Counting the distance traveled horizontally

Unity Engine
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1

There are many similar topics (like this one), but I have found nothing about this specific part - horizontally.

I need to simulate a spending some travel points, but according to slopes angle. So, I have a normal points spending when I am traveling on a surface completely horizontal, and additional penalty spending when I am climbing on a slope. Therefore, I need to calculate it separately:

  1. Calculate the distance traveled horizontally.
  2. Calculate a penalty according to angle of the surface, which I am traveling on.

The first one is what I am interested in now. So, my approach is the following, with saving the last position:

Vector3 lastPosition;

void Update()
{
	var difference = transform.position - lastPosition;
	var horizontalProjection = Vector3.ProjectOnPlane(difference, Vector3.up);
	var traveledHorizontally = horizontalProjection.magnitude;

	lastPosition = transform.position;
}

And it seems like it works - traveledHorizontally is the actual path traveled horizontally every frame. The question is - is there any more compact and cheap calculations, or perhaps even a built-in solution?

I tried distance, but this is actually the distance in 3D space, not horizontally. And I can’t get a projection from this distance, because it is scalar.

var distance = Vector3.Distance(transform.position, lastPosition);

It seems like this distance is the magnitude of the difference between two vectors as in my approach, but I can’t get access to this vector to project it on a horizontal plane.

I tried also a magnitude of some horizontal movement acquired from the old input system, but the magnitude of the resulting vector is too big - it definitely not the actual distance traveled horizontally.

var horizontalMovement = transform.forward * Input.GetAxis("Vertical");
horizontalMovement += transform.right * Input.GetAxis("Horizontal");
var traveledHorizontally = horizontalMovement.magnitude;
2

Project on the plane is the technically correct answer and will correctly handle any normal.

An optimization for flatness: If you have a Vector3 that is the difference between this and last frame’s positions, just flatten it on the y and sum its magnitude:

Vector3 distanceThisFrame = currentPosition - previousPosition;

distanceThisFrame.y = 0;

totalDistanceFloat += distanceThisFrame.magnitude;
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3

just flatten it on the y
distanceThisFrame.y = 0;

Thank you. I just didn’t realize this. This trick works for projections to the planes of the main axes only, isn’t it?

4

Correct. The Vector3.Project() method is just handy matrix math to give you it in any arbitrary plane.

While running that project-on-plane method is unlikely to ever be an actual bottleneck (you would need to have a LOT of these things, and if you have a lot of them, it’s likely that something else would bottleneck you sooner than some extra math), but I do appreciate the simplicity of setting y to zero if you only care about x and z :slight_smile:

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