Calculate the color at a given point on a gradient between two colors?

So this is essentially the method I would like to write (in Objective-C/Cocoa, using UIColors, but I'm really just interested in the underlying math):

+ (UIColor *)colorBetweenColor:(UIColor *)startColor andColor:(UIColor *)endColor atLocation:(CGFloat)location;

So as an example, say I have two colors, pure red and pure blue. Given a linear gradient between the two, I want to calculate the color that's at, say, the 33% mark on that gradient: Example
So if I were to call my method like so:

UIColor *resultingColor = [UIColor colorBetweenColor:[UIColor redColor] andColor:[UIColor blueColor] atLocation:0.33f];

I would get the resulting color at 'B', and similarly, passing 0.0f as the location would return color 'A', and 1.0f would return color 'C'.

So basically my question is, how would I go about mixing the RGB values of two colors and determining the color at a certain 'location' between them?


Solution 1:

You simply linearly interpolate the red, the green, and the blue channels like this:

double resultRed = color1.red + percent * (color2.red - color1.red);
double resultGreen = color1.green + percent * (color2.green - color1.green);
double resultBlue = color1.blue + percent * (color2.blue - color1.blue);

where percent is a value between 0 and 1 (location in your first method prototype).

Solution 2:

RGB color space is like a circle. With highest saturation along the outer border, and grey in the middle. Traveling from one color to another, you'd preferably want to be doing that, along the same radius (circle) from the middle; so as to where saturation and value stay the same. In that case, the hue is changing in a linear fashion. You will not cross into more grey area than your left and right colors initially are. You can travel from an inner ring to an outer ring, simple go up (or down) the saturation; again linearly. See here for color circle (try it in e.g. paint.net)

Apple's (iOS) objective classes allow you to work with other spectra than RGB.