• Mathematical Calculation puzzle? (Excel – ANY)

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    #397696

    I have an interesting puzzle/riddle to solve. I am using this forum simply because I want the result in Excel. I need to determine the area of an odd shape given two parameters. The formulas/method to determine this would be helpful in an Excel spreadsheet as there are many values of the two variables. I am attaching a GIF of the shape for reference. The only two variables supplied are “T” or thickness, and “W” or width. The radius of the edge is simply T*3/4.

    Any help would be most appreciated!!

    Drew

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    • #755014

      (Edited by sdckapr on 09-Dec-03 12:35. OOPs made a mistake, Corrected it here)

      Given your “W”, “T” and your “RF” (radius factor) of 0.75 you can calculate:

      x = SQRT((RF^2-0.5^2)*T^2) = The distance from the center of “radius” to the end of rectangular section.
      alpha =2*ASIN(0.5/RF) = the angle of the “PieSections”
      AreaPie =alpha/2*(RF*T)^2 = Area of each end “pie segment”
      AreaTriangle =x*T/2 = Area of each “triangle” in the pie (the pie without the circular cap)
      AreaCap = AreaPie – AreaTriangle
      AreaRectangle =T*(W-2*(RF*T-x))= area of the rectangular area

      Total area = AreaRectangle +2*AreaCap

      In one formula this is:
      Area =T*(W-2*T*RF+SQRT((RF^2-0.5^2)*T^2))+2*ASIN(0.5/RF)*(RF*T)^2

      Steve

      • #755033

        Hi!! Thanks for the help!! You know, I copied the data into an Excel worksheet, began testing it and it didn’t come out right. I looked it over step by step figuring, “Something must be wrong, but where. Hmmmm..” So, when i reviewed the last formula (AreaRectangle) I realized there was an error. I corrected my worksheet and all worked well. So, I thought I’d come on here and let you know there was a minor error, and LO-and-BEHOLD, you ALREADY CORRECTED IT!!!

        Thanks for following up!! I appreciate the help!! And thanks for forcing me to walk through it rather than simply taking it for granted. I am better off for it.

        Drew

        • #755038

          Yes, I had removed the 2 “x”s originally from the length. I wanted to subtract the radius then add the “x” back.

          I am glad it worked out.

          Steve

        • #755039

          Yes, I had removed the 2 “x”s originally from the length. I wanted to subtract the radius then add the “x” back.

          I am glad it worked out.

          Steve

      • #755034

        Hi!! Thanks for the help!! You know, I copied the data into an Excel worksheet, began testing it and it didn’t come out right. I looked it over step by step figuring, “Something must be wrong, but where. Hmmmm..” So, when i reviewed the last formula (AreaRectangle) I realized there was an error. I corrected my worksheet and all worked well. So, I thought I’d come on here and let you know there was a minor error, and LO-and-BEHOLD, you ALREADY CORRECTED IT!!!

        Thanks for following up!! I appreciate the help!! And thanks for forcing me to walk through it rather than simply taking it for granted. I am better off for it.

        Drew

    • #755015

      (Edited by sdckapr on 09-Dec-03 12:35. OOPs made a mistake, Corrected it here)

      Given your “W”, “T” and your “RF” (radius factor) of 0.75 you can calculate:

      x = SQRT((RF^2-0.5^2)*T^2) = The distance from the center of “radius” to the end of rectangular section.
      alpha =2*ASIN(0.5/RF) = the angle of the “PieSections”
      AreaPie =alpha/2*(RF*T)^2 = Area of each end “pie segment”
      AreaTriangle =x*T/2 = Area of each “triangle” in the pie (the pie without the circular cap)
      AreaCap = AreaPie – AreaTriangle
      AreaRectangle =T*(W-2*(RF*T-x))= area of the rectangular area

      Total area = AreaRectangle +2*AreaCap

      In one formula this is:
      Area =T*(W-2*T*RF+SQRT((RF^2-0.5^2)*T^2))+2*ASIN(0.5/RF)*(RF*T)^2

      Steve

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