点投影到yx旋转平面(point projection into yx rotated plane)

点投影到yx旋转平面(point projection into yx rotated plane)

我想在2D空间中模拟深度，如果我有一个点P1，我想我需要将给定点P1投影到平面x轴顺时针旋转“theta”rads，得到P1'

似乎P1'.x coord必须与P1.x和P1'相同。它必须比P1.y短。 在3D世界中：

cosa = cos(theta) sina = sin(theta) P1'.x = P1.x P1'.y = P1.y * cosa - P1.z * sina P1'.z = P1.y * sina + P1.z * cosa

我的P1.z = 0吗？ 我试过了，P1'.y = P1.y * cosa没有按预期结果

任何回应将不胜感激，谢谢！

编辑：我想要的，现在我旋转相机并翻译矩阵

编辑2：具有start1点和end1点的单行示例（它是水平线，只要倾斜角度增加，预期结果是到“floor”的下降线）

我认为这是一个符号错误或需要的偏移量（java画布图（0,0）位于左上角），因为我的新线倾斜度为0，是一个低于所有值且值为90º的新线和原来的一个匹配

I want to simulate depth in a 2D space, If I have a point P1 I suppose that I need to project that given point P1 into a plane x axis rotated "theta" rads clockwise, to get P1'

It seems that P1'.x coord has to be the same as the P1.x and the P1'.y has to b shorter than P1.y. In a 3D world:

cosa = cos(theta) sina = sin(theta) P1'.x = P1.x P1'.y = P1.y * cosa - P1.z * sina P1'.z = P1.y * sina + P1.z * cosa

Is my P1.z = 0? I tried it and P1'.y = P1.y * cosa doesn't result as expected

Any response would be appreciated, Thanks!

EDIT: What I want, now I rotate camera and translate matrix

EDIT 2: an example of a single line with a start1 point and a end1 point (it's an horizontal line, result expected is a falling line to the "floor" as long as tilt angle increases)

I think it's a sign error or an offset needed (java canvas drawing (0,0) is at top-left), because my new line with a tilt of 0 is the one below of all and with a value of 90º the new line and the original one match

最满意答案

如果要围绕x轴顺时针旋转，则执行的计算是正确的。 如果您将线条视为一张纸，则直接观察线条时旋转0度。

对于示例，您已经给出了与x轴水平的线。 这不会改变围绕x轴的旋转（线和它旋转的轴彼此平行）。 当你在0到90度之间旋转时，线的y坐标将随着P1.y * cos（theta）在90度下降到0而减小（考虑一下我们围绕它的底边旋转的那张纸， x轴，90度纸张是平的，y轴垂直于页面，因此页面的两个边缘具有相同的y坐标，两侧是“x轴”和相反的平行方将有y = 0）。

因此，您可以看到您的示例，这已正常工作。

编辑：乘以90度的原因并不能给出完全零答案，只是浮点舍入

The calculation you are performing is correct if you would like to perform a rotation around the x axis clockwise. If you think of your line as a sheet of paper, a rotation of 0 degrees is you looking directly at the line.

For the example you have given the line is horizontal to the x axis. This will not change on rotation around the x axis (the line and the axis around which it is rotating are parallel to one another). As you rotate between 0 and 90 degrees the y co-ordinates of the line will decrease with P1.y*cos(theta) down to 0 at 90 degrees (think about the piece of paper we have been rotating around it's bottom edge, the x axis, at 90 degrees the paper is flat, and the y axis is perpendicular to the page, thus both edges of the page have the same y co-ordinate, both the side that is the "x-axis" and the opposite parallel side will have y=0).

Thus as you can see for your example this has worked correctly.

EDIT: The reason that multiplying by 90 degrees does not give an exactly zero answer is simply floating point rounding