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 Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora 10.6 Feet Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Paul R. Martin on September 7, 2002 04:47:03 UTC

Hi Aaron,

So we can talk about it, suppose the moat is square with the world. Go to the southwest corner and place a board across the corner at a 45 degree angle. Then gingerly walk to the center of that board and place the second board from that point to the inside corner of the moat.

The first board forms a right triangle with the grass ledge with legs of length 10/sqrt(2). Complete the square by drawing a congruent triangle to the north of this one with a common hypotenuse. The second board will have its center at the right angle corner of this triangle. There will be five more feet of that board sticking up to the northwest to touch the inside corner of the grass ledge on the other side. The north-south (or east-west) distance covered by that five feet is 5/sqrt(2). If you sketch the diagram out, it is easy to see that the width of the moat is 10/sqrt(2) + 5/sqrt(2) which is just over 10.6 feet.

Warm regards,

Paul

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