当给出开始点、结束点、速度和移动开始时间时，有没有更简单的方法来计算 2D 中的对象当前坐标？

Question

抱歉，几何从来不是我在学校最喜欢的科目，所以...

目前我正在编写 2D MMORPG 服务器模拟器，想改进我当前的坐标检测算法。

客户端向服务器发送带有 start point coords 和 finish point coords 的数据包。服务器接收它并将 movement start 变量分配给 DateTime.Now。服务器知道角色在 1 秒内可以移动多远（即 speed）。如何计算他开始移动后 some 秒后的角色位置？

因为我不擅长几何，所以我根据行进距离的百分比编写了这个……垃圾……。正在运行，但是计算量太大了。

        public ushort X;
        public ushort Y;
        public ushort TargetX;
        public ushort TargetY;

        public DateTime MovementStartTime;
        public ushort CalculatedX
        {
            get
            {
                if (this.X == this.TargetX && this.Y == this.TargetY)
                    return this.X;

                float total_dst = CF.GetEuclideanDistanceBetween(this.X, this.Y, this.TargetX, this.TargetY);
                float seconds_since_movement_start = (float)(DateTime.Now - this.MovementStartTime).TotalSeconds;
                float passed_dst = Limits.CharacterSpeed * seconds_since_movement_start;

                if (passed_dst > total_dst)
                    return this.TargetX;

                float passed_dst_in_normalized_percents = (passed_dst * 100 / total_dst) / 100;

                return (ushort)(this.X - ((this.X - this.TargetX) * passed_dst_in_normalized_percents));
            }
        }

        public ushort CalculatedY
        {
            get
            {
                if (this.X == this.TargetX && this.Y == this.TargetY)
                    return this.Y;

                float total_dst = CF.GetEuclideanDistanceBetween(this.X, this.Y, this.TargetX, this.TargetY);
                float seconds_since_movement_start = (float)(DateTime.Now - this.MovementStartTime).TotalSeconds;
                float passed_dst = Limits.CharacterSpeed * seconds_since_movement_start;

                if (passed_dst > total_dst)
                    return this.TargetY;

                float passed_dst_in_normalized_percents = (passed_dst * 100 / total_dst) / 100;

                return (ushort)(this.Y - ((this.Y - this.TargetY) * passed_dst_in_normalized_percents));
            }
        }

想法是根据它的速度确定已经通过了多少百分比的线字符，然后将它与start X和end X之间的平坦距离结合起来。

也许有人知道一种不那么丑陋的方法？..

Answer 1

物体匀速运动时，X-coordinate对时间的依赖性为

X = X0 + Vx * t

其中 Vx 是速度的 x-component：

Vx = CharacterSpeed * dx / sqrt(dx*dx + dy*dy)

其中 dx, dy 是沿轴的距离。在你的情况下是这样的：

dx = this.TargetX - this.StartX

Answer 2

正如我在评论中提到的，一个简单的向量 class 将大大简化这一过程。如果您使用的是更高版本的 .Net，那么您可以在 System.Numerics 中使用 Vector2 class。如果你不这样做，那么你可以只添加我在这个 post.

末尾包含的简单向量 class (Vec2d)

然后您可以像这样将两个属性重写为一个属性：

Vec2d calculatedXY
{
    get
    {
        Vec2d source = new Vec2d((float)this.X, (float)this.Y);
        Vec2d target = new Vec2d((float)this.TargetX, (float)this.TargetY);

        Vec2d diff = target - source;

        if(source == target)
            return source;

        float total_dst = diff.length;
        float seconds_since_movement_start = (float)(DateTime.Now - this.MovementStartTime).TotalSeconds;
        float passed_dst = Limits.CharacterSpeed * seconds_since_movement_start;

        if(passed_dst > total_dst)
            return target;

        return source + ((diff/total_dst) * passed_dst);
    }
}

如果您需要，这里是 Vec2d 结构（也可以写成 class）：

public struct Vec2d
{
    public float X;
    public float Y;

    public Vec2d(float x_, float y_)
    {
        X = x_;
        Y = y_;
    }

    // calculated properties:
    public float length
    {
        get { return (float)Math.Sqrt(X * X + Y * Y); }
    }


    // class operator definitions:

    // equality
    public static bool operator ==(Vec2d left, Vec2d right)
    {
        return (left.X == right.X) && (left.Y == right.Y);
    }
    public static bool operator !=(Vec2d left, Vec2d right)
    {
        return !(left == right);
    }

    // addition
    public static Vec2d operator +(Vec2d left, Vec2d right)
    {
        return new Vec2d(left.X + right.X, left.Y + right.Y);
    }
    // subtraction
    public static Vec2d operator -(Vec2d left, Vec2d right)
    {
        return new Vec2d(left.X - right.X, left.Y - right.Y);
    }

    // unary negation
    public static Vec2d operator -(Vec2d value)
    {
        return new Vec2d(-value.X, -value.Y);
    }

    // scalar multiplication
    public static Vec2d operator *(float scalar, Vec2d value)
    {
        return new Vec2d(scalar * value.X, scalar * value.Y);
    }
    public static Vec2d operator *(Vec2d value, float scalar)
    {
        return new Vec2d(scalar * value.X, scalar * value.Y);
    }

    // multiplication (vector dot product, NOT scalar dot product)
    public static Vec2d operator *(Vec2d left, Vec2d right)
    {
        return new Vec2d(left.X * right.X, left.Y * right.Y);
    }

    // scalar division
    public static Vec2d operator /(Vec2d value, float scalar)
    {
        return new Vec2d(value.X / scalar, value.Y / scalar);
    }
}

当给出开始点、结束点、速度和移动开始时间时，有没有更简单的方法来计算 2D 中的对象当前坐标？

Any less ugly way to calculate object current coords in 2D when start pt, end pt, speed and movement start time are given?

c#

geometry

linear-algebra