I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres.

No you haven’t. The length of your line will tell you the distance you have travelled (vertical plus horizontal). Your line would look like the one on the right. If you straighten it out, it will be

*longer* than the one on the left and

*more* than ¼ of the circumference. You’ve dropped ¼ of the circumference, but you’ve travelled more distance (vertical plus horizontal)

Never mind on average its a fact all the way round; the curve is the same all the way round - it doesn't drop away any more or less than anywhere else on the circle.

Wrong. Curvature can be measured by how much it deviates from straight. Here’s a graph of a circle.

Notice how the circumference hits the grid lines at irregular intervals. That means it deviates from straight a different amount each time it hits the grid and the rate of drop varies. That’s why there is 8in of drop at 1 mile, but 32 in of drop at two, like I already explained. Compare that to the graph of a straight line where each coordinate hits the graph line in equal intervals. That’s a constant rate of drop

And I can't understand why no one agrees with this.

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Because this is pretty basic well understood geometry and math. Even Rowbotham accepted the 8in. m^2 equation, which works pretty good up to a point. But it’s the equation for a parabola not a sphere so eventually the errors start compounding.