Visual Line of Sight Calculations dependent on Earth's Curvature

by David Senesac

How does one calculate a visual line of sight for objects at a given height that are distant enough that the Earth's curvature needs to be considered? Why might a photographer find such information useful? Some examples of what one might want to know: Whether or not a distant mountain can be seen from the top of another mountain. By how many degrees the rising full moon will be blocked by a distant mountain range in order to estimate whether it might be bathed in warm sunset light. Whether an alpine lake will catch sunset alpenglow or be blocked by a nearby ridge to the west.

The Earth has a radius of approximately 3965 miles. Using the Pythagorean theorem, that calculates to an average curvature of 7.98 inches per mile or approximately 8 inches per mile (squared). The distance to the horizon in miles from height of an observer is approximately equal to 1.23 times the square root of the height in feet. For example 1.23 times the square root of 8 divided by 12 equals 1 mile. Inversely given the horizon distance in miles, the height in feet required to be visible equals the distance in miles squared divided by 1.513. The second example above concerning the Moon rising over a distant range also requires some topographic map calculations using the tan trigonometric function. Thus if a peak rises up 1844 feet at a distance of 10.0 miles or 52,800 feet, it will form an angle of 2 degrees with a theoretical flat horizon. The tan is 1844/52800=0.0349 or 2 degrees. However due to the Earth's curvature, it would appear as though it was only 1778 feet tall with the lowest 66 feet below the horizon.

Disregarding refraction, on a perfectly flat plain like the Bonneville Salt Flats in Utah, if one's eyes are 9 inches above the ground, one would be able to see at night a flashlight one mile distant laying on the surface but not if one lowers their eyes to 7 inches. For example per the above diagram, that might be between the tangent point and Object B.

If a 64 inch tall person's eyes are at a height of 60 inches or 5 feet, they might be able to see at night, a flashlight laying on the ground at 1.23 the root of 5 = 2.75 miles. They would also be able to see another like standing person's headlamp twice that distance or at 5.5 miles distance, since each person would be able to see the midway tangent point. For example per the above diagram, that might be between the Object A and Object B. Likewise one of those person's would be able to see the last inch of a 9 inch tall object 2.75+1.0=3.75 miles distant. Lake Tahoe is 12 miles wide and 22 miles long. For a person's eyes at a height of 5 feet above the water on the south shore, a streetlight at night on the north shore needs to be up on a nearby hill at least 370 feet above the shore, else it is below the horizon.

Mount Shasta rises 14,262 feet about 50 miles north of the end of the Sacramento Valley. According to the below chart, one ought to be able to see the mountain from about 147 miles away. Elevations south in the valley at that distance are a relatively insignificant 200 feet in elevation. By checking a map, one will see that Shasta ought to be visible on clearest days once one has gone just a few miles north of the town of Colusa along the Sacramento River. Near Colusa are the Sutter Buttes that rise to as high as 2117 feet. Thus the Buttes might be seen from the top of Mount Shasta. A bit less than 100 miles south rising above San Francisco Bay to the east, Mount Diablo rises to 3849 feet. From the chart, one can see that volcanic peak is about 2800 feet too short to see Shasta.

In the real world there are however atmospheric effects of mainly ray refraction that tend to cause objects beyond the theoretical horizon to sometimes be visible. Thus the visible setting sun is usually a little below the theoretical horizon. In like manner, the effect is to increase the apparent height of distant peaks. Some approximate heights and distance in the below table:. ...David Senesac

Distance Height Miles Feet -------------------------- 1.0 0.67 1.23 1.0 3.0 5.95 3.9 10 6.0 23.8 10 66 12.3 100 20 264 30 595 39 1000 40 1060 50 1650 60 2380 70 3240 87 5000 100 6610 110 8000 120 9520 123 10000 129 11000 135 12000 140 13000 146 14000 148 14496

David Senesac Photography