# Answer to Riddle #67: How Far is the Horizon

67. How far is the Horizon?

This is not a typical 'Puzzle' but it is some common sense geometry. I saw some puzzles of the form '*John is stood on the beach...*' I thought these all come down, basically, to working out the distance to the horizon. So let's just do that.

Before reading the answer can I interest you in a clue?

There are a number of ways to make this calculation, the important thing is that you can justify any assumptions you make.

AOB is a right angel triangle where from O the opposite side '

The algebra to now should be fairly easy to follow. Note the ± is simply because in our 2D model the horizon is in both directions, not strictly necessary, but good for completeness.

On the right we have

Given

Since the limiting factor in how far we can see is the hump of the earth in the way as it just disappears ABC will be a straight line. We can simply add the tow horizon distances together. As in

As you would expect both AI's did well at this.

If you're curious what

If you're curious what

There are a number of ways to make this calculation, the important thing is that you can justify any assumptions you make.

AOB is a right angel triangle where from O the opposite side '

*d*' is the distance to the horizon, the adjacent side '*R*' is the radius of the earth and the hypotenuse is '_{E}*R*' + '_{E}*h*' where '*h*' is the height of the observer. We will apply Pythagoras Theorem*X*^{2}+ Y^{2}= Z^{2}*R*_{E}^{2}+ d^{2}= (R_{E}+ h)^{2}*R*_{E}^{2}+ d^{2}= R_{E}^{2}+ 2R_{E}h + h^{2}*d*^{2}= 2R_{E}h + h^{2}*d*^{2}= h(2R_{E}+ h)*d = ±√h(2R*_{E}+ h)The algebra to now should be fairly easy to follow. Note the ± is simply because in our 2D model the horizon is in both directions, not strictly necessary, but good for completeness.

On the right we have

*±√h(2R*and we will be approximating this to_{E}+ h)*√2hR*. Some people would argue that the approximation is less accurate, but the earth is not a sphere, it's radius varies by ±15km and so using 2R_{E}_{E}+ h seems like an accuracy we can't claim.*d = √2hR*_{E}Given

*R*= 6,371,000m for me at least whose eyes are 6ft2 from the ground or 1.8796m the horizon is 4,894m away._{E}## Two Objects Above Sea-level

So, to now, we have considered an observer for example stood on a beach looking at the horizon and calculating that distance. What if you have someone on the shore looking for the top of a ship's mast?Since the limiting factor in how far we can see is the hump of the earth in the way as it just disappears ABC will be a straight line. We can simply add the tow horizon distances together. As in

*d = d*_{1}+ d_{2}= √2h_{1}R_{E}+ √2h_{2}R_{E}As you would expect both AI's did well at this.

If you're curious what

**Bard**made of this puzzle...If you're curious what

**ChatGPT**made of this puzzle...© Nigel Coldwell 2004 - – The

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