In a comment on the post, bernieT36 asked a good question: whether the vertical component of any walk would account for this. In other words, the hypotenuse of a triangle (the ground really walked) is longer than the some route when measured on a map.

I thought I'd write a quick post to say why I think the effect is negligible.

Firstly, although there are some very hilly days on the coast (an example being between Bude and Hartland on the South West Coast Path), many are also fairly flat with little ascent or descent to be done. Indeed, some walks in Suffolk, Essex and Kent had total ascent and descent of a few feet over many miles.

### The long, steady drag

Now for some maths (yay!).As an example, let us take a fictitious walk up a mountain from sea level, with a total ascent of 2 kilometres over 10 kilometres horizontal distance.

For simplicity, this post assumes that all ascent is cumulative; that is, an ascent followed by a short descent and another ascent, equates to the sum of the ascents and descents.

The vertical component is not done in one go, as if the horizontal distance is walked, and then a ladder climbed. If this was the case, it would make a total distance of 12km. Instead, the vertical component is shared with the horizontal component.

Using basic trigonometry, the distance walked (the hypotenuse of the triangle) can be calculated as the square root of the horizontal distance squared plus the vertical distance squared.

Therefore: walked distance = sqrt ((10^2)+(2^2)).

This comes out as 10.19 km. Therefore the vertical distance accounted for not even a fifth of a kilometre on such an extreme walk.

### Arcs

Ah, I hear people cry, but hills are not a straight line!So what happens if we walk a curve instead of a straight line? For simplicity I will choose an impossibly steep and long walk - a walk with 10 kilometres of ascent over 10 kilometres of horizontal distance. Using the formula above, the straight-line distance between the two points is sqrt ((10^2)+(10^2)), or 14.14 km. So even in this extreme case, the vertical component has added far less than half the horizontal component.

If we do this as an arc of a quarter of a circle, we get (2*π*10) * (90/360), which is 15.7 km.

Even in this artificially harsh example, the vertical component adds only just over half of the original distance.

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