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  • Octagon inside a circle

    What's the formula for determining the exact distance across the flats of an octagon inside a circle of a given diameter, in this case it is an odd size of 1.033.

  • #2
    Sqrt(trig)
    Amount of experience is in direct proportion to the value of broken equipment.

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    • #3
      From google W=.9239D, where W = width of the octagon and D = the circle diameter. If I understand what you are looking for correctly,
      1.033 * 0.9239 = .9543887.

      Edit - this is for an octagon with the points (vertices) touching the inside of the circle.
      Last edited by Ridgerunner; 06-12-2011, 05:21 PM.

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      • #4
        Hmmmmm...... here we go...

        http://buster2058.netfirms.com/octag...ayout_calc.htm

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        • #5
          It's only ink and paper

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          • #6
            Originally posted by radkins
            What's the formula for determining the exact distance across the flats of an octagon inside a circle of a given diameter, in this case it is an odd size of 1.033.
            You know that the base angle of each section facing the flats is 360 divided by 8 or 45 degrees. Dividing that to get a right triangle for trig gives 22.50 degrees. Then you know the length of the hypotenuse is 1/2 the circle diameter so half the distance across the flats is (1.033/2) * cos 22.5, and the across flats distance is twice that or .95436.

            EDIT: Which I see does not agree with the book. Hmmm.
            .
            "People will occasionally stumble over the truth, but most of the time they will pick themselves up and carry on" : Winston Churchill

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            • #7
              Thanks guys that one sure had me confounded big time! I not only got my problem solved but I leaned something new, thanks again!

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              • #8
                Isn't it sin(22.5).

                Phil

                Originally posted by TGTool
                You know that the base angle of each section facing the flats is 360 divided by 8 or 45 degrees. Dividing that to get a right triangle for trig gives 22.50 degrees. Then you know the length of the hypotenuse is 1/2 the circle diameter so half the distance across the flats is (1.033/2) * cos 22.5, and the across flats distance is twice that or .95436.

                EDIT: Which I see does not agree with the book. Hmmm.

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                • #9
                  Originally posted by philbur
                  Isn't it sin(22.5).

                  Phil
                  Some People Have, Curly Black Hair, Through Persistent Brushing

                  sine = p/h
                  cosine = b/h
                  tangent = p/b

                  One thing I was missing was not multiplying the book constant by 1.033 so I think I'm out of the woods now.
                  .
                  "People will occasionally stumble over the truth, but most of the time they will pick themselves up and carry on" : Winston Churchill

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                  • #10
                    The distance across the flats is just diameter x cos 22.5 which is .92387. The book said .9239.

                    I don't understand what your 1.033 is Tgtool.

                    Philbur, I always work everything out from cos. In this case, sin will give you the length of the flats.
                    Richard - SW London, UK, EU.

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                    • #11
                      Originally posted by TGTool
                      Some People Have, Curly Black Hair, Through Persistent Brushing

                      sine = p/h
                      cosine = b/h
                      tangent = p/b.
                      ?????? p h b ??????
                      don't look/sound like any parts of a triangle I've ever seen .
                      I guess h may be hypotenues, but p and b for adjacent and opposite?????????
                      ...lew...

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                      • #12
                        We referenced Chief soh cah toa to remember this.

                        Sine= Opposite/Hypotenuse
                        Cosine= Adjacent/Hypotenuse
                        Tangent= Opposite/Adjacent
                        Guaranteed not to rust, bust, collect dust, bend, chip, crack or peel

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                        • #13
                          Originally posted by Lew Hartswick
                          ?????? p h b ??????
                          don't look/sound like any parts of a triangle I've ever seen .
                          I guess h may be hypotenues, but p and b for adjacent and opposite?????????
                          ...lew...
                          Right, oPposite, Base, and Hypotenuse

                          Originally posted by rohart
                          I don't understand what your 1.033 is Tgtool.
                          That's the size of the OP's circle. Factor times actual circle size. I knew you knew that.
                          .
                          "People will occasionally stumble over the truth, but most of the time they will pick themselves up and carry on" : Winston Churchill

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                          • #14
                            Originally posted by radkins
                            What's the formula for determining the exact distance across the flats of an octagon inside a circle of a given diameter, in this case it is an odd size of 1.033.

                            You could download a free copy of Draftsight 2D CAD. Very handy for these and other type problems of this sort.

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                            • #15
                              Originally posted by KIMFAB
                              We referenced Chief soh cah toa to remember this.

                              Sine= Opposite/Hypotenuse
                              Cosine= Adjacent/Hypotenuse
                              Tangent= Opposite/Adjacent
                              What is "Chief?"

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