鏈狀碰撞

Question

我正在通過使用鏈狀來面對新問題。兩個鏈狀體之間的碰撞不像正常人那樣發生。所以這是正常行爲，還是我爲此做了任何錯誤？

以下是用於此目的的代碼。

 ChainShape mChainShape = new ChainShape(); 
     Vector2[] mVector2 = new Vector2[lineList.size()]; 

     for (int i = 0; i < lineList.size(); i++) { 
       mVector2[i] = new Vector2(lineList.get(i).getX1() 
           /PhysicsConstants.PIXEL_TO_METER_RATIO_DEFAULT, 
           lineList.get(i).getY1() 
           /PhysicsConstants.PIXEL_TO_METER_RATIO_DEFAULT); 
     } 

     mChainShape.createChain(mVector2); 
     FixtureDef mFixtureDef = new FixtureDef(); 
     Body mChainBody; 
     BodyDef mBodyDef = new BodyDef(); 
     mBodyDef.type = BodyType.DynamicBody; 
     mChainBody = mPhysicsWorld.createBody(mBodyDef); 
     mFixtureDef.shape = mChainShape; 
     mFixtureDef.density = 1f; 
     mFixtureDef.friction = 0.5f; 
     mFixtureDef.restitution = 0.5f; 
     mChainBody.createFixture(mFixtureDef); 
     mChainShape.dispose(); 

如果我想創建基於觸摸座標的身體，那麼這件事可能與否？

請給出任何指導。

Answer 1

BOX 2D不支持鍊形狀碰撞。

但是，如果您的身體是多邊形，您可以爲此創建一個三角近似值，並從b2PolygonShape組成您的身體。

b2Vec2 triVerts[3]; 
// for this example polygon should have "center" point, from that it can be observed around. 
triVerts[0] = b2Vec2(0,0); // center 
for(int idx = 1; idx < lineList.size(); idx++) 
{ 
    b2PolygonShape triangle; 
    fixtureDef.shape = &triangle; 
    // Assumes the vertices are going around 
    triVerts[1] = mVector2[idx-1]; 
    triVerts[2] = mVector2[idx]; 
    triangle.Set(triVerts,3); 
    mChainBody.CreateFixture(&fixtureDef); 
} 

for b2PolygonShape collision will be fine fine。