Need Math! Got Point A and B Need....
Author
I have a game with sound and I need to figure out right and left volume for the sound source to the players camera.
Ok.. In 3D space I just need 2D cordinates (x,y) of point 1 and 2. I can figure out the distance of the two points.
sqr((x1 - x2) ^ 2 + (y1 - y2) ^ 2) = distance between X and Y
What I need to do is with point 1 and 2 I have to figure out witch side is closer. What would be the best way??? Get a right and left point next to the players location, then calc. the distance from the R and L points to the sound location and figure out the difference for the sound balance?
- Joe
Have you looked into using DirectSound? I believe it handles 3D sounds pretty good. Even implementing things like the doplar effect for you.
In 3D space the distance is sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)...
To get the right volume, calculate the unit vector from the player to the sound source, then calculate the scalar product x of the unit right vector of the player with this player-sound source vector. Then the right volume should be (1 + x)/2 (if 1 is the whole volume), the left volume is 1 - right volume.
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GA
Edited by - ga on June 13, 2000 6:09:13 PM
To get the right volume, calculate the unit vector from the player to the sound source, then calculate the scalar product x of the unit right vector of the player with this player-sound source vector. Then the right volume should be (1 + x)/2 (if 1 is the whole volume), the left volume is 1 - right volume.
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GA
Edited by - ga on June 13, 2000 6:09:13 PM
Author
I am using DX7a. I didn''t see anything about this in Direct Sound. But I''m a newbi. 
Well, you should avoid the sqr function at all times. If you ONLY need to find which one is closer, you simply do this
if (x1*x1+y1*y1) < (x2*x2+y2*y2) then
p1 is closer
else
p2 is closer
end if
See how easy it was to just avoid the sqr function? Well, I got this from "The good-looking textured light-sourced bouncy fun smart and stretchy page."
------------------------
Captured Reality.
if (x1*x1+y1*y1) < (x2*x2+y2*y2) then
p1 is closer
else
p2 is closer
end if
See how easy it was to just avoid the sqr function? Well, I got this from "The good-looking textured light-sourced bouncy fun smart and stretchy page."
------------------------
Captured Reality.
Author
There is a 3D sample in the DX7 SDK. But its in VC++ and I''m working in VB6 
Don''t know much about VC.
Don''t know much about VC.
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