# Converting relative velocities

If $\displaystyle \vec{v}_{\text{XY}}$ represents the velocity of X with respect to Y, then

$\displaystyle \vec{v}_{\text{AB}} + \vec{v}_{\text{BC}} = \vec{v}_{\text{AC}}$.

In other words, we can eliminate the B and go straight from A to C, provided we know the velocity of B relative to C. Sometimes you will have to rearrange this equation to solve for a different velocity—that’s about as complicated as these problems get.

## Example

An airplane has a heading of 245 km/h [N 17º E], and the wind is coming from [S 35º W] at 89.0 km/h. What is the course?

First, we need to familiarize ourselves with airplane terminology:

- heading ($\displaystyle \vec{v}_{\text{PA}}$)
- velocity of the plane with respect to the air
- wind ($\displaystyle \vec{v}_{\text{AE}}$)
- velocity of the air with respect to the Earth
- course/groundspeed ($\displaystyle \vec{v}_{\text{PE}}$)
- velocity of the plane with respect to the Earth

The velocity of the wind is a vector pointing in the direction in which it is going. Since the wind is coming *from* [S 35º W], it is going *to* [N 35º E].

Since $\displaystyle \vec{v}_{\text{PE}} = \vec{v}_{\text{PA}} + \vec{v}_{\text{AE}}$, the course is

$\displaystyle \vec{v}_{\text{PE}} = 245 \, \text{km/h [N 17º E]} + 89.0 \, \text{km/h [N 35º E]} \\ = 331 \, \text{km/h [N 22º E]}$.