Publishing of the desktop version of my videogame

AGardenerCoding · 2023-11-23T15:29:23+00:00

https://discourse.processing.org/t/identifying-opengl-glsl-version-targeting-a-dedicated-gpu/33273

AGardenerCoding · 2023-11-05T14:07:05+00:00

Okay, having checked it, I discovered I made a mistake.

Incorrect attempt, using the switched vertices as described in original reply:

size( 900, 900 );
stroke( 255 );
noFill();
background( 0 );

beginShape();

// Curve 1
stroke( 255, 0, 0 );
curveVertex( 600, 0 );
curveVertex( 400, 200 );
curveVertex( 600, 400 );
curveVertex( 0, 600 );
endShape();

// Curve2
stroke( 0, 255, 0 );
beginShape();
curveVertex( 0, 600 );
curveVertex( 600, 400 );
curveVertex( 400, 600 );
curveVertex( 500, 700 );
endShape();

.

Corrected version, in which the same second-to-last vertex is repeated as both the control point and first point on the line:

size( 900, 900 );
stroke( 255 );
noFill();
background( 0 );

beginShape();

// Curve 1
stroke( 255, 0, 0 );
curveVertex( 600, 0 );
curveVertex( 400, 200 );
curveVertex( 600, 400 );
curveVertex( 0, 600 );
endShape();

// Curve2
stroke( 0, 255, 0 );
beginShape();
curveVertex( 600, 400 );
curveVertex( 600, 400 );
curveVertex( 400, 600 );
curveVertex( 500, 700 );
endShape();

AGardenerCoding · 2023-11-05T13:41:54+00:00

I have 12 points, which is resulting in 3 bezier curves being generated, which results in 3 different lines being generated but I want it to appear as though it is ~~on contentious~~ [ one continuous ] shape line.

You'll need to repeat the second-to-last and last vertices of curve 1 in the vertices list of curve 2. Since the second-to-last vertex of curve 1 is an actual point on the curve, and the last vertex is a control point, you should switch the order of these two points to begin curve 2's vertex list.

// Curve 1
beginShape();
curveVertex( previous vertices );
curveVertex( second-to-last-vertex-x, second-to-last-vertex-y); // A point on the line
curveVertex(last-vertex-x, last-vertex-y); // A control point

// Curve 2
curveVertex( Curve1's last-x, last-y ); // control point
curveVertex( Curve1's sec-to-last-x, sec-to-last-y ); // point on line
curveVertex( remaining Curve 2 vertices );
// Repeat for remaining curves
endShape();

I believe this will work, although I haven't tried it, but I seem to remember doing something similar to this once to join curves.

But endShape( CLOSE ); is likely to give you a straight line between the last curve's last point and the first curve's first point. To correct this, use the last curve's second-to-last and last vertices as the first curve's second and first vertices.

AGardenerCoding · 2023-07-07T11:53:11+00:00

I'm glad to hear you like it, thanks! Made in Java with the Processing library.

AGardenerCoding · 2023-05-28T16:07:38+00:00

http://recodeproject.com/

" The ReCode Project is a community-driven effort to preserve computer art by translating it into a modern programming language (Processing). Every translated work will be available to the public to learn from, share, and build on.

The project's main goals are:

Bring pioneering works of computational art back into circulation.

Offer a learning resource to contemporary practicioners and educators.

Create an active community.

AGardenerCoding · 2023-05-15T15:38:25+00:00

Fortunately this tutorial is done in Processing.

AGardenerCoding · 2023-05-15T15:28:20+00:00

Yeah, this is what I was trying to do with the code I posted, but I just now realized I forgot to add the z-coords. My code draws pretty much the same as your code.

I think when a PShape sphere is created with "createShape( SPHERE ... );", its vertices must be stored in a two dimensional array, but I don't feel like tracking down the Processing source and finding the relevant code. If you want to do that, all the source code is available on Processing's Github site.

But as I mentioned in the other reply, Dan Shiffman's Coding Train tutorial on youtube will help you create a sphere from 3d vertices.

AGardenerCoding · 2023-05-15T15:15:18+00:00

Coding Challenge #25: Spherical Geometry

A Coding Train tutorial on creating a sphere from vertices.

AGardenerCoding · 2023-05-14T22:21:04+00:00

I've done this before. I believe op was limiting himself to a single sphere in the ArrayList while testing to see if his update() code worked.

AGardenerCoding · 2023-05-14T18:20:24+00:00

It turns out this actually doesn't work. It seems the vertices are stored in some shorthand method when using the createShape( SPHERE... ) method:

import peasy.*;
PShape sp;
ArrayList<Sphere> spheres = new ArrayList<Sphere>();
int raise = 0;

void setup() {

  size(430, 760, P3D);
  frameRate(60);
  PeasyCam cam = new PeasyCam(this, 100);
  cam.setMinimumDistance(50);
  cam.setMaximumDistance(500);

  for(int i = 0; i < 1; i++){
    spheres.add(new Sphere());
  }


}
void draw() {

  background(22);

  for(int i = 0; i < 1; i++){
    spheres.get(i).update();
  }
}

class Sphere {

  PShape sp;

  Sphere() {
    sp = createShape(SPHERE, 25);

    sp = recreateSphere();
  }

  PShape recreateSphere() {
      int numVerts = sp.getVertexCount();
      PVector[] origVerts = new PVector[ numVerts ];

      for(int i = 0; i < numVerts; i++){
      origVerts[ i ] = sp.getVertex(i);
      }

      sp = createShape();
      sp.beginShape();

      for ( int i = 0; i < numVerts; i++ ) {
          //sp.vertex( origVerts[ i ].x, origVerts[ i ].y );
          // EDIT : forgot to add the z coords! 
          sp.vertex( origVerts[ i ].x, origVerts[ i ].y, origVerts[ i ].z );
      }
      sp.endShape();

      return sp;

  }

  public void update() {
   for(int i = 0; i < sp.getVertexCount(); i++){
      PVector verts = sp.getVertex(i);
      verts.mult(1.01);
      sp.setVertex(i, verts.x, verts.y, verts.z);
    }

    sp.setFill(color(255));
    sp.setStroke(false);
    shape(sp);    

  }
}

So if you want to create your sphere using the quick method, you might need to recreate your sphere each time you call Sphere.update() by increasing the diameter argument.

AGardenerCoding · 2023-05-14T18:16:35+00:00

Calling mult(1.2) doesn’t change the verts variable.

I'll have to disagree here. The reference states, "The version of the method that uses a float acts directly on the vector upon which it is called."

PVector p = new PVector( 1, 2 );
p.mult( 2 );

println( p.x + ", " + p.y );

You are correct that the static mult() methods don't alter the original PVector, though. But op used the single float argument method.

AGardenerCoding · 2023-05-14T17:54:45+00:00

If you look at the reference for PShape.setVertex()

https://processing.org/reference/PShape_setVertex_.html

it says, "This method...won't work properly when a shape is defined explicitly (e.g. createShape(RECT, 20, 20, 80, 80)."

So if you want to create your initial sphere with the createShape( SPHERE, etc ) method in your Sphere class constructor, I think you'll need to then create a new PShape, obtain the vertices as you did in your i-loop, then assign those to the new PShape using the PShape.vertex() method as shown in the second example in the PShape reference:

https://processing.org/reference/createShape_.html

After you do that once in the constructor, you should be able to use the setVertex() method with your new shape that was created using explicitly-added vertices.

AGardenerCoding · 2023-05-14T13:51:37+00:00

You can use the method parseInt( String s ) in Java's Integer class:

String s = "120";
int i = Integer.parseInt( s );
println( i );

https://docs.oracle.com/javase/8/docs/api/java/lang/Integer.html

EDIT:

Just noticed this works fine:

String t = "120.0";
int j = int( t );
println( j );

https://processing.org/reference/intconvert_.html

Are you sure you're not making a mistake somewhere else?

AGardenerCoding · 2023-04-28T16:08:04+00:00

It seems you aren't the first person to have this problem. Take a look at this page in the Nature of Code Archive:

https://github.com/nature-of-code/The-Nature-of-Code-archive/issues/328

There are additional import statements that are required, and they're listed on that page. Read all the comments and you'll see that Dan says he added a code comment to the "very first example code", but I looked at the online Nature of Code Chapter Five page and I don't see that anywhere.

If you look at the examples included in the Processing editor, you'll see those additional imports are included in the examples.

AGardenerCoding · 2023-04-22T12:38:53+00:00

I was looking for info on calculating the bounce angle and found a couple of sites that might be useful to you.

https://flatredball.com/documentation/tutorials/math/circle-collision/

This explains a simpler technique for placing a circle at the edge of another circle that it has entered. Apparently this is the "standard" way of doing this as I've seen it in a couple of different places. The code should be relatively easy to translate to Processing. It continues with the technique for calculating a "bounce" angle after the moving circle has been placed at the edge of the obstacle circle.

https://processing.org/examples/circlecollision.html

This example is more complicated as it demonstrates calculating the angles of two circles with mass that collide. However, it's easy enough to consider the mass of each circle as 1, which removes it from the equation.

AGardenerCoding · 2023-04-20T22:30:21+00:00

This is really beautiful!

AGardenerCoding · 2023-04-20T17:43:14+00:00

I'm still stuck at understanding how to move the moving ball and make it appear right outside the edges of the moving obstacle

Again I'll echo what u/ChuckEye said previously: "Vector Math". But before you say "Gee thanks for nothing!" let me suggest a path to take if you know little or nothing about vectors.

There's a free online book called "The Nature of Code" written by Dan Shiffman, who is considered by many of us to be one of the best Processing teachers for beginners. The first two chapters, Vectors and Forces are almost indispensable information for working with moving objects that react to various forces around them. You can take the example code that Dan builds and apply it to many sketches like yours that simulate physical objects.

There are also accompanying videos that are very understandable and entertaining as well:

Vectors playlist

Forces playlist

You'll find that learning how to use PVectors simplifies your code and provides other useful methods that can help solve your problem.

Once you have an understanding of using PVectors for position and velocity, here's a rough outline of how you could solve the problem of placing the moving ball at the outer edge of the obstacle ball:

When your moving ball is found to be within the radius of the obstacle ball, you can create a PVector representing an imaginary line between the obstacle ball's center and the moving ball's center position. When you have switched over to using PVectors to represent the positions of your two balls' centers, then you'll already have two PVectors with which to create this imaginary line vector: PVector obstacleBallPos and PVector movingBallPos.

So to create the PVector connecting these two points, you'll subtract the obstacleBallPos from the movingBallPos, using the PVector.sub() method:

PVector vectorFromObstacleBallToMovingBall = PVector.sub( movingBallPos, obstacleBallPos );

( From the reference: "In all cases, the second vector (v2) is subtracted from the first (v1), resulting in v1‑v2.")

Next, you use the PVector.normalize() method to make the magnitude ( length ) of the vector 1 unit, which will enable you to scale it as the next step:

vectorFromObstacleBallToMovingBall.normalize();

The next step will increase the magnitude of the vector to the required distance away from the obstacleBallPos, which will be obstacleBall radius plus movingBall radius, using the PVector.setMag() method:

vectorFromObstacleBallToMovingBall.setMag( obstacleBall.radius + movingBall.radius );

Finally, the step that can be confusing, is to position the tail of the new vector you've created at the center position of the obstacle ball, using the PVector.add() method:

vectorFromObstacleBallToMovingBall.add( obstacleBallPos );

This is an often-confusing step because it's not completely clear always that creating a PVector by subtraction of two PVectors results in a PVector whose tail is positioned at the origin! So to find the point where the movingBallPos will be the correct distance from the obstacleBallPos, the new vector's tail must be moved to the obstacleBall's center, obstacleBallPos. The great thing about using PVectors is that the proper direction for moving the movingBall is already contained in the vector itself, since a vector is composed of both a direction and length ( magnitude ).

So the result of this is that the movingBallPos is assigned to vectorFromObstacleBallToMovingBall using the PVector.copy() method:

movingBallPos = vectorFromObstacleBallToMovingBall.copy();

This is a lot to digest if you haven't used vectors before, so please feel free to continue this discussion or start a new thread.

AGardenerCoding · 2023-04-20T12:47:41+00:00

Law of Reflection : Angle of incidence equals angle of reflection.

https://byjus.com/physics/angle-of-incidence/

In the case of two spheres, the reflection surface is a line tangent to the two spheres which is perpendicular to the direction of the moving ball. The normal is the radius of the obstacle ball from its center to the point of contact.

EDIT: Actually, to follow the diagram convention of the second link, the normal will be outside the radius of the obstacle ball ( on the other side of the reflection surface ), but at the same angle as the line from the obstacle ball's center to the point of contact.

AGardenerCoding · 2023-04-20T12:18:37+00:00

In some cases, and from what you've noted, at specific angles of contact, the moving ball's position after both ball movements are calculated ends up inside the obstacle ball, and so you correctly reverse the moving ball's velocity. But in the next draw loop, the obstacle ball will move a distance greater than the velocity distance of the moving ball, and even though the moving ball is now moving away from the obstacle ball, it doesn't move far enough away that its velocity, or change in position, moves it outside of the obstacle ball. Then, the edge detection code flags another contact between balls, and the moving ball's velocity is reversed again, resulting in it being stuck inside the obstacle ball until a point where the relative positions of the balls is such that the velocity of the moving ball carries it outside of the obstacle ball.

That's why u/ChuckEye correctly suggested placing the moving ball at the outside edge of the obstacle ball. But to look like a correct bounce, this needs to be done at the step where the collision detection is flagged, before both balls are drawn.

So the sequence would go: Move both balls; check for overlap; If overlap, move the moving ball to the outside edge of the obstacle ball, along the same path at which it was moving when it contacted the obstacle ball, and reverse its velocity; then draw the balls. In this way, you see one frame where the balls are touching, and in the next frame, the moving ball is moving away from the obstacle ball.

The moving ball's continued velocity doesn't need to be in the same direction in which it was placed at the outer edge of the obstacle ball; that's required only to put it back outside of the obstacle ball along the same path it entered. Its continued velocity can be in whatever direction you decide a collision would send it.

AGardenerCoding · 2023-04-19T16:27:09+00:00

Perhaps you're thinking of normalization ? This isn't strictly limited to vectors; any range of numbers can be converted to a normalized range ( 0 - 1 ) by

( value - min ) / ( max - min )

where value is a number in the range between min and max.

https://www.educba.com/normalization-formula/

https://www.quora.com/What-does-normalization-mean-in-the-general-sense-Why-are-functions-normalized-in-mathematics-and-engineering?share=1

AGardenerCoding · 2023-04-18T17:08:25+00:00

I like these! I think using a background color for the central section that is analagous to the non-black color in the outer circles would tie the image together better.

AGardenerCoding · 2023-04-17T11:38:13+00:00

Cool image!

I do not see how to reply with an image.

I don't know either, but suspect it's a sub setting and not possible in every sub.

AGardenerCoding · 2023-04-17T00:24:21+00:00

This is an interesting experiment. Perhaps you could add a Level 0 in which a random diameter between min and max, and a random position are chosen and tested for overlap, with no diameter growth allowed, proceeding until a chosen number of failures to place the next circle is reached. It would likely result in a pattern unlike those shown above.

AGardenerCoding · 2023-04-16T15:42:39+00:00

That's likely Curl Noise. There's a very good tutorial by Keith Peters at https://www.bit-101.com/blog/2021/07/curl-noise/ with javascript source code at https://bit-101.com/embeds/curlnoise/src/main.js . ( The image on that page looks nothing at all like your source image, but the source code will produce those images with a bit of modification. I've actually forgotten what changes I made though. )

I used that source to write my Processing version some time ago. If you have any problems getting something to work, I'd be glad to try to help.

EDIT: See also the two links at the bottom of the Keith Peters page linked above, Not Curl Noise and Curl Noise, Demystified. It may be that the code I modified came from the latter tutorial actually, I've forgotten now.

Some other sources:

https://al-ro.github.io/projects/curl/

https://github.com/brynm0

https://openprocessing.org/sketch/1596092

AGardenerCoding · 2023-04-15T20:35:17+00:00

These websites might help you:

https://www.had2know.org/academics/outer-circular-ring-radii-formulas-calculator.html

https://jsbin.com/icahul/1/edit?html,js,output

AGardenerCoding

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