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Parabola and catenary

ValeryOchkov
24-Ruby I

Parabola and catenary

There are three random points on the plane through which an arc of a parabola and a similar arc of a catenary pass. What is the probability that the length of the parabola's arc will be greater than the length of the catenary's arc??

Catenary-Parabola.png

11 REPLIES 11

Since a catenary describes a condition of minimum potential energy, I would guess that the probability is quite low.

About 20...25 %.

First run over 100 iterationsFirst run over 100 iterations

No, it's more towards 30 %. The next run is over 200 iterations:

LM_20210514_CatenaryParabole200.gif

And finally over 1000 iterations, but showing only 1 in 5:

LM_20210514_CatenaryParabole1000.gif

Luc

Is this a new math constant?

 

Nah... don't think so.

You would have something if you could prove that a catenary line through 3 points is always x % longer than a Parabola line through the same three points. Since that is not the case....Keep searching.

 

I've got a puzzle for you: try to find what my select_dhm() function is... (The function I use to provide guess values for the solve block of finding the parameters for the catenary function)

 

Success!

Luc

what my select_dhm() function is...

rnd, error and cycle?

 

No, it's not trial and error. I have an algorithm (a simple formula) that spits out a single set of guess values, with a success rate of over 99 % (that using those guess values, the solve block will find the catenary function parameters. If you look carefully at the animations, you can see the miss shots every now and then...).

That's better than a Covid19 vaccine...

 

Luc

Rnd, on error and cycle (trial and error) - 99.999%

That's better than a Sputnik V vaccine...

000100000100010000-0000011010000-00---110000100010000100-011001-0101101-01100000--1101--110101-010-010000101-001-010-0-0001-1-00000--0-10011101000-0-0-00-01010000-0-01-10-0001000-010100-0-00001101011100000110110111--0101-01-101010---000000001000100-000001100111000-10 (Tomorrow I will continue my research)

1 - Sc > Sp

0 - Sc < Sp

- (hiccup) - no solution

* - Sc = Sp (I am waiting for)

It's like a hiccup. You don't know when you will hiccup next time - when next time there will be no solution of the problem of catenaries and parabolas.
By the way, there is a very interesting study of the frequency and strength (- or ---) of hiccups.

Catenary longer than a parabola about (a bit more) in 26 cases out of a hundred.

p-c.png

With prorgramming

p-c-Prog.png

p-c-Prog-3.png

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