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I am trying to plot points around a center point using spherical coordinates. I know this is by far not the most efficient way to plot a sphere in OpenGL but i want to do it as an excersive to understand spherical coordinates better.
I want to step through each point by a certain angle so for this i have a nested for loop itterating through theta 0 - 360 and phi 0-360 and i am attempting to get the Cartesian coordinates of each of these steps and display it as a single point.
so far i have this:
float r = 1.0;
for( float theta = 0.0; theta < 360.0; theta += 10.0){
for(float phi = 0.0; phi < 360.0; phi += 10.0){
float x = r * sin(theta) * cos(phi);
float y = r * sin(theta) * sin(phi);
float z = r * cos(theta);
}
}
i store these points a display them. the display function works fine as i have used it to display other point structure before but for some reason i can't get this to work.
I have also tried converting the angles from degrees to radians:
float rTheta = theta * M_PI * 180.0;
float rPhi = phi * M_PI * 18.0;
as sin() and cos() both use radians but it yields the same results.
Am i doing something wrong and badly misunderstanding something?
In the conversion from degrees to radians of angle x, the correct formula is x * M_PI / 180..
I'm using processing, and I'm trying to create a circle from the pixels i have on my display.
I managed to pull the pixels on screen and create a growing circle from them.
However i'm looking for something much more sophisticated, I want to make it seem as if the pixels on the display are moving from their current location and forming a turning circle or something like this.
This is what i have for now:
int c = 0;
int radius = 30;
allPixels = removeBlackP();
void draw {
loadPixels();
for (int alpha = 0; alpha < 360; alpha++)
{
float xf = 350 + radius*cos(alpha);
float yf = 350 + radius*sin(alpha);
int x = (int) xf;
int y = (int) yf;
if (radius > 200) {radius =30;break;}
if (c> allPixels.length) {c= 0;}
pixels[y*700 +x] = allPixels[c];
updatePixels();
}
radius++;
c++;
}
the function removeBlackP return an array with all the pixels except for the black ones.
This code works for me. There is an issue that the circle only has the numbers as int so it seems like some pixels inside the circle won't fill, i can live with that. I'm looking for something a bit more complex like I explained.
Thanks!
Fill all pixels of scanlines belonging to the circle. Using this approach, you will paint all places inside the circle. For every line calculate start coordinate (end one is symmetric). Pseudocode:
for y = center_y - radius; y <= center_y + radius; y++
dx = Sqrt(radius * radius - y * y)
for x = center_x - dx; x <= center_x + dx; x++
fill a[y, x]
When you find places for all pixels, you can make correlation between initial pixels places and calculated ones and move them step-by-step.
For example, if initial coordinates relative to center point for k-th pixel are (x0, y0) and final coordinates are (x1,y1), and you want to make M steps, moving pixel by spiral, calculate intermediate coordinates:
calc values once:
r0 = Sqrt(x0*x0 + y0*y0) //Math.Hypot if available
r1 = Sqrt(x1*x1 + y1*y1)
fi0 = Math.Atan2(y0, x0)
fi1 = Math.Atan2(y1, x1)
if fi1 < fi0 then
fi1 = fi1 + 2 * Pi;
for i = 1; i <=M ; i++
x = (r0 + i / M * (r1 - r0)) * Cos(fi0 + i / M * (fi1 - fi0))
y = (r0 + i / M * (r1 - r0)) * Sin(fi0 + i / M * (fi1 - fi0))
shift by center coordinates
The way you go about drawing circles in Processing looks a little convoluted.
The simplest way is to use the ellipse() function, no pixels involved though:
If you do need to draw an ellipse and use pixels, you can make use of PGraphics which is similar to using a separate buffer/"layer" to draw into using Processing drawing commands but it also has pixels[] you can access.
Let's say you want to draw a low-res pixel circle circle, you can create a small PGraphics, disable smoothing, draw the circle, then render the circle at a higher resolution. The only catch is these drawing commands must be placed within beginDraw()/endDraw() calls:
PGraphics buffer;
void setup(){
//disable sketch's aliasing
noSmooth();
buffer = createGraphics(25,25);
buffer.beginDraw();
//disable buffer's aliasing
buffer.noSmooth();
buffer.noFill();
buffer.stroke(255);
buffer.endDraw();
}
void draw(){
background(255);
//draw small circle
float circleSize = map(sin(frameCount * .01),-1.0,1.0,0.0,20.0);
buffer.beginDraw();
buffer.background(0);
buffer.ellipse(buffer.width / 2,buffer.height / 2, circleSize,circleSize);
buffer.endDraw();
//render small circle at higher resolution (blocky - no aliasing)
image(buffer,0,0,width,height);
}
If you want to manually draw a circle using pixels[] you are on the right using the polar to cartesian conversion formula (x = cos(angle) * radius, y = sin(angle) * radius).Even though it's focusing on drawing a radial gradient, you can find an example of drawing a circle(a lot actually) using pixels in this answer
The Problem
I am making a game where enemies appear at some point on the screen then follow a smooth curvy path and disappear at some point. I can make them follow a straight path but can't figure out the way to make them follow the paths depicted in the image.
Attempts
I started with parabolic curve and implemented them successfully. I just used the equation of parabola to calculate the coordinates gradually. I have no clue what is the equation for desired paths supposed to be.
What I want
I am not asking for the code.I just want someone to explain me the general technique.If you still want to show some code then I don't have special preference for programming language for this particular question you can use C,Java or even pseudo-code.
First you need to represent each curve with a set of points over time, For example:
-At T(0) the object should be at (X0, Y0).
-At T(1) the object should be at (X1, Y1).
And the more points you have, the more smooth curve you will get.
Then you will use those set of points to generate two formulas-one for X, and another one for Y-, using any Interpolation method, like The La-grange's Interpolation Formula:
Note that you should replace 'y' with the time T, and replace 'x' with your X for X formula, and Y for Y formula.
I know you hoped for a simple equation, but unfortunately this is will take from you a huge effort to simplify each equation, and my advise DON'T do it unless it's worth it.
If you are seeking for a more simple equation to perform well in each frame in your game you should read about SPline method, In this method is about splitting your curve into a smaller segments, and make a simple equation for every segment, for example:
Linear Spline:
Every segment contains 2 points, this will draw a line between every two points.
The result will be some thing like this:
Or you could use quadratic spline, or cubic spline for more smooth curves, but it will slow your game performance. You can read more about those methods here.
I think linear spline will be great for you with reasonable set of points for each curve.
Please change the question title to be more generic.
If you want to generate a spiral path you need.
Total time
How many full rotations
Largest radius
So, total time T_f = 5sec, rotations R_f = 2.5 * 2 * PI, the final distance from the start D_f = 200px
function SpiralEnemy(spawnX, spawnY, time) {
this.startX = spawnX;
this.startY = spawnY;
this.startTime = time;
// these will change and be used for rendering
this.x = this.startX;
this.y = this.startY;
this.done = false;
// constants we figured out above
var TFinal = 5.0;
var RFinal = -2.6 * 2 * Math.PI;
var RStart = -Math.PI / 2;
var DFinal = 100;
// the update function called every animation tick with the current time
this.update = function(t) {
var delta = t - this.startTime;
if(delta > TFinal) {
this.done = true;
return;
}
// find out how far along you are in the animation
var percent = delta / TFinal;
// what is your current angle of rotation (in radians)
var angle = RStart + RFinal * percent;
// how far from your start point should you be
var dist = DFinal * percent;
// update your coordinates
this.x = this.startX + Math.cos(angle) * dist;
this.y = this.startY + Math.sin(angle) * dist;
};
}
EDIT Here's a jsfiddle to mess with http://jsfiddle.net/pxb3824z/
EDIT 2 Here's a loop (instead of spiral) version http://jsfiddle.net/dpbLxuz7/
The loop code splits the animation into 2 parts the beginning half and the end half.
Beginning half : angle = Math.tan(T_percent) * 2 and dist = Speed + Speed * (1 - T_percent)
End half : angle = -Math.tan(1 - T_percent) * 2 and dist = **Speed + Speed * T_percent
T_percent is normalized to (0, 1.0) for both halfs.
function LoopEnemy(spawnX, spawnY, time) {
this.startX = spawnX;
this.startY = spawnY;
this.startTime = time;
// these will change and be used for rendering
this.x = this.startX;
this.y = this.startY;
this.last = time;
this.done = false;
// constants we figured out above
var TFinal = 5.0;
var RFinal = -2 * Math.PI;
var RStart = 0;
var Speed = 50; // px per second
// the update function called every animation tick with the current time
this.update = function(t) {
var delta = t - this.startTime;
if(delta > TFinal) {
this.done = true;
return;
}
// find out how far along you are in the animation
var percent = delta / TFinal;
var localDelta = t - this.last;
// what is your current angle of rotation (in radians)
var angle = RStart;
var dist = Speed * localDelta;
if(percent <= 0.5) {
percent = percent / 0.5;
angle -= Math.tan(percent) * 2;
dist += dist * (1 - percent);
} else {
percent = (percent - 0.5) / 0.5;
angle -= -Math.tan(1 - percent) * 2;
dist += dist * percent;
}
// update your coordinates
this.last = t;
this.x = this.x + Math.cos(angle) * dist;
this.y = this.y + Math.sin(angle) * dist;
};
}
Deriving the exact distance traveled and the height of the loop for this one is a bit more work. I arbitrarily chose a Speed of 50px / sec, which give a final x offset of ~+145 and a loop height of ~+114 the distance and height will scale from those values linearly (ex: Speed=25 will have final x at ~73 and loop height of ~57)
I don't understand how you give a curve. If you need a curve depicted on the picture, you can find a curve is given analytically and use it. If you have not any curves you can send me here: hedgehogues#bk.ru and I will help find you. I leave e-mail here because I don't get any messages about answers of users from stackoverflow. I don't know why.
If you have some curves in parametric view in [A, B], you can write a code like this:
struct
{
double x, y;
}SPoint;
coord = A;
step = 0.001
eps = 1e-6;
while (coord + step - eps < B)
{
SPoint p1, p2;
p1.x = x(coord);
p1.y = y(coord);
coord += step;
p2.x = x(coord);
p2.y = y(coord);
drawline(p1, p2);
}
Can you suggest an algorithm that can draw a sphere in 3D space using only the basic plot(x,y,z) primitive (which would draw a single voxel)?
I was hoping for something similar to Bresenham's circle algorithm, but for 3D instead of 2D.
FYI, I'm working on a hardware project that is a low-res 3D display using a 3-dimensional matrix of LEDs, so I need to actually draw a sphere, not just a 2D projection (i.e. circle).
The project is very similar to this:
... or see it in action here.
One possibility I have in mind is this:
calculate the Y coordinates of the poles (given the radius) (for a sphere centered in the origin, these would be -r and +r)
slice the sphere: for each horizontal plane pi between these coordinates, calculate the radius of the circle obtained by intersecting said plane with the sphere => ri.
draw the actual circle of radius ri on plane pi using Bresenham's algorithm.
FWIW, I'm using a .NET micro-framework microprocessor, so programming is C#, but I don't need answers to be in C#.
The simple, brute force method is to loop over every voxel in the grid and calculate its distance from the sphere center. Then color the voxel if its distance is less than the sphere radius. You can save a lot of instructions by eliminating the square root and comparing the dot product to the radius squared.
Pretty far from optimal, sure. But on an 8x8x8 grid as shown, you'll need to do this operation 512 times per sphere. If the sphere center is on the grid, and its radius is an integer, you only need integer math. The dot product is 3 multiplies and 2 adds. Multiplies are slow; let's say they take 4 instructions each. Plus you need a comparison. Add in the loads and stores, let's say it costs 20 instructions per voxel. That's 10240 instructions per sphere.
An Arduino running at 16 MHz could push 1562 spheres per second. Unless you're doing tons of other math and I/O, this algorithm should be good enough.
I don't believe running the midpoint circle algorithm on each layer will give the desired results once you reach the poles, as you will have gaps in the surface where LEDs are not lit. This may give the result you want, however, so that would be up to aesthetics. This post is based on using the midpoint circle algorithm to determine the radius of the layers through the middle two vertical octants, and then when drawing each of those circles also setting the points for the polar octants.
I think based on #Nick Udall's comment and answer here using the circle algorithm to determine radius of your horizontal slice will work with a modification I proposed in a comment on his answer. The circle algorithm should be modified to take as an input an initial error, and also draw the additional points for the polar octants.
Draw the standard circle algorithm points at y0 + y1 and y0 - y1: x0 +/- x, z0 +/- z, y0 +/- y1, x0 +/- z, z0 +/- x, y0 +/- y1, total 16 points. This forms the bulk of the vertical of the sphere.
Additionally draw the points x0 +/- y1, z0 +/- x, y0 +/- z and x0 +/- x, z0 +/- y1, y0 +/- z, total 16 points, which will form the polar caps for the sphere.
By passing the outer algorithm's error into the circle algorithm, it will allow for sub-voxel adjustment of each layer's circle. Without passing the error into the inner algorithm, the equator of the circle will be approximated to a cylinder, and each approximated sphere face on the x, y, and z axes will form a square. With the error included, each face given a large enough radius will be approximated as a filled circle.
The following code is modified from Wikipedia's Midpoint circle algorithm. The DrawCircle algorithm has the nomenclature changed to operate in the xz-plane, addition of the third initial point y0, the y offset y1, and initial error error0. DrawSphere was modified from the same function to take the third initial point y0 and calls DrawCircle rather than DrawPixel
public static void DrawCircle(int x0, int y0, int z0, int y1, int radius, int error0)
{
int x = radius, z = 0;
int radiusError = error0; // Initial error state passed in, NOT 1-x
while(x >= z)
{
// draw the 32 points here.
z++;
if(radiusError<0)
{
radiusError+=2*z+1;
}
else
{
x--;
radiusError+=2*(z-x+1);
}
}
}
public static void DrawSphere(int x0, int y0, int z0, int radius)
{
int x = radius, y = 0;
int radiusError = 1-x;
while(x >= y)
{
// pass in base point (x0,y0,z0), this algorithm's y as y1,
// this algorithm's x as the radius, and pass along radius error.
DrawCircle(x0, y0, z0, y, x, radiusError);
y++;
if(radiusError<0)
{
radiusError+=2*y+1;
}
else
{
x--;
radiusError+=2*(y-x+1);
}
}
}
For a sphere of radius 4 (which actually requires 9x9x9), this would run three iterations of the DrawCircle routine, with the first drawing a typical radius 4 circle (three steps), the second drawing a radius 4 circle with initial error of 0 (also three steps), and then the third drawing a radius 3 circle with initial error 0 (also three steps). That ends up being nine calculated points, drawing 32 pixels each.
That makes 32 (points per circle) x 3 (add or subtract operations per point) + 6 (add, subtract, shift operations per iteration) = 102 add, subtract, or shift operations per calculated point. In this example, that's 3 points for each circle = 306 operations per layer. The radius algorithm also adds 6 operations per layer and iterates 3 times, so 306 + 6 * 3 = 936 basic arithmetic operations for the example radius of 4.
The cost here is that you will repeatedly set some pixels without additional condition checks (i.e. x = 0, y = 0, or z = 0), so if your I/O is slow you may be better off adding the condition checks. Assuming all LEDs were cleared at the start, the example circle would set 288 LEDs, while there are many fewer LEDs that would actually be lit due to repeat sets.
It looks like this would perform better than the bruteforce method for all spheres that would fit in the 8x8x8 grid, but the bruteforce method would have consistent timing regardless of radius, while this method will slow down when drawing large radius spheres where only part will be displayed. As the display cube increases in resolution, however, this algorithm timing will stay consistent while bruteforce will increase.
Assuming that you already have a plot function like you said:
public static void DrawSphere(double r, int lats, int longs)
{
int i, j;
for (i = 0; i <= lats; i++)
{
double lat0 = Math.PI * (-0.5 + (double)(i - 1) / lats);
double z0 = Math.Sin(lat0) * r;
double zr0 = Math.Cos(lat0) * r;
double lat1 = Math.PI * (-0.5 + (double)i / lats);
double z1 = Math.Sin(lat1) * r;
double zr1 = Math.Cos(lat1) * r;
for (j = 0; j <= longs; j++)
{
double lng = 2 * Math.PI * (double)(j - 1) / longs;
double x = Math.Cos(lng);
double y = Math.Sin(lng);
plot(x * zr0, y * zr0, z0);
plot(x * zr1, y * zr1, z1);
}
}
}
That function should plot a sphere at the origin with specified latitude and longitude resolution (judging by your cube you probably want something around 40 or 50 as a rough guess). This algorithm doesn't "fill" the sphere though, so it will only provide an outline, but playing with the radius should let you fill the interior, probably with decreasing resolution of the lats and longs along the way.
Just found an old q&a about generating a Sphere Mesh, but the top answer actually gives you a short piece of pseudo-code to generate your X, Y and Z :
(x, y, z) = (sin(Pi * m/M) cos(2Pi * n/N), sin(Pi * m/M) sin(2Pi * n/N), cos(Pi * m/M))
Check this Q&A for details :)
procedurally generate a sphere mesh
My solution uses floating point math instead of integer math not ideal but it works.
private static void DrawSphere(float radius, int posX, int poxY, int posZ)
{
// determines how far apart the pixels are
float density = 1;
for (float i = 0; i < 90; i += density)
{
float x1 = radius * Math.Cos(i * Math.PI / 180);
float y1 = radius * Math.Sin(i * Math.PI / 180);
for (float j = 0; j < 45; j += density)
{
float x2 = x1 * Math.Cos(j * Math.PI / 180);
float y2 = x1 * Math.Sin(j * Math.PI / 180);
int x = (int)Math.Round(x2) + posX;
int y = (int)Math.Round(y1) + posY;
int z = (int)Math.Round(y2) + posZ;
DrawPixel(x, y, z);
DrawPixel(x, y, -z);
DrawPixel(-x, y, z);
DrawPixel(-x, y, -z);
DrawPixel(z, y, x);
DrawPixel(z, y, -x);
DrawPixel(-z, y, x);
DrawPixel(-z, y, -x);
DrawPixel(x, -y, z);
DrawPixel(x, -y, -z);
DrawPixel(-x, -y, z);
DrawPixel(-x, -y, -z);
DrawPixel(z, -y, x);
DrawPixel(z, -y, -x);
DrawPixel(-z, -y, x);
DrawPixel(-z, -y, -x);
}
}
}
I'm currently writing a program to generate really enormous (65536x65536 pixels and above) Mandelbrot images, and I'd like to devise a spectrum and coloring scheme that does them justice. The wikipedia featured mandelbrot image seems like an excellent example, especially how the palette remains varied at all zoom levels of the sequence. I'm not sure if it's rotating the palette or doing some other trick to achieve this, though.
I'm familiar with the smooth coloring algorithm for the mandelbrot set, so I can avoid banding, but I still need a way to assign colors to output values from this algorithm.
The images I'm generating are pyramidal (eg, a series of images, each of which has half the dimensions of the previous one), so I can use a rotating palette of some sort, as long as the change in the palette between subsequent zoom levels isn't too obvious.
This is the smooth color algorithm:
Lets say you start with the complex number z0 and iterate n times until it escapes. Let the end point be zn.
A smooth value would be
nsmooth := n + 1 - Math.log(Math.log(zn.abs()))/Math.log(2)
This only works for mandelbrot, if you want to compute a smooth function for julia sets, then use
Complex z = new Complex(x,y);
double smoothcolor = Math.exp(-z.abs());
for(i=0;i<max_iter && z.abs() < 30;i++) {
z = f(z);
smoothcolor += Math.exp(-z.abs());
}
Then smoothcolor is in the interval (0,max_iter).
Divide smoothcolor with max_iter to get a value between 0 and 1.
To get a smooth color from the value:
This can be called, for example (in Java):
Color.HSBtoRGB(0.95f + 10 * smoothcolor ,0.6f,1.0f);
since the first value in HSB color parameters is used to define the color from the color circle.
Use the smooth coloring algorithm to calculate all of the values within the viewport, then map your palette from the lowest to highest value. Thus, as you zoom in and the higher values are no longer visible, the palette will scale down as well. With the same constants for n and B you will end up with a range of 0.0 to 1.0 for a fully zoomed out set, but at deeper zooms the dynamic range will shrink, to say 0.0 to 0.1 at 200% zoom, 0.0 to 0.0001 at 20000% zoom, etc.
Here is a typical inner loop for a naive Mandelbrot generator. To get a smooth colour you want to pass in the real and complex "lengths" and the iteration you bailed out at. I've included the Mandelbrot code so you can see which vars to use to calculate the colour.
for (ix = 0; ix < panelMain.Width; ix++)
{
cx = cxMin + (double )ix * pixelWidth;
// init this go
zx = 0.0;
zy = 0.0;
zx2 = 0.0;
zy2 = 0.0;
for (i = 0; i < iterationMax && ((zx2 + zy2) < er2); i++)
{
zy = zx * zy * 2.0 + cy;
zx = zx2 - zy2 + cx;
zx2 = zx * zx;
zy2 = zy * zy;
}
if (i == iterationMax)
{
// interior, part of set, black
// set colour to black
g.FillRectangle(sbBlack, ix, iy, 1, 1);
}
else
{
// outside, set colour proportional to time/distance it took to converge
// set colour not black
SolidBrush sbNeato = new SolidBrush(MapColor(i, zx2, zy2));
g.FillRectangle(sbNeato, ix, iy, 1, 1);
}
and MapColor below: (see this link to get the ColorFromHSV function)
private Color MapColor(int i, double r, double c)
{
double di=(double )i;
double zn;
double hue;
zn = Math.Sqrt(r + c);
hue = di + 1.0 - Math.Log(Math.Log(Math.Abs(zn))) / Math.Log(2.0); // 2 is escape radius
hue = 0.95 + 20.0 * hue; // adjust to make it prettier
// the hsv function expects values from 0 to 360
while (hue > 360.0)
hue -= 360.0;
while (hue < 0.0)
hue += 360.0;
return ColorFromHSV(hue, 0.8, 1.0);
}
MapColour is "smoothing" the bailout values from 0 to 1 which then can be used to map a colour without horrible banding. Playing with MapColour and/or the hsv function lets you alter what colours are used.
Seems simple to do by trial and error. Assume you can define HSV1 and HSV2 (hue, saturation, value) of the endpoint colors you wish to use (black and white; blue and yellow; dark red and light green; etc.), and assume you have an algorithm to assign a value P between 0.0 and 1.0 to each of your pixels. Then that pixel's color becomes
(H2 - H1) * P + H1 = HP
(S2 - S1) * P + S1 = SP
(V2 - V1) * P + V1 = VP
With that done, just observe the results and see how you like them. If the algorithm to assign P is continuous, then the gradient should be smooth as well.
My eventual solution was to create a nice looking (and fairly large) palette and store it as a constant array in the source, then interpolate between indexes in it using the smooth coloring algorithm. The palette wraps (and is designed to be continuous), but this doesn't appear to matter much.
What's going on with the color mapping in that image is that it's using a 'log transfer function' on the index (according to documentation). How exactly it's doing it I still haven't figured out yet. The program that produced it uses a palette of 400 colors, so index ranges [0,399), wrapping around if needed. I've managed to get pretty close to matching it's behavior. I use an index range of [0,1) and map it like so:
double value = Math.log(0.021 * (iteration + delta + 60)) + 0.72;
value = value - Math.floor(value);
It's kind of odd that I have to use these special constants in there to get my results to match, since I doubt they do any of that. But whatever works in the end, right?
here you can find a version with javascript
usage :
var rgbcol = [] ;
var rgbcol = MapColor ( Iteration , Zy2,Zx2 ) ;
point ( ctx , iX, iY ,rgbcol[0],rgbcol[1],rgbcol[2] );
function
/*
* The Mandelbrot Set, in HTML5 canvas and javascript.
* https://github.com/cslarsen/mandelbrot-js
*
* Copyright (C) 2012 Christian Stigen Larsen
*/
/*
* Convert hue-saturation-value/luminosity to RGB.
*
* Input ranges:
* H = [0, 360] (integer degrees)
* S = [0.0, 1.0] (float)
* V = [0.0, 1.0] (float)
*/
function hsv_to_rgb(h, s, v)
{
if ( v > 1.0 ) v = 1.0;
var hp = h/60.0;
var c = v * s;
var x = c*(1 - Math.abs((hp % 2) - 1));
var rgb = [0,0,0];
if ( 0<=hp && hp<1 ) rgb = [c, x, 0];
if ( 1<=hp && hp<2 ) rgb = [x, c, 0];
if ( 2<=hp && hp<3 ) rgb = [0, c, x];
if ( 3<=hp && hp<4 ) rgb = [0, x, c];
if ( 4<=hp && hp<5 ) rgb = [x, 0, c];
if ( 5<=hp && hp<6 ) rgb = [c, 0, x];
var m = v - c;
rgb[0] += m;
rgb[1] += m;
rgb[2] += m;
rgb[0] *= 255;
rgb[1] *= 255;
rgb[2] *= 255;
rgb[0] = parseInt ( rgb[0] );
rgb[1] = parseInt ( rgb[1] );
rgb[2] = parseInt ( rgb[2] );
return rgb;
}
// http://stackoverflow.com/questions/369438/smooth-spectrum-for-mandelbrot-set-rendering
// alex russel : http://stackoverflow.com/users/2146829/alex-russell
function MapColor(i,r,c)
{
var di= i;
var zn;
var hue;
zn = Math.sqrt(r + c);
hue = di + 1.0 - Math.log(Math.log(Math.abs(zn))) / Math.log(2.0); // 2 is escape radius
hue = 0.95 + 20.0 * hue; // adjust to make it prettier
// the hsv function expects values from 0 to 360
while (hue > 360.0)
hue -= 360.0;
while (hue < 0.0)
hue += 360.0;
return hsv_to_rgb(hue, 0.8, 1.0);
}