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Limit the max derivate of 1D Perlin Noise

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#1 Makers_F   Members   

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Posted 15 April 2012 - 07:15 AM

I've already wrote the code to produce a "cavern" like the one attached using 1 dimensional perlin noise.
I'll have an object fly through the tunnel, and this object has a maximum velocity along the Y axis.
I want to grant the possibility to travel all the tunnel, but i need to limit the derivate of both the upped and lower part of the cave to a specific maximum.
How can i do this?

Here the code i use to generate the noise

public class PerlinNoise {

private static final int X_NOISE_GEN = 1619;
private static final int SEED_NOISE_GEN = 1013;

private double mPersistence = 1/4;
private int mOctaves = 6;
private int mSeed;

public PerlinNoise() {
  mSeed = (int) System.currentTimeMillis();
}

PerlinNoise(double persistence, int octaves, int seed) {
  mPersistence = persistence;
  mOctaves = octaves;
  mSeed = seed;
}

private double IntNoise_1D(int x) {
  int n = (X_NOISE_GEN * x + SEED_NOISE_GEN * mSeed )& 0x7fffffff;
  n = (n<<13) ^ n;
  return ( 1.0 - ( (n * (n * n * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0);
}

private double SmoothedNoise_1D(double x) {
	 return IntNoise_1D((int) x)/2  +  IntNoise_1D((int) (x-1))/4  +  IntNoise_1D((int) (x+1))/4;
}

private double InterpolatedNoise_1D(double x) {
	 int integer_X    = (int) x;
	 double fractional_X = x - integer_X;

	 double v1 = SmoothedNoise_1D(integer_X);
	 double v2 = SmoothedNoise_1D(integer_X + 1);

	 return linearInterp(v1 , v2 , SCurve5(fractional_X));
}

public double Noise_1D(float x) {
  double total = 0;
	 for(int i = 0; i< mOctaves - 1; i++) {
	  int frequency = (int) Math.pow(2, i);
	  double amplitude = (int) Math.pow(mPersistence, i);
	  total = total + InterpolatedNoise_1D(x * frequency) * amplitude;
	 }
    return total;
}

private static double linearInterp (double n0, double n1, double a)
    {
	  return (((1.0 - a) * n0) + (a * n1));
    }

private static double SCurve5 (double a)
    {
	  double a3 = a * a * a;
	  double a4 = a3 * a;
	  double a5 = a4 * a;
	  return (6.0 * a5) - (15.0 * a4) + (10.0 * a3);
    }
}

Attached Thumbnails

  • Immagine.jpg





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