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You can find the magnitude (the length) using the Pythagorean theorem. In a car game, you might need to find this value show show the speed of an object on the HUD display.
c = (4, 5)
sqrt(4^2 + 5^2))
The PVector class in processing has a method that returns this value: mag();
Normalising
A normalised vector or a unit vector has had its magnitude set to one, with the direction left unchanged. A unit vector shows us a direction alone, without a magnitude. Sometimes we are only interested in direction, and removing magnitude can make many calculations much simpler.
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PVector vec1 = new PVector(1,2);
PVector vec2 = new PVector(3,4);
// read out components of vec1
println(vec1.x);
println(vec1.y);
println(vec1.z);
println(vec1);
println(vec2);
// set new components of vector
vec1.set(10,20);
println(vec1);
// vector addition when creating a new vector
println("-- Vec Add --");
PVector resVec = PVector.add(vec1,vec2);
println(resVec);
// adding a vector to an existing vector
vec1.add(vec2);
println(vec1);
// vector subtraction
println("-- Vec Sub --");
resVec = PVector.sub(vec1,vec2);
println(resVec);
vec1.sub(vec2);
println(vec1);
// vector multiplication
println("-- Vec Mult --");
resVec = PVector.mult(vec1,2);
println(resVec);
vec1.mult(2);
println(vec1);
// find vector magnitude
println("-- Vec Length --");
println(vec1.mag());
// find angle between two vectors
println("-- Vec Angle Between --");
println(degrees(PVector.angleBetween(vec1, vec2)));
// normalise a vector
println("-- Vec Normalise --");
vec1.normalize();
println(vec1);
println(vec1.mag()); |
Example 1: Gravity
Here we use vectors to create a force of attraction. The force's effect is accumulative, always adding to the velocity of our ellipse. But since the force is sometimes negative, depending on if the ellipse is above, below, left, or right of the mouse, the ellipse is subject to an oscillating effect.
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PVector pos = new PVector(200,200);
PVector velocity = new PVector(1,0);
void setup() {
size(600,600);
fill(255);
}
void draw() {
background(0);
PVector direction = new PVector(mouseX, mouseY);
direction.sub(pos); // by subtracting the pos from mouse coordinates, we end up with a vector between the two points
//
direction.normalize(); // now we have a vector with the length of 1, this tells us the direction we want to push out ellipse in
direction.mult(.1); // we shorten the vector (magnitude), this gives us our speed
velocity.add(direction); //we add the direction to our velocity, so the changes accumulate over time
pos.add(velocity); // pos + velocity give us our new position!
ellipse(pos.x, pos.y,30,30);
}; |
Example 2: Bouncy ball
Here we use vectors to simulate the effect of bouncing. When the ball hits the edges of the canvas, it's vector is inverted in the appropriate axis, sending it back to where it came from.
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PVector pos = new PVector(200,200); PVector velocity = new PVector(1,0); float damper = 0.99; void setup() { size(600,600); fill(255); } void draw() { background(0); velocity.mult(damper); // always slow the ball down to simulate resistance pos.add(velocity); // pos + velocity give us our new position! // if the ball is above or below the screen, invert the y value of the velocity to send it back where it came from if (pos.y<0 || pos.y>height) { velocity.y = -velocity.y; pos.add(velocity); } // if the ball is left or right of the screen, invert the x value if (pos.x<0 || pos.x>width) { velocity.x = -velocity.x; pos.add(velocity); } ellipse(pos.x, pos.y,30,30); if(mousePressed) { // show us our "line of power" stroke(255,0,0); line(pos.x,pos.y, mouseX,mouseY); noStroke(); } }; void mouseReleased() { // when the mouse is released we give a ball a big push PVector direction = new PVector(mouseX, mouseY); direction.sub(pos); // by subtracting the pos from mouse coordinates, we end up with a vectore between the two points direction.mult(.2); // we shorten the magnitude to reduce the power of the push velocity.set(direction); //now we have our new velocity }; |
Exercise:
Modify example 2 to include a gravity force. The circles should fall towards the bottom of the screen, where it will bounce up again.
(Harder) Further modify the example so that each mouse press adds a new circle, and releasing the mouse sets the new circles velocity.
(Very Hard) Further modify the example so that each ball bounces off any ball it collides with.