Cartesian-to-polar in the pixel array

Programming Questions

amnon.owed

Cartesian-to-polar in the pixel array

in Programming Questions • 2 years ago

Hey everyone,

I want to do a cartesian-to-polar transformation on an image (sequence) using the pixel array. I've created a first implementation of this idea, which solves some of the challenges: cartesian nature of the pixel array itself, low density in the outer polar rings at stepSize of 1 pixel. My solution is the inputBasedPolar. It starts from the input xy and via qr ends up at the output xy.

It works, but it's inefficient. There may be ways to improve it, but what I would really like is the outputBasedPolar. This should start from the output xy and via qr end up at the input xy. However my math is a little rusty. I would appreciate some help with 'inverting' the formulas. Or hear other suggestions on cartesian-to-polar transformations of course.

Thanks!

This is what I got...

Code-in-progress

String url = "http://newtech.aurum3.com/images/polar-bear.jpg";
PImage input, output;
boolean original;
void setup() {
input = loadImage(url);
size(input.width, input.height, P2D);
inputBasedPolar(input, 0.25, 0.25, 0.5); // try stepSizes of 1 to see the density issue
}
void draw() {
if (original) { image(input, 0, 0); }
else { image(output, 0, 0); }
}
void keyPressed() {
if (key == ' ') { original = !original; }
}
void inputBasedPolar(PImage input, float stepSizeX, float stepSizeY, float factor) {
output = createImage(input.width, input.height, RGB);
for (float y=0; y<input.height; y += stepSizeY) {
float r = y * factor;
for (float x=0; x<input.width; x += stepSizeX) {
float q = map(x, output.width, 0, 0, TWO_PI);
int polarX = int(r * cos(q)) + output.width/2;
int polarY = int(r * sin(q)) + output.height/2;
polarX = constrain(polarX, 0, output.width-1);
polarY = constrain(polarY, 0, output.height-1);
int outputIndex = polarX + polarY * output.width;
int inputIndex = int(x) + int(y) * input.width;
output.pixels[outputIndex] = input.pixels[inputIndex];
}
}
}
void outputBasedPolar(PImage input, float factor) {
output = createImage(input.width, input.height, RGB);
for (int y=0; y<output.height; y++) {
for (int x=0; x<output.width; x++) {
// ? HOW? ;-)
int inputIndex = 0;
int outputIndex = x + y * output.width;
output.pixels[outputIndex] = input.pixels[inputIndex];
}
}
}

Replies(5)

allonestr..

Re: Cartesian-to-polar in the pixel array

2 years ago

I've deliberately not looked at your code, but at the results. I wonder if you're making it harder by not having the transformation symmetrical.

My diagram below shows the result of a pole-transform plug-in that I have for Paint-Shop Pro. Unlike yours, it doesn't fit the resulting circle into the origainal image size.

I've tried to illustrate how I see the transformation working. I can see that the bottom corner pixels, say, are moved up and round to the centre top.

I've included the results of inverse transforming each image.

To untransform the symmetrical version, I can see that I need to take all the pixels along the circumference of the circle and rearrange them to form the bottom row of the image ... and so on. There's going to be some loss as you'll have only one pixel in the centre of the circle to map to the top row.

Is this enough, or do I need to start really thinking?

string

allonestr..

Re: Re: Cartesian-to-polar in the pixel array

2 years ago

Sorry - (un)symmetry irrelevant!

string

amnon.owed

Re: Cartesian-to-polar in the pixel array

2 years ago

I like your diagrams. They give me a different perspective. And also thanks for the plugin-tip. I just downloaded it. The options in Mura Meister's version gave me some ideas about polar and inverse-polar transformations. It looks like he is using color interpolation for low density areas, which is something I've also considered. I'll keep experimenting with the code!

We can disregard the 90 degree (quarter image) difference though. That's just because the plugin's 0 degree is at 12 o'clock and Processing's 0 degree is at 3 o'clock. Or something like that, you know what I mean. In the adapted code below, it's similar to the plugin (where the plugin uses an endRadius of 120 so that it also fits the circle within the original image).

Adapted Code

// use these input settings
inputBasedPolar(input, 0, input.height, 0, input.width, 0.25, 0.25, 360, 0.5);
// with this polar function
void inputBasedPolar(PImage input, int startY, int endY, int startX, int endX, float stepSizeX, float stepSizeY, int degrees, float factor) {
output = createImage(input.width, input.height, RGB);
for (float y=startY; y<endY; y += stepSizeY) {
float r = y * factor;
for (float x=startX; x<endX; x += stepSizeX) {
float q = map(x, output.width, 0, 0, radians(degrees))-HALF_PI;
int polarX = int(r * cos(q)) + output.width/2;
int polarY = int(r * sin(q)) + output.height/2;
polarX = constrain(polarX, 0, output.width-1);
polarY = constrain(polarY, 0, output.height-1);
int outputIndex = polarX + polarY * output.width;
int inputIndex = int(x) + int(y) * input.width;
output.pixels[outputIndex] = input.pixels[inputIndex];
}
}
}

Just found out that if I want polar-to-cartesian, I can just switch the outputIndex and inputIndex in the function I have. That's also a pretty cool effect. And inverting the x and/or y also gives some interesting visuals. So this ongoing investigation is already paying off. However, my original outputBasedPolar() request was something else though.

amnon.owed

Re: Cartesian-to-polar in the pixel array

2 years ago

I stumbled upon this post from the old forum. It seems davbol stole my idea... 5 years ago!

All kidding aside, his pseudocode helped me solve this problem and create the outputBasedPolar I was after.

Now I'll be looking into lookup-tables.

amnon.owed

Re: Cartesian-to-polar in the pixel array

2 years ago

For possible future dwellers looking for more information on this topic...

I have written a blog post that among other things covers this and includes three working code examples to do cartesian-to-polar transformations in the pixel array: input-based, output-based and output-based using lookup tables.

For those interested in the end result as well, check out the two videos I made here and here.

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