Octagon Shaped Pond

TreeBay

From the wiki


I'm assuming your eight foot measurement is measured across the parallel sides and not the corners. It is not clear if by formula you are looking for volume, or surface area of the water or surface area of the pond.

If the octagon is going to have equilateral sides (each side the same length - like a stopsign) and it's 8 feet across then the length of each side will be about 3.314 feet.

As a formula you can use the side length to be equal to the width (measured across the parallel sides) divided by 2.414

My calculation is that the surface area of the regular octagon pond will be equal to the square of the diameter minus the area of one side squared.

In the case of the 8 foot diameter with the 3.13 foot side:
8x8=64 and 3.314x3.314=11; and 64-11 is about 53 square feet.

Volume is surface area times depth. With a depth of 2 you'd have 106 cubic feet.

Graph paper is a very handy way of physically checking your calculations. You can sketch out the pond design to scale and then count squares.

Regards,

Matt

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