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在x秒内预测刚体对象的位置

rigid-bodies unity3d c#

Programmer · 技术社区 · 9 年前

Rigidbody Rigidbody.AddForce 或 Rigidbody.velocity

Extrapolation 但在这种情况下,几乎不可能使用一些公式来获得物体在 x

Physics.autoSimulation Physics.Simulate 来解决这个问题。对于2D物理,这是 Physics2D.autoSimulation 和 Physics2D.Simulate 若为false,则调用 物理。模拟 作用

刚体 将在 4 1 5 更重要的是 Simulate 精确的太离谱了。

为什么会发生这种情况,我该如何解决?这个问题在Android设备上更严重。

Unity 2017.2.0b5

下面是我正在使用的示例代码。这个 guide GameObject只是用来显示/显示预测的位置。

public GameObject bulletPrefab;
public float forceSpeed = 50;

public GameObject guide;

// Use this for initialization
IEnumerator Start()
{
    //Disable Physics AutoSimulation
    Physics.autoSimulation = false;

    //Wait for game to start in the editor before moving on(NOT NECESSARY)
    yield return new WaitForSeconds(1);

    //Instantiate Bullet
    GameObject obj = Instantiate(bulletPrefab);

    Rigidbody bulletRigidbody = obj.GetComponent<Rigidbody>();

    //Calcuate force speed. (Shoot towards the x + axis)
    Vector3 tempForce = bulletRigidbody.transform.right;
    tempForce.y += 0.4f;
    Vector3 force = tempForce * forceSpeed;

    //Addforce to the Bullet
    bulletRigidbody.AddForce(force, ForceMode.Impulse);

    //yield break;
    //Predict where the Rigidbody will be in 4 seconds
    Vector3 futurePos = predictRigidBodyPosInTime(bulletRigidbody, 4f);//1.3f
    //Show us where that would be
    guide.transform.position = futurePos;
}

Vector3 predictRigidBodyPosInTime(Rigidbody sourceRigidbody, float timeInSec)
{
    //Get current Position
    Vector3 defaultPos = sourceRigidbody.position;

    Debug.Log("Predicting Future Pos from::: x " + defaultPos.x + " y:"
        + defaultPos.y + " z:" + defaultPos.z);

    //Simulate where it will be in x seconds
    Physics.Simulate(timeInSec);

    //Get future position
    Vector3 futurePos = sourceRigidbody.position;

    Debug.Log("DONE Predicting Future Pos::: x " + futurePos.x + " y:"
        + futurePos.y + " z:" + futurePos.z);

    //Re-enable Physics AutoSimulation and Reset position
    Physics.autoSimulation = true;
    sourceRigidbody.velocity = Vector3.zero;
    sourceRigidbody.useGravity = false;
    sourceRigidbody.position = defaultPos;

    return futurePos;
}

2 回复 | 直到 9 年前

PassetCronUs 9 年前

你甚至幸运地发现 1 根本不起作用。你 不应该 0.03 到 Physics.Simulate Physics2D.Simulate

当该值高于 0.03 ,你必须把它切成碎片,然后使用 Simulate x 检查时间是否仍然大于或等于 Time.fixedDeltaTime 应该这样做。

代替

Physics.Simulate(timeInSec);

具有

while (timeInSec >= Time.fixedDeltaTime)
{
    timeInSec -= Time.fixedDeltaTime;
    Physics.Simulate(Time.fixedDeltaTime);
}

predictRigidBodyPosInTime 函数应该如下所示:

Vector3 predictRigidBodyPosInTime(Rigidbody sourceRigidbody, float timeInSec)
{
    //Get current Position
    Vector3 defaultPos = sourceRigidbody.position;

    Debug.Log("Predicting Future Pos from::: x " + defaultPos.x + " y:"
        + defaultPos.y + " z:" + defaultPos.z);

    //Simulate where it will be in x seconds
    while (timeInSec >= Time.fixedDeltaTime)
    {
        timeInSec -= Time.fixedDeltaTime;
        Physics.Simulate(Time.fixedDeltaTime);
    }

    //Get future position
    Vector3 futurePos = sourceRigidbody.position;

    Debug.Log("DONE Predicting Future Pos::: x " + futurePos.x + " y:"
        + futurePos.y + " z:" + futurePos.z);

    //Re-enable Physics AutoSimulation and Reset position
    Physics.autoSimulation = true;
    sourceRigidbody.velocity = Vector3.zero;
    sourceRigidbody.useGravity = false;
    sourceRigidbody.position = defaultPos;

    return futurePos;
}

thatGuyDown 4 年前

{
Vector3 PredictionPos(Rigidbody _TargetRb, float _predictiontime){

//get the rigidbodies velocity
Vector3 _targvelocity = _TargetRB.velocity;
//multiply it by the amount of seconds you want to see into the future
_targvelocity *= _predictiontime;
//add it to the rigidbodies position
_targvelocity += _TargetRB.position;
//Return the position of where the target will be in the amount of seconds you want to see into the future
Return _targvelocity;
}
}