- #1

rede96

- 663

- 15

x = sin((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*(d/2)

x = cos((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*d/2)*2

My math is pretty crap, so any help would be appreciated.

Thanks

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- Thread starter rede96
- Start date

- #1

rede96

- 663

- 15

x = sin((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*(d/2)

x = cos((cos(((1-(cos(θ/2)^2))*360)/2)+1)/2*θ)*d/2)*2

My math is pretty crap, so any help would be appreciated.

Thanks

- #2

mfb

Mentor

- 36,291

- 13,366

- #3

rede96

- 663

- 15

Oh well, will have to use them in the current form. Thanks anyway.

If you don't mind me asking (as I really am crap at math) why is the contour plot interesting?

- #4

- 17,777

- 18,916

It looks like right from a seismograph or a piece of modern art.Oh well, will have to use them in the current form. Thanks anyway.

If you don't mind me asking (as I really am crap at math) why is the contour plot interesting?

- #5

It looks like right from a seismograph or a piece of modern art.

Tbh, I didn't see anything special in it either.

- #6

- 17,777

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- #7

Edgardo

- 705

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It has a simpler looking form for x, d and theta real.

- #8

DrClaude

Mentor

- 8,109

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It has an interesting contour plot, however.

My guess is that this is incorrect, as the 360 smells of degrees, leading to a much too rapid oscillation and a crappy contour plot. Change the 360 to 2*pi and you get something much more banal.

This is the same result as

It has a simpler looking form for x, d and theta real.

##1-\cos^2(x) = \sin^2(x)##

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