Compound tangents of equal length

Hi all,

I’m looking to solve this piece of geometry, basically, I’m looking to make it so that each red curve is of equal length, while being perpendicular to the convergence point. So far I’ve only managed to do it using Galapagos, but I was wondering there was a theorem that could be applied instead? I’ve tried to search for it but so far nothing

190306_Tangent_arcs01.gh (19.5 KB)

Are red segments “r” orthogonal to the gray dashed lines?

Probably kangaroo is the fastest way for now.
Maybe there is a pure math solution, but kangaroo would certainly work it out and it’s way… lazier XD

A note: having one segment “r” being co-linear and parallel with segment “a” make the whole structure somehow “rigid”… maybe to make it work you should just use a distance for vertex “b-c” from “a” , instead of using direct values for “b” and “c”.

In the meantime someone find a theorem, this is a possible way of solving this with kangaroo:

2019-03-07%2017_28_18-Window
what
what.3dm (40.4 KB)
what.gh (13.3 KB)

Note how left point move; as said, having one segment “r” co-linear with “a” make this impossible to have a rigid triangle “abc”. One of the segments need to be “free” to change length.
(in my solution both b and c are “free” and i’ve set a vertical anchor to left point)

This would be way easier with software like Inventors, using “bonds” and such.

Thanks for this, looks promising! Ideally I’d like to keep the red segments perpendicular to the focal point, while keeping the focal point in the same place, also leaning towards a trigonometric solution to make it quicker to calculate

x is divided to 5 ? or they have different angles?

they all have different angles because the distance of the vertical curve from the point changes as the segments go up the curve

i analyse the image and find something interesting maybe that help you to find the final solution

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As i said… I think that’s impossible.
Again referring to your first picture, If you want one red segment being parallel and co-linear with “a”, you cannot use any triangle abc , you have to let a or b or c change its length.

If you don’t need one red segment being co-linear and parallel to a, then it’s a different case.

Thanks for your response, there’s no need for any of the red segments to be co-linear/parallel with line “a”.

The only requirement is that the focal point stays in the same place and that the red segments are of equal length.

Ok… maybe that will make everything even mor complex. :rofl:
Even on just kangaroo, the anchors would need to be changed in a smart way to make them work.

What @seghierkhaled pointed out is really interesting.
I did a fast check and maybe the "1,2,3,4 … " proportionality is theoretically exact with infinite-diameter circles, so with smaller circles you have to implement non-linear proportion between them.

I’ll look on this further later…

Definitely, I’m convinced there’s a fairly simple equation than can be used to describe the relative radii of the red curves

hi
i create this no equation or complex math

math.gh (15.9 KB)

Hi Seghier, thanks for this but ideally I’d need to subdivide a length exactly and using a fixed point

what you mean by subdivide a length exactly

as in, if you refer to the original image, length a is of a fixed dimension

ok i understand you i will check it again

if you can solve this you will solve the problem:
we know : x , y , ∝ and we need find γ