The curvature of Earth’s gravity field is approximately 8 centimeters per kilometer. This produces horizon effects even within a single large building.

—J. Andrew Rogers, “Why Are Geospatial Databases So Hard To Build?”, March 2, 2015

http://www.jandrewrogers.com/2015/03/02/geospatial-databases-are-hard/?h

That sounds reasonable. Can we verify it?

Of course! A little geometry gives the deflection as \(\sqrt{x^2 + R^2} - R\) where \(R\) is the radius of the earth and \(x\) is the distance traveled parallel to the surface by an object that does not follow the curve of the earth.

You have: earthradius
You want:
        Definition: 6371.01 km = 6371010 m
You have: sqrt((1km)^2 + earthradius^2) - earthradius
You want: cm
        * 7.8480491
        / 0.1274202

Yup, about 8 cm.

Of course, we tend to care more about deflection as a function of \(s\), the distance travelled along the surface by an object that follows the curve of the earth. This is \(R( \sec(s/R)-1 )\).

Not surprisingly, it gives nearly the same answer:

You have: earthradius ( 1/cos(1km/earthradius) - 1)
You want: cm
        * 7.8480493
        / 0.1274202

Only when \(s\) or \(x\) are roughly the order of magnitude of the radius of earth does it begin to matter.

You have: sqrt((earthradius)^2 + earthradius^2) - earthradius
You want: km
        * 2638.9587
        / 0.00037893734
You have: earthradius ( 1/cos(earthradius/earthradius) - 1)
You want: km
        * 5420.5554
        / 0.00018448294