Hi. Can you please show me how can I mathematically reach the measurment of x in the fish base of the following image, when I know the measurement of y?
Thank you in advance .
How to reach the measurement of this crease on a fish base?
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How to reach the measurement of this crease on a fish base?
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Re: How to reach the measurement of this crease on a fish ba
It comes from trigonometry and geometry. The bisector of the angle at the bottom forms a 22.5deg angle from the edge, the cosine of which is 1/(1+sqrt2). It also happens that the distance from the left corner of the square to the left end of x is the same as the length of the square's edge (y). Since the main diagonal is sqrt2 times the side length, just multiply y by sqrt2 and subtract it from the result.
The equation is x = (sqrt2)y - y.
The equation is x = (sqrt2)y - y.
Re: How to reach the measurement of this crease on a fish ba
I didn't get everything you wrote, but I did get that it's possible to present the problem as the Pythagorean theorem:
y²+y² = (y + x)² right?
Was that what you did?
You mentioned x = √(2)y - y
I can take it one step further x = y(√(2) - 1)
If y = 1 then x = 0.41421356237
One more question, is there a shortish fraction close to that number?
y²+y² = (y + x)² right?
Was that what you did?
You mentioned x = √(2)y - y
I can take it one step further x = y(√(2) - 1)
If y = 1 then x = 0.41421356237
One more question, is there a shortish fraction close to that number?
Last edited by Gerardo on September 18th, 2018, 1:48 am, edited 1 time in total.
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Re: How to reach the measurement of this crease on a fish ba
http://www.mindspring.com/~alanh/fracs.html
This website might be helpful. I typed in the number and it gave a few fractions:
2/5 = 0.4
5/12 = 0.4167
12/29 = 0.41379
29/70 = 0.41428
70/169 = 0.414201
This website might be helpful. I typed in the number and it gave a few fractions:
2/5 = 0.4
5/12 = 0.4167
12/29 = 0.41379
29/70 = 0.41428
70/169 = 0.414201
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Re: How to reach the measurement of this crease on a fish ba
Awesome webpage! Thanks NeverCeaseToCrease .
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