POV-Ray : Newsgroups : povray.general : Math help Server Time
30 Jul 2024 08:19:40 EDT (-0400)
  Math help (Message 1 to 8 of 8)  
From: SharkD
Subject: Math help
Date: 31 Aug 2009 04:48:08
Message: <4a9b8e48$1@news.povray.org>
I have the following formula:

	tan(theta) * R1 = sin(theta) * R2

R1 and R2 are known. How do I solve for theta? Thanks!

-Mike


Post a reply to this message

From: Stephen
Subject: Re: Math help
Date: 31 Aug 2009 05:45:06
Message: <bl6n95dkdie52jvh33cjo1uporhc4rf1a5@4ax.com>
On Mon, 31 Aug 2009 04:48:08 -0400, SharkD <mik### [at] gmailcom> wrote:

>I have the following formula:
>
>	tan(theta) * R1 = sin(theta) * R2
>
>R1 and R2 are known. How do I solve for theta? Thanks!
>
>-Mike

It has been a very, very long time since  I did anything like this and I could
be far out. So don't take this for gospel ;)

tan(theta) * R1 = sin(theta) * R2

Therefore: tan(theta)  = sin(theta)  * r2/r1

But tan(theta)  = sin(theta) /cos(theta)  

Therefore: sin(theta) /cos(theta)  = sin(theta)  * r2/r1

Divide both sided by sin(theta) /cos(theta)  

1 = (sin(theta) * cos(theta) *r2)/(r1 * sin(theta) )

1 = cos(theta)  *r2/r1

Therefore: cos(theta)  = r2/r1

-- 

Regards
     Stephen


Post a reply to this message

From: scott
Subject: Re: Math help
Date: 31 Aug 2009 05:57:30
Message: <4a9b9e8a@news.povray.org>
> 1 = cos(theta)  *r2/r1
> 
> Therefore: cos(theta)  = r2/r1

You messed up at the very last step :-)

Easier if at this step:

sin(theta)/cos(theta)  = sin(theta)  * r2/r1

You just cancel the sin(theta) term on each side, leaves you with:

1/cos(theta) = r2/r1

Taking the reciprocal of both sides then leaves you with the answer.


Post a reply to this message

From: Stephen
Subject: Re: Math help
Date: 31 Aug 2009 06:06:36
Message: <c28n951ifrhp35atk14vjunlm97naboq5t@4ax.com>
On Mon, 31 Aug 2009 11:57:28 +0200, "scott" <sco### [at] scottcom> wrote:

>
>You messed up at the very last step :-)

LOL I inverted the "r1/r2" when I was checking.
I said it was a long time ago.

But it at least got someone to check it ;)
-- 

Regards
     Stephen


Post a reply to this message

From: SharkD
Subject: Re: Math help
Date: 31 Aug 2009 06:37:26
Message: <4a9ba7e6$1@news.povray.org>
scott wrote:
> You just cancel the sin(theta) term on each side, leaves you with:
> 
> 1/cos(theta) = r2/r1
> 
> Taking the reciprocal of both sides then leaves you with the answer.
> 

Thank you! I forgot that tan = sin/cos.

-Mike


Post a reply to this message

From: Jellby
Subject: Re: Math help
Date: 31 Aug 2009 11:40:11
Message: <vm4tm6-30n.ln1@badulaque.unex.es>
Among other things, SharkD saw fit to write:

>> You just cancel the sin(theta) term on each side, leaves you with:
>> 
>> 1/cos(theta) = r2/r1
>> 
>> Taking the reciprocal of both sides then leaves you with the answer.
>> 
> 
> Thank you! I forgot that tan = sin/cos.

You all missed the last step:

theta = arccos(R1/R2)    ;-)


-- 
light_source{9+9*x,1}camera{orthographic look_at(1-y)/4angle 30location
9/4-z*4}light_source{-9*z,1}union{box{.9-z.1+x clipped_by{plane{2+y-4*x
0}}}box{z-y-.1.1+z}box{-.1.1+x}box{.1z-.1}pigment{rgb<.8.2,1>}}//Jellby


Post a reply to this message

From: "Jérôme M. Berger"
Subject: Re: Math help
Date: 31 Aug 2009 13:15:31
Message: <4a9c0533@news.povray.org>
SharkD wrote:
> I have the following formula:
> 
>     tan(theta) * R1 = sin(theta) * R2
> 
> R1 and R2 are known. How do I solve for theta? Thanks!
> 
	theta = 0 ;)

-- 
mailto:jeb### [at] freefr
http://jeberger.free.fr
Jabber: jeb### [at] jabberfr


Post a reply to this message


Attachments:
Download 'us-ascii' (1 KB)

From: Slime
Subject: Re: Math help
Date: 1 Sep 2009 01:08:32
Message: <4a9cac50$1@news.povray.org>
> Easier if at this step:
>
> sin(theta)/cos(theta)  = sin(theta)  * r2/r1
>
> You just cancel the sin(theta) term on each side, leaves you with:
>
> 1/cos(theta) = r2/r1

However, now that you've divided both sides by sin(theta), you're only 
solving the equation over values where sin(theta) is not 0. You have to 
check that possibility separately, and it looks like the equation is 
trivially true when sin(theta) = 0, or when theta = n*pi. (As well as the 
solution you gave.)

 - Slime
 [ http://www.slimeland.com/ ]


Post a reply to this message

Copyright 2003-2023 Persistence of Vision Raytracer Pty. Ltd.