PDA

View Full Version : How to find the length of helix


mranji1
Nov 10, 2010, 12:43 AM
A helix is defined as,
x(s)=a(s), y(s)=-R(s) cos(2*pi*np*s/l), z(s)=R(s) sin(2*pi*np*s/l) where 0.le.R(s).le.Rh and is given by,
R(s)=Rh[1/pi *arctan(beta*(s/l -gama))+0.5] where beta and gama are constants and np is the no.of turns on the helix. Find
the expression for a(s) such that the tangent vector [x'(s),y'(s),z'(s)] has unit length




Where the amplitude satisfies and is given by,

Where and are constants and is the no. of turns in a helix.

Find the expression for such that the tangent vector has unit length.

harum
Nov 12, 2010, 11:23 AM
Are you sure that x(s) = a(s) and not a*s? In general, the length L is the integral over the parameter s of this function: sqrt (x'(s)^2 + y'(s)^2 + z'(s)^2).

mranji1
Nov 12, 2010, 09:41 PM
It just means x(s) is a function of s. I used the above formula u mentioned and I tried. But the final integration becoming complex since here R(s) is a function of (s). Can u help me in that... Thanks for your comment

harum
Nov 12, 2010, 10:05 PM
Are you sure you need to calculate the helix length? There is no mention of helix length here. What is asked here is to find a(s) such that [x'(s),y'(s),z'(s)] is a unity vector. You said your integration was complex; what a(s) did you use for integration? Read you original post -- it hasn't come completely through. The last sentences are with gaps and meaningless.

If you select a(s) such that [x'(s),y'(s),z'(s)] has unit length everywhere then sqrt( (dx/ds)^2 + (dy/ds)^2 + (dz/ds)^2 ) = 1 irrespective of parameter s. Then your integration becomes trivial.