blocks|key|1529599|text|除非你选择的子弹在一定范围内引爆(离物体的距离很重要)，那么你只需要你与飞机的交点来确定你的角度。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1529600|限制你的光标躺在一个垂直于你看方向的平面上，离你有一段距离。然后从枪管出口的位置到光标的位置，计算出你的角度。|1529601|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Unless your bullet of choice detonates at some range (so distance to the thing is important) then you only need the intersection point of your boresight with a plane to determine your angles.

Constrain your cursor to lie in a plane that's perpendicular to your look direction, and some distance away from you. Then take the displacement from the location of the exit of the gun barrel to the location of the cursor, and calculate your angles from there.

blocks|key|1886067|text|简单的答案是，渲染场景的一部分涉及到将世界坐标转换为查看坐标--基本上，相对于我的眼球而言，所有的内容都在哪里。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1886068|视图平面是屏幕，这就是鼠标所处的位置。如果你的鼠标坐标是(xx，yy)，那么它可能位于(xx，yy，1)或(xx，yy，-1)的视图坐标。您可能需要使用一个系统，其中(0，0)是您的视口在屏幕上的中心-不是鼠标协调通常的工作方式，但不变的偏移是很容易的。|1886069|zz=0是玩家所在的位置，而视图平面是前面、玩家和他能看到的东西之间的一段短距离。这就是为什么视图平面被作为(xx，yy，1)或(xx，yy，-1)在这里-如果你使用%2B1+(我认为是正常情况)，并且视图坐标为正z的物体就在你前面。|1886070|若要将此转换回世界协调，请应用相同的转换，将您的世界协调器转换为查看协调，但向后转换。然后你得到你的鼠标指针的位置在世界协调。要得到方向矢量，减去玩家的位置。|1886071|向后转换在每一步都使用逆矩阵，并向后应用步骤。原则上，所有步骤都可以组合成一个矩阵，并且可以为整个变换计算一个逆矩阵--但这不是最好的方法。例如，计算旋转的逆比计算一般矩阵的逆要容易得多。如果您从世界协调到查看协调的转换是根据需要从组件组成的(位置、方向.)您只需要计算反转组件并将它们组合起来。而不是30度左(从世界到视图)，你使用30度-右(从视图到世界)等。|1886072|如果你幸运的话，你可以完全放弃玩家位置的整个问题--计算鼠标指针在视图空间中的角度(欧拉角就足够了)，然后只应用视点到世界转换的方向部分。|1886073|不，这是公式，抱歉，部分原因是世界对世界的转换取决于你如何进行从世界到视图的转换，部分原因是我没有一个简单的例子。|1886074|虽然问题不同，但你可以在这里找到一些相关的参考资料.|1886075|矩阵乘法-视图/投影、世界/投影等|offset|length|1886076|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1283865/matrix-multiplication-view-projection-world-projection-etc^0|0|0|0|0|0|0|0|0|0|H|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|13|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|14|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|15|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|16|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|17|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|18|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|19|8|@]|9|@$R|1A|S|1B|1|1C]]|A|$]]|$1|T|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|U|$V|$5|W|X|Y|A|$Z|10]]]]

The simple answer is that part of rendering a scene involves converting world co-ordinates to view co-ordinates - where's all that stuff relative to my eyeballs, basically.

The view plane is the screen, which is what the mouse sits on. If your mouse co-ordinates are (xx, yy) then it is probably positioned at either (xx, yy, 1) or (xx, yy, -1) in view co-ordinates. You'll probably need to use a system where (0, 0) is the centre of your viewport on screen - not how mouse co-ordinates normally work, but constant offsets are easy enough.

zz=0 is where the player is, whereas the view plane is a short distance in front, between the player and what he can see. That's why the view plane is taken as either (xx, yy, 1) or (xx, yy, -1) here - if you use +1 (the normal case, I think) objects with view co-ordinates that have positive z are in front of you.

To convert this back to world co-ordinates, apply the same transform you do to transform your world co-ordinates to view co-ordinates, but backwards. You then get your mouse pointer position in world co-ordinates. To get the direction vector, subtract your player position.

The backwards conversion uses the inverse matrix at each step, and applies the steps backwards. In principle, all steps can be combined into a single matrix, and an inverse matrix can be calculated for the whole transform - but that's not the best approach. It's much easier, for instance, to calculate the inverse of a rotation than to calculate a general matrix inverse. If your transform from world co-ordinates to view co-ordinates is composed as needed from components (postion, orientation...) you really only need to calculate inversed components and combine them. Instead of 30-degrees-left (from world to view) you use 30-degrees-right (from view to world) etc.

If you're lucky, you might be able to drop the whole issue of player position completely - calculate the angle for the mouse pointer in view space (Euler angles are enough for this), then apply the orientation part only of the view-to-world transform.

No here's-the-formula, sorry, partly because the view-to-world transform depends on how you do the world-to-view transform and partly because I don't have an easy example to hand.

Though the question is different, you can find some relevant references in the answers here...

<a href="https://stackoverflow.com/questions/1283865/matrix-multiplication-view-projection-world-projection-etc">Matrix multiplication - view/projection, world/projection, etc</a>

blocks|key|333298|text|如果你认为你的整个场景适合一个金字塔，你的眼睛在顶峰(面向底部)，你的屏幕是一个横截面平行于底部，整个场景是投影(到2D空间)。想必，你知道屏幕的尺寸，以及光标在2D空间的位置。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|333299|你可以从X面看那个金字塔，然后从Y边再看一遍，把问题变成两个直角三角形。三角形的底部由从“眼睛”到屏幕中心的直线定义，低音由从“眼睛”到光标的直线定义。|333300|当光标位于2D显示的极端角时，您可以知道角度(给定的透视信息)，因此您可以确定“眼睛”离2D投影平面(或三角形底部的长度)有多远(以像素为单位)。使用这个长度，以及三角形的另一边的长度(光标的x坐标或y坐标)，您可以确定角度。所有这些都使用基本的trig。|333301|如果有帮助，可以画出所有相关的三角形：|333302|由(0,+0)；(base,+0)；(base,+Y_MAX)定义的三角形，其中Y_MAX是光标(屏幕上边缘)的最大值。base+=+Y/tan(max_angle)|unordered-list-item|offset|length|style|CODE|333303|由(0,+0)；(base,+0)；(base,+X_MAX)定义的三角形。|333304|由(0,+0)；(base,+0)；(base,+Y)定义的三角形，其中Y是光标的y位置.angle+=+(tan%5E(-1))(Y/base)|333305|由(0,+0)；(base,+0)；(base,+X)定义的三角形。|333306|entityMap^0|0|0|0|0|1|6|8|9|I|D|14|5|1O|N|0|1|6|8|9|I|D|0|1|6|8|9|I|9|10|1|19|Q|0|1|6|8|9|I|9|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Z|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@$K|11|L|12|M|N]|$K|13|L|14|M|N]|$K|15|L|16|M|N]|$K|17|L|18|M|N]|$K|19|L|1A|M|N]]|9|@]|A|$]]|$1|O|3|P|5|J|7|1B|8|@$K|1C|L|1D|M|N]|$K|1E|L|1F|M|N]|$K|1G|L|1H|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|J|7|1I|8|@$K|1J|L|1K|M|N]|$K|1L|L|1M|M|N]|$K|1N|L|1O|M|N]|$K|1P|L|1Q|M|N]|$K|1R|L|1S|M|N]]|9|@]|A|$]]|$1|S|3|T|5|J|7|1T|8|@$K|1U|L|1V|M|N]|$K|1W|L|1X|M|N]|$K|1Y|L|1Z|M|N]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|20|8|@]|9|@]|A|$]]]|V|$]]

If you consider your entire scene to fit within a pyramid, with your eye at the peak (looking toward the base), your screen is a cross-section parallel to the base, on which the entire scene is projected (into 2D space). Presumably, you know the dimensions of the screen, and the location of the cursor in 2D-space.

You can look at that pyramid from the X side, and then again from the Y side, to turn the problem into two right triangles. The base of the triangle is defined by the line from the "eye" to the center of the screen, and the hypotenuse is defined by the line from the "eye" to the cursor.

With the cursor at the extreme corners of the 2D display, you know the angle (given perspective information), so you can determine how far the "eye" is from the 2D projection plane (or the length of the base of your triangles), in pixels. Using that length, and the length of the other side of the triangle (the x- or y-coordinate of the cursor), you can determine the angle. All of this uses basic trig.

If it helps, draw out all the triangles in question:

<ul>
<li>The triangle defined by <code>(0, 0)</code>; <code>(base, 0)</code>; <code>(base, Y_MAX)</code> where <code>Y_MAX</code> is the maximum value for the cursor (the top edge of the screen). <code>base = Y/tan(max_angle)</code></li>
<li>The triangle defined by <code>(0, 0)</code>; <code>(base, 0)</code>; <code>(base, X_MAX)</code>.</li>
<li>The triangle defined by <code>(0, 0)</code>; <code>(base, 0)</code>; <code>(base, Y)</code>, where <code>Y</code> is the y-position of the cursor. <code>angle = (tan^(-1))(Y/base)</code></li>
<li>The triangle defined by <code>(0, 0)</code>; <code>(base, 0)</code>; <code>(base, X)</code>.</li>
</ul>

blocks|key|1529730|text|gluUnProject|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1529731|entityMap|0|LINK|mutability|MUTABLE|url|http://www.opengl.org/sdk/docs/man/xhtml/gluUnProject.xml^0|0|C|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

<a href="http://www.opengl.org/sdk/docs/man/xhtml/gluUnProject.xml" rel="nofollow noreferrer">gluUnProject</a>

I only need to know which angle (x,y,z vector) my cursor is pointing at. So i can shoot any particle in the direction i am pointing at.

I cant use depth testing, so the typical picking methods are not an option.

Is this possible with pure math only? if i know the view port aspect ratio and perspective etc?

Edit: I need to get the coordinate from empty scene. Totally empty, no drawing at all!

OpenGL: How to calculate the angle where my mouse is pointing at in the 3d space?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我只需要知道我的光标指向哪个角度(x，y，z矢量)。所以我可以朝我所指的方向发射任何粒子。我不能使用深度测试，所以典型的采摘方法不是一种选择。只有数学才能做到这一点吗？如果我知道视图，端口，纵横比和透视等？编辑：，我需要从空场景中得到坐标。完全是空的，一点都不画！

问OpenGL:如何计算我的鼠标在3d空间中指向的角度？
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问OpenGL:如何计算我的鼠标在3d空间中指向的角度？EN