沿其方向将 3D(x,y,z) 转换为 2D(x,y)（正交）
我尽可能地浏览了互联网上所有可用的学习资源，这些资源都是简单方程、向量或三角方程的形式。 我找不到执行以下操作的方法：
假设 Y 在 3D 世界中。 我需要为 3D 轨迹正交绘制两个 2D 轨迹（不是投影），例如，XY 平面作为轨迹的侧视图相对于轨迹本身和 XZ 平面对于相同的顶视图。
我有 3D 轨迹的所有 3D 点、初始速度、两个角度都可以通过向量数学计算。
我应该如何进一步进行？
参考： 在不同角度的曲线下方，如果沿 XY 平面投影，则可能会失去其意义。我想要的只是沿着自身转换红色曲线，沿着绿色曲线转换绿色曲线等等。以及进一步我如何将侧视图映射到飞机上。顶视图相对简单，只需获取每个点的 X 和 Z 坐标即可完成。
我的意思是这个要求。 :)
I have gone through all available study resources in the internet as much as possible, which are in form of simple equations, vectors or trigonometric equations.
I couldn't find the way of doing following thing:
Assuming Y is up in a 3D world.
I need to draw two 2D trajectories orthogonally (not the projections) for a 3D trajectory, say XY-plane for side view of the trajectory w.r.t. the trajectory itself and XZ-plane for top view for the same.
I have all the 3D points of the 3D trajectory, initial velocity, both the angles can be calculated by vector mathematics.
How should I proceed further?
refer:
Below a curve in different angles, which can loose its significance if projected along XY-plane. All I want is to convert the red curve along itself, the green curve along green curve and so on. and further how would I map side view to a plane. Top view is comparatively easy and done just by taking X and Z ordinates of each points.
I mean this the requirement. :)
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
我认为我不明白这个问题，但无论如何我都会回答我的解释。
您有一个由一系列点 p0, ..., pN 描述的 3D 轨迹。我们给定一个平行于 Y 轴的平面 P 的角度 v，并希望计算点 p< 的 2D 坐标 (di, hi) sub>i 投影到该平面上，其中 hi 是 Y 方向的高度坐标，di 是 < em>距离 v 方向上的坐标。假设 p0 = (0, 0, 0) 或从所有向量中减去 p0。
设 pi = (xi, yi, zi)。高度坐标为 hi = yi。假设角度 v 是相对于 Z 轴给出的。 v 方向的向量为 r = (sin(v), 0, cos(v))，距离坐标变为 di = dot(pi ，r）。
I don't think I understand the question, but I'll answer my interpretation anyway.
You have a 3D trajectory described by a sequence of points p0, ..., pN. We are given an angle v for a plane P parallel to the Y-axis, and wish to compute the 2D coordinates (di, hi) of the points pi projected onto that plane, where hi is the height coordinate in the direction Y and di is the distance coordinate in the direction v. Assume p0 = (0, 0, 0) or else subtract p0 from all vectors.
Let pi = (xi, yi, zi). The height coordinate is hi = yi. Assume the angle v is given relative to the Z-axis. The vector for the direction v is then r = (sin(v), 0, cos(v)), and the distance coordinates becomes di = dot(pi, r).