Re: Spirals and arc length
- To: mathgroup at smc.vnet.net
- Subject: [mg41340] Re: Spirals and arc length
- From: "Narasimham G.L." <google.news.invalid at web2news.net>
- Date: Thu, 15 May 2003 04:03:03 -0400 (EDT)
- References: <b9n9ki$8sl$1@smc.vnet.net>
- Reply-to: "Narasimham G.L." <mnoos1p+-a3mmathma18 at hotmail.com>
- Sender: owner-wri-mathgroup at wolfram.com
> Simply put, I wish to find the polar coordinates of a > point that has been moved along a spiral arc. First draw a differential triangle consisting of ds,dr and r dth opposite to the fundamental angle si. Divide out each side by dth to get s' , r' , and r ; tan(si)=r/r' ; sin(si)= r/s' ; this is common approach for any spiral in polar coordinates. For Logarithmic spiral b=cot(al); al=si = constant at all points; r= a E^ b.th ; r'= r. cot(al); 1/s'= sin(al) E^(-cot(al).th)/a ; integrate to get s= a sec(al)[E^(cot(al).th-1] with proper boundary condns As an exercise try sin(si)/r =1/a =1/s', integrate to get all circles through the origin. If you succeed, I believe you got what u wanted. HTH -- Posted via http://web2news.com To contact in private, remove noos1p+-a3mm