POV-Ray : Newsgroups : povray.animations : Trig for Lohmueller chain animation : Re: Trig for Lohmueller chain animation Server Time
20 Apr 2024 08:18:02 EDT (-0400)
  Re: Trig for Lohmueller chain animation  
From: Le Forgeron
Date: 16 Jul 2014 16:24:27
Message: <53c6df7b$1@news.povray.org>
Le 16/07/2014 22:10, Bald Eagle nous fit lire :
> 
>> Just a small point or two:
> 
> Yeah, I think Error Checking got toggled to OFF today.
> Maybe it's the heat.  Or intellectual sloth.   :O
> 
>> The expression of total length based on alpha is missing.
>> It should be
> 
> 2*sqrt[P^2 - (r1-r2)^2)]      // Twice the linear length
> +
> 2*pi*r2                       // that gives the whole circumference of circle 2
>      Should it be r2 x [(pi/2) - (2 x alpha)] ?

No, the full circumference is intended.
> +
> (pi+2*asin((r1-r2)/P))*(r1-r2)
>          =alpha * difference

I made a sign error when r1<r2, see below.
But yes, it's the difference*(half-circumference + twice adjustment of
alpha)

> 
> ------------------------------
> I worked out what you did there.
> You have a very different way of thinking than I do.
> 
> I got:
> Twice the linear length  +   Sector length of Large Pulley    +   Sector length
> of Small Pulley
> 
> 2*sqrt[P^2 - (r1-r2)^2)] + [r1 x (pi/2)] + [r1 x (2 x alpha)] + [r2 x (pi/2)] -
> [r2 x (2 x alpha)]

you have an issue with pi/2 (you were to write pi), otherwise it's ok.

> 
> which if rewritten to use (r1-r2) would give (r1 - r2) x [(pi/2) + (2 x alpha)],
> which is then subtracting [r2 x (pi/2)], so then you have to add back in twice
> that, 2 x [r2 x (pi/2)], which gives your 2 x pi x r2 term.
> 
> Included in Rev 3.0   :)
> 
>> The interesting point of that formula is that it is invariant about the
>> value of r1 & r2. Whereas the Wiki formula has to have r1 > r2 or it fails.
> 
> If I'm understanding you, that's because it gives a negative value for the
> dimension of that side of the right triangle.  Taking the absolute value of the
> difference ought to rectify that, correct?
> 
> 
> 
> 
Yes.

My bad, it should have been :

2*sqrt(P^2 - (r1-r2)^2)) +2*pi*r2 + (pi+2*asin((r1-r2)/P))*(abs(r1-r2))

-- 
IQ of crossposters with FU: 100 / (number of groups)
IQ of crossposters without FU: 100 / (1 + number of groups)
IQ of multiposters: 100 / ( (number of groups) * (number of groups))


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