• This topic is empty.
Viewing 4 reply threads
  • Author
    • #37841

      Is there a formula or chart for detemining the length of a piece of wood that’s used as a segment in a circle?

      Example: If I cut eight pieces of wood equal lengths and miter the ends at 22.5° and glue them together, I will end up with a semicircle. The question is, to make a semicircle that’s 30 inches in diameter, what is the length that I need to cut each segment?

      I would assume that there would be an existing chart or formula somewhere to figure this out…

      Or, who’s good at geometry?

    • #188710

      today, but I am curious how you ended up with your compressor motor repairs. ????

      Luck, Kizmet

    • #188733

      Do a google search on “chord length”.

      I don’t have the formula handy but it would be there.

    • #188767

      I’ve never had a geometry class, but am pretty good at figuring these things out. I hope this does not need to exact as I will be rounding to the hundreths and a true circle is round whereas what you are talking about is made from straight pieces:

      We know that: Pi R2 = circumference

      We know: R = 1/2 Diameter,

      So we can say Pi D = circumference

      So circumference = 3.1416 x 30 = 94.25

      We know the circumference of a semi-circle is half of a circle, so the outside of this semi-circle is 47.125.

      You want to use 8 straight pieces to miter together, so 47.125/8 = 5.890625, or 5 57/64.

      For ease, round to 5 56/64, or 5 7/8, and you will have a semi-circle with an approximate 29.92 inch diameter.

      I’m sure I do not need to say this, but if you want the id to be approx 30, the 5 7/8 needs to be on the mitered length (short side), if you want an approx 30 in od, the unmitered length (long side), needs to be 5 7/8.

    • #188791

      If the diameter is 30″, the length of each cut will be 15″.

      The radius (which happens to be the exact length of each side) is 1/2 the diameter.

Viewing 4 reply threads
  • You must be logged in to reply to this topic.