Is this code correct ?

[johnnash]

In my project I need to shoot rays from the source.I know you cannot
follow all the rays from the source as you can actually shoot rays at
infinite angles,therefore my plan is to launch rays from the
trasmitter in all directions spaced eg. x degrees or something. This
is the logic I have used for a C program -

consider the source to be centre of the unit sphere.

0<=theta<=180(zenith)
0<=phi<=360 (azimuth)

Then run a for loop like this -

for( theta = 0 ; theta <=180; theta = theta + 5 ) /* taking some 5
degree difference or there will be infinite rays */
for( phi = 0; phi <=360; phi = phi + 5 )
{
/*Calculate the cartesian coordinates of a point p(a struct) on
sphere*/
P.x = r*sin(theta)*cos(phi);

P.y = r * sin(theta) *sin(phi);

P.z = r * cos(theta);

direction[index] = VectorSubtract(P, Center_of_Sphere);

VectorNormalize(direction[index]);

/*direction is just a list that will store ray directions */

[Micah Cowan]

johnnash wrote:

On Feb 28, 11:55 pm, Eric Sosman <[e-mail]:

>johnnash wrote:

>>for( phi = 0; phi <=360; phi = phi + 5 )

> 2) Your phi steps from 0 degrees up to and including
360 degrees. Since 0 and 360 are the same angle, each
zenith has one ray that's shot twice.

Thank you very much. However I did not understand the 2nd point.

There are 360 degrees to a circle. 0 through 360 is 361. Shooting at phi
== 0 and again at phi == 360 is exactly the same as shooting at phi == 0
twice.

You want that for statement to begin like:
for (phi = 0; phi < 360; phi += 5)
^^^
rather than with <=.

--
Micah J. Cowan
Programmer, musician, typesetting enthusiast, gamer...
http://micah.cowan.name/

[Coos Haak]

Op Thu, 28 Feb 2008 10:31:20 -0800 (PST) schreef johnnash:

In my project I need to shoot rays from the source.I know you cannot
follow all the rays from the source as you can actually shoot rays at
infinite angles,therefore my plan is to launch rays from the
trasmitter in all directions spaced eg. x degrees or something. This
is the logic I have used for a C program -

consider the source to be centre of the unit sphere.

0<=theta<=180(zenith)

0<=theta<=90 is enough.
--
Coos

[Thad Smith]

johnnash wrote:

In my project I need to shoot rays from the source.I know you cannot
follow all the rays from the source as you can actually shoot rays at
infinite angles,therefore my plan is to launch rays from the
trasmitter in all directions spaced eg. x degrees or something. This
is the logic I have used for a C program -

consider the source to be centre of the unit sphere.

0<=theta<=180(zenith)
0<=phi<=360 (azimuth)

Then run a for loop like this -

for( theta = 0 ; theta <=180; theta = theta + 5 ) /* taking some 5
degree difference or there will be infinite rays */
for( phi = 0; phi <=360; phi = phi + 5 )
{
/*Calculate the cartesian coordinates of a point p(a struct) on
sphere*/

Be aware that the distribution you get with this approach will not be
uniform. You will have a hot spot at 90 degrees zenith darkening as your
angle approaches 0 and 180 degrees.

I am not a graphics programmer, but I think to get a uniform distribution,
you should aim each vector at an equal-size portion of a sphere at a given
distance. You could do this, for each zenith angle, by going around a
circle on a unit sphere at equal spacing.

--
Thad

[johnnash]

On Feb 29, 10:29 am, Thad Smith <[e-mail]:

johnnash wrote:

In my project I need to shootraysfrom the source.I know you cannot
follow all theraysfrom the source as you can actually shootraysat
infinite angles,therefore my plan is to launchraysfrom the
trasmitter in all directions spaced eg. x degrees or something. This
is the logic I have used for a C program -

consider the source to be centre of the unit sphere.

0<=theta<=180(zenith)
0<=phi<=360 (azimuth)

Then run a for loop like this -

for( theta = 0 ; theta <=180; theta = theta + 5 ) /* taking some 5
degree difference or there will be infiniterays*/
for( phi = 0; phi <=360; phi = phi + 5 )
{
/*Calculate the cartesian coordinates of a point p(a struct) on
sphere*/

Be aware that the distribution you get with this approach will not be
uniform. You will have a hot spot at 90 degrees zenith darkening as your
angle approaches 0 and 180 degrees.

I am not a graphics programmer, but I think to get a uniform distribution,
you should aim each vector at an equal-size portion of a sphere at a given
distance. You could do this, for each zenith angle, by going around a
circle on a unit sphere at equal spacing.

--
Thad

I thought what would actually happen over here is that I would get a
thicker density of rays near the poles as opposed to when zenith
becomes 90

[johnnash]

There are a number of ways to approach this. One would be to scale the
steps in phi by the sine of theta,

for ( phi = 0; phi < 360; phi += 5.0 / sin( theta )) ...

The poles, where theta = 0 or 180, are treated as a special case, since
there's no need to test different increments of phi.

why increment phi in steps of sin(theta) ?? if we do this, then when
zenith is 90, the rays will far apart whereas near the poles, they
would be bundled closely and there will be a very thick density.

[Ernie Wright]

johnnash wrote:

>>There are a number of ways to approach this. One would be to scale the
steps in phi by the sine of theta,

for ( phi = 0; phi < 360; phi += 5.0 / sin( theta )) ...

The poles, where theta = 0 or 180, are treated as a special case, since
there's no need to test different increments of phi.

why increment phi in steps of sin(theta) ?? if we do this, then when
zenith is 90, the rays will far apart whereas near the poles, they
would be bundled closely and there will be a very thick density.

It depends on how you define your angles. If sine doesn't work, you can
use cosine (which is actually more common).

- Ernie http://home.comcast.net/~erniew

[johnnash]

On Mar 5, 7:09 pm, Ernie Wright <[e-mail]:

johnnash wrote:

>There are a number of ways to approach this. One would be to scale the
steps in phi by the sine of theta,

for ( phi = 0; phi < 360; phi += 5.0 / sin( theta )) ...

>The poles, where theta = 0 or 180, are treated as a special case, since
there's no need to test different increments of phi.

why increment phi in steps of sin(theta) ?? if we do this, then when
zenith is 90, the rays will far apart whereas near the poles, they
would be bundled closely and there will be a very thick density.

It depends on how you define your angles. If sine doesn't work, you can
use cosine (which is actually more common).

- Ernie http://home.comcast.net/~erniew

ok i think i understood. may be what you are trying to say is that the
rays tend to bundled along the poles(i.e. theta = 0 and 180) so what
we do is we increment phi by greater values near pole. because when
theta is closer to zero or 180, the sine value will be very low
resulting in a very high value of 5 / sin(theta) ?? but at theta = 90
which represents the equatior, increment is only 5 degree..