I just redrawed, with de 2 circles in the middle farther away from each other, and then you see that the circles are not on the hyperboloïd. Maybe I explained wrong, the 2 small circles does not represent the circle of the center of the hyperboloïd, they are just symmetrical circles that are on the surface, just like the 2 big ones.
You’re right. Your initial example seems to work because it must have been taken from a pre-exisitng hyperboioid or I was just “lucky”. I’m working on better solution but Laurent’s will take into account your changes.
Yes, when you intersect the circles from the first model with the straight line, there is no intersection point. damn 
OK. Here’s a new approach and probably a little more involved than either of us was hoping. Play around with it and let me know. Assuming that everything lies on an hyperboloid, I used some analytic geometry to solve for the circle that intersects the xy plane, using the radii and heights of the two circles. Thereafter, it’s similar to my first post… If you want four circles, just mirror the first two, but its purely cosmetic.Sorry for the unwarranted confidence in my previous reply.
hyperboloid from circles 2.gh (13.2 KB)
Nice, thanks!! I try this out and let you know 
@akilli This is really awesome what you did! I wonder if, could we change the equation, to add 3 more circles on the hyperboloïd? I don’t know which rule you applied, so I could change it for more circles… Maybe I am asking the impossible? 
… As 5 points define a hyperbola, it’s maybe easier, to draw a hyperbola defined by 4 points, and then I can move it with the 5 th point. And rotate the hyperbola around the center line. I am trying to make a hyperboloïd the same in some points as an onther one, and I was using the circles to obtain it.
Connect just two circles, not four…
hyperboloïd_2018Mar16a.gh (9.4 KB)
Intersect plane(s) (white group) to get circle, ellipse, parabola, hyperbola curves (yellow).
hyperboloïd_2018Mar16b.gh (18.3 KB)
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@Joseph_Oster hi, it is important all the circles are on the hyperbola, that’s why the 4… I have drawn a hyperboloïd, and I am drawing a second one, with some parts the same of the first one.
As alternative, we can use Kangaroo as an approximation tool.
Line from 4 circles with K2.gh (12.9 KB)
@Dani_Abalde Hi, thanks for the drawing! Could I use this method using 5 circles above the middle? Actually, being occupied with my drawing, I am now in the point, that it would be nice, if i have the hyperbola curve going true 5 points (in 1 plane - as if I would make a verticale cut true the hyperboloïd), so that I can move one point, and the 4 others stay fix. A hyperbola is defined by 5 points mathematically, so technical it can work. I just don’t find a function to draw a hyperbola in grasshopper using points on the curve.
The general equation of an hyperboloid requires three parameters, but since you are requiring that the cross sections be circles, that leaves two degrees of freedom. That’s why it is possible to use just two circles to define it here. The five points you mention for an hyperbola refer to an arbitrary hyperbola in the plane that can be oriented in any way. Lets say your five circles intersect the xz plane in five points. Then you could solve five equations in five unknowns and come up with the equation for an hyperbola but it would probably have its major axis not pointing in the the x direction and if you rotated it about the z axis, what resulted would not be an “hyperboloid” according to the definition. https://en.wikipedia.org/wiki/Hyperboloid. If you want something that “looks” like an hyperboloid but is not mathematically exact, then that is a different matter.In that case, you could just do it by eyeballing it.
@akilli Thanks very much for this explanation! I just have some questions.
Are the cross sections not always circles on an hyperboloïd? I was thinking the way it is formed, turned around an axis that it always result into a circle as cross section.
Do you mean that if rotating the hyperbola about the z axis, it would not be an hyperboloïd if the axis of the hyperbola is not perpendicular to the z axis? (my englisch is not that well, thats why I ask to be shure I understand) So if my hyperbola’s axis is perpendicular to the z, there would be no issue. My four points I pick, are lying on a hyperboloid drawed with rotating of a straight line. I am trying to draw a second hyperboloid that has some points the same as my original hyperboloid. Taking 4 points the same, would let me try to see the difference how far I need to change the 5th point to see a real difference between the two surfaces. (I am drawing a double structure, and on the circles points need to connect to each other…) I would think because of taking 4 points from a real hyperboloid, it will still be one after changing the 5th point, but I am not a mathematician to be sure of this, only quiet good and passionate about it… Is it possible to make the equation of the hyperboloid with grasshopper, so I can try it to see if it is an hyperboloïd? (my 4 points are fix, available, so 1 point should be able to change) I would love I can make a hyperboloïd of it 
what can I understand by ‘eyeballing’?
Thanks!
It’s clear you have something in mind that requires five points but, sorry, its not clear to me. Now you’re talking about two surfaces. I think it would be extremely helpful if you drew a picture of what it is you’re trying to do.
To answer your first question, you are looking for an “hyperboloid of revolution”, which has circular cross sections but a general hyperboloid can have elliptical cross sections.
@akilli sorri for making it unclear. I was trying to explain more but didn’t work. 
I made 2 pictures. All what I was trying know with the circles for the hyperboloid, was for the surface on the inside you see on the picture. So, I am drawing a double structure, were both surfaces are lying on a ‘circular’ hyperboloïd. I draw a technique out to divide them into triangles. The inside surface on this image is just a 1D scaling of the hyperboloid to show how I want a surface inside another. Because of the scaling it is not a circular hyperboloïd anymore, so I cannot use my technique for division, which is necessary. I would like in my triangulation method, that the eindpoints at the border are on both hyperboloids, so I can connect my triangles of the two layers at the end of the border. (don’t know if this is clear). So to draw the inside hyperboloid, I first wanted to secure the two outside circles, so they would have the same ‘range’. That was my question I started with here. Looking further how it could be better, I start thinking making circles true each endpoint on the border, and look at the intersection point on the curve on the bottom (=were the plane is drawn). Then I wanted to use this points, to define a new hyperbola with one other point, so my curves would be different, but still coming together nice at the end. (as if you would not see 2 layers when you look at it, but 1 line going open in 2 layers) Using more points, give me more points at the end of my curve that will be the same… The points are marked in yellow on the picture. On the picture the lines come together in the bottem, I want to try make them together at the border. Is this explanation more clear? If not just say and I try to make it more clear on another drawing… Thanks for the effort!
- Yes, i had totally the hyperboloid of revolution in mind -
Two questions: Why do the surfaces have to be hyperboloids and if they are, do they both need to have circular cross sections?
@akilli I am working with ‘basic’ forms like a sphere (dome), a cylinder, a cone, and a hyperboloïd. Those are selfsupporting structures, that I start cutting through to connect them to each other. The other one there structure is ok, but the hyperboloïd needed the second structure underneath to be strong enough. If it is not possible that the turned hyperbola is a hyperboloïd, I am fine with the option to rotate the hyperbola, without being a real hyperboloïd after.
The cross section is important for my division method. I make big triangles in there, and if it is not a circle, there is no way to make some lengths the same as others (thats another story to explain why )
this is the first step in the division method, and all points of the triangles are lying on the surface.
Working with points exactly on the surface is important, because the form of a hyperboloïd make the ‘forces in a structure’ go nice down from the top til the bottom…
If you are willing to allow the inner hyperboloid to have elliptical cross sections, then you can get the bases to align perfectly. You accomplish this by scaling the outer structure down in one direction. The bases still coincide but the roofs never touch The panels come from the Lunchbox plugin
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hyperboloid from circles 3.gh (18.4 KB)
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@akilli, waw, this looks really nice. I am drawing for the moment a surface of revolution, based on a curve I draw with the points of the circles I needed on. I draw it in gh, so I can change it if i would get a better curve that would be a hyperboloid function I have quiet a big grasshopper file for the moment with my division method of triangles I make, so I am afraid if I open now your file, my Rhino will crash… But I am very curious to see it and thank you very much spending time on my drawing! I am in my last step of my gh for the moment, and I am missing just 1 command to finish it… I don’t find the ‘curve-> normal to surface’ command form Rhino in gh. Do you know this?
Thanks again!
@akilli Hi, opened your file, installed the plugin… A new world is opening to me! This toolbox is great! I tried to do use the panels on a piece of a surface of an hyperboloïd I divide with my method to see the difference in division, but he is making the panels like the hyperboloïd is still complete. Can I do something to change that, that he divides only my piece into triangles?
I don’t think I can give you a reasonable reply at this point without seeing your code. Could you upload the file and also internalize whatever geometry you’re importing from Rhino? Thanks.
@akilli hi, here are both files:
test verdeling driehoek plugin.gh (29.8 KB)
test verdeling driehoek plughin.3dm (6.0 MB)
I tried to mark were the issues are and what I tried, It’s a bit messy in my rhino file as I tried things out and copies… What I was thinking if the gh maybe only want to triangulate the revolved line, and any changes on that surface he doesn’t look at. Thanks