I’m working on a project with p5.js where I draw a circle, draw arcs (straight red lines) to separate the circle then another arc between each of the red lines (blue lines). The idea looks like the included image below:
What I'm confused about is how to position the labels in the circle drawing so that they're positioned in each segment inside the circle but outside the blue arcs. My question is how do I add text labels to this figure so that it looks like the image below?
Here is the shortened code to produce the first image (circle without the labels) so far:
function setup() {
createCanvas(400, 400);
}
function draw() {
background(255);
let startX = 50;
let startY = 50;
let data = [1, 2, 3, 4];
let width = 80;
let angle = -Math.PI / 2;
let radianPer = Math.PI * 2 / Object.keys(data).length;
noStroke();
fill(255);
ellipse(startX, startY, width, width);
Object.keys(data).forEach(i => {
fill(255);
stroke(255, 0, 0);
arc(startX, startY, width, width, angle, angle + radianPer, PIE);
fill(255);
stroke(0, 0, 255);
arc(startX, startY, width / 2, width / 2, angle, angle + radianPer, PIE);
angle += radianPer;
// add label here
});
}
Edit (02/05/22): updated code to match the screenshot image example.
Displaying a label in the middle of a segment of an arc involves using the angle for the middle of that arc with the sine and cosine functions to find the X and Y coordinates. For more information see the trigonometric functions article on wikipedia.
function setup() {
createCanvas(400, 400);
// Text settings
textAlign(CENTER, CENTER);
}
function draw() {
background(255);
let startX = 50;
let startY = 50;
let data = [1, 2, 3, 4];
let width = 80;
let angle = -Math.PI / 2;
let radianPer = (Math.PI * 2) / Object.keys(data).length;
noStroke();
fill(255);
ellipse(startX, startY, width, width);
Object.keys(data).forEach((i) => {
fill(255);
stroke(255, 0, 0);
arc(startX, startY, width, width, angle, angle + radianPer, PIE);
fill(255);
stroke(0, 0, 255);
arc(startX, startY, width / 2, width / 2, angle, angle + radianPer, PIE);
// add label here
let textAngle = angle + radianPer / 2;
// Use sine and cosine to determine the position for the text
// Since sine is opposite / hypotenuse, taking the sine of the angle and
// multiplying by distance gives us the vertical offset (i.e. the Y
// coordinate).
// Likewise with cosine for the X coordinate
noStroke();
fill(0);
text(
data[i].toString(),
startX + cos(textAngle) * width / 2 * 0.75,
startY + sin(textAngle) * width / 2 * 0.75
);
// Don't update angle until after calculating the angle for the label
angle += radianPer;
});
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/1.4.0/p5.js"></script>
Related
I'm trying to animate a spiral using a line, but can only seem to get it to work using ellipses.
Does anyone know how to replace the ellipse() with line()?
here is the code:
var angle = 0.0;
var offset = 60;
var scalar = 10;
var speed = 0.05;
function setup() {
createCanvas(600, 120);
fill(0);
}
function draw() {
var x = offset + cos(angle) * scalar;
var y = offset + sin(angle) * scalar;
ellipse( x, y, 2, 2);
angle += speed;
scalar += speed;
}
Assuming you would like to draw the entire spiral instantaneously using line segments, the you simply need a for loop that calculates the x and y coordinates for the current and next point in the spiral for some increment of change, and then draw lines between each pair of points. There are certainly numerous ways to write such a for loop, depending on what the constrains are (do you want a specific number of rings in your spiral? a specific number of degrees of rotation?), but importantly the bigger your increment of change the less smooth your spiral will look. Here is an example that uses the mouse position to determine the number of rings and the size of the change increments:
function setup() {
createCanvas(windowWidth, windowHeight);
stroke(0);
strokeWeight(4);
textAlign(LEFT, TOP);
}
function draw() {
background(255);
// let the horizontal mouse position indicate the
// size of the steps
let speed = map(mouseX, 0, width, 0.01, 1, true);
// let the vertical mouse position indicate the
// total amount of rotation
let maxRotation = map(mouseY, 0, height, TWO_PI, TWO_PI * 50, true);
push();
noStroke();
fill('red');
text(`Rings: ${(maxRotation / TWO_PI).toFixed(1)}, Speed: ${speed.toFixed(2)}`, 10, 10);
pop();
translate(width / 2, height / 2);
let scalar = 10;
if (speed <= 0) {
console.error('cannot have <= 0 speed');
return;
}
for (let angle = 0; angle < maxRotation; angle += speed, scalar += speed) {
const x = cos(angle) * scalar;
const y = sin(angle) * scalar;
const x2 = cos(angle + speed) * (scalar + speed);
const y2 = sin(angle + speed) * (scalar + speed);
line(x, y, x2, y2);
}
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/1.4.0/p5.js"></script>
In processing, when you apply a matrix transformation, you can draw on your canvas without worrying of the "true" position of your x y coordinate.
I thought that by the same logic, I could copy a section of the canvas by using ParentApplet.get(x, y, width, height) and that it would automatically shift the x and y, but it does not, it uses the coordinates as raw inputs without applying the matrix stack to it.
So the easiest way I see to deal with the problem would be to manually apply the matrix stack to my x, y, width, height values and using the results as input of get(). But I cannot find such a function, does one exist ?
EDIT : As requested, Here's an example of my problem
So the objective here is to draw a simple shape, copy it and paste it. Without translate, there is no problem:
void settings(){
size(500, 500);
}
void draw() {
background(255);
// Fancy rectangle for visibility
fill(255, 0 ,0);
rect(0, 0, 100, 100);
fill(0, 255, 0);
rect(20, 20, 60, 60);
// copy rectangle and paste it elsewhere
PImage img = get(0, 0, 101, 101);
image(img, 200, 200);
}
Now if I applied a translate matrix before drawing the shape, I wish that I could use the same get() code to copy the exact same drawing:
void settings(){
size(500, 500);
}
void draw() {
background(255);
pushMatrix();
translate(10, 10);
// Fancy rectangle for visibility
fill(255, 0 ,0);
rect(0, 0, 100, 100);
fill(0, 255, 0);
rect(20, 20, 60, 60);
// copy rectangle and paste it elsewhere
PImage img = get(0, 0, 101, 101);
image(img, 200, 200);
popMatrix();
}
But it doesn't work that way, The get(0, 0, ..) doesn't use the current transformation matrix to copy pixels from origin (10, 10):
Can you please provide a few more details.
It is possible to manipulate coordinate systems using pushMatrix()/PopMatrix() and you can go further and manually multiply matrices and vectors.
The part that is confusing is that you're calling get(x,y,width,height) but no showing how you render the PImage section. It's hard to guess the matrix stack you're mentioning. Can you post an example snippet ?
If you render it at the same x,y you call get() with it should render with the same x,y shift:
size(640, 360);
noFill();
strokeWeight(9);
PImage placeholderForPGraphics = loadImage("https://processing.org/examples/moonwalk.jpg");
image(placeholderForPGraphics, 0, 0);
int x = 420;
int y = 120;
int w = 32;
int h = 48;
// visualise region of interest
rect(x, y, w, h);
// grab the section sub PImage
PImage section = placeholderForPGraphics.get(x, y, w, h);
//filter the section to make it really standout
section.filter(THRESHOLD);
// display section at same location
image(section, x, y);
Regarding the matrix stack, you can call getMatrix() which will return a PMatrix2D if you're in 2D mode (otherwise a PMatrix3D). This is a copy of the current matrix stack at the state you've called it (any prior operations will be "baked" into this one).
For example:
PMatrix m = g.getMatrix();
printArray(m.get(new float[]{}));
(g.printMatrix() should be easier to print to console, but you need to call getMatrix() if you need an instance to manipulate)
Where g is your PGraphics instance.
You can then manipulate it as you like:
m.translate(10, 20);
m.rotate(radians(30));
m.scale(1.5);
Remember to call applyMatrix() it when you're done:
g.applyMatrix(m);
Trivial as it may be I hope this modified version of the above example illustrates the idea:
size(640, 360);
noFill();
strokeWeight(9);
// get the current transformation matrix
PMatrix m = g.getMatrix();
// print to console
println("before");
g.printMatrix();
// modify it
m.translate(160, 90);
m.scale(0.5);
// apply it
g.applyMatrix(m);
// print applied matrix
println("after");
g.printMatrix();
PImage placeholderForPGraphics = loadImage("https://processing.org/examples/moonwalk.jpg");
image(placeholderForPGraphics, 0, 0);
int x = 420;
int y = 120;
int w = 32;
int h = 48;
// visualise region of interest
rect(x, y, w, h);
// grab the section sub PImage
PImage section = placeholderForPGraphics.get(x, y, w, h);
//filter the section to make it really standout
section.filter(THRESHOLD);
// display section at same location
image(section, x, y);
Here's another example making a basic into PGraphics using matrix transformations:
void setup(){
size(360, 360);
// draw something manipulating the coordinate system
PGraphics pg = createGraphics(360, 360);
pg.beginDraw();
pg.background(0);
pg.noFill();
pg.stroke(255, 128);
pg.strokeWeight(4.5);
pg.rectMode(CENTER);
pg.translate(180,180);
for(int i = 0 ; i < 72; i++){
pg.rotate(radians(5));
pg.scale(0.95);
//pg.rect(0, 0, 320, 320, 32, 32, 32, 32);
polygon(6, 180, pg);
}
pg.endDraw();
// render PGraphics
image(pg, 0, 0);
}
This is overkill: the same effect could have been drawn much simpler, however the focus in on calling get() and using transformation matrices. Here a modified iteration showing the same principle with get(x,y,w,h), then image(section,x,y):
void setup(){
size(360, 360);
// draw something manipulating the coordinate system
PGraphics pg = createGraphics(360, 360);
pg.beginDraw();
pg.background(0);
pg.noFill();
pg.stroke(255, 128);
pg.strokeWeight(4.5);
pg.rectMode(CENTER);
pg.translate(180,180);
for(int i = 0 ; i < 72; i++){
pg.rotate(radians(5));
pg.scale(0.95);
//pg.rect(0, 0, 320, 320, 32, 32, 32, 32);
polygon(6, 180, pg);
}
pg.endDraw();
// render PGraphics
image(pg, 0, 0);
// take a section of PGraphics instance
int w = 180;
int h = 180;
int x = (pg.width - w) / 2;
int y = (pg.height - h) / 2;
PImage section = pg.get(x, y, w, h);
// filter section to emphasise
section.filter(INVERT);
// render section at sampled location
image(section, x, y);
popMatrix();
}
void polygon(int sides, float radius, PGraphics pg){
float angleIncrement = TWO_PI / sides;
pg.beginShape();
for(int i = 0 ; i <= sides; i++){
float angle = (angleIncrement * i) + HALF_PI;
pg.vertex(cos(angle) * radius, sin(angle) * radius);
}
pg.endShape();
}
Here's a final iteration re-applying the last transformation matrix in an isolated coordinate space (using push/pop matrix calls):
void setup(){
size(360, 360);
// draw something manipulating the coordinate system
PGraphics pg = createGraphics(360, 360);
pg.beginDraw();
pg.background(0);
pg.noFill();
pg.stroke(255, 128);
pg.strokeWeight(4.5);
pg.rectMode(CENTER);
pg.translate(180,180);
for(int i = 0 ; i < 72; i++){
pg.rotate(radians(5));
pg.scale(0.95);
//pg.rect(0, 0, 320, 320, 32, 32, 32, 32);
polygon(6, 180, pg);
}
pg.endDraw();
// render PGraphics
image(pg, 0, 0);
// take a section of PGraphics instance
int w = 180;
int h = 180;
int x = (pg.width - w) / 2;
int y = (pg.height - h) / 2;
PImage section = pg.get(x, y, w, h);
// filter section to emphasise
section.filter(INVERT);
// print last state of the transformation matrix
pg.printMatrix();
// get the last matrix state
PMatrix m = pg.getMatrix();
// isolate coordinate space
pushMatrix();
//apply last PGraphics matrix
applyMatrix(m);
// render section at sampled location
image(section, x, y);
popMatrix();
save("state3.png");
}
void polygon(int sides, float radius, PGraphics pg){
float angleIncrement = TWO_PI / sides;
pg.beginShape();
for(int i = 0 ; i <= sides; i++){
float angle = (angleIncrement * i) + HALF_PI;
pg.vertex(cos(angle) * radius, sin(angle) * radius);
}
pg.endShape();
}
This is an extreme example, as 0.95 downscale is applied 72 times, hence a very small image is rendered. Also notice the rotation is incremented.
Update Based on your update snippet it seems the confusion is around pushMatrix() and get().
In your scenario, pushMatrix()/translate() will offset the local coordinate sytem: that is where elements are drawn.
get() is called globally and uses absolute coordinates.
If you're only using translation, you can simply store the translation coordinates and re-use them to sample from the same location:
int sampleX = 10;
int sampleY = 10;
void settings(){
size(500, 500);
}
void draw() {
background(255);
pushMatrix();
translate(sampleX, sampleY);
// Fancy rectangle for visibility
fill(255, 0 ,0);
rect(0, 0, 100, 100);
fill(0, 255, 0);
rect(20, 20, 60, 60);
// copy rectangle and paste it elsewhere
PImage img = get(sampleX, sampleY, 101, 101);
image(img, 200, 200);
popMatrix();
}
Update
Here are a couple more examples on how to compute, rather than hard code the translation value:
void settings(){
size(500, 500);
}
void setup() {
background(255);
pushMatrix();
translate(10, 10);
// Fancy rectangle for visibility
fill(255, 0 ,0);
rect(0, 0, 100, 100);
fill(0, 255, 0);
rect(20, 20, 60, 60);
// local to global coordinate conversion using PMatrix
// g is the global PGraphics instance every PApplet (sketch) uses
PMatrix m = g.getMatrix();
printArray(m.get(null));
// the point in local coordinate system
PVector local = new PVector(0,0);
// multiply local point by transformation matrix to get global point
// we pass in null to get a new PVector instance: you can make this more efficient by allocating a single PVector ad re-using it instead of this basic demo
PVector global = m.mult(local,null);
// copy rectangle and paste it elsewhere
println("local",local,"->global",global);
PImage img = get((int)global.x, (int)global.y, 101, 101);
image(img, 200, 200);
popMatrix();
}
To calculate the position of a vector based on a transformation matrix, simply multiply the vector by that matrix. Very roughly speaking what's what happens with push/pop matrix (a transformation matrix is used for each push/pop stack, which is then multiplied all the way up the global coordinate system). (Notice the comment on efficienty/pre-allocating matrices and vectors as well).
This will be more verbose in terms of code and may need a bit of planning if you're using a lot of nested transformations, however you have finer control of which transformations you choose to use.
A simpler solution may be to switch to the P3D OpenGL renderer which allows you use screenX(), screenY() to do this conversion. (Also checkout modelX()/modelY())
void settings(){
size(500, 500, P3D);
}
void draw() {
background(255);
pushMatrix();
translate(10, 10);
// Fancy rectangle for visibility
fill(255, 0 ,0);
rect(0, 0, 100, 100);
fill(0, 255, 0);
rect(20, 20, 60, 60);
// local to global coordinate conversion using modelX,modelY
float x = screenX(0, 0, 0);
float y = screenY(0, 0, 0);
println(x,y);
PImage img = get((int)x, (int)y, 101, 101);
image(img, 200, 200);
popMatrix();
}
Bare in mind that you want to grab a rectangle which simply has translation applied. Since get() won't take rotation/scale into account, for more complex cases you may want to convert local to global coordinates of not just the top left point, but also the bottom right one with an offset. The idea is to compute the larger bounding box (with no rotation) around the transformed box so when you call get() the whole area of interest is returned (not just a clipped section).
I have this code that basically reads each pixel of an image and redraws it with different shapes. All shapes will get faded in using a sin() wave.
Now I want to rotate every "Pixelshape" around its own axis (shapeMode(CENTER)) while they are faded in and the translate function gives me a headache in this complex way.
Here is the code so far:
void setup() {
size(1080, 1350);
shapeMode(CENTER);
img = loadImage("loremipsum.png");
…
}
void draw() {
background(123);
for (int gridX = 0; gridX < img.width; gridX++) {
for (int gridY = 0; gridY < img.height; gridY++) {
// grid position + tile size
float tileWidth = width / (float)img.width;
float tileHeight = height / (float)img.height;
float posX = tileWidth*gridX;
float posY = tileHeight*gridY;
// get current color
color c = img.pixels[gridY*img.width+gridX];
// greyscale conversion
int greyscale = round(red(c)*0.222+green(c)*0.707+blue(c)*0.071);
int gradientToIndex = round(map(greyscale, 0, 255, 0, shapeCount-1));
//FADEIN
float wave = map(sin(radians(frameCount*4)), -1, 1, 0, 2);
//translate(HEADACHE);
rotate(radians(wave));
shape(shapes[gradientToIndex], posX, posY, tileWidth * wave, tileHeight * wave);
}
}
I have tried many calculations but it just lets my sketch explode.
One that worked in another sketch where I tried basically the same but just in loop was (equivalent written):
translate(posX + tileWidth/2, posY + tileHeight/2);
I think I just don't get the matrix right? How can I translate them to its meant place?
Thank you very much #Rabbid76 – at first I just pasted in your idea and it went of crazy – then I added pushMatrix(); and popMatrix(); – turned out your translate(); code was in fact right!
Then I had to change the x and y location where every shape is drawn to 0,0,
And this is it! Now it works!
See the code here:
float wave = map(sin(radians(frameCount*4)), -1, 1, 0, 2);
pushMatrix();
translate(posX + tileWidth/2, posY + tileHeight/2);
rotate(radians(wave*180));
shape(shapes[gradientToIndex], 0, 0, tileWidth*wave , tileHeight*wave );
popMatrix();
PERFECT! Thank you so much!
rotate defines a rotation matrix and multiplies the current matrix by the rotation matrix. rotate therefore causes a rotation by (0, 0).
You have to center the rectangle around (0, 0), rotate it and move the rotated rectangle to the desired position with translate.
Since translate and rotate multiplies the current matrix by a new matrix, you must store and restore the matrix by pushMatrix() respectively popMatrix().
The center of a tile is (posX + tileWidth/2, posY + tileHeight/2):
pushMatrix();
translate(posX + tileWidth/2, posY + tileHeight/2);
rotate(radians(wave));
shape(shapes[gradientToIndex],
-tileWidth*wave/2, -tileHeight*wave/2,
tileWidth * wave, tileHeight * wave);
popMatrix();
I'm currently trying to make a Ying and Yang symbol spin using a circular path. SO far I have made the medium and smaller ones rotate just fine. However, the stationary arc's are wrecking the illusion. Here is an open link to see my current code.
https://editor.p5js.org/Nathan65bmx/sketches/PAu3xx6Bd
Just looking for someone to help me make it look like it is rotating properly.
Draw all shapes from a common central point, then use the rotate() function. https://p5js.org/reference/#/p5/rotate Here's the link.
Do ask if you need help modifying the code.
[EDIT]
Here's the working version
function setup() {
createCanvas(600, 600);
angleMode(DEGREES);
a = 0;
x = 180;
}
let ANGLE = 0
let a;
let x;
function draw() {
background(180, 13, 123);
//Big Circle
noStroke();
//Change starts from here
push();
translate(300, 300);
rotate(a);
fill("black");
arc(0, 0, 300, 300, 0, x);
fill("white")
arc(0, 0, 300, 300, x,0);
pop();
a+=2;
//Till here
// Medium Circles
fill("black");
let CENTRE_X4 = width / 2;
let CENTRE_Y4 = height / 2;
let RADIUS4 = 75;
let X4 = RADIUS4 * cos(ANGLE);
let Y4 = RADIUS4 * sin(ANGLE);
ellipse(CENTRE_X4 + X4, CENTRE_Y4 + Y4, 150);
fill("white");
let CENTRE_X3 = width / 2;
let CENTRE_Y3 = height / 2;
let RADIUS3 = 75;
let X3 = RADIUS3 * cos(ANGLE);
let Y3 = RADIUS3 * sin(ANGLE);
ellipse(CENTRE_X3 - X3, CENTRE_Y3 - Y3, 150);
// Small Circles
fill("white");
let CENTRE_X = width / 2;
let CENTRE_Y = height / 2;
let RADIUS = 75;
let X = RADIUS * cos(ANGLE);
let Y = RADIUS * sin(ANGLE);
ellipse(CENTRE_X + X, CENTRE_Y + Y, 50);
fill("black");
let CENTRE_X2 = width / 2;
let CENTRE_Y2 = height / 2;
let RADIUS2 = 75;
let X2 = RADIUS2 * cos(ANGLE);
let Y2 = RADIUS2 * sin(ANGLE);
ellipse(CENTRE_X2 - X2, CENTRE_Y2 - Y2, 50);
ANGLE = ANGLE + 2;
}
All the edits have been done using the push() & pop() and rotate() functions.
Hope this has helped!
My answer is not adding anything new to Ruskin's great answer suggesting rotate() as well as push()/pop(), but wanted to mention that you could isolate the drawing instructions into a re-usable function and additionally simply reduce some of complexity and repetition (see D.R.Y):
function setup() {
createCanvas(600, 600);
angleMode(DEGREES);
}
function draw() {
background (200, 13, 123);
// isolate coordinate system
push();
// move everything to the center
translate(width / 2, height / 2);
// rotate everything from the center
rotate(frameCount % 360);
// draw ying Yang
drawYingYang(300);
// return to the original coordinate system (0,0 = top left)
pop();
}
function drawYingYang(outerDiameter){
let innerYOffset = outerDiameter / 4;
let outerRadius = outerDiameter / 2;
let innerDiameter = innerYOffset / 1.5;
// Big Circle
noStroke();
fill("black");
arc(0, 0, outerDiameter, outerDiameter, -90, -270);
fill("white")
arc(0, 0, outerDiameter, outerDiameter, 90, 270);
// Medium Circles
fill("black");
ellipse(0, innerYOffset, outerRadius);
fill("white");
ellipse(0, - innerYOffset, outerRadius);
// Small Circles
fill("white");
ellipse(0, innerYOffset, innerDiameter);
fill("black");
ellipse(0, - innerYOffset, innerDiameter);
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/1.1.9/p5.min.js"></script>
If that's the only thing you want to draw, removing push()/pop() won't make a difference visually, however, if you want to draw other shapes it will much easier to have independent control over where each shape is drawn
How can I curve a sheet (cube)? I'd like to control the angle of the bend/curve.
e.g.
cube([50,50,2]);
You can rotate_extrude() an rectangle with the parameter angle. This requires the openscad version 2016.xx or newer, see documentation.
It is necessary to install a development snapshot, see download openscad
$fn= 360;
width = 10; // width of rectangle
height = 2; // height of rectangle
r = 50; // radius of the curve
a = 30; // angle of the curve
rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);
looks like this:
The curve is defined by radius and angle. I think it is more realistic, to use other dimensions like length or dh in this sketch
and calculate radius and angle
$fn= 360;
w = 10; // width of rectangle
h = 2; // height of rectangle
l = 25; // length of chord of the curve
dh = 2; // delta height of the curve
module curve(width, height, length, dh) {
// calculate radius and angle
r = ((length/2)*(length/2) - dh*dh)/(2*dh);
a = asin((length/2)/r);
rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);
}
curve(w, h, l, dh);
Edit 30.09.2019:
considering comment of Cfreitas, additionally moved the resulting shape to origin, so dimensions can be seen on axes of coordinates
$fn= 360;
w = 10; // width of rectangle
h = 2; // height of rectangle
l = 30; // length of chord of the curve
dh = 4; // delta height of the curve
module curve(width, height, length, dh) {
r = (pow(length/2, 2) + pow(dh, 2))/(2*dh);
a = 2*asin((length/2)/r);
translate([-(r -dh), 0, -width/2]) rotate([0, 0, -a/2]) rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);
}
curve(w, h, l, dh);
and the result:
Edit 19.09.2020: There was a typo in the last edit: In the first 'translate' the local 'width' should be used instead of 'w'. Corrected it in the code above.
I can do it this way but it would be better if you could specify the bend/curve in #degrees as an argument to the function:
$fn=300;
module oval(w, h, height, center = false) {
scale([1, h/w, 1]) cylinder(h=height, r=w, center=center);
}
module curved(w,l,h) {
difference() {
oval(w,l,h);
translate([0.5,-1,-1]) color("red") oval(w,l+2,h+2);
}
}
curved(10,20,30);
Using the concept used by a_manthey_67, corrected the math and centered (aligned the chord with y axis) the resulting object:
module bentCube(width, height, length, dh) {
// calculate radius and angle
r = (length*length + 4*dh*dh)/(8*dh);
a = 2*asin(length/(2*r));
translate([-r,0,0]) rotate([0,0,-a/2])
rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);}
Or, if you just want something with a fixed length, and a certain bent angle do this:
module curve(width, height, length, a) {
if( a > 0 ) {
r = (360 * (length/a)) / (2 * pi);
translate( [-r-height/2,0,0] )
rotate_extrude(angle = a)
translate([r, 0, 0])
square(size = [height, width], center = false);
} else {
translate( [-height/2,0,width] )
rotate( a=270, v=[1,0,0] )
linear_extrude( height = length )
square(size = [height, width], center = false);
}
}
The if (a > 0) statement is needed to make an exception when the bending angle is 0 (which, if drawing a curved surface, would result in an infinite radius).
Animated GIF here