棘手的问题，关于投影变换

snhun

对了，还少了一些。如果
分辨率是 800*600 ，
屏幕宽高比是 16:10 = 1.6，
视野是 60度，那么

s=
(
400+ 400*x/ [z*tan(30度)*1.6]
,
300- 300*y/ [z*tan(30度)]
)

liurunfeng

Projects a 3D vector from object space into screen space.

D3DXVECTOR3 * D3DXVec3Project(
  D3DXVECTOR3 * pOut,
  CONST D3DXVECTOR3 * pV,
  CONST D3DVIEWPORT9 * pViewport,
  CONST D3DXMATRIX * pProjection,
  CONST D3DXMATRIX * pView,
  CONST D3DXMATRIX * pWorld
);
Parameters
pOut
[in, out] Pointer to the D3DXVECTOR3 structure that is the result of the operation.
pV
[in] Pointer to the source D3DXVECTOR3 structure.
pViewport
[in] Pointer to a D3DVIEWPORT9 structure, representing the viewport.
pProjection
[in] Pointer to a D3DXMATRIX structure, representing the projection matrix.
pView
[in] Pointer to a D3DXMATRIX structure, representing the view matrix.
pWorld
[in] Pointer to a D3DXMATRIX structure, representing the world matrix.
Return Values
Pointer to a D3DXVECTOR3 structure that is the vector projected from object space to screen space.

Remarks
The return value for this function is the same value returned in the pOut parameter. In this way, the D3DXVec3Project function can be used as a parameter for another function.

Requirements
Header: Declared in D3dx9math.h.

See Also
D3DXVec3Unproject

以上摘自DirectX文档

lights

I 服了you ，dx里有一个直接的方法取。
这不光是投影矩阵要参与，视矩阵和世界矩阵都乘上。
D3DXVECTOR3 * D3DXVec3Project(
  D3DXVECTOR3 * pOut,
  CONST D3DXVECTOR3 * pV,
  CONST D3DVIEWPORT9 * pViewport,
  CONST D3DXMATRIX * pProjection,
  CONST D3DXMATRIX * pView,
  CONST D3DXMATRIX * pWorld
);
(不好意思，没注意楼上的兄弟已经发了)

jjjyes

13楼正解，可以看下MSDN，关于D3DXProject的函数，同理还有D3DXUnproject，但是具体用法似乎不像楼上说的那么简单，算出的点还需要自己考量一下。

Chiyer

D3DXVec3Project
得到的应该是你在 backbuffer 里的位置，如果你的 backbuffer 大小跟窗口不一致，那你还需要做相应的变换，

bracelet

简单得很，乘上视矩阵和投影矩阵不就完了

tonykee

我用纯几何方法已经算出来了。呵呵。。。。还不知道，DX里面有这个函数？呵呵。。
谢谢各位

千里马肝

有种顶点格式叫做XYZRHW

gb

呵呵