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What is the formula for finding the height of a circle perpendicular to a
given point on the radius?
Thank you,
-Shay
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On Wed, 18 Apr 2001 15:57:26 -0500, Shay wrote:
>What is the formula for finding the height of a circle perpendicular to a
>given point on the radius?
Your question doesn't make any sense. Can you draw a picture?
--
Ron Parker http://www2.fwi.com/~parkerr/traces.html
My opinions. Mine. Not anyone else's.
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Ron Parker <ron### [at] povray org> wrote in message
news:slr### [at] fwicom...
> On Wed, 18 Apr 2001 15:57:26 -0500, Shay wrote:
> >What is the formula for finding the height of a circle perpendicular to a
> >given point on the radius?
>
> Your question doesn't make any sense. Can you draw a picture?
>
> --
> Ron Parker http://www2.fwi.com/~parkerr/traces.html
> My opinions. Mine. Not anyone else's.
Post a reply to this message
Attachments:
Download 'drawing.jpg' (8 KB)
Preview of image 'drawing.jpg'
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On Wed, 18 Apr 2001 16:49:10 -0500, Shay wrote:
>
>Ron Parker <ron### [at] povrayorg> wrote in message
>news:slr### [at] fwicom...
>> On Wed, 18 Apr 2001 15:57:26 -0500, Shay wrote:
>> >What is the formula for finding the height of a circle perpendicular to a
>> >given point on the radius?
>>
>> Your question doesn't make any sense. Can you draw a picture?
>
>begin 666 drawing.jpg
I guess I kinda asked for that one, didn't I? The picture should have gone
in .binaries.images, but since it's small, I won't complain too loudly.
--
Ron Parker http://www2.fwi.com/~parkerr/traces.html
My opinions. Mine. Not anyone else's.
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On Wed, 18 Apr 2001 15:57:26 -0500, Shay wrote:
>What is the formula for finding the height of a circle perpendicular to a
>given point on the radius?
Ah. The picture helps. But we also need to know what you do know.
Obviously if you knew the location of the center and the location of
the point, the answer would be trivial, so you must have something
else in mind.
So... what information do you have?
--
Ron Parker http://www2.fwi.com/~parkerr/traces.html
My opinions. Mine. Not anyone else's.
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Shay wrote:
>
> [Image]
Umm... Perhaps
2*sin(acos(Distance/Radius))*Radius
where Distance is the distance of the point from the centre.
This is provided I don't get my sin and cos mixed up as per usual.
--
Margus Ramst
Personal e-mail: mar### [at] peakeduee
TAG (Team Assistance Group) e-mail: mar### [at] tagpovrayorg
Home page http://www.hot.ee/margusrt
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nothing more complex than that. I do know the location of the point and of
the center. This is an easy question with an easy answer, just one with
which I am not familiar. What I have in mind is an arrangement of objects
beneath a dome, but not touching it. I am going to test out Margus Ramst's
formula. I'm sure he got it right. Thank you.
-Shay
Ron Parker <ron### [at] povrayorg> wrote in message
news:slr### [at] fwicom...
> On Wed, 18 Apr 2001 15:57:26 -0500, Shay wrote:
> >What is the formula for finding the height of a circle perpendicular to a
> >given point on the radius?
>
> Ah. The picture helps. But we also need to know what you do know.
> Obviously if you knew the location of the center and the location of
> the point, the answer would be trivial, so you must have something
> else in mind.
>
> So... what information do you have?
>
> --
> Ron Parker http://www2.fwi.com/~parkerr/traces.html
> My opinions. Mine. Not anyone else's.
Post a reply to this message
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On Wed, 18 Apr 2001 19:50:32 -0500, Shay and/or Carrie Ann wrote:
>nothing more complex than that. I do know the location of the point and of
>the center. This is an easy question with an easy answer, just one with
>which I am not familiar. What I have in mind is an arrangement of objects
>beneath a dome, but not touching it. I am going to test out Margus Ramst's
>formula. I'm sure he got it right. Thank you.
If you know the location of the point and the location of the center, the
height is just Point.y-Center.y for a circle. For a dome, you can use
sqrt( Radius^2 - vlength(<Point.x,Point.y,0>-<Center.x,Center.y,0>)^2 )
--
Ron Parker http://www2.fwi.com/~parkerr/traces.html
My opinions. Mine. Not anyone else's.
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Margus Ramst wrote:
>
> 2*sin(acos(Distance/Radius))*Radius
You're right.
Though when the sin is >0, sin(acos x) = sqrt(1-x^2).
So a prettier formula would be :
2*sqrt(Radius^2-Distance^2)
--
JM
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Shay wrote:
> What is the formula for finding the height of a circle perpendicular to a
> given point on the radius?
sqrt((radius)^2 - (distance_from_center)^2)
--
David Fontaine <dav### [at] faricynet> ICQ 55354965
My raytracing gallery: http://davidf.faricy.net/
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