相机的位置取决于型号吗？ [英] Camera position based on model size?

问题描述

在C＃WPF中如何自动根据 3D模型 尺寸设置相机 位置 >。

in C# WPF how to automatically set camera Position based on 3DModel size.

例如，如果我加载较小型号的相机，位置应更近，而如果加载较大型号的相机位置则应更长。

For example if I load small model,position be closer and if I load large model camera position be longer.

我想从模型的各个侧面获取 Max 和 Min Point3D 。

What I thought was to get the Max and Min Point3D from model from each side.

-通过X值查找最大和最小point3D。

-Find Max and Min point3D by X value.

-通过Y值查找最大和最小point3D。

-Find Max and Min point3D by Y value.

这是通过 foreach 循环遍历所有3D点来完成的。

this is done by foreach loop through all 3D points.

现在我在边缘有4个点，但是如何计算相机位置？

Now I have 4 points at the edges, but how to calculate camera position?

是否有任何公式可以获取正确的位置，并为位置增加一些额外的长度？

Is there any formula to get the right position and the add some additional length to position?

推荐答案

所有取决于您使用的框架/库，但是我想您必须使用摄像机的FOV来计算宽和高gh相机可以在一定距离内看到。

All depends on the framework / library you're using, but I'd guess you'd have to use the camera's FOV to calculate how wide and high the camera can 'see' at a certain distance.

如果您看摄像机的示意图，则FOV是摄像机看到东西的角度。它可能是Camera对象的成员变量。将三角形一分为二将为您提供两个具有已知角度（1/2 FOV）的直角三角形。基本数学应该让您计算窗口的距离和大小。

If you look at the schematic of the camera, the FOV is the angle at which the camera sees stuff. It is probably a member variable of your Camera object. Splitting the triangle in two will give you two right triangles with a know angle ( 1/2 FOV ). Basic mathematics should let you calculate the distance and size of the viewwindow.

由于您已经有4个点，因此您已经知道 L（在示意图中）应该有多大

Since you already have 4 points you already know how big 'L' (in the schematic) should be.

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