Re: How do you measure cloud speed?

Without more detail I am going to have to make some reasonable guesses as to exactly how you were taking the measurements. But even if I do not guess the exact method or configuration I think there will be enough information in this answer to allow you to reconstruct your method.

You know the width of the mirror, and I assume you know the distance from your eye to the mirror. If you knew the height of the clouds then you could compute the distance in the sky that the clouds move during the time interval, and from that the speed is easily computed.

One of the ways you can determine the height of the clouds is to go to this site (which is a Unisys Weather site) and try to find a city that is close to your location. Another way to find the cloud height would be to call your local National Weather Service office and ask for the cloud ceiling height.

Okay, let's assume you now know the width of the mirror, which we will call "w", the distance from your eye to the mirror, which we will call "e", the time interval, which we will call "t", and the height of the clouds, which we will call "h". To a good approximation, the angle, A, that the image of the clouds makes (with your eye at the center of the angle) from one edge of the mirror to the other is A =arctan(w/e). Using that angle, the distance, D, that the clouds actually move across the sky is D = h tan(A). An alternative with small error, using similar triangles, is D = hw/e.

The speed is then simply D/t.

There are some simplifying steps taken in these formulae. For instance, it is impossible to get your eye directly over the mirror and see the clouds above you because your head gets in the way! So the perpendicular height of your eye above the mirror is only approximately e. If the angle is small then the perpendicular height is roughly e, but not exactly. In the same way, the perpendicular distance of the clouds is only roughly h. Also, the distance of the clouds' image across the mirror is not a perpendicular distance, and the actual distance across the sky is also not a perpendicular distance, so the calculated distance, D, is not exact, but it is close enough.

John Link, MadSci Physicist