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Am I thinking correctly math wise?

Am I thinking correctly math wise?

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Old 05-16-2014, 08:40 AM
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Default Am I thinking correctly math wise?

I have been saving my son's jeans and I am going to make one of these quilts for my couch. http://www.equilters.com/library/jea...gallbaros.html Am I thinking correctly that if I cut a 5" square for inside the circle the square will finish at 5"? Because the jean circles will be sewn together first and then fold over the 5" circle. I am just wondering how many squares I will need. Please let me know your thoughts. Thank you
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Old 05-16-2014, 08:59 AM
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Never saw this before, and wow, thanks for sharing. I've been saving Blue Jeans for several years wanting to make a quilt and I absolutely love it. After reading the instructions 2x's. Yes you are thinking correctly. Your block should finish 5 inches, however think about it, when you sew on the line you will loose probably 2 threads width due to the thickness of the fabric. She mentions you can cut the squares a scant or two shorter to allow for that. I would probably go with 4 7/8's or even 4 3/4. do a test block first. So to figure how many squares you need Take your width and length and multiply together to get your square inches of the total quilt. Say 70x90= 6300, divide by 25 ( since a 5x5 block is 25 square inches) You will need 252 5 inch squares for a 70x90 quilt. I've bookmarked the page to do one ( 1 of these days). Good luck.

Last edited by Gannyrosie; 05-16-2014 at 09:05 AM.
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Old 05-16-2014, 09:25 AM
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Thank you! Good point, they may actually finish at 4-7/8. I think it looks like a really neat quilt.. no batting or backing needed. I am wondering.. but I think it would be self binding as well
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Old 05-16-2014, 10:03 AM
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the diameter of the circle is the length of the square's diagonal. so half of the square is a right triangle. the pythagorean theorem states, the square of the diagonal is the sum of the squares of the other two sides. so if D is the diagonal and the other two equal sides are A, then D [5] squared = 2 x A squared... 25=2x[AxA] ... [AxA] = 25/2 ... 12.5 = [AxA] ... A = square root of 12.5 ...A = 3.5355". so the inside square is 3.5355"
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Old 05-16-2014, 10:05 AM
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This is basically the pattern for Rings That Bind quilt only without the square of batting inside the blocks. It is easy to do and you can easily change your mind about what size as you go.
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Old 05-16-2014, 10:10 AM
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QuiltnNan is correct. The corner to corner measurement of the square is 5" so the sides of the square are 3.535". As also mentioned, you area going to loose a bit in the seam, so they will wind up pretty close to 3.5".
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Old 05-16-2014, 11:32 AM
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oh my! I need to get my college student to help me with this!!
the pythagorean theorem states, the square of the diagonal is the sum of the squares of the other two sides. so if D is the diagonal and the other two equal sides are A, then D [5] squared = 2 x A squared... 25=2x[AxA] ... [AxA] = 25/2 ... 12.5 = [AxA] ... A = square root of 12.5 ...A = 3.5355". so the inside square is 3.5355"
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Old 05-16-2014, 08:00 PM
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No, the squares will not finish at 5", if the circles at 5". The diagonal of the square will be 5", so the sides of the square will be 1/2 of the square root of 25. Remember that the 2 right angle sides of a triangle, when squared and added together, will equal the square of the hypotenuse of the triangle.

x squared + x squared = 5 squared
2 (x squared) = 25
x squared = 12.5
x = square root of 12.5 = 3.5 (same as an earlier commenter came up with)
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