an airplane moves with constant velocity ${\bf v}_a$ with respect to the air, and that the air moves with constant velocity ${\bf u}$ with respect to the ground. What is the vector velocity ${\bf v}_g$ of the plane with respect to the ground? In principle, the answer to this question is very simple:

$${\bf v}_g = {\bf v}_a + {\bf u}.$$ (81)

In other words, the velocity of the plane with respect to the ground is the vector sum of the plane's velocity relative to the air and the air's velocity relative to the ground. See Fig. 18. Note that, in general, ${\bf v}_g$ is parallel to neither ${\bf v}_a$ nor ${\bf u}$. Let us now consider how we might implement Eq. (81) in practice.

Figure 18: Relative velocity

As always, our first task is to set up a suitable Cartesian coordinate system. A convenient system for dealing with 2-dimensional motion parallel to the Earth's surface is illustrated in Fig. 19. The $x$-axis points northward, whereas the $y$-axis points eastward. In this coordinate system, it is conventional to specify a vector ${\bf r}$ in term of its magnitude, $r$, and its compass bearing, $\phi$. As illustrated in Fig. 20, a compass bearing is the angle subtended between the direction of a vector and the direction to the North pole: i.e., the $x$-direction. By convention, compass bearings run from $0^\circ$ to $360^\circ$. Furthermore, the compass bearings of North, East, South, and West are $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$, respectively.

Figure 19: Coordinates for relative velocity problem

According to Fig. 20, the components of a general vector ${\bf r}$, whose magnitude is $r$ and whose compass bearing is $\phi$, are simply

$${\bf r} = (x, y) = (r \cos\phi, r \sin\phi).$$ (82)

Note that we have suppressed the $z$-component of ${\bf r}$ (which is zero), for ease of notation. Although, strictly speaking, Fig. 20 only justifies the above expression for $\phi$ in the range $0^\circ$ to $90^\circ$, it turns out that this expression is generally valid: i.e., it is valid for $\phi$ in the full range $0^\circ$ to $360^\circ$.

Figure 20: A compass bearing

As an illustration, suppose that the plane's velocity relative to the air is $300 {\rm km/h}$, at a compass bearing of $120^\circ$, and the air's velocity relative to the ground is $85 {\rm km/h}$, at a compass bearing of $225^\circ$. It follows that the components of ${\bf v}_a$ and ${\bf u}$ (measured in units of km/h) are

$${\bf v}_a = (300 \cos 120^\circ, 300 \sin 120^\circ) = (-1.500\times 10^2, 2.598\times 10^2),$$ (83)

$${\bf u} = (85 \cos 225^\circ, 85 \sin 225^\circ) =(-6.010\times 10^1, -6.010\times 10^1).$$ (84)

According to Eq. (81), the components of the plane's velocity ${\bf v}_g$ relative to the ground are simply the algebraic sums of the corresponding components of ${\bf v}_a$ and ${\bf u}$. Hence,

$${\bf v}_g = (-1.500\times 10^2 - 6.010\times 10^1, 2.598\times 10^2 - 6.010\times 10^1)$$

$$= (-2.101\times 10^2, 1.997\times 10^2).$$ (85)

Our final task is to reconstruct the magnitude and compass bearing of vector ${\bf v}_g$, given its components $(v_{g x}, v_{g y})$. The magnitude of ${\bf v}_g$ follows from Pythagoras' theorem [see Eq. (35)]:

$$v_g = \sqrt{(v_{g x})^2 + (v_{g y})^2}$$

$$= \sqrt{(-2.101\times 10^2)^2 + (1.997\times 10^2)^2} = 289.9 {\rm km/h}.$$ (86)

In principle, the compass bearing of ${\bf v}_g$ is given by the following formula:

$$\phi = \tan^{-1} \left(\frac{v_{g y}}{v_{g x}}\right).$$ (87)

This follows because $v_{g x}=v_g \cos\phi$ and $v_{g y}=v_g \sin\phi$ [see Eq. (82)]. Unfortunately, the above expression becomes a little difficult to interpret if $v_{g x}$ is negative. An unambiguous pair of expressions for $\phi$ is given below:

$$\phi = \tan^{-1} \left(\frac{v_{g y}}{v_{g x}}\right),$$ (88)

if $v_{g x}\geq 0$; or

$$\phi = 180^\circ - \tan^{-1} \left(\frac{v_{g y}}{\vert v_{g x}\vert}\right),$$ (89)

if $v_{g x}< 0$. These expressions can be derived from simple trigonometry. For the case in hand, Eq. (89) is the relevant expression, hence

$$\phi = 180^\circ - \tan^{-1} \left(\frac{1.997\times 10^2}{2.101\times 10^2}\right) = 136.5^\circ.$$ (90)

Thus, the plane's velocity relative to the ground is $289.9 {\rm km/h}$ at a compass bearing of $136.5^\circ$.