Such as, if you can see a light house flashing, and it should be far below the horizon (you have to see through the earth to see it),
As far as I can obseerve, lighthouses that are meant to be seen from the distance are usually built at the top of towers, and even local lighthouses are typically raised above the sea.
In the sea map there may be information on how far the light house can be seen, like, one of our strongest light houses here in the Oslo fjord is Færder Fyr, the light is situated 47 metres above the sea level (on high water), and is supposed to be visible 19 nm away.
Now there is 60 nautic miles for every degree of curvature, meaning that at the distances where the light is supposed to either be too faint to be seen by sailors or below the horizon, the difference in curvature is less than half a degree.
How many metres of elevation is there between two points on a globe separated by half a degree? Long time since I've been doing geometry, I will try to find a pen and a paper and see if I can figure it out a bit later today.
After thinking for a while, I came to this approximation being "good enough". On big scales, it will give too big numbers:
h(l) = tan(l)*l/2where h is the elevation between two points due to the curvature, and l is the length between two points, measured as an angle in the tangens function.
So, let's calculate h(19 nm). In radians, 19 nm = 192pi/360/60 = 0.0055269. In metres, 19 nm = 19*1852m = 35188 m
Hence, h(19 nm) = 15.47m
Meaning that the light is supposed to be too faint to be seen long before it disappears due to the curvature. At least if my formula is right, which I believe it is.
Around 33 nm the light should become below the horizon.
A typical relatively big sailing boat will stretch 2 metres up from the water, meaning that at 7 nautic miles, only the mast should be seen. I will try to do some binocular observations next time I have the chance.