# Local Cartesian System and Earth`s curvature

Hello,

Yes you are right. I should have written "almost equal".

Okay.

But in my example the difference is huge - 1250mm for a 4000m line.

Yes, that is right. If you estimate the influence of the curvature of the Earth, the result is

$\frac{4000^2}{2R} = 1.256~\mathrm{m}$

This is the different in the up component you already mention. The equation (approximately) corrects the zenith angle to the Earth surface.

However, if the ellipsoidal approach is applied, the coordinates are related to a local planar system defined by the principal point, cf. the figures in the documentation. That is the different. In your case, the principal point is identical to the station position.

Kind regards
Micha

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