Display temperature as a color with C#?

Question

Someone knows an algorithm that gets temperatue in Kelvin/Celsius and returns RGB?

Like in thermal cameras.

I found some links :

http://www.brucelindbloom.com/index.html?Eqn_XYZ_to_T.html

http://www.fourmilab.ch/documents/specrend/specrend.c

But i cant figure what XYZ color is ?

I only have temperature in Celsius..

i can convert it to any temperature Temperature Conversion Formulas

UPDATE: Blackbody Color Datafile I have found this.. but those Kelvin degrees are impossible.. i mean red suppose to be hot.. so why 8000k is blue and 1000k is red...

Klopstock

The second link contains a complete implementation. Reading the code should be all that's necessary. – Bypath

Please forgive my ignorance: But isn't temperature a single value which can be mapped directly to a single point on a simple pre-defined gradient? – Cristinecristiona

@Cristinecristiona thats exectly what i'm trying to do.. But where can i get ot how can i make a pre-defined gradient ? – Klopstock

8000K -> blue and 1000K -> red are color temperatures. And blue is indeed higher energy than red. – Dishonest

I think it may be worth pointing out that the colors shown by thermal cameras are not in fact "real colors". They are are chosen somewhat arbitrarily and are only intended to visualize, not measure. – Bloodcurdling

Answer 1

The best option is to use an image with GetPixel:

private void UpdateTemp()
{
    Bitmap temps = (Bitmap)Properties.Resources.temp;
    if (curTemp >= 0)
    {
        int i = curTemp;
        if (i < 0)
            i = 0;
        if (i > temps.Width-1)
            i = temps.Width-1;
        this.BackColor = temps.GetPixel(i, 10);
    }
}

Or building an array. Source

    private static Color[] colors = 
    {
        Color.FromArgb(155, 188, 255), //    40000
        Color.FromArgb(155, 188, 255), //    39500
        Color.FromArgb(155, 188, 255), //    39000
        Color.FromArgb(155, 188, 255), //    38500
        Color.FromArgb(156, 188, 255), //    38000
        Color.FromArgb(156, 188, 255), //    37500
        Color.FromArgb(156, 189, 255), //    37000
        Color.FromArgb(156, 189, 255), //    36500
        Color.FromArgb(156, 189, 255), //    36000
        Color.FromArgb(157, 189, 255), //    35500
        Color.FromArgb(157, 189, 255), //    35000
        Color.FromArgb(157, 189, 255), //    34500
        Color.FromArgb(157, 189, 255), //    34000
        Color.FromArgb(157, 189, 255), //    33500
        Color.FromArgb(158, 190, 255), //    33000
        Color.FromArgb(158, 190, 255), //    32500
        Color.FromArgb(158, 190, 255), //    32000
        Color.FromArgb(158, 190, 255), //    31500
        Color.FromArgb(159, 190, 255), //    31000
        Color.FromArgb(159, 190, 255), //    30500
        Color.FromArgb(159, 191, 255), //    30000
        Color.FromArgb(159, 191, 255), //    29500
        Color.FromArgb(160, 191, 255), //    29000
        Color.FromArgb(160, 191, 255), //    28500
        Color.FromArgb(160, 191, 255), //    28000
        Color.FromArgb(161, 192, 255), //    27500
        Color.FromArgb(161, 192, 255), //    27000
        Color.FromArgb(161, 192, 255), //    26500
        Color.FromArgb(162, 192, 255), //    26000
        Color.FromArgb(162, 193, 255), //    25500
        Color.FromArgb(163, 193, 255), //    25000
        Color.FromArgb(163, 193, 255), //    24500
        Color.FromArgb(163, 194, 255), //    24000
        Color.FromArgb(164, 194, 255), //    23500
        Color.FromArgb(164, 194, 255), //    23000
        Color.FromArgb(165, 195, 255), //    22500
        Color.FromArgb(166, 195, 255), //    22000
        Color.FromArgb(166, 195, 255), //    21500
        Color.FromArgb(167, 196, 255), //    21000
        Color.FromArgb(168, 196, 255), //    20500
        Color.FromArgb(168, 197, 255), //    20000
        Color.FromArgb(169, 197, 255), //    19500
        Color.FromArgb(170, 198, 255), //    19000
        Color.FromArgb(171, 198, 255), //    18500
        Color.FromArgb(172, 199, 255), //    18000
        Color.FromArgb(173, 200, 255), //    17500
        Color.FromArgb(174, 200, 255), //    17000
        Color.FromArgb(175, 201, 255), //    16500
        Color.FromArgb(176, 202, 255), //    16000
        Color.FromArgb(177, 203, 255), //    15500
        Color.FromArgb(179, 204, 255), //    15000
        Color.FromArgb(180, 205, 255), //    14500
        Color.FromArgb(182, 206, 255), //    14000
        Color.FromArgb(184, 207, 255), //    13500
        Color.FromArgb(186, 208, 255), //    13000
        Color.FromArgb(188, 210, 255), //    12500
        Color.FromArgb(191, 211, 255), //    12000
        Color.FromArgb(193, 213, 255), //    11500
        Color.FromArgb(196, 215, 255), //    11000
        Color.FromArgb(200, 217, 255), //    10500  
        Color.FromArgb(204, 219, 255), //    10000
        Color.FromArgb(208, 222, 255), //    9500
        Color.FromArgb(214, 225, 255), //    9000
        Color.FromArgb(220, 229, 255), //    8500
        Color.FromArgb(227, 233, 255), //    8000
        Color.FromArgb(235, 238, 255), //    7500
        Color.FromArgb(245, 243, 255), //    7000
        Color.FromArgb(255, 249, 253), //    6500
        Color.FromArgb(255, 243, 239), //    6000
        Color.FromArgb(255, 236, 224), //    5500
        Color.FromArgb(255, 228, 206), //    5000
        Color.FromArgb(255, 219, 186), //    4500
        Color.FromArgb(255, 209, 163), //    4000
        Color.FromArgb(255, 196, 137), //    3500
        Color.FromArgb(255, 180, 107), //    3000
        Color.FromArgb(255, 161,  72), //    2500
        Color.FromArgb(255, 137,  18), //    2000
        Color.FromArgb(255, 109,   0), //    1500 
        Color.FromArgb(255,  51,   0), //    1000
    };

Klopstock

Answer 2

I realize this is a two-year old thread, but I had the same predicament.

I took the data from the color table and did applied piece-wise 5th order polynomial fitting using Numpy.polyfit in Python. From those coefficients I was able to come up with the C# function below. R-squared values for the fits are close to or exceed 0.999. It has less than .01% error through most of its domain, but it does have a couple of points where it is closer to 3%. Should be good enough for most situations though.

private Color blackBodyColor(double temp)
{
    float x = (float)(temp / 1000.0);
    float x2 = x * x;
    float x3 = x2 * x;
    float x4 = x3 * x;
    float x5 = x4 * x;

    float R, G, B = 0f;

    // red
    if (temp <= 6600)
        R = 1f;
    else
        R = 0.0002889f * x5 - 0.01258f * x4 + 0.2148f * x3 - 1.776f * x2 + 6.907f * x - 8.723f;

    // green
    if (temp <= 6600)
        G = -4.593e-05f * x5 + 0.001424f * x4 - 0.01489f * x3 + 0.0498f * x2 + 0.1669f * x - 0.1653f;
    else
        G = -1.308e-07f * x5 + 1.745e-05f * x4 - 0.0009116f * x3 + 0.02348f * x2 - 0.3048f * x + 2.159f;

    // blue
    if (temp <= 2000f)
        B = 0f;
    else if (temp < 6600f)
        B = 1.764e-05f * x5 + 0.0003575f * x4 - 0.01554f * x3 + 0.1549f * x2 - 0.3682f * x + 0.2386f;
    else
        B = 1f;

    return Color.FromScRgb(1f, R, G, B);
}

Mayne

Can you use Color.FromArgb instead? Apparently the mapping from ScRgb to Argb is not linear: https://mcmap.net/q/752059/-fromargb-vs-fromscrgb – Gurl

Answer 3

If get you right you are looking for a theoretical background on XYZ color space

Enjoin

Answer 4

Color temperature is based on the actual color of light emitted from something (theoretically, an "ideal black body") that emits light based solely on its temperature.

Some examples of this kind of light source: if you have an electric stove element that is glowing red, it might be around 1000K. A regular incandescent bulb filament is around 2700K, and the sun is roughly 5700K. All three are fair approximations of a "black body"; they emit a particular spectrum of light based on their actual temperature.

Many artificial light sources are not actually the "temperature" of the light they emit (and their spectra are generally not black-body spectra, either...). Instead, their "temperature" rating is the temperature a theoretical black body would have to be in order to emit light of that color. There are also colors that cannot be generated by a black body: light which is greenish or purplish compared to a more "natural"-looking black body illumination.

As mentioned in one of the comments, the kind of thermal camera displays you are probably thinking about are all false-color. In a false-color display, the colors are chosen for convenience only: so, for a thermal camera they might choose a "hot"-looking red for warm, and "cold"-looking blue for cold. But, they could just as easily choose a range from black to white, or fuschia to green.

Because false-color displays are arbitrary, you really need to check the color key to a particular image or system if you want to estimate the temperature (scientific images should generally have some kind of color key for this purpose). If you have no color key, and no documentation on how the image was generated, you are out of luck.

Apophyllite

Answer 5

The above function overestimate red color when temp > 10000 K. Colors turn to purple when temp>14000. I refitted the data with 7th order polynomials. The coefficients of should be:

def temp_to_rgb(temp):
    t = temp/1000.

    # calculate red
    if t < 6.527:
        red = 1.0
    else:
        coeffs = [  4.93596077e+00,  -1.29917429e+00,
                    1.64810386e-01,  -1.16449912e-02,
                    4.86540872e-04,  -1.19453511e-05,
                    1.59255189e-07,  -8.89357601e-10]
        tt = min(t,40)
        red = poly(coeffs,tt)
    red = max(red,0)
    red = min(red,1)

    # calcuate green
    if t < 0.85:
        green = 0.0
    elif t < 6.6:
        coeffs = [ -4.95931720e-01,   1.08442658e+00,
                   -9.17444217e-01,   4.94501179e-01,
                   -1.48487675e-01,   2.49910386e-02,
                   -2.21528530e-03,   8.06118266e-05]
        green = poly(coeffs,t)
    else:
        coeffs = [  3.06119745e+00,  -6.76337896e-01,
                    8.28276286e-02,  -5.72828699e-03,
                    2.35931130e-04,  -5.73391101e-06,
                    7.58711054e-08,  -4.21266737e-10]
        tt = min(t,40)
        green = poly(coeffs,tt)
    green = max(green,0)
    green = min(green,1)

    # calculate blue
    if t < 1.9:
        blue = 0.0
    elif t < 6.6:
        coeffs = [  4.93997706e-01,  -8.59349314e-01,
                    5.45514949e-01,  -1.81694167e-01,
                    4.16704799e-02,  -6.01602324e-03,
                    4.80731598e-04,  -1.61366693e-05]
        blue = poly(coeffs,t)
    else:
        blue = 1.0
    blue = max(blue,0)
    blue = min(blue,1)

    return (red,green,blue)

Here poly(coeffs,x) = coeffs[0] + coeffs[1]*x + coeffs[2]*x**2 + ...

Sorry I am not familiar with C# but you can easily read the codes.

The error is within only 0.5% for most cases and at most 1.2% for red in temp = 6600 K. High order polynomials are adopted here so red and green must keep constant for temp > 40000 K, otherwise strange things will happen.

Fraxinella

Sorry to see that 6500K is not maximum (1,1,1) for R,G,B. It ends up as (1, 0.955, 0.985). To make each max out would be a nice feature if you have the time to make the appropriate adjustments. I have converted your pseudo code to C# if anyone is interested by the way. – Gurl

This answer is the bottom here while the top is this (2000-7000K). This answer doesn't compare favorably (unless I goofed it, Java code here). – Boylan

@DanW Try 6600K for 1,1,1 RGB. – Boylan