Parabola and catenary

Forum|Forum|4 years ago
May 14, 2021
3 replies
4445 views

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??

1_Catenary-Parabola.zip

Calculus_Derivatives

F

Fred_Kohlhepp

23-Emerald I

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

L

LucMeekes

23-Emerald IV

About 20...25 %.

First run over 100 iterations

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

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

Luc

V

ValeryOchkov

Author

24-Ruby IV

Is this a new math constant?

L

LucMeekes

23-Emerald IV

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

V

ValeryOchkov

Author

24-Ruby IV

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.