Segmented using multiple size wedges

[BobCoates]

I have been making segmented bowls and vases using wedges cut at 4 different angles and different number of wedges of each size (), 30(3), 221/2(4), 18(5), and 15(6) degrees(number of each). Per math friend there is over 7 million arrangements for each ring. I have found that you can get interesting patterns by mixing the wedges. Each group makes up ¼ of the ring. You can group them in any arrangements to form the rings. They make nice “Ikea” type packages to give to friends or “sa” and let them form the rings, return them when glued to their liking and then give rings back to assemble and turn the rings. Neighbor 85 years old love the doing one. I use Excel or LibreOffice Calc to calculate the wedge sizes. Here are two samples.

 

[BobCoates]

Here is a attempt to explain the spread sheet used to create these.

This sheet can be used in excel or LibreOffice Cal to find the wedge length to cut. Row1 is the number of wedges for a normal ring of degrees in row 2.

Ie 12 for 30 degrees. Row 3 shows the degrees and how many [ number ] of that degree is used in a ring for this project.

Column 1 is the diameter of the ring you wish to make. And the result in columns 2 - 5 is the estimated wedge size. It is the actual arc length, but is a good estimate for wedge size.

The math is circumference ( Pie * diameter)/ number of wedges.

To use, paste the four rows below into either spreadsheet. Select one cell from row 3 and copy it to row 4 and remove the " both front and back. Repeat for other columns. With the 4 in column 1 you should see the same numbers. You can also format the cells to 2 decimals or fractions.

Place your desired diameters in successive rows of column 1 and then use the copy feature in both. (Hint: Highlight the columns 2 -5 for row 4 and the rows that you entered data and then press ctrl and d key). This should have copied all the formulas and used your diameters from your numbers entered in column 1.

	12	16	20	24
	30 [3]	22.5 [4]	18 [5]	15 [6]
	"=PI()*RC[-1]/R1C2”	"=PI()*RC[-2]/R1C3"	"=PI()*RC[-3]/R1C4'	"=PI()*RC[-4]/R1C5"
4	1.0471975511966	0.785398163397448	0.628318530717959	0.523598775598299