Lake Tahoe is 12 miles wide and 22 miles long. 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. For example per the above diagram, that might be between the Object A and Object B. 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. 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. For example per the above diagram, that might be between the tangent point and Object B. 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.ĭisregarding 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. 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 second example above concerning the Moon rising over a distant range also requires some topographic map calculations using the tan trigonometric function. 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. For example 1.23 times the square root of 8 divided by 12 equals 1 mile. 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. Using the Pythagorean theorem, that calculates to an average curvature of 7.98 inches per mile or approximately 8 inches per mile (squared).
The Earth has a radius of approximately 3965 miles. Whether an alpine lake will catch sunset alpenglow or be blocked by a nearby ridge to the west.
Horizon line calculator full#
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. 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. Visual Line of Sight Calculations dependent on Earth's Curvature by David Senesac David Senesac Visual Line of Sight Calculations dependent on Earth's Curvature return to home page
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