help with equation in function component - catenary from two points - where is arcsinh function?

I want to calculate these equation in a function component:
sinh^(-1)[h/sqrt(s^2 - h^2)] - L/(2a) but it's not working...
Can someone help me with the correct syntax for the "expression editor" in grasshopper?

PD: "h", "L" and "a" are inputs variables.

I'm trying to do a catenary from two points like in these video: http://www.grasshopper3d.com/video/catenaria-1

The equation it's suppose to be the "A" constant in the catenary equation: y(x) = a cosh(x/a + A) + B
The "B" constant is B = -a cosh A

Replies to This Discussion

Permalink Reply by Bojan Koncarevic on November 30, 2011 at 6:03am

ps it the same question image as Artur Mani had at this forum page ..

Permalink Reply by Bradley Bowers on July 12, 2013 at 2:54pm

I have been using an older definition (attached to this message) in an attempt to make a semi Catenary dome . . . but at the moment I am unable to give 2 DIFFERENT Catenary curves the SAME EXACT apex coordinates. I have attached an image depicting what I get. I am unable to find a solution. When I set the z-height to the same input, I still get two different apex heights . . . is this happening for a particular reason? And is there a way to rectify it?

Attachments:

cantenary_halfdome.png, 108 KB
img04 Jul. 12, 2013 04.45 Mr.Bowers.jpg, 131 KB
catenary_dome.ghx, 167 KB

Permalink Reply by Benedetta Gargiulo on April 29, 2010 at 4:13am

DId you find a way???To have the right funciton in GH?Xxx

Permalink Reply by Benedetta Gargiulo on April 29, 2010 at 4:14am

DId you find a way???To have the right funciton in GH?Xxx

Permalink Reply by Pep Tornabell on April 29, 2010 at 5:55am

There are several definitions in this post to draw the catenary, each of them shows/reveals the complexity of the problem in different ways. I posted one definiton some time ago that only works when the 2 points are (-1,0,0),(1,0,0), I imagine its just a matter of mapping values to make it work for any pair of points. Giulio posted a really cool one that works for any number of points (and any position) using Ralph-Newton aproximation (correct me if I'm wrong, Giulio) for the catenary problem. And sure there are more posible solutions.

Depending on what you need (defined the catenary lenght, defined catenary high) you sould use one or another.

The mathematical function works fine in GH, but notice that you need to map values (and I might be very wrong about that) when working with sinh and cosh (hiperbólic functions)...

Hope this is usefull, anything just reply, lol

Btw, if somebody could explain to all of us something about hiperbólic functions (hard maths) I'd appreciate it very much

Cheers

Permalink Reply by Bojan Koncarevic on November 30, 2011 at 6:01am

Dear David

I have a problem with the catenary gh component..

It does outcomes in a curve as You have attached the set of curves creating the arched tunnel..

but actually it gives out the simple triangle connection of the 2 base point and height ending point

What i am doing wrong (since i do not have curve (actualy i do but it is not the curved one) ; ) )

I hope i will get the reply asap your schedule can estimate..

I did not upload any file since the question is very simple.

I just simply make 2 points and catch them to the Catenary icon (or line end points)

best

Bojan