# Parallax Cloud Displacement - Angular Distance from Geostationary Satellite

When observing clouds from a geostationary satellite, they appear displaced because of the parallax effect. The satellite looks at the clouds "at an angle" and it projects them on a point of the earth that is not exactly "under" the cloud.

This image shows the Normalized Cloud Offset as a function of the Angular Distance from Nadir.

$\text{Normalized Cloud Offset} \times \text{Cloud Height} = \text{Offset}$ (because of displacement)

Given a geostationary satellite at a given longitude at 35,800 km over the equator, how do I calculate the Angular Distance to a generic point on the earth (given its latitude/longitude)?

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If this doesn't get an answer here, I'm guessing math.stackexchange.com could rustle one up for you pretty easily –  Rory Alsop Oct 13 '11 at 13:51
Not to dodge the question, but wouldn't the parallax at ~36 x 10^3km be negligible for clouds, whose height is measured in single-or-at-most-two digit kms, even for clouds at extreme latitudes? –  Larry OBrien Oct 14 '11 at 0:08
Given a cloud that is at the same longitude of the satellite, the Angular Distance is equal to the cloud latitude. Most of Europe for instance falls roughly between 35N and 65N with offsets between 0.8 and 3. For a 10km high cloud (ignoring the cloud width/height) this translates in a displacement between 8km and 24km. The above considering satellite and cloud at the same longitude only. I'm afraid that with different longitudes (and there are only a small number of geostationary satellites out there) things can get pretty bad faster and at lower latitudes as well... –  Gevorg Oct 14 '11 at 14:31
@Gevorg You can write that up into an answer. I am pretty confident of what you say too. –  Cheeku Mar 2 '13 at 8:40