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I am given 7 categories and 2 colors: blue (#02A7E3) and orange (#F6A800). There are numerous online tools to interpolate between these two, for instance:

Chroma.js example

But what if I don't want my given colors to be at the ends? More precisely:

  1. unknown
  2. blue (#02A7E3)
  3. unknown
  4. unknown
  5. unknown
  6. orange (#F6A800)
  7. unknown

I think reasonable to calculate the colors 3 to 5 with the same tool:

Chroma.js 2

So:

  1. unknown
  2. blue (#02A7E3)
  3. #8ba7b5
  4. #bba787
  5. #dba858
  6. orange (#F6A800)
  7. unknown

How could I interpolate the colors at both ends?

Ideally (but I don't know whether these requirements are compatible), I would all the resulting colors to be dark enough to serve as background against a white text (with a bold font).

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  • 5
    Yes you could do this, no problem there its just 3 linear extrapolations after all. However, consider that while this make sense mathematically it does not really make sense as to what color actually IS. – joojaa Feb 22 '18 at 8:58
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    Are you sure you want to do this? As joojaa says, while mathematical interpolation might be a scientifically sound way of creating variant colours, it might not be the sensible answer. Are you limited to these intermediate shades, or is it conceivable that you'd use other ways of finding harmonising tints? – Vincent Feb 22 '18 at 9:54
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    I have to agree with @Vincent and joojaa, these colours don't look particularly inviting/pleasant. They're what I'd call muddy. – Billy Kerr Feb 22 '18 at 10:51
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    One of the issues you are struggling with here is that blue and orange are exactly opposite each other on the HSB/HSV colour wheel. You could get from one of those colours to the other in just about any conceivable route. – Vincent Feb 22 '18 at 10:55
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    I second Vincent: besides, if you try to use interpolation you get saturation of the red and blue channels at both ends, so you won't get what you expect – clabacchio Feb 22 '18 at 13:10
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Do 3 transitions, and discard the colours you don't need.

For the sake of an example, here I have chosen MidnightBlue at one end, and a Yellow colour for the other transitions at either side:

enter image description here

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  • Thanks, but what I want is to "prolongate" the staircase interpolation given on my second screenshot. – Aristide Feb 22 '18 at 9:53
  • @Aristide updated my answer, but basically you can choose any colours you want. – Billy Kerr Feb 22 '18 at 10:56
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    @Aristide Actually this is not worse than my messings with linearinterpolation, Opposite: it opens a wide range of possblities and is simple.. Mr.Kerr has shown a practical way to pick visible more examples from the sea of zillion possibliies . Definitely an excellent receipe once again! – user287001 Feb 22 '18 at 11:17
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    @user287001 actually I wouldn't normally go down this route to make colour schemes myself. To be honest the colours in the middle are somewhat muddy and not very pleasant I much prefer to use tools like Adobe Color, where you can select different colour harmony options, like complementary colours or triads, etc. – Billy Kerr Feb 22 '18 at 11:23
  • Yes, but as an answer how to continue the already existing, series (muddy or not, good for easy to watch website or not)) the receipe is fine. It shows literally that there's possiblities infinitely. – user287001 Feb 22 '18 at 11:30
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In Photoshop and many other graphic software you have layer blending modes Add and Subtract. With subtract you can calculate the step size (=B-A) between two colors A and B. You can add that step to B to get as distant new color C or subtract it from A to see what belongs in front of A to that series.

A nice theory, but it works poorly in normal RGB mode with 0 to 255 channel value range. We have unfortunately no negative colors. You get better results in 32 bit RGB mode. See an example:

enter image description here

I have taken 5th and 6th colors of your series and calculated the 7th. For reference the original 7th color is shown as the right half of the image, the calculated (=extrapolated) version is the left half.

Another example: I have subtracted and merged 2nd and 6th color of your series. Then I subtract it with 50% opacity from your 2nd color. I get your first color. You see that the first color in the reference series has nearly sunken to the calculated color. :

enter image description here

Actually you do not need graphic software. You know the RGB values and you can calculate the colors with a calculator or spreadsheet. It's elementary, if you use linear interpolation. Calculate X1 plus or minus N*(X2-X1)/M where X is R, G and B in their turn, N is an integer "how many steps you want to advance", M is another integer "how many steps there already is from color1 to color2, use plus and minus for different directions. Do not make any roundings until you are ready. If you happen to get a final R,G or B number below zero or over 255, then you have wanted something impossible in 8 bit RGB range.

NOTE: This is not exact even with 32 bit colors because we used elementary linear interpolation with no idea how the series was actually formed. A series between two colors can be formed via arbitary routes in the color space. Linear interpolation is only one possiblity and it happened to give quite the same result than your color series generator just in this case. Another source for inaccuracy is that I used as starting values numbers which are already rounded to 8 bit. Finally the system color management and the trip via Imgur distort colors.

linear interpolation between RGB numbers is only one of the zillion possiblities to advance from color 1 to color 2 and create some intermediate steps. Somehow visually optimal result needs surely something advanced which takes into the account the limitations of the devices and the human color perception. Artists do all that intuitively.

If you want to get a series without radically differently saturated intermediate colors, you can try interpolating between hue, saturation and luminance values. Note that you must select do you want to go clockwise or CCW along the hue circle.

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  • So, to find the color 1, I should do: #02A7E3 + #02A7E3 - #8ba7b5? I have no graphic software. Do you know any online tool where I can perform such calculations? – Aristide Feb 22 '18 at 9:51
  • @Aristide Actually you do not need graphic software. You know the RGB values and you can calculate the colors with a calculator or spreadsheet. It's elementary, if you use linear interpolation. Calculate X1 plus or minus N*(X2-X1)/M where X is R, G and B in their turn and N is an integer "how many steps you want to advance", M is another integer "how many steps there already is from 1 to 2, use plus and minus for different directions. Do not make any roundings until you are ready. I add that also to my answer. – user287001 Feb 22 '18 at 10:17
  • Thanks. I just discovered that WolframAlpha can do this wolframalpha.com/input/?i=%2302A7E3+%2B+%2302A7E3+-+%238ba7b5. I am experimenting with it right now. – Aristide Feb 22 '18 at 10:32
  • According to WolframAlpha, c2 - c3 ≠ c3 - c4 (more precisely, #02A7E3 - #8ba7b5 = #3DCFFF but #8ba7b5 - #bba787 = #BFF9F9). Hence, it seems that there is no constant step I could add to c2 to obtain c1. – Aristide Feb 22 '18 at 10:48
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Let me clarify my comment a bit. Color is not a very simple concept. While color has 3 components it is not really a straightforward way to say what the vector means like say physical location that also has 3 variables.

First RGB color values are not linear. Secondly while we have information on how to equally space colors in certain limited sets of colors its not the same thing as lineary spacing RGB values. So we know in fact that this is not linear interpolation of color. It is linear interpolation of RGB values, while a perfectly valid request not the same as intermediate color.

Now, it does not particularly help us that there can be two perfectly valid and scientifically correct and incompatible definitions of color. Simply inetremidate is badly defined and equally spaced is HARD, can not be done in a RGB space not even in LAB space. One can do it with quite a lot of fiddling and uncertainty as to whether or not you actually did it.

But yeah you can treat RGB values as linear color coordinates, and it works out as you have show. But then who is tio say RGB is the space to use LAB gives different results, XYZ gives different results a space consisting of Munsel colors would admittedly give you well spaced results an and again you get different results in HSV, HSB and XYZ spaces

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