Chords From Inscribed Polygons
Date: 07/11/2002 at 22:36:31 From: Edmond Clay Subject: Unknown Chord Known Radius 8 parts I am a carpenter and would like the formula for deriving the length of a chord of a circle with a known radius. The chord here is defined as equal parts of the circle and in this particular case is an octagon. I need to know this in order to cut the blocks (spacers) that fall between the trusses of a turret that I am building. Thank you for your time and effort.
Date: 07/12/2002 at 03:01:48 From: Doctor Jeremiah Subject: Re: Unknown Chord Known Radius 8 parts Hi Edmond, To find the side length of an octagon inscribed in a circle with a known radius, notice first that the distance from the center to the corner of the octagon is the same as the radius of the circle. This means that the octagon is really eight triangles with three sides, one of which is the chord length, and two of which are the radius. +------L------+ + \ / + + \ / + + R R + + \ / + | + \ / + | | + \a/ + | | + | | + / \ + | | + / \ + | + / \ + + / \ + + / \ + + / \ + +-------------+ But we don't know L, so we need some other piece of data. It turns out that angle 'a' is 360/n, where n is the number of sides. For an octagon, of course, n = 8. We can cut one of the triangles in half, +----L/2--+ \ | \ | \ | R | \ | \a/2| \ | \ | \| + and use the definition of the sine of an angle: sine of angle = length of opposite side / length of longest side which translates to sin(a/2) = (L/2) / R R sin(a/2) = L/2 2R sin(a/2) = L But since an octagon has 8 sides, the value of a would be 360/n, and we have: 2R sin( (360/n)/2 ) = L 2R sin(180/n) = L So in the case of an octagon with a radius of 10: 20 sin(180/8) = L 7.654 = L Let me know if you need more details. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/
Date: 07/13/2002 at 18:57:13 From: Edmond Clay Subject: Thank you (Unknown Chord Known Radius 8 parts) Bingo! The good news is that I applied the calculation to the frieze block cuts and installed away. Everything fit perfectly. It never ceases to amaze me how a very accurate calculation always makes the woodwork look good. I'm in production framing and quite often "fuzzy" math is the "go to" method for in the field applications. I had hoped to avoid carrying around a trig table, but I suppose I'll have to move my craft to the next level now...Now about building the compound arch....Thanks Dr. Math!
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