在 Unity 2D 自上而下游戏中击退玩家 [英] Knocking back the Player in a Unity 2D TopDown Game

问题描述

当有东西击中玩家时，可能是敌人或尖刺等，我希望他被击退到与敌人相反的方向.

When something hits the Player, maybe an enemy or spikes etc., I want him to get knocked back in the oppisite direction of the enemy.

所以我已经得到了:

public void ChangeHealth(float healthToAdd, Vector2 objectPosition) // Change the players health, objectPosition = enemies position or something else
    {
        if (healthToAdd < 0) // incoming damage
        {
            // ... other stuff

            Knockback(objectPosition);
        }

        // ... other stuff
    }

    void Knockback(Vector2 objectPosition) // Knockback routine
    {
        Vector2 knockbackPosition = new Vector2( , ); // calculation is missing here! Calculate the new position by the knockback direction

        rigid.MovePosition(Vector2.MoveTowards(playerPos, knockbackPosition, 2 * Time.deltaTime)); // the knock back
    }

    private void Update() // !! TEST !!
    {
        if (Input.GetKeyDown(KeyCode.E)) // TEST routine
        {
            ChangeHealth(-7, new Vector2(10,10)); // decrease players health by 7 and knock him back
        }
    }

缺少什么:

Vector2 knockbackPosition = new Vector2( , );

我正在寻找像这张图所示的计算:

I am looking for a calculation like this picture is showing:

推荐答案

Vector2 knockbackPosition = 
    transform.position + (transform.position - objectPosition).normalized *a

<小时>

要理解为什么它等于你必须通读.

---

To understand why is it equal to that you have to read through.

三个点:E、P、K(敌人、玩家、击退)

There are three points: E,P,K (Enemy, Player, Knockback)

还有一个标量:a(这个值越大，你的击退就越多)

And one scalar number: a (the greater this value is, the more knockback you'll have)

现在从你的照片:

PK = EP*a

将向量扩展为两点之间的距离:

expand vectors into distances between two points:

(K-P) = (P-E)*a

计算K的位置:

K = P + (P-E)*a

不过有一个问题.(感谢 Rotem)使用这个公式:

There is one problem though. (thanks to Rotem) with this formula as it is:

你会期待距离更近的对手更大的击退.

you'd expect a bigger knockback from a closer opponent.

我们不希望击退取决于 P 和 E 之间的距离.

We don't want the knockback to be dependent on the distance between P and E.

为了消除对前者的依赖，在乘以 a 之前对向量进行归一化

To remove the dependency on the former, normalize the vector before multiplying by a

所以我们添加 .normalized 以仅使用 (P-E) 的方向而不是它的原始向量

So we add .normalized to use just the direction of (P-E) instead of its original vector

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