Looking up at say a cumulus cloud at a height of 2km which covers say 10 degrees of the visible sky, what is the easiest way to estimate its rough size at its base?

Simply put, how large is the arc of visible sky viewed from the surface?

  • $\begingroup$ The 10 degrees of visible sky- is that supposed to be in steradians? $\endgroup$ Nov 11, 2015 at 16:24
  • $\begingroup$ That would give the size in square meters I guess, so yes that would also be good. Basically I just want to satisfy my curiosity when I look at a cloud to know how big it is as I have no idea - 100kms, 10km, 10meters who knows? From horizon to horizon, how far is that? How long will it take that air-plane to cross that cloud, etc. $\endgroup$ Nov 12, 2015 at 18:05

1 Answer 1


Three methods spring to mind:

1) If you can really estimate both the width in degrees and distance to the cloud, you can draw a triangle and use trigonometry to calculate the width.

2) The distance to the cloud you may estiamte from the cloud type. But both height and width in degrees are uncertain numbers. Instead you can estimate cloud sizes by looking at planes passing through them. If you time how long it takes a plane to pass through a cloud, you can multiply that with its cruising speed to get the width of the cloud. The cruising speed (typically 500km/h) can be found via say, the Flightradar24 app.

3) Using that app, you can also track the plane in real-time to map out not only the length of the cloud, but what land it is covering. No need to time the plane or know its speed.

By following a plane on the app until it goes under the horizon at your location, you get a feel for "how far away the horizon is", for clouds at that plane's altitude. high-altitude clouds and aurora at the horizon will be much farther away than low-altitude clouds.

  • $\begingroup$ Thanks, method 1 seems the quickest and easiest. From what I can find on the internet, an observer of height 2meters, the horizon is 5km away and base of cumulus clouds another 1km further (average height). Therefore the total distance from horizon to horizon (point to point, 180 degrees), at that height is 19km (6*pi). So for every 10 degrees is about 1km. At cirrus level about 8km high, the arc from horizon to horizon is 41km (13*pi), so each 10 degrees is 2.3km. Is that correct? $\endgroup$ Nov 15, 2015 at 5:51

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