Hello, Does any body how to represent a curve in Processing, I have the equation of the curve but I don’t know how to represent it. If somebody could help me and telling me how to draw it I would thank it a lot
2 Likes
Hello,
and welcome to the forum!
Great to have you here!
Here is a medium advanced graph. Not by me but from the forum.
I hope this helps!
Warm regards,
Chrisir
// the max x for our test
int maxX = 100; // (in neg and pos direction)
// ---------------------------------------------------------------
void setup() {
// init
size(900, 600);
background(111);
drawCoordSystem();
// draw graph
float y;
for (float x= -maxX; x < maxX; x+=.01) { // from - maxX to + maxX with step 0.01 (very fine)
// formulas:
// y = x*x;
y = (3*(x*x))+(2*x)-1;
// apply to screen coordinate system
PVector getScreenCoords = matchToScreenCoordinates(x, y);
// draw point
pointPVector(getScreenCoords);
}//for
} // func
void draw() {
// text box upper left corner
fill(111);
noStroke();
rect (0, 0, 100, 100);
fill(255);
text(" Plot Graph.", 30, 30);
} // func
// ---------------------------------------------------------------
void drawCoordSystem() {
// draw coord system in white
// (apply to screen coordinate system)
PVector getScreenCoord1 = matchToScreenCoordinates(-maxX, 0);
PVector getScreenCoord2 = matchToScreenCoordinates(maxX, 0);
linePVectors(getScreenCoord1, getScreenCoord2);
getScreenCoord1 = matchToScreenCoordinates(0, height);
getScreenCoord2 = matchToScreenCoordinates(0, -height);
linePVectors(getScreenCoord1, getScreenCoord2);
for (int x = -10; x<10; x++) {
getScreenCoord1 = matchToScreenCoordinates(x, -.3);
getScreenCoord2 = matchToScreenCoordinates(x, .3);
linePVectors(getScreenCoord1, getScreenCoord2);
}//for
for (int y = -10; y<10; y++) {
getScreenCoord1 = matchToScreenCoordinates( -.3, y);
getScreenCoord2 = matchToScreenCoordinates( .3, y);
linePVectors(getScreenCoord1, getScreenCoord2);
}//for
}//func
// -------------------------------------------------------------------
// convert the data from formula to screen appropriate format
PVector matchToScreenCoordinates (float x, float y) {
// convert to screen coordinate system
x*=40.0;
y*=40.0; //y=y*0.019;
return new PVector (x+width/2, height/2-y-30);
}
// -------------------------------------------------------------------
// PVector functions
void pointPVector(PVector pv) {
// draw point at pv
stroke(255, 2, 2); // red
point(pv.x, pv.y);
}
void linePVectors(PVector pv1, PVector pv2) {
// draw line from pv1 to pv2
stroke(255); // white
line(pv1.x, pv1.y,
pv2.x, pv2.y);
}
//
1 Like
Show us the equation here.
You could also try one either the QScript or Jasmine libraries to evaluate the equation and graphica to plot it.
1 Like
How can I do this??
could you text me in whatts app and help me please??
+34 667307423
THANK YOU SO MUCH
Thank you so much, but I triet to descifrate what it puts but I didn’t discovered anything hahaha
Instead of this you can also use ellipse
Not sure what you mean though
Did the program run?
What did you see?
The red dots are small, look closely
Need to know what the equation looks like.
The libraries I mentioned are all available through the Contributions Manager (Use the Sketch > Add Library menu option) and they all include examples for you to look at.
The equation is x^4+y^3-((x^2)*y) = 0
I’d say this is a 3D formula, not a 2D formula ?
Do you have a graph how this looks like?
can you transform your formula in a way that instead of ......= 0 we have
y = .......................
please?
I have this:
x=t-(t^3)
y=(t^2)-t^4)
what is the parameterized function
This formulae cannot be plotted as a simple function because it would have to be rearranged as a function i.e. y = f(x) where the function f(x) only has 1 unknown x and that is not a trivial task.
This is a pair of parametric equations where the two unknowns x and y are both functions of a shared variable t (the parametric variable) i.e.
x = f(t) = t - t^2 and
y = g(t) = t^2 - t^4
Since you want to plot this parametric function I recommend Jasmine over QScript as it’s faster.
The sketch below uses the Jasmine library evaluate your parametric equations over a user defined range of t giving this output.
t x y -50.0 -2550.0 -6247500.0 -40.0 -1640.0 -2558400.0 -30.0 -930.0 -809100.0 -20.0 -420.0 -159600.0 -10.0 -110.0 -9900.0 0.0 0.0 0.0 10.0 -90.0 -9900.0 20.0 -380.0 -159600.0 30.0 -870.0 -809100.0 40.0 -1560.0 -2558400.0 50.0 -2450.0 -6247500.0
The sketch code
import org.quark.jasmine.*;
String xt_expr, yt_expr;
Expression x_expr, y_expr;
void setup() {
xt_expr = "t - t^2";
yt_expr = "t^2 - t^4";
x_expr = Compile.expression(xt_expr, false);
y_expr = Compile.expression(yt_expr, false);
float minT = -50, maxT = 50, deltaT = (maxT - minT) / 10;
float t = minT;
System.out.println("t \t\tx\t\ty = ");
while(t <= maxT){
float x = x_expr.eval(t).answer().toFloat();
float y = y_expr.eval(t).answer().toFloat();
println(t + "\t\t" + x + "\t\t" + y);
t += deltaT;
}
}
Bezier curves are probably the most famous types of parametric equations 
2 Likes
Thanks for explaining!
What are those functions called?
2D functions
2D functions receive one parameter x and the result is y. In a 2D coordinate system enter the two values as one point x,y.
3D functions
3D functions receive two parameters x and y and the result is z. Think of a grid where x,y gives you a cell and z is the value (the height) of this cell (above the cell in 3D space).
Further Readings
see also : Math Projects · Kango/MathProjects Wiki · GitHub
and GitHub - Kango/MathProjects: Math Projects in Processing
Further Readings on Cartesian coordinate system
see Cartesian coordinate system - Wikipedia
see Cartesian coordinate system - Wikipedia
This could be a nice tutorial.
Chrisir
1 Like
I should point out that I finished studying maths at 18 so I am not a qualified mathematician . Although I am quite good at maths and do find the subject interesting I do have my limits
I don’t know if there is a technical / mathematical name for this type of equation, I generally just call them bastards LOL
I am also not sure if there is such a thing as a “3D formulae” in mathematics, but it appears to be a big thing in spreadsheets, according to google.
As I said Bezier curves are parametric curves and are great fun and very useful in computing.
I am currently working on a Bezier library based on this primer and using EJML to do some of the maths.
Although we are familiar with seeing 2D Bezier curves where x = f(t) and y = g(t) we can have 3D Beziers by adding a third expression for z where z = h(t)
If you want more detail on parametric curves and Bezier curves in particular here is a lot of information on the internet.
3 Likes
A challenging polynomial.
:)
1 Like