On Jan 23, 4:46 pm, "Graham Easterling"
wrote:
Atmospheric refraction bends the rays of light downwards and
effectively increases the radius of the earth by about a fifth. Since
the horizon distance is inversely proportional to the square root of
the earth's radius all the calculated distances should be increased by
about 10%. Don't forget that if you see the sun sitting right on the
horizon it is actually just entirely below it on a geometrical basis.
If the atmosphere were about five times denser you could see
right round the earth. One consequence would be that the sun would
never set though it may well look a bit dim and sausage-shaped at
"night".

Tudor Hughes, Warlingham, Surrey.So this means that, all other things being equal, the horizon distance
is greater with high pressure.

So if the horizon distance was say 30 miles at 940mb, what would it be
at 1040mb?

Graham
Penzance

The figure I gave of one-fifth for the curvature is a little
out; it should be one-sixth, nearly, at 1013 mb, 15°C and lapse rate
6.5 deg/km. So a rise of 100 mb (10% of atmospheric pressure) would
put the horizon further away by about one sixtieth, half a mile in your
case.
The lapse rate makes some difference. The larger it is the
smaller the curvature of a ray of light and if it exceeds 34.2 deg/km
(easily possible close to a heated surface) the ray actually bends
upwards, the density *increasing* with height. On the other hand, a
strong inversion bends the light more than normal, giving rise to
mirages of a different kind. This all gets a bit complicated to
explain without diagrams but I hope that's OK.

Tudor Hughes, Warlingham, Surrey.