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abeylin

need an algorithm for following a path (curved line)?

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abeylin    304

I am wondering how the game https://play.google.com/store/apps/details?id=com.namcowireless.flightcontrol&hl=en makes planes follow the path drawn by player.

 

Does anyone know of a simple algorithm to do this?

 

I am thinking of a game where there are predefined paths, which is kind of similar to the game I linked, but more like this:

sample.jpg

 

So I want to go from one box to another, but following the paths. Please help. The paths would not be dynamic in my game.

 

edit (not sure if it's better to add post or update original, so doing both):

The paths will not be in terms of any function. The paths are going to represent roads from one point to another. During level creation, I'd define them as makes sense for the level.

Edited by abeylin

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TheComet    3896

What you're looking for is an appropriate interpolation algorithm, if the path you wish to follow can be represented mathematically.

 

The Bezier Curve is a good place to start. Maybe you could even use cubic interpolation.

 

If the path cannot be represented mathematically, then I suggest trying to approximate it as best as possible using the interpolation algorithms mentioned above, or create waypoints along the paths for your characters to follow. You could even use interpolation between waypoints to make transitioning ultra smooth.

Edited by TheComet

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Paragon123    620

Here is an explaination of legrange interpolation

http://mathforum.org/library/drmath/view/63984.html

 

Here's another that does a better job of describing how to actually compute the function

https://brilliant.org/assessment/techniques-trainer/lagrange-interpolation-formula/

 

 

These are slightly differen't than bezier curves, Bezier curves use control points, for which the final line does not pass through.

Legrange interpolation ensures that each point in the data set is in the final line.

 

These are good for lines where x is always increasing.

 To apply it to a line where x is not always increasing you have to convert your data set into a parameterized data set and interpolate the x and y components independently by t. 

I.E: f(t)=(fx(t),fy(t))

Get the lagrange interpolation formula for fx(t) and fy(t).

Edited by Paragon123

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abeylin    304

Hi, I didn't go through links yet, but they are greately appreciated.

 

The paths will not be in terms of any function. The paths are going to represent roads from one point to another. During level creation, I'd define them as makes sense for the level.

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Paragon123    620

So, I got pretty interested in this topic b/c I was looking to do something similar.

So, it turns out that Interpolating the x and y components separately actually made the calculation simpler because I could choose which values of T to use.

So given a data set of at least four points (x1,y1,) (x2,y2) (x3,y3) (x4,y4), we separate it 

(0,x1) (1,x2) (2,x3) (3,x4)  and (0,y1) (0,y2) (0,y3) and (0,y4)

 

The function created with four points can always be represented as F(x)=ax^3+bx^2+cx+d, so we can store the function as a set of four values (a,b,c,d). 

To get a value at T=t

double Px(Tuple<double, double, double, double> coefficents, double t)
{
  double ret = coefficents.Item1 * Math.Pow(t, 3);
  ret += coefficents.Item2 * Math.Pow(t, 2);
  ret += coefficents.Item3 * t;
  ret += coefficents.Item4;
  return ret;
}

so the full polynomail is

    y1*1/((x1-x2)(x1-x3)(x1-x4))*(x-x2)(x-x3)(x-x4)

+  y2*1/((x2-x1)(x2-x3)(x2-x4))*(x-x1)(x-x3)(x-x4)

+  y3*1/((x3-x1)(x3-x2)(x3-x4))*(x-x1)(x-x2)(x-x4)

+ y4*1/((x4-x1)(x4-x2)(x4-x3))*(x-x1)(x-x2)(x-x3)

 

But, we already choose the X(T) values (0,1,2,3) so we can calculate that part and put it as a constant

   y1*-1/6*(x-1)(x-2)(x-3)

+ y2*1/2*(x-0)(x-2)(x-3)

+ y3*-1/2*(x-0)(x-1)(x-3)

+ y4*1/6*(x-0)(x-1)(x-2)

 

And we can expand to get

  y1* -1/6 *( x^3-6x^2+11x-6)

+y2* 1/2*(x^3-5x+6x)

+y3* -1/2*(x^3-4x^2+3x

+y4* 1/6*(x^3-3x^2+2x)

 

Then distribute to get

    -(1/6)y1x^3+y1x^2 -(11/6)y1x+1y1

+ (y2/2)x^3-(5/2)y2*x^2+3y2x

+ -(y3/2)x^3+2y3x^2-(3/2)y3x

+ (y4/6)x^3-(y4/2)x^2+(y4/3)x

 

Now we can collect the terms and it leaves us with the following (v1,v2,v3, and v4 are the points to parameterize by t):

Tuple<double, double, double, double> GetCoeffInT(double v1, double v2, double v3, double v4)
{
  double a = (-v1 / 6.0) + (v2 / 2.0) + (-v3 / 2.0) + (v4 / 6.0);
  double b = v1 - (5.0 / 2.0) * v2 + 2 * v3 - v4 / 2.0;
  double c = (-11.0 / 6.0) * v1 + 3.0 * v2 - (3.0 / 2.0) * v3 + v4 / 3.0;
  double d = v1;
  return new Tuple<double, double, double, double>(a, b, c, d);
}

Now we take our four X/Y cords and convert them to two sets of co-efficents 

Tuple<Tuple<double, double, double, double>,Tuple<double, double, double, double>> GetCoefficents(Point p1, Point p2, Point p3, Point p4)
{
  Tuple<double, double, double, double> xCoeff = GetCoeffInT(p1.X, p2.X, p3.X, p4.X);
  Tuple<double,double,double,double> yCoeff=GetCoeffInT(p1.Y,p2.Y,p3.Y,p4.Y);
  return new Tuple<Tuple<double, double, double, double>, Tuple<double, double, double, double>>(xCoeff, yCoeff);
}

The last part is to draw a line of n>4 data points

double step = .01;
for (int i = 3;i < DataPoints.Count; i++) 
{
int ii = i;
  var v = GetCoefficents(DataPoints[ii-3], DataPoints[ii-2], DataPoints[ii-1], DataPoints[ii]);
  //On the first segment draw from T=0 to T=2
  //On the last segment draw from T=1 to T=3
  //On every other segment draw from T=1 to T=2
  for (double t = (i==3?0:1); t <= (i==DataPoints.Count-1?3.0:2.0); t += step) 
  {
    Point p1 = new Point((int)Px(v.Item1, t - step), (int)Px(v.Item2, t - step));
    Point p2 = new Point((int)Px(v.Item1, t), (int)Px(v.Item2, t));
    g.DrawLine(Pens.Green, p1, p2);
   }
}

Even more interesting is that is it trivial to expand this to any number of dimentions. 

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ferrous    6137
For comparison, Catmull Rom, taken from here
/// <summary>
/// Return a point on the curve between P1 and P2 with P0 and P4 describing curvature, at
/// the normalized distance t.
/// </summary>
 
public static Vector3 PointOnCurve(Vector3 p0, Vector3 p1, Vector3 p2, Vector3 p3, float t)
{
	Vector3 result = new Vector3();
 
	float t0 = ((-t + 2f) * t - 1f) * t * 0.5f;
	float t1 = (((3f * t - 5f) * t) * t + 2f) * 0.5f;
	float t2 = ((-3f * t + 4f) * t + 1f) * t * 0.5f;
	float t3 = ((t - 1f) * t * t) * 0.5f;
 
	result.x = p0.x * t0 + p1.x * t1 + p2.x * t2 + p3.x * t3;
	result.y = p0.y * t0 + p1.y * t1 + p2.y * t2 + p3.y * t3;
	result.z = p0.z * t0 + p1.z * t1 + p2.z * t2 + p3.z * t3;
 
	return result;
}


 

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Paragon123    620

What is the difference between Catmull-Rom and legrange? I read somewhere that Catmull-Rom gives you "Local control" while lerange does not, but I am not 100% certain what they mean by that? Also, The artical you pointed out mentions "The location can be calulated... assuming uniform spacing of control points", does this mean that D(p1,p2)=D(p2,p3)=D(p3,p4) is a requirement?  Is it able to create curves that double back on it's self?

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ferrous    6137

I am not sure what he means by that statement, it isn't true, the points do not need to be equidistant from one another.  I'm not sure what you mean by a curve that doubles back on itself, but Catmull-Rom is local in that if you have multiple points on your spline, if you move the 4th point, it will not affect the curve anywhere but with the lines that are attached to that point.  I'm not that familiar with legrange, but I'm guessing moving one point would require a recalculation of the entire curve?

 

(This has a nice interactive example: http://www.ibiblio.org/e-notes/Splines/cardinal.html)

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ryt    294

I would also go with Catmull-Rom. Its a cubic interpolation curve that is connected of more Hermite curves. I think its similar to Lagrange curve. You can get the same Lagrange curve by manipulating coefficients.

 

Bezier curve does not pass trough all of control points or nodes. But if you choose to go with it maybe it would be better to go with B-Splines instead or with NURBS (Non-Uniform Rational Splines)

Edited by ryt

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