Truncating a Square to Get an OctagonDate: 10/13/2003 at 18:29:10 From: Joe Subject: carpentry How do I figure the meaurements to lay out an octagon? I tried starting with a square and cutting 45 degrees off the corners but it's still not working out. Date: 10/14/2003 at 09:30:53 From: Doctor Luis Subject: Re: carpentry Hi Joe, You're right. You do have to saw off 45 degree angles from the square, since octagons have external angles of 45 degrees. The trick is figuring *where* to start sawing off. What distance from the corner is that? Well, don't worry. Mr. Pythagoras will come to the rescue. In the diagram below, let 'a' be the length of the octagon's side, let 'L' be the length of the square, and let 'b' be the distance from the corner of the square to a corner of the octagon. (Before we start our analysis, let me just say that this is an approximate drawing. It's probably not an octagon of equal sides, but it will be sufficient to establish the relationships between our desired quantities a,b,L for the case of the regular octagon.) Looking at the diagram, we can conclude a couple of things right away L = a + 2b a^2 = b^2 + b^2 Of these equations, the first one comes just from adding the lengths that make up L, and the second one comes from the Pythagorean theorem. So, a^2 = 2*b^2, or a = sqrt(2)*b. Plugging this in the equation for L, we get L = sqrt(2)*b + 2*b = (sqrt(2) + 2)*b Since we want b as a function of L, b = L / (2 + sqrt(2)) we can get 'a' as a function of L from the equation a=sqrt(2)*b a = L * sqrt(2) / (2 + sqrt(2)) As percentages of the side L, we find that b/L = 1/(2 + sqrt(2)) a/L = sqrt(2)/(2 + sqrt(2)) These are exact values. Approximate values are b/L = 0.29289.. a/L = 0.41421.. (or L/a = 2.41421..) Therefore, given a square of side L, your octagon's side will come out to about 41.4% of L, and you must start sawing off the 45 degree wedge a distance 29.3% of L from the corner of the square Also, if your target is an octagon of side 'a', you must start with a square of side L that is 241.4% of the length 'a'. I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
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