Trigonometric riddle. Obtaining Wind Speed based on two vectors of Ground Speed of an Airplane

Question

The question:

An airplane flies, and we know its ground speed and direction. There is a wind of unknown speed and direction. The airplane changes its heading, so that we have a new ground speed and direction. The airplane speed with respect to air does not change. Can we calculate the wind speed and direction based on this information?

What we have is:

Vector $C_1$: Ground Speed 1 and Direction 1, all known

Vector $C_2$: Ground Speed 2 and Direction 2, all known

Vector $A_1$: Air Speed (same) Direction $B_1$ (unknown)

Vector $A_2$: Air Speed (same) Direction $B_2$ (unknown)

Vector $B_1$: Wind Speed and Direction (same)

Vector $B_2$: Wind Speed and Direction (same)

This question has practical use for determining wind speed from UAVs. One easy way to measure wind speed is to make a $360$ degree circle and then obtain the wind speed which is half the difference between minimum and maximum ground speed. I wounder if it is possible to obtain the same result without doing a full circle, but just change course by a smaller angle ($60 - 90$ degrees) and use the changes in ground speed to determine the wind speed

I am not sure how to tackle it, but I feel that since we have the same airspeed, we probably could solve it...

Answer 1

Buy an E6B flight computer. It is a device created explicitly for this purpose.

If you want to learn the math.

Method 1, law of cosines / sin.

$C = A^2 + B^2 - 2AB\cos c$

In this case, $c$ the deviation between heading and course.

$A$ is you airspeed. $B$ is your ground-speed. $C$ will be the wind-speed.

law of sines

$\frac {\sin a}{A} = \frac {\sin b}{B} = \frac {\sin c}{C}$

$A,B, C, c$ are the same as above.

$a$ will be the difference between heading and the wind direction. $b$ will be the difference between ground-track and the wind direction.

Method 2 convert to Cartesian.

$\text{Heading} (v, \theta) = v\cos\theta, v\sin\theta\\ \text {Course} (s, \phi) = s\cos\phi, s\sin\phi\\ \text{wind} = \text {Course - Heading} = s\cos\phi - v\cos\theta , s\sin\phi - s\sin\theta$

Wind speed $= \sqrt {(s\cos\phi - v\cos\theta)^2 + (s\sin\phi - s\sin\theta)^2}$

Wind direction $= \arctan \frac {s\sin\phi - s\sin\theta}{s\cos\phi - v\cos\theta}$

Wind direction is measured on a $0$ to $360$ scale, and $\arctan$ will return a number in $-90$ to $90$ so you will need to adjust, and you might find your indicated direction by the calculation to be $180$ degrees off. So, you will need to run a sanity check to see if you have a headwind or a tailwind.

Answer 2

If you simply enable the measurement of heading (the direction of the airspeed vector) as well as ground course (the direction of the ground speed vector) then one measurement is enough to deduce the wind speed and direction.

If you insist that we know absolutely nothing about the heading of the aircraft when each ground speed measurement is taken, then we can at least draw the vector diagram a little differently (representing the exact same vector sum):

Knowing the magnitude and direction of each of the vectors $c_1$ and $c_2,$ as well as knowing the magnitude of $a_1$ (and that this is also the magnitude of $a_2$), we can determine the exact positions of the tips of the arrows for $c_1$ and $c_2,$ and then plot the common starting point of the vectors $a_1$ and $a_2.$ (That point will be on the perpendicular bisector of the segment joining the tips of $c_1$ and $c_2$ at a suitable distance from both those points; an easy way to construct the vectors is to take half the vector difference $c_2 - c_1$ as one component, and use the Pythagorean theorem and the known length of the vectors $a_1$ and $a_2$ to compute the length of the component perpendicular to the first component.)

Once you have determined at least one of the vectors $a_1$ or $a_2$ in this way, you can use a vector difference to find the wind vector $b.$

A weakness of this approach is that nothing in the solution so far tells us that the correct vector diagram is the one shown above rather than the one shown below.

Here we have the exact same direction and magnitude as before for each of the vectors $c_1$ and $c_2,$ and the same magnitude as before for each of the vectors $a_1$ and $a_2,$ but we get a very different wind vector. And if you really know nothing about the heading of the aircraft, you have no way (other than guessing) which of these is the correct figure for the actual wind.

If you can distinguish a $90$-degree left turn from a $90$-degree right turn (and from a $270$-degree left turn), however, you can tell which of the two diagrams is correct. (It's the first diagram for a left turn, the second for a right turn.)