Math question???

[jpummill]

Hi All,

I am trying to figure out how many objects can be seen by my camera. From a previous email that I sent out yesterday, a couple of you have pointed me at tutorials on frustum culling (Thank you).

I am still having problems with the math though and I was wondering if someone would help.

What I know:

Given A and B, I can use atan(B/A) to find the angle in radians.
Given C, and the angle, I can find A with (C * cos(angle)) and I can find B with (C * sin(angle)).

what I don't know is:
Given A, and the angle in radians, How do I find B, and C.

Ex. A = 10, angle in radians = .698132 (through trial and error, I figure out that B = appox. 8.5)

I know C = sqrt( sqr(A) + sqr(B)).

Thanks a lot for any replies,
jpummill

P.S. Sorry for being so long winded... err, long handed

[reubenhawkins]

what is A, B and C? is it a right triangle? and where is the angle? between A and C or B and C or is it elsewhere?

[satan]

http://www.gametutorials.com/Tutoria...OpenGL_Pg4.htm

good frustum culling tutorial with source code (source is only in c or c++ (nearly the same), but well documented)
it should answer you questions

[jpummill]

I don't remember the proper math terms but A would be a line on the X axis. B is perpendicular to A so it is on the Y axis. So A, and B form a right triangle.

I know A=10, and the angle is 40 degrees between A and C.

I want to find the length of B.

Any suggestions are very appreciated,
jpummill

Also, satan (I'm assuming not the one and only), I appreciate the responses you have given to both this and my previous emails. I did look up frustum culling on the web and in the Red Book. Found a lot of useful information but not an example that matched exactly what I was trying to do (I probably just didn't understand it correctly.)

I will check out the examples on gametutorials.com and see if they can explain it in simple enough terms that I can understand it (as I am still relearning all the math that I slept through in high school and college).


[This message has been edited by jpummill (edited 04-08-2002).]

[imported_JML]

I'm assuming you're talking about the edges of a triangle when you say A, B and C (given the formula C*C = A*A + B*B you are talking about a right triangle).

c
|\
| \
+--
a   b
triangle abc, angles alpha (=PI/2), beta, gamma, edges ab, ac, bc

IIRC some goniometric formula's are (PI some well known constant somewhere in the neighbourhood of 3.14 ):

sin(beta) = ac/bc
cos(beta) = ab/bc
tan(beta) = ac/ab --> beta=atan(ac/ab)
sin(gamma) = ab/bc
cos(gamma) = ac/bc
tan(gamma) = ab/ac --> gamma=atan(ab/ac)
alpha + beta + gamma = PI

BTW, I'm not sure how this helps you determining how many objects are visible to your camera...

HTH

Jean-Marc

[ioquan]

How are your objects stored internally? I assume you are storing an origin, and the vertices are relative to that. You can do a simple (not precise) test to see if an object is on screen by perspective projecting the origin and testing if its on screen. To perspective project a point in 3d space, you first translate and rotate it relative to the viewpoint, then divide the x and y coordinates by the z coordinate, multiply those values by your field of view, and add half of the screen width to the x component and half of the height to the y component. It sounds complicated, but it's actually very simple. Look up 3d perspective projection on the internet, or open up a good textbook on 3d graphics. Once you have the origin perspective projected, just see if its within the boundaries of your viewport( i.e. x < 800 && x > 0, y<600 && y>0).

[jpummill]

>>BTW, I'm not sure how this helps you determining how many objects are visible to your camera...

What I am really trying to do is draw lines and functions that "grow" across the screen on a frame by frame basis (ie. I don't want the full line to instantly apear, I want it to take a second or two to grow to its full length.)
Based on the camera position, FOV, and size of my window, I want to figure out how to start the line just off screen so I don't waste time looping to draw the pieces that are not seen.

By the way, thanks again to everyone for helping with this little problem. I have still not found exactly what I am looking for but with the leads you have given, I am sure I will soon.

jpummill

[jpummill]

I have found the basic formula that I needed from the post by JML. Here is is if anyone is interested.

To find the length of a line oposite an angle:
linelength = tan(rad(fov)/2) * distance_from_line

Thanks again to everyone who helped. I am going to do more reading on frustum culling because it appears to be an extension of what I was wanting to do.

[imported_JML]

Originally posted by jpummill:
What I am really trying to do is draw lines and functions that "grow" across the screen on a frame by frame basis (ie. I don't want the full line to instantly apear, I want it to take a second or two to grow to its full length.)
Based on the camera position, FOV, and size of my window, I want to figure out how to start the line just off screen so I don't waste time looping to draw the pieces that are not seen.

That approach will only work when you have a stationary camera position and orientation. It may be more efficient to just either cull the entire line or not, instead of trying to determine what part of the line can be drawn.

Jean-Marc