My son did an extra credit problem in Geomety (with my idea)…

Being in Southern California, the tallest mountain is San Gorgonio (~11,500’). The problem I presented my son for extra credit is: Can you see Catalina Island from San Gorgonio? First we determined the statute miles between the peak and closest point on the island relative to the peak. Turned out it was about 106(?) miles using a distance calculator from the internet.

Next we applied the equation:

A square of tangent (line of sight) line segment is equal to a product of a secant line segment (earth’s diameter) by the secant line external part (elevation of San Gorgonio).

LOS[sup]2[/sup] = (ED)(ESG)

LOS[sup]2[/sup] = (7912 mi.)(2.18 mi.)

LOS[sup]2[/sup] = 17248.2 mi[sup]2[/sup]

LOS = 131.3 miles to the horizon (from peak to tangent point)

Since Catalina was 106 miles away, then it is visible from San Gorgonio (weather permitting).

If there are two peaks that are 10,500’ in elevation that are ~260 miles away, then I guess you could possibly see them by doubling the LOS distance where the midpoint must be at sea-level (and all land between is below the LOS) so there is no visual obstruction. Of course, I’m not taking atmospheric conditions in consideration, but this site does… (hey it was just a 10th grade extra credit project!)

For Mt. Everest, the LOS is 208.56 to sea-level horizon, so a theoretically you can see (w/o atmospheric conditions) a 24,600’ peak from a distance of 400 miles (if there actually is one, and the LOS is not obstructed). That’s about the limit, I’m guessing.

A more detailed approach…