Hard Math Problem!

[wossname]

I have a small universe, with the 3 standard dimensions: X,Y and Z.

Given these things:
- The (x,y,z) coordinates of the "camera"
- the direction that the camera is pointing
- the (x,y,z) coordinate of a point in space

Can someone tell me the maths expression that will resolve this info into an (x,y) coordinate, so I can plot it on a picturebox.

My brain is aching, please help!

[HarryW]

You need to use some linear algebra for this kind of thing, what you're basically looking for is a projection matrix (I think that's what it's called). I don't know too much about it myself but there are a lot of articles on this kind of translation at www.gamedev.net , since it's the sort of thing that 3D games use a lot of. www.flipcode.com has some good maths resources for this sort of stuff too.

[Zaei]

What are you programming this in? GDI, OGL, D3D, Other? If its D3D8, there is a function in the D3DX library called D3DXVec3Project(...) that will do what you want. OGL has a similar function, although i dont know what its called. Search for "Project". If its GDI, you are going to have some fun playing with vector calculus.

Z.

[wossname]

Oh dear, I knew someone would ask me that. I'm using PocketC on the PalmOS platform! All I have to work with are the standard math expressions: Cos() Atn() etc...

Its a black/white LCD display on my Palm IIIx handheld, so I'm looking for simple solutions really.

[parksie]

This may be of use: http://www.dai.ed.ac.uk/CVonline/LOC...psrc_ssaz.html

[Rebis]

Depends if the camera is always upright, or whether it rolls as well.

The maths for a rolling camera is well tricky, but I can give you the bits for upright no sweat.

Assuming x for horizontal, y for vertical & z for lateral,

Using X,Y for final screen,
x1,y1,z1 for camera pos,
x2,y2,z2 for object pos,
and ang for camera angle

X = (x2 - x1) * Cos(ang) + (z2 - z1) * Sin(ang)
Y = (y2 - y1)

This assumes no perpective

With perspective,

Z = (x2 -x2) * Cos(ang) + (z2 - z1) * Sin(ang)

X = ((x2 - x1) * Cos(ang) + (z2 - z1) * Sin(ang)) * Log(Z)
Y = (y2 - y1) * Log(Z)


This is in maths speak and not code, gotta leave some work for you to do