Box2D 中的链形

ChainShape in Box2D(Box2D 中的链形)

问题描述

我最近开始学习 libgdx，但遇到了 Box2D 的 CainShape 的问题.

I recently began to learn libgdx and I am stuck at a problem with the CainShape of Box2D.

我的第一个目标是简单地创建一个带有 ChainShape 的盒子.

My first goal is to simply create a box with a ChainShape.

为了实现这一点，我将四个 Vector2 添加到一个数组中并使用它们创建一个循环.

In order to achieve that, I added four Vector2 to an array and use them to create a loop.

结果取决于阵列中的排列，要么是沙漏形的东西(左上角与右下角相连，右上角与左下角相连)或错误

The result is depending on the arrangement in the array either an hourglass shaped thing (top left is connected with bottom right and top right is connected with bottom left) or the error

表达式:b2DistanceSquared(v1, v2) > 0.005f * 0.005f

Expression: b2DistanceSquared(v1, v2) > 0.005f * 0.005f

这是我目前使用的代码:

This is the code I used so far:

Vector2[] box = new Vector2[4];

    box[1] = new Vector2(0 -    bounds.getWidth() / 2 / Main.PPM, 0 -   bounds.getHeight() / 2 / Main.PPM);


    box[0] = new Vector2(       bounds.getWidth() / 2 / Main.PPM, 0 -   bounds.getHeight() / 2 / Main.PPM);


    box[2] = new Vector2(0 -    bounds.getWidth() / 2 / Main.PPM,       bounds.getHeight() / 2 / Main.PPM);


    box[3] = new Vector2(       bounds.getWidth() / 2 / Main.PPM,       bounds.getHeight() / 2 / Main.PPM);

    ChainShape chainShape = new ChainShape();
    chainShape.createLoop(box);

    fdef.shape = chainShape;
    fixture = body.createFixture(fdef);

我希望有人能告诉我我错过了什么.
提前致谢！

I hope somebody can tell me what I am missing.
Thanks in advance!

推荐答案

调整您传递的参数并放入此代码中.这段代码肯定会跑

Adjust the parameters you are passing and put in this code. This code will definitely run

ChainShape chain = new ChainShape();

createChain().Vec2[] vertices = new Vec2[2];
vertices[0] = box2d.coordPixelsToWorld(0,150); 
vertices[1] = box2d.coordPixelsToWorld(width,150);
chain.createChain(vertices, vertices.length);

FixtureDef fd = new FixtureDef();
fd.shape = chain; 
fd.density = 1;
fd.friction = 0.3;
fd.restitution = 0.5;
body.createFixture(fd);

希望这能回答这个问题，如果不是，请询问.

Hope this answers the question if not please ask.

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