在 x 秒内预测刚体对象的位置

Question

假设您有一个 Rigidbody 移动的对象。通过 Rigidbody.AddForce 或 Rigidbody.velocity 向此对象添加力。该对象可以滚动撞击另一个对象并改变方向。

我知道 Extrapolation 但在这种情况下，几乎不可能使用某些公式在 x 秒内获得对象的位置，因为对象可以击中另一个对象并在此过程中更改speed/direction。

引入了 Unity 2017 Physics.autoSimulation and Physics.Simulate to solve this problem. For 2D physics, that is Physics2D.autoSimulation and Physics2D.Simulate。我所做的只是首先将 Physics.autoSimulation 设置为 false 然后调用 Physics.Simulate 函数。

在我的示例中，我想知道 Rigidbody 在对其施加力后 4 秒内的位置，它似乎在像 1 这样的微小秒内工作正常。问题是，当我将 5 及以上的更大数字传递给 Simulate 函数时，预测位置不准确。它的路途遥远。

为什么会发生这种情况，我该如何解决？此问题在 Android 台设备上更为严重。

我当前的Unity版本是Unity 2017.2.0b5.

下面是我正在使用的示例代码。 guide 游戏对象仅用于 display/show 预测位置。

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;
}

Answer 1

您甚至很幸运 1 的值完全有效。您不应该 将任何高于 0.03 的值传递给 Physics.Simulate 或 Physics2D.Simulate 函数。

当值高于0.03时，您必须将其分成几部分，然后在循环中使用Simulate函数。减少 x 时间，同时检查它是否仍然大于或等于 Time.fixedDeltaTime 应该这样做。

替换

Physics.Simulate(timeInSec);

和

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

您新的完整 predictRigidBodyPosInTime 函数应如下所示：

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;
}

Answer 2

我知道这是一个迟到的回复，但是，以防万一有人想知道如何解决这个问题。所以，我阅读了关于 rigidbody.velocity 的文档，如果你将它添加到刚体的位置，你将在 1 秒内得到刚体的位置，所以如果你得到速度，将它乘以你想要的秒数see in the future 然后你将它添加到刚体的位置你将在 x 秒内得到刚体所在的位置，像这样

{
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;
}
}

在 x 秒内预测刚体对象的位置

Predict the position of a Rigidbody Object in x second

c#

unity3d

rigid-bodies