Hyperbolic Spiral

Replies to This Discussion

Permalink Reply by David Rutten on November 14, 2016 at 10:43am

Just to make sure we're on the same page... you're looking for a curve shape which would result in a flat graph rather than the bumpy one shown here right?

ps. I took the liberty of simplifying the file somewhat and also remove one of the sample points which resulted in an awkward lump at the top of the curve.

Attachments:

• HyperbolicRamp.gh, 25 KB

Permalink Reply by David Rutten on November 14, 2016 at 10:49am

You could also use the Modulo operator to wrap your samples:

Permalink Reply by Toni Rubio Soler on November 14, 2016 at 11:20am

Exactly David!

I need the graph to be flat resulting a equiangular spiral.

Any suggestion?

Thanks a lot!

ps: Thanks for cleaning up the mess as well :-)

Permalink Reply by David Rutten on November 14, 2016 at 2:40pm

I can't seem to figure it out. I'm trying to manually tweak a sampling curve, hoping that by controlling how fast the UV points most along the surface I could adjust the tangent distribution, but it's like pushing air-bubbles underneath tape...

The exact mathematics are well beyond me so I definitely can't furnish you with an analytic solution.

Permalink Reply by Toni Rubio Soler on November 15, 2016 at 3:04am

Thanks for your help David!

Permalink Reply by Ethan Gross on November 14, 2016 at 8:30pm

Hi Toni,

This was a very interesting question that proved much harder than I thought. I assumed a logarithmic spiral projected from the xy plane onto the hyperboloid might work and it looked pretty good, except at the hyperboloid's "waist", where it deviated from constant slope. Using calculus, I came up with differential equation for the radius of the spiral versus the angle and it turned out that the radius is the hyperbolic cosine (cosh)of the angle. Coincidence???? You be the judge... Hyperbolic cosine looks like an exponential at large +/- values but it deviates near zero.

I generated a cosh spiral in the xy plane that fit between the large and small radii of the hyperbola and wound a specified number of times. The points of this spiral were essentially projected onto the hyperbola to generate a spiral about it. Using David's method you can see the deviation from constant slope is quite small.

Attachments:

• HyperbolicRamp2.gh, 20 KB

Permalink Reply by Toni Rubio Soler on November 15, 2016 at 3:04am

Yesss!!!

Is brilliant to generate the hyperbolic spiral in the xy plane!!

That tiny variation on the slope will do the job perfectly!

Thousands thanks Ethan!

Best,

Toni

Permalink Reply by David Rutten on November 15, 2016 at 3:44am

Gorgeous!

Permalink Reply by Ethan Gross on November 15, 2016 at 6:21am

Toni and David, congratulations were premature. I went over my calculations again, and discovered that, in my eagerness to find an neat analytic expression, I left out a term which would have made my integral intractable instead of one where everything cancelled out so obligingly. Thus, my solution is really a first-order approximation -a pretty good one, but an approximation nonetheless. If you play around with some extreme values like a very narrow waist and a steep spiral you'll see it goes off somewhat. So, it's not gorgeous but it is good enough for government work.

Permalink Reply by Toni Rubio Soler on April 2, 2019 at 8:30am

Hi Ethan and David,

I want to thank you again for your brilliant help 2 years ago when trying to figure out the definition of a hyperbolic spiraling ramp with constant slope.

Also to share with you the result of that studies:

Is this 45 meter tall observation tower located in Denmark for Camp Adventure and designed by EFFEKT Architects. It was inaugurated last Saturday.

As you can see the ramp is a sculptural element on itself, the constant change of diameter of the hyperbolic shape, makes the ramp fluctuate on every turn making it incredibly dynamic.

Please let me know if you ever visit Copenhagen :-)

Best,

Toni

(photo: Rasmus Hjortshøj)

Permalink Reply by Ethan Gross on April 2, 2019 at 8:52am

Hi Toni,

Thanks for the follow up; I'm glad I was able to play a small role in this, and it looks great! What's even better, is that actually exists and not just on someone's hard drive. How did you arrange the supports. or rulings, so they overlap and don't intersect? That's another area where reality intrudes over theory.

Best,

Ethan Gross

ewgart.com

Permalink Reply by Toni Rubio Soler on April 2, 2019 at 9:28am

Hi Ethan,

Glad you like it!

The supporting diagrid structure is in an exterior plane while the spiraling ramp beam is offsetted to the interior. The connection between them happens every second diagrid member through "offset connections".

That makes the ramp appear as a separate element and feels almost like levitating!

Best,

Toni