
September 29th, 2011, 05:04 PM
#1
rotating a point by xaxis is the formula valid for any theta
Dear Friends
I am trying to sweep a set of points by some degree theta. So I am using this formula
newx = x;
newy = y * cos(theta)  z * sin(theta);
newz = y * sin(theta) + z * cos(theta);
Is this formula valid for any degree theta.
Because when theta=360 degree, then sin(theta) = 1.47e05
Is this okay. Please give me some idea.

October 3rd, 2011, 04:10 PM
#2
Re: rotating a point by xaxis is the formula valid for any theta
Originally Posted by sujan.dasmahapatra
Because when theta=360 degree, then sin(theta) = 1.47e05
Floating point operations are never mathematically exact. You expected an answer of zero, but instead got 1.47e05, which is probably within expected errors for floating point operations.
How close to zero do you need it to be?

October 5th, 2011, 03:53 AM
#3
Re: rotating a point by xaxis is the formula valid for any theta
Last edited by nuzzle; October 5th, 2011 at 04:23 AM.

October 5th, 2011, 04:16 AM
#4
Re: rotating a point by xaxis is the formula valid for any theta
Originally Posted by sujan.dasmahapatra
Is this okay. Please give me some idea.
You should get 0.0 for sin(2.0*PI).
Maybe you aren't specifying PI with enougth precision (digits). You can try this,
const double PI = 3.141592653589793238;
Also if you're using floats try double throughout. If you're using doubles already make sure you don't have a narrowing conversion to float somewhere.
Or maybe you simply are specifying theta in degrees while sin() expects radians.
Last edited by nuzzle; October 5th, 2011 at 04:27 AM.
Posting Permissions
 You may not post new threads
 You may not post replies
 You may not post attachments
 You may not edit your posts

Forum Rules

Click Here to Expand Forum to Full Width
OnDemand Webinars (sponsored)
