Question

An airplane is flying on a compass heading(bearing) of 340 degrees art 325 mph. A wind is blowing with the bearing 320 degrees at 40 mph, how do you find the component form of the velocity of the airplane?

Answer 1

Below are the formulas for X- and Y-components of the resulting movement as well as the amplitude and angle of this movement (all angles are measured according to trigonometric standard).

Explanation:

In navigation the angle of the course (on a compass) is counted clockwise from the North (so, the direction to the North is `0^o`, to the East is `90^0`, to the South is `180^o` and to the West is `270^o`).
The North on most maps is a vertically up direction.
In coordinate Geometry and Trigonometry, which we will use, angles are measured counterclockwise from the positive direction of the horizontal X-axis (the East on most maps).

Let's make a simple transformation into Trigonometric standard using the direction to the East as an X-axis.:
`340^o` on a compass is `90^o +(360^o -340^o)=110^o` counterclockwise from the X-axis.
`320^o` on a compass is `90^o +(360^o -320^o)=130^o` counterclockwise from the X-axis.

This is a problem on addition of two vectors. Each is defined by its amplitude and angle of direction:
airplane (vector `A`) has amplitude `325` (`mph`) and angle `110^o`;
wind (vector `W`) has amplitude `40` (`mph`) and angle `130^o`.

To add these two vectors, we represent both as sums of X-component and Y-component:
`A_X = 345*cos(110^o)`
`A_Y = 345*sin(110^o)`
`W_X = 40*cos(130^o)`
`W_Y = 40*sin(130^0)`

Both X-components act along the same direction, both Y-components act along the same direction. So, we can add X-components to get an X-component of the resulting movement and add Y-components to get a Y-component of the resulting movement.

`(A+W)_X = A_X + W_X = 345*cos(110^o)+40*cos(130^o)`
`(A+W)_Y = A_Y + W_Y = 345*sin(110^o)+40*sin(130^o)`

Knowing two components of the resulting vector of movement, we can easily determine the amplitude `|A+W|` and direction `/_(A+W)` of the resulting vector:

`|A+W| = sqrt((A+W)_X^2+(A+W)_Y^2)`

`/_(A+W) = arctan[(A+W)_Y/(A+W)_X]`

We leave the calculations to the person who suggested the problem.

If it's necessary to express the angle `/_(A+W)` in compass format (`/_alpha`) from its standard trigonometric format (`/_beta`), use the conversion formula
`/_alpha = 360^o-(/_beta-90^o) = 450^o-/_beta`
(if the result is greater than `360^o`, subtract `360^o`)