On a clear day: solution

R is the radius of the Earth, H is your height and D is how far you can see. The walking distance across the Earth's surface is the length of the arc s.
Solution to problem 1: In Issue 54's Outer Space you were asked to work out the "walking" distance over the Earth's surface to the horizon. This is equal to the length
of the arc of the circle from the point where you are to the point where your line of sight hits the surface of the Earth, which is
where

is the angle as shown in the diagram and

is the radius of the Earth. Now bearing in mind that we have a right-angled triangle we have
so
If

is much larger than

then the straight line distance to the horizon is approximately equal to

, as we worked out in the article. Moreover, in this case the straight line distance is approximately equal to the walking distance over the Earth's surface. This gives that for

Solution to problem 2: You see something, like a cloud or a hot air balloon, high in the sky beyond the horizon and you know its maximum possible height above the ground. Can you work out how far away it could be?
Let's assume that the object is at the maximum possible height
above the surface of the Earth. So in our two-dimensional view it lies somewhere on the circle of radius
about the centre
of the Earth. Since it is beyond the horizon, the object lies beyond (from your point of view) the straight line through
and the horizon point
. And since it is visible over the horizon, the object lies "above" the tangent line passing through your view point and the horizon point
. Thus the object lies somewhere on the arc of the circle between points
and
(see the diagram on the right).
Using Pythagoras' theorem, we see that the distance from you to point
is
We have already worked out that for a person of height

, we have

and we know that

So for a maximum height of

we get that the object is at least

away from you.
Now let
be the distance from
to
. Using Pythagoras' theorem on the right-angled triangle
, we get
Noting that

, so that we can ignore the

term, we have
Now the maximum distance to the object from the observer is approximately equal to
Using

,

and

we get that the maximum distance to the object is approximately

Back to Outer space