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我正在尝试创建一个定位器网格，这些定位器用作从 Maya 中的相机在指定深度的平行有限平面上的投影点。网格应与指定的分辨率对齐，以匹配渲染输出。

目前我的计算已经关闭，我正在寻找一些帮助来确定我用于确定投影点的公式是如何不正确的。

我有一个自包含的 python 脚本和图像，显示了作为示例生成的定位器的当前位置。

显示当前生成的定位器在 y 和 z 轴上关闭的图像

import maya.cmds as mc
import maya.OpenMaya as om

res = [mc.getAttr('defaultResolution.width'), 
        mc.getAttr('defaultResolution.height')]

print res
grid = [5, 5]    


def projectedGridPoint(camera, coord, depth, res):


    selList = om.MSelectionList()
    selList.add(camera)
    dagPath = om.MDagPath()
    selList.getDagPath(0,dagPath)
    dagPath.extendToShape()
    camMtx = dagPath.inclusiveMatrix()

    fnCam = om.MFnCamera(dagPath)
    mFloatMtx = fnCam.projectionMatrix()
    projMtx = om.MMatrix(mFloatMtx.matrix)

    #center of camera
    eyePt = fnCam.eyePoint()

    #offset position
    z = eyePt.z - depth

    #calculated xy positions
    x = (2 * z * coord[0] / res[0]) - z
    y = (2 * z * coord[1] / res[1]) - z

    return om.MPoint(x,y,depth) * camMtx * projMtx.inverse()

for y in range(grid[1] + 1):
    for x in range(grid[0] + 1):
        coord = ( x / float(grid[0]) * res[0], y / float(grid[1]) * res[1] )
        pt = projectedGridPoint('camera1', coord, 10, res)

        mc.spaceLocator(a=1, p=[pt.x, pt.y, pt.z])
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2 回答 2

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一旦我调整了 Theodox 的答案以考虑所有可能的网格划分，使得 ndc_x 和 ndc_y 始终在 -1 和 1 的范围内。我能够得到一个可行的解决方案。

import maya.api.OpenMaya as om
import maya.cmds as cmds


def projectedGridPoint(camera, coord, depth):
    selList = om.MGlobal.getSelectionListByName(camera)
    dagPath = selList.getDagPath(0)
    dagPath.extendToShape()
    view = dagPath.inclusiveMatrix()

    fnCam = om.MFnCamera(dagPath)
    projection = om.MMatrix(fnCam.projectionMatrix())

    viewProj =  projection * view 


    r =    om.MPoint(coord[0],coord[1], -1 * depth)  * projection.inverse()
    return r.homogenize() * view


xx, yy = (6, 6) 

for y in range(yy + 1):
    for x in range(xx + 1):
        ndc_x = -1
        ndc_y = -1

        if x > 0:
            ndc_x = (x / float(xx) * 2) - 1
        if y > 0:    
            ndc_y = (y / float(yy) * 2) - 1

        coord = ( ndc_x, ndc_y)
        print coord
        pt = projectedGridPoint('camera1', coord, 0)

        c,_ = cmds.polyCube(w = 0.1, d = 0.1, h = 0.1)
        cmds.xform(c, t = (pt[0], pt[1], pt[2]))
于 2018-05-04T03:16:53.117 回答
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我认为您想要更像这样的东西（注意，我将其转换为 API 2 以减少样板）

    import maya.api.OpenMaya as om
    import maya.cmds as cmds


    def projectedGridPoint(camera, coord, depth):
        selList = om.MGlobal.getSelectionListByName(camera)
        dagPath = selList.getDagPath(0)
        dagPath.extendToShape()
        view = dagPath.inclusiveMatrix()

        fnCam = om.MFnCamera(dagPath)
        projection = om.MMatrix(fnCam.projectionMatrix())

        viewProj =  projection * view 


        r =    om.MPoint(coord[0],coord[1], -1 * depth)  * projection.inverse()
        return r.homogenize() * view


    xx, yy = (2, 2) 

    for y in range(yy):
        for x in range(xx):

            ndc_x =  2.0 * x  / float(xx - 1) - 1
            ndc_y =  2.0 * y  / float(yy - 1) - 1
            coord = ( ndc_x, ndc_y)
            pt = projectedGridPoint('camera1', coord,0)

            c,_ = cmds.polyCube(w = 0.1, d = 0.1, h = 0.1)
            cmds.xform(c, t = (pt[0], pt[1], pt[2]))

坐标以标准化设备坐标的形式提供（从 -1,-1 到 1, 1 在视图的角落），深度从近到远剪辑平面 - 深度 1 正好在近平面上，并且深度为 0 位于远平面上。我认为在实践中我会将深度锁定在 0 并使用相机上的剪辑平面设置来设置深度

编辑 我合理化了将索引值转换为 NDC 坐标的原始、不稳定的方法

于 2018-05-04T02:32:22.163 回答