algorithmic modeling for Rhino
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- Comment on: Topic 'Oloid - Curve division - any math expert?'
- he next step was to define the points on the two curves in terms of t and s. Point A for the lying arc and point B for the one standing. A=r*{-cos(t), -sin(t),0} 0<t<2/3*pi B=r*{1-cos(s),0,sin(s)} pi/3<s<pi When t=0, s=pi/3 [that's the line from the middle of the lying arc to the end of the one standing upright] This can be used to calculate the distance between the two points. Some basic geometry gives that |AB|=r*sqrt(3) This can be used to get an expression for s in terms of t. That means that we can express the position of point B in terms of t. Let's first find the function s(t). Distance in 3 dimensions is given by sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2). When we apply this to the coordinates of A and B we get |AB|=r*sqrt{(1-cos(s)+cos(t))^2+(sin(t))^2+(sin(s))^2} The part in the root has to be equal to 3 [see |AB| above]. When we get rid of all the brackets we get: 3-2cos(s)+2cos(t)-2cos(s)cos(t)=3 [we used: cos^2(t)+sin^2(t)=1] so cos(s)=cos(t)/(1+cos(t)) or: s=arccos(cos(t)/(1+cos(t)) Now you can get the mapping from A to B, both depending on parameter t. Hopefully this is what you were looking for :D By the way, this is only for one quarter of the shape, but the rest is trivial due to symmetry.…
- Added by Reinier de Graaf at 1:59pm on January 17, 2013
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