Of course, that is just plain nuts. Nevertheless, it raised the question, “Just how far away could a person see an object if nothing interfered? Let’s consider the answer to that question.

## Conditions

First, we need to set conditions or ground rules.Earth is sufficiently round to call it a sphere, so we treat it as such. In fact, we assume it is perfectly smooth even to an ant. Further, we assume the atmosphere is perfectly clear, and the observer has perfect eyes.

When we look out toward Earth’s horizon, we are actually looking along one of Earth’s tangents. If we look below the horizon, we see only the ground, not our object at maximum distance, d.

If we assume we have zero height, and we draw a line down through our body to the center of the earth, the second line is a radius, r, of Earth.

Finally, at the point of maximum distance we can see our object, we see only the very tip of the object. We draw a line through the object down to the center of Earth, a second Earth’s radius.

## The Math

Since we have described a triangle, and that triangle is a right triangle, we can employ the very simplest geometry and trigonometry. We write two equations…**r² + d² = (h + r)²**

**sin (θ) = d/(r + h)**

Rearranging the first equation, the maximum distance away on the earth an object can be (as the crow flies) and it can still be seen is,

**d = √ h(h +2r)**

For instance, if the object is of infinite height, the furthest distance it is still possible to see it is infinity! To a point… If the object is at an angle θ of 90º away from the observer, no matter what the height of the object, the observer can never see it.

## The Eiffel Tower

The object of our choice, the Eiffel Tower, is 1,063′ tall. This converts to almost exactly 1/5 mile. Additionally, the median radius of Earth is 3,959 miles.Substituting these values in our equation, we get,

**d = √ 0.2(0.2 +7918)**

**d = 39.8 miles**

**sin (θ) = 39.8 / (3959.2) = 0.01005**

**θ = 0.58 degrees**

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