I have an RGB LED and have pins to 9, 10, 11 and a pin to ground.
Resistors have been provided for R, G, and B.
When I do:
analogWrite(r, 255); // I see a red color
analogWrite(g, 0);
analogWrite(b, 0);
analogWrite(r, 0);
analogWrite(g, 255); // I see a green color
analogWrite(b, 0);
analogWrite(r, 0);
analogWrite(g, 0); // I see a blue color
analogWrite(b, 255);
When I do:
analogWrite(r, 153);
analogWrite(g, 102);
analogWrite(b, 51);
it does not look brown to me, more like a blue color.
Am I missing something I need to do?
Brown is a very difficult color to achieve. (It's actually dark red. Brown isn't in the rainbow.)
Make sure that your colors are balanced: write a dim white / gray, 128, 128, 128 and make sure this looks white. Then write a 255, 255,255 and make sure this looks white. If these don't look white, adjust your resistors to reduce current through the component that is too bright (be careful not to allow too much current and burn out the LED.
Its weird, but blue color seems exactly opposite to brown color, that you are trying to achieve. Moreover, when I've tried to invert the color (255 - x), I've got blue colors. Maybe something is wrong with PWM configuration?
is the RGB LED common anode or common cathode,
usually RGB LEDs are common anode.
this means: lowwer the analogWrite value, higher brightness (more emitted light)
http://www.hertaville.com/wp-content/uploads/2011/07/rgb.jpg
you also need resistors, notice that blue emits more light with the same value of resistans, than green or red, so you need to set a higher resistor value.
Related
I'm wirtting a program that changes all the image pixels to grayscale except for the red ones. At first, i thought it would be easier, but I'm having trouble trying to find the best way to determine if a pixel is red or not.
The first method I tried was a formula: Green < Red/2 && Blue < Red/1.5
results:
michael jordan
goldhill
Michael Jordan's image shows some not red pixels that pass the formula, like #7F3222 and #B15432. So i tried a different method, hue >= 345 || hue <= 9, trying to limit only the red part of the color wheel.
results:
michael jordan 2
goldhill 2
Michael Jordan's image now has less not red pixels and goldhill's image has more red pixels than before but still not what I want.
My methods are incorrect or just some adjustments are missing? if they're incorrect, how can I solve this problem?
Your question "How to identify 'real' red pixels", begs the question "what a red pixel actually is, especially if it has to be 'real'".
The RGB (red, green, blue) color model is not well suited to answer that question, therefore you should use the HSV (hue, saturation, value) model.
Hue defines the color in degrees (0 - 360 degrees)
Saturation defines the intensity of the color (0 - 100 %)
Value or Brightness defines the luminosity (0 - 100 %)
Steps:
convert RGB to HSV
if the H value is not red (+/- 30 degrees, you'll have to define a threshold range of what you consider to be red, 'real' red would be 0 degrees)
set S to 0 (zero), by doing so we remove the saturation of the color, which results in a gray shade
leave the brightness (V) as it is (or play around with it and see how it effects the results)
convert HSV to RGB
Convert from RGB to HSV and vice versa:
RGB to HSV
HSV to RGB
More info on HSV:
https://en.wikipedia.org/wiki/HSL_and_HSV
"All cats are gray in the dark"
Implement a dynamic color range. Adjust the 'red' range based on the brightness and/or saturation of the current pixel. Put a weight scale (on how much they affect the range in %) on the saturation and brightness values to determine your range ... play around to achieve the best results.
You used RGB, and HSV method, which it is good, and both are ok.
The problem is about defining red. Hue (or R) is not enough: it contains many other colours (in the broader sense): browns are dark/unsaturated reds (or oranges). Pink is also a tint of red (so red + white, so unsaturated).
So in your first method, I would add a condition: R > 127 (you must check yourself a good threshold). And possibly change the other conditions with a higher ratio of R to G and B and possibly adding also the ration R to (G+B). The first new added condition is about reds (and not "dark reds/browns), and brightness. Your two conditions are about "hue" (hue is defined by the top two values), and the last condition I wrote is about saturation.
You can do in a similar way for HSV: filter H (as you did), but you must filter also V (you want just bright reds), and also an high saturation, so you must filter all channels.
You should test yourself the saturation levels. The problem is that eyes adapt quickly to colours, so some images with a lot of redish colours are seen normally (less redish) by humans, but more red by above calculation. Etc. (so usually for such works there is some sliders to modify, e.v. you can try to automatize, but you need to find overall hue and brightness of image, and possibly complex methods, see CIECAM).
I would like to have a gradient which goes from black to transparent (not white). How can I achieve this?
From my attempt below I assume the gradient style color's alpha value is not considered:
gui_Footer.allStyles.apply {
backgroundType = Style.BACKGROUND_GRADIENT_LINEAR_VERTICAL
border = RoundRectBorder.create().topOnlyMode(true).cornerRadius(1f)
backgroundGradientEndColor = ColorUtil.BLACK
backgroundGradientStartColor = ColorUtil.argb(0, 255, 255, 255)
}
Gradients in Codename One ignore the alpha byte. While we could technically add support for alpha gradients it's not something that's planned at this time. You can probably generate such an image by manipulating the RGB data but it would be more efficient to just generate an RGB image of a gradient and draw it scaled.
Notice that this is generally the most efficient approach since the GPU works by drawing textures very efficiently. If an image is a power of 2 (e.g. 256x128 pixels) it can fit perfectly in a texture and it's drawn very fast. Much faster than our builtin gradients.
I want to detect an Eye, I have some code where I can detect blue color object, so if I made changes(how I can?) then it would be possible for me to detect an eye. As the below color has its own specific range value so, if I specify the eye color HSV value then can I detect EYE with this method.
In this below code I am going to detect BLUE Color Object, please tell me that where I do changes in my code so that I could get EYE using Open CV.
IplImage* GetThresholdedImage(IplImage* img)
{
// Convert the image into an HSV image
IplImage* imgHSV = cvCreateImage(cvGetSize(img), 8, 3);
cvCvtColor(img, imgHSV, CV_BGR2HSV);
IplImage* imgThreshed = cvCreateImage(cvGetSize(img), 8, 1);
//For detecting BLUE color i have this HSV value,
cvInRangeS(imgHSV, cvScalar(112, 100, 100), cvScalar(124, 255, 255), imgThreshed);//this will not recognize the yellow color
cvReleaseImage(&imgHSV);
return imgThreshed;
}
Eye detection is much easier with Haar classifier.
link here
Such a simple method may work at extracting a blue object using some thresholding but even if it could be adapted using a different colour black? blue? green? Everyone has different eye colours. I don't see a non hacky method working for you using blob extraction like this based on a HSV threshold value. This method works well on large blocks of the same colour, i.e. removing a blue background.
Look more at shape, everyone has different coloured eyes but the shape is circular/ellipse ish. There are varients of the Hough Transform for detecting circles.
...the Hough transform has been extended to identifying positions of
arbitrary shapes, most commonly circles or ellipses.
I have a basic png file with two colors in it, green and magenta. What I'm looking to do is to take all the magenta pixels and make them transparent so that I can merge the image into another image.
An example would be if I have an image file of a 2D character on a magenta background. I would remove all the magenta in the background so that it's transparent. From there I would just take the image of the character and add it as a layer in another image so it looks like the character has been placed in an environment.
Thanks in advance.
That's the code i would use,
First, load your image :
IplImage *myImage;
myImage = cvLoadImage("/path/of/your/image.jpg");
Then use a mask like this to select the color, you should refer to the documentation. In the following, I want to select a blue (don't forget that in OpenCV images are in BGR format, therefore 125,0,0 is a blue (it corresponds to the lower bound) and 255,127,127 is blue with a certain tolerance and is the upper bound.
I chose lower and upper bound with a tolerance to take all the blue of your image, but you can select whatever you want...
cvInRangeS(image,
cvScalar(125.0, 0.0, 0.0),
cvScalar(255.0, 127.0, 127.0),
mask
);
Now we have selected the mask, let's inverse it (as we don't want to keep the mask, but to remove it)
cvNot(mask, mask);
And then copy your image with the mask,
IplImage *myImageWithTransparency; //You may need to initialize it before
cvCopy(myImage,myImageWithTransparency,mask);
Hope it could help,
Please refer to the OpenCVDocumentation for further information
Here it is
Julien,
I have a project that would classify the color of a pixel. Whether it is red,violet, orange or simply any color in the color wheel. I know that there are over 16 million color combination for pixels. But I was able to read a web page that says its possible for me to do my project using the wavelengths of color. Please give me the formula to compute for the wavelength using RGB values. Thanks!
A pure color has a wavelength (any single color LED will have a specific wavelength).
Red, green and blue each have a range of wavelength. However, when you make an RGB color, you add these wavelengths together, which will NOT give you a new wavelength. The eye can't distinguish a yellow composed of one wavelength from that of adding red and green (just how the eye works).
I'd recommend reading up on color theory
http://en.wikipedia.org/wiki/RGB_color_model
Well RGB for a monitor maps to 3 independant levels of Red Green and Blue light, so there are (mostly) 3 distinct wavelengths present of any one percieved colour.
BUT If you can convert your RGB colour value to its equivilent HSL, the H part (Hue) is the dominant colour in so far as wavelength goes if you are prepared to ignore the saturation (think of it as whiteness).
Based on that you could approximate the dominante wavelength of a colour based on its H value.
Red light is roughly 630–740nm wavelength, Violet is roughly 380–450nm.
Working out wavelength is a bit tricky, and as Goblin mentioned, not always possible (another example is the colour obtained by mixing equal amounts of red and blue light. That purple has no single wavelength).
But if all you want to do is identify the colour by name, then the HSV model would be a good one to use. HSV is Hue (where the colour is around the colour wheel), Saturation (how much colour there is as opposed to being a shade of black/grey/white) and Value (how bright or dark the pixel is). In this case Hue is probably exactly what you want.
If you are using a .NET language, then you're in luck. See the Color.GetHue Method which does all the work for you.
Otherwise, see HSV at Wikipedia for more details.
In essence, if you have R, G and B as floats ranging from 0.0 to 1.0 (instead of ints from 0 to 255 for example), then:
M = max(R, G, B)
m = min(R, G, B)
C = M-m
if M = m then H' is undefined (The pixel is some shade of grey)
if M = R then H' = (G-B)/C mod 6
if M = G then H' = (B-R)/C + 2
if M = B then H' = (R-G)/C + 4
When converting RGB to HSV you then multiply H' by 60 degrees, but for your purposes H' is probably fine. It will be a float ranging from 0 to 6 (almost). 0 is Red (as is 6). 1 is Yellow, with values between 0 and 1 being shaded between Red and Yellow. So 0.5 would be Orange. The important landmarks are:
0 - Red
1 - Yellow
2 - Green
3 - Cyan
4 - Blue
5 - Purple
6 - Red (again)
Hope that helps.
http://en.wikipedia.org/wiki/Visible_spectrum
It is possible. See above. Gray background apparently makes it easier. You might get something like that on your own, and even improve on it. But to do it accurately will cost major dollars. U will need a colorimetry expert, a calibrated monitor and viewing environment (since what the dominant wavelength of your pixel is just means what monochromatic wavelength it approximates on your calibrated monitor in your calibrated viewing environment). All this will be a few thousand dollars. The work done at the above link, shown on wikipedia, does not seem that accurate but it is probably what you want.
Just convert the RGB to HSV then get the HSV value to degrees and this is the answer:
650 - 250 / 270 * D
where D is the degrees.
Considering...
Violet has a 380–450 nm wavelength, &
Blue has a 450–495 nm wavelength, &
Green has a 495–570 nm wavelength, &
Yellow has a 570–590 nm wavelength, &
Orange has a 590–620 nm wavelength, &
Red has a 620–750 nm wavelength,
then you just need to check if it is in those ranges, then you can classify it.
Hope this helps!