Math question: Does a 4.25" square cut diagonally yield a 3" pinwheel?
#1
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Join Date: Dec 2012
Posts: 538
Math question: Does a 4.25" square cut diagonally yield a 3" pinwheel?
My pattern says cut (68) 3" squares, then cut diagonally for pinwheel parts (times 2 fabrics)
I will end up with a total of 17 pinwheel blocks which are ultimately 4" square.
If I were to use Jennie Doan's easy pinwheel method of sewing all the way around two pieces, then cutting diagonal both ways, do I need to start with 4.25" squares?
I've drawn this out on a piece of paper, and measured, and I get the whole pythagorean theorem, but just need to confirm. And the second part of the question is: if it is the 4.25, do I need to be concerned about the decimals? (square root of 18 is 4.24264068712).
Should I bag it and just cut the flippin triangles?
I will end up with a total of 17 pinwheel blocks which are ultimately 4" square.
If I were to use Jennie Doan's easy pinwheel method of sewing all the way around two pieces, then cutting diagonal both ways, do I need to start with 4.25" squares?
I've drawn this out on a piece of paper, and measured, and I get the whole pythagorean theorem, but just need to confirm. And the second part of the question is: if it is the 4.25, do I need to be concerned about the decimals? (square root of 18 is 4.24264068712).
Should I bag it and just cut the flippin triangles?
#4
My pattern says cut (68) 3" squares, then cut diagonally for pinwheel parts (times 2 fabrics)
I will end up with a total of 17 pinwheel blocks which are ultimately 4" square.
If I were to use Jennie Doan's easy pinwheel method of sewing all the way around two pieces, then cutting diagonal both ways, do I need to start with 4.25" squares?
I've drawn this out on a piece of paper, and measured, and I get the whole pythagorean theorem, but just need to confirm. And the second part of the question is: if it is the 4.25, do I need to be concerned about the decimals? (square root of 18 is 4.24264068712).
Should I bag it and just cut the flippin triangles?
I will end up with a total of 17 pinwheel blocks which are ultimately 4" square.
If I were to use Jennie Doan's easy pinwheel method of sewing all the way around two pieces, then cutting diagonal both ways, do I need to start with 4.25" squares?
I've drawn this out on a piece of paper, and measured, and I get the whole pythagorean theorem, but just need to confirm. And the second part of the question is: if it is the 4.25, do I need to be concerned about the decimals? (square root of 18 is 4.24264068712).
Should I bag it and just cut the flippin triangles?
Try it with scrap fabric. Then you'll know for sure.
#5
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Join Date: May 2011
Location: New Jersey (North)
Posts: 32
I had bookmarked this site before and found it very helpful. I rounded my squares up and then trimmed to size.
edited to add the link. oops.
http://www.moonlightquilters.org/tipHSTfav.htm
edited to add the link. oops.
http://www.moonlightquilters.org/tipHSTfav.htm
#6
Pinwheels are made from 4 - HSTs. For a finished 4" pinwheel then each corner of the HST should finish at 2".
When doing HSTs, I use method 2 in the following tutorial
http://www.thatgirlthatquilt.com/201...es-3-ways.html
The rule of thumb for assembling HSTs this way is "finished size plus 7/8". So a 3" square will be adequate using method 2 - giving you a little wiggle room for trimming the final pinwheel into a perfect square.
My only concern about Jenny Donan's method (method 1 in the first link) is that the resulting edges are not on the straight of the goods and can stretch, If you do want to do that method, the following site even has a chart for measurements
http://whipup.net/2011/03/17/guest-b...are-triangles/
When doing HSTs, I use method 2 in the following tutorial
http://www.thatgirlthatquilt.com/201...es-3-ways.html
The rule of thumb for assembling HSTs this way is "finished size plus 7/8". So a 3" square will be adequate using method 2 - giving you a little wiggle room for trimming the final pinwheel into a perfect square.
My only concern about Jenny Donan's method (method 1 in the first link) is that the resulting edges are not on the straight of the goods and can stretch, If you do want to do that method, the following site even has a chart for measurements
http://whipup.net/2011/03/17/guest-b...are-triangles/
#7
I have to respectfully disagree entirely with the math in that whipup link above. The relationship is NOT proportional as the blogger states.
The math would be the same as for QST's. You're just sewing the first seam differently to get to the same result. For any finished size HST, add 1.25" and you have the size of the squares to start with. Your finished pinwheel will be twice the size of the finished HSTs, so for 4" finished pinwheels you need 2" finished HST's, which requires two 3.25" squares.
The math would be the same as for QST's. You're just sewing the first seam differently to get to the same result. For any finished size HST, add 1.25" and you have the size of the squares to start with. Your finished pinwheel will be twice the size of the finished HSTs, so for 4" finished pinwheels you need 2" finished HST's, which requires two 3.25" squares.
#8
Super Member
Join Date: Aug 2012
Location: Highland, CA
Posts: 1,407
I had bookmarked this site before and found it very helpful. I rounded my squares up and then trimmed to size.
edited to add the link. oops.
http://www.moonlightquilters.org/tipHSTfav.htm
edited to add the link. oops.
http://www.moonlightquilters.org/tipHSTfav.htm
#9
Thanks for this, I just printed it to hang on my bulletin board.
I had bookmarked this site before and found it very helpful. I rounded my squares up and then trimmed to size.
edited to add the link. oops.
http://www.moonlightquilters.org/tipHSTfav.htm
edited to add the link. oops.
http://www.moonlightquilters.org/tipHSTfav.htm
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