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Topic: Time Gradients (Serrated Images) (Read 857 times) |
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TomC
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Time Gradients (Serrated Images)
« on: Dec 17th, 2003, 11:57am » |
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NB:- self-moved/repost from 'beyond categories' following subtle renaming of this category 
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Inspired by www.serratedimage.com, I've been playing with my new webcam and I came up with these little ditties...
Code:
//
// Tom Carden - web (at) tom (dash) carden (dot) co (dot) uk
// Time Tiles - 5th December 2003
// Takes the central column from the webcam image and tiles it from left to right.
//
int xpos = 0;
int sliceWidth= 4;
boolean newFrame = false;
void setup() {
size(640, 480);
beginVideo(320, height, 30);
}
void videoEvent() {
newFrame = true;
}
void loop() {
if(newFrame) {
xpos += sliceWidth;
if(xpos > width - sliceWidth) {
//saveFrame();
xpos = 0;
}
for (int y = 0; y < height; y++) {
for (int x = 0; x < sliceWidth; x++) {
pixels[y*width + xpos + x] = video.pixels[y*video.width + video.width/2 + x];
}
}
newFrame = false;
}
}
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The first one looks good with size(screenWidth,480). Anyone know if there's a way to avoid hardcoding the screensize in processing? Is there a screen size constant somewhere?
Code:
//
// Tom Carden - web (at) tom (dash) carden (dot) co (dot) uk
// Time Gradient - 5th December 2003
// Delaying video across the screen (left hand side is 10 seconds older than right hand side).
// NB - needs stacks of RAM and a Webcam.
//
int currentLine = 0;
BImage[] buffer;
void setup() {
size(320,240); // the most my 512MB laptop will cope with!
beginVideo(width, height, 30);
buffer = new BImage[width]; // a buffer for each column of pixels
for (int j = 0; j < buffer.length; j++) {
buffer[j] = new BImage(width-j, height); // remember "width" (e.g. 320) columns for the left hand side, and 1 for the right hand side
}
}
void loop() {
for (int x = 0; x < width; x++) {
for (int y = 0; y < height; y++) {
buffer[x].pixels[(currentLine % buffer[x].width) + (y * buffer[x].width)] = video.pixels[x + (y * video.width)]; // record each column of pixels in new column of buffer
pixels[(y * width) + x] = buffer[x].pixels[(y * buffer[x].width) + ((currentLine + 1) % buffer[x].width)]; // draw the oldest column (1 before the one we just remembered)
}
}
currentLine++;
}
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If anyone has a powerful enough machine (ie enough RAM) to run the second one high-res, I'd be interested in seeing it. Any hints on optimisation would be greatfully received. I think it needs width * (width/2 * height) pixels storing, which if BImage is 32bit will mean e.g. ~100MB for 320x240 but ~750MB for 640x480. Am I right? Can BImage (or an alternative) use compression on the fly (and how quickly can images be compressed and uncompressed to memory?).
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flight404
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Re: Time Gradients (Serrated Images)
« Reply #1 on: Dec 17th, 2003, 3:38pm » |
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I am rather fond of the second execution. Very underwater-trippy. Thought I had more RAM than I obviously do because I couldnt do more than 320x240 as well.
r
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Jerronimo
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Re: Time Gradients (Serrated Images)
« Reply #2 on: Dec 17th, 2003, 8:28pm » |
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The second one is awesome!
The first one reminds me of my linear strip sketches (first two in the bottom section of this page: http://www.cis.rit.edu/~jerry/Software/p5/ )
Neat stuff!
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| « Last Edit: Dec 18th, 2003, 5:58am by Jerronimo » |
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Andrew
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Re: Time Gradients (Serrated Images)
« Reply #3 on: Dec 18th, 2003, 1:15am » |
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yes the second one is quite strange. I love it
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charmhaste
Guest
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Re: Time Gradients (Serrated Images)
« Reply #4 on: Apr 11th, 2004, 7:41am » |
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I also dig teh second one
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