Calculating position to position B with speed

Question

I am trying to find out a formula that calculates the actual position of a player using his speed and a destination position.

These are the informations I have:

• Vector3 (X,Y,Z) : Current position
• Vector3 (X,Y,Z) : Destination position
• float : Speed

I can manage to calculate the angle between these two vectors, but I can't find an efficient formula to calculate the exact position using the speed at each loop. Do you have an idea?

EDIT*

I finally found a formula, but it's not really working as I want...

private void WalkNew()
{
    float speed = 0.1f;
    float distX = this.DestinationPosition.X - this.Position.X;
    float distZ = this.DestinationPosition.Z - this.Position.Z;
    float distance = (float)Math.Sqrt((distX * distX) + (distZ * distZ)); // distance between A and B positions

    float time = distance / speed;
    float currentTime = Time.GetTickFrom(this.lastMoveTime);
    this.lastMoveTime = Time.GetTick();
    float percentage = currentTime / time;

    if (this.Position.IsInCircle(this.DestinationPosition, 0.1f))
    {
        // We arrived
    }
    else
    {
        this.Position.X += distX / percentage;
        this.Position.Z += distZ / percentage;
    }
}

Any thoughts about this formula ?

Thanks

Answer 1

You subtract the current position from the destination position, this way you get a vector pointing to the destination.

You need to normalize this vector (divide it with it's length) and multiple with the speed. You get a velocity vector this way.

Finally you add this to the current position to get the new position.

In pseudo code:

currPos += normalize(dest - currPos) * speed

This way you don't need to uses that many costy function, such as atan2, cos or sin, only a square root.