, in this case $\sin \alpha \; \sin \beta = \frac{1}{2}\left(\cos (\alpha-\beta) - \cos (\alpha+\beta)\right)$ applied to an input signal $x(t) = \sin(f t)$, you'll notice that for $x^2 = \sin^2(ft)= \sin(ft)\sin(ft) = \frac12 \left(\cos(ft-ft)-\cos(ft+ft)\right)$ you see a signal with twice the input frequency.

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>> >> >> >> function x(r, theta, rho) = r * sin(theta) * cos(rho); >> function y(r, theta, rho) = r * sin(theta) * sin(rho); >> function z(r, theta, rho) = r * cos(theta); >> >> sphereD = 5; >> numRots = 7; >> numSpheres = numRots * 360; >> r = 100; >> for (rho = [0:5:numSpheres]) >> let (theta = (180 * (rho / numSpheres))) >> translate([x(r, theta

Using > "max(triangle)" [1,0] is returned (so I can extract the max "x" value), is > there a way to get the max "y" value or should I just index through each > "y" and store the max value?

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-------- I have a DNR officer on my dock who asked if he could take a picture of my "Y" valve being secured so he could share it with a legal class he's teaching to a new batch of DNR officers. -------- In my case I will invite him to: - Raise the hatch in the hall outside the head and stow it in the guest bed room.

Here is my solution from ~8.5 yrs ago https://www.thingiverse.com/thing:1098806 On Mon, Feb 26, 2024 at 11:59 PM Jordan Brown via Discuss < discuss@lists.openscad.org> wrote: > [ This was a math puzzle. It's almost certainly not the *right* way to > generate a helix.

Idea is in this code: include sec=m_points_so(cr(pts1([[0,10],[0,5],[12,0,5],[10,-5],[23,0]]),5),20,.1); max_x=max(sec*[1,0]); p_line(sec,.1); echo(max_x); sec1=[for(i=[0:len(sec)-1])[13*cos(sec[i].x/max_x*360),13*sin(sec[i].x/max_x*360),sec[i].y]]; sec2=[for(i=[0:len(sec)-1])[10*cos(sec[i].x/max_x*360),10*sin(sec[i].x/max_x*360),sec[i].

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