沿X和Z轴旋转相机时出现问题

Question

我正在实现一个能够响应鼠标位置变化的相机。 与编码相比，这是一个数学问题，但我也想知道如何使用它。

我有一个Camera对象，当鼠标更改其X位置时，该对象沿Y轴旋转。 这可以按预期工作，并且可以绕着我正在绘制的立方体旋转。 现在，我想实现由鼠标垂直触发的向上和向下查找，但是X和Z轴是相对于相机对象的，所以我不仅可以沿X轴旋转，还必须结合X和Z轴以流畅的动作做到这一点。

public class Camera {

    public float moveSpeed = 0.05f;

    private Vector3f position, rotation;
    private float oldMouseX, oldMouseY, newMouseX, newMouseY, mouseSensitivity;

    public Camera () {
        position = new Vector3f(0f, 0f, 0f);
        rotation = new Vector3f(0f, 0f, 0f);

        mouseSensitivity = 0.1f;
        oldMouseX = 0.0f;
        oldMouseY = 0.0f;
        newMouseX = 0.0f;
        newMouseY = 0.0f;
    }

    public Camera (Vector3f pos, Vector3f rot) {
        this.position = pos;
        this.rotation = rot;

        mouseSensitivity = 0.1f;
        oldMouseX = 0.0f;
        oldMouseY = 0.0f;
        newMouseX = 0.0f;
        newMouseY = 0.0f;
    }

    public void setCursor (int x, int y) {
        oldMouseX = x;
        oldMouseY = y;
        newMouseX = x;
        newMouseY = y;
    }

    public Matrix4f getViewMatrix () {
        Matrix4f rotateX = new Matrix4f().rotate(rotation.x * (float)Math.PI / 180f, new Vector3f(1f, 0f, 0f));
        Matrix4f rotateY = new Matrix4f().rotate(rotation.y * (float)Math.PI / 180f, new Vector3f(0f, 1f, 0f));
        Matrix4f rotateZ = new Matrix4f().rotate(rotation.z * (float)Math.PI / 180f, new Vector3f(0f, 0f, 1f));

        Matrix4f rotation = MatrixMath.mul(rotateX, MatrixMath.mul(rotateZ, rotateY));

        Vector3f negPosition = new Vector3f(-position.x, -position.y, -position.z);
        Matrix4f translation = new Matrix4f().translate(negPosition);

        return MatrixMath.mul(translation, rotation);
    }
            public Vector3f getPosition() {
        return position;
    }

    public Vector3f getRotation() {
        return rotation;
    }

    public void update (Window window) {

        if (window.isKeyDown(GLFW.GLFW_KEY_W)) {
            position.x += Math.sin(Math.PI * rotation.y / 180) * -moveSpeed;
            position.z += Math.cos(Math.PI * rotation.y / 180) * moveSpeed;
        }

        if (window.isKeyDown(GLFW.GLFW_KEY_S)) {
            position.x -= Math.sin(Math.PI * rotation.y / 180) * -moveSpeed;
            position.z -= Math.cos(Math.PI * rotation.y / 180) * moveSpeed;
        }

        if (window.isKeyDown(GLFW.GLFW_KEY_D)) {
            position.x += Math.sin(Math.PI * (rotation.y - 90) / 180) * -moveSpeed;
            position.z += Math.cos(Math.PI * (rotation.y - 90) / 180) * moveSpeed;
        }

        if (window.isKeyDown(GLFW.GLFW_KEY_A)) {
            position.x -= Math.sin(Math.PI * (rotation.y - 90) / 180) * -moveSpeed;
            position.z -= Math.cos(Math.PI * (rotation.y - 90) / 180) * moveSpeed;
        }

        if (window.isKeyDown(GLFW.GLFW_KEY_SPACE)) {
            addPosition(0f, moveSpeed, 0f);
        }

        if (window.isKeyDown(GLFW.GLFW_KEY_LEFT_SHIFT)) {
            addPosition(0f, -moveSpeed, 0f);
        }

        newMouseX = (float)window.getMouseX();
        newMouseY = (float)window.getMouseY();

        float dx = newMouseX - oldMouseX;
        float dy = newMouseY - oldMouseY;

        if (window.isMouseButtonDown(GLFW.GLFW_MOUSE_BUTTON_LEFT)) {
            rotation.y += dx * mouseSensitivity;
        }

        //unPos = unPos.rotateAxis(dy * mouseSensitivity, (float)Math.cos(Math.PI * rotation.y / 180), 0f, (float)Math.sin(Math.PI * rotation.y / 180));

//      rotation.x += (float)Math.cos(rotation.y * Math.PI / 180) * (dy * mouseSensitivity);
//      rotation.z += (float)Math.sin(rotation.y * Math.PI / 180) * (dy * mouseSensitivity);

        oldMouseX = newMouseX;
        oldMouseY = newMouseY;
    }
}

我认为没有必要向您展示我的Window类，因为这些功能是不言而喻的。 如您所见，我在底部注释掉的部分是我解决问题的方法，乍一看似乎可行，但轮换略有偏离。

我期望流体上下运动（即相对于相机），但是会收到奇怪的滚动运动。

任何帮助是极大的赞赏！

Answer 1

我解决了我的问题。 这很奇怪，但是我必须将Y旋转矩阵乘以X旋转矩阵。 这对我来说没有意义，但可以。 谢谢你的帮助！