POV-Ray : Newsgroups : povray.general : Fitting a gothic trefoil into an equilateral triangle : Re: Fitting a gothic trefoil into an equilateral triangle Server Time
4 Aug 2024 14:21:55 EDT (-0400)
  Re: Fitting a gothic trefoil into an equilateral triangle  
From: LinuxLibrarian
Date: 25 Jun 2003 20:43:23
Message: <3efa41ab@news.povray.org>

> Hi Mark,
> 
> obviously *I* am dense and you found my 3 errors. The section about
> the centers should begin as follows:
> 
> [ centers ]
> Now look at the midpoint 'Center2' of a circle touching the sides
> emanating from Corner2. Due to symmetry of this circle and the triangle
> with respect to the line through Corner2 and Center2, the angle between
> this line and the x-axis is 30 degrees (60/2). Now imagine a new point:
> the mirror image 'M' of Center2 with respect to the x-axis. The triangle
>    Corner2 -- Center2 -- M
> has 30+30 degrees at Corner2 (x-axis symmetry), its other two angles are
> <continues as before>
> 
> Sorry for creating this confusion.
> 
> 
> 
>>If I make point 'M' a mirror image
>>(with respect to the x-axis) of Center1, I end up with an isosceles
>>triangle whose longest side is the vertical line Center1 -- M .  This
>>leaves me with point 'M' having a y-value which is the negative of

> 
> 
> If this "angle at Corner2" is meant to be
>    angle ( Center1, Corner2, x-axis mirror image of Center1 )

>    M1 = x-axis mirror image of Center1
> Because Center1 is inside the triangle Corner1 -- Corner2 -- Corner3,
>    angle ( Center1, Corner2, x-axis)
>    < angle ( Corner1, Corner2, x-axis)

> Now
>    "angle at Corner2"
>    = 2 * angle ( Center1, Corner2, x-axis) [x-axis symmetry]


> case the "longest" side has the same length as the other sides!
> 
> 
> Is the rest of my previous post understandable (with the above
> corrections)?
> 
>    Sputnik
> 
> 
> P.S.
> Why "INVALID_ADDRESS" at the beginning of your last post?
> 
I don't know, it's how *your* address appears in the postings on the 
web-based interface to news.povray.org (see for yourself at 
http://news.povray.org  -- maybe you never read the newsgroups that way, 
so you didn't realize your address was being (or rather, not) displayed 
that way.

BTW, I also wondered about how you solved for the height.  I quote:

"The y-coordinate is the Height of the triangle, found by the law of
pythagoras in the right-angled triangle
    top point -- bottom left point -- midpoint of base line
so we get
    Height^2 + (Side/2)^2 = Side^2
Solving for Height gives
    Height = sqrt( Side^2 + Side^2/4 ) = sqrt(3)/2*Side"

Shouldn't the last line read:
Height = sqrt( Side^2 - (Side/2)^2)  ?
Or is that really the same thing?  I'm a little fuzzy on my exponents 
where fractions & diving are concerned.

But for the rest of it, I'm going off line now to study your
"revised edition." ;-)  THanks for all your indulgence!  Sorry to be
such a persnickety pain-in-the-ass!

--Mark


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