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Creating a Linear Gradient with Java for a drop down menu

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#1 Hankenstien   Members   -  Reputation: 128

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Posted 05 August 2014 - 12:56 PM

 
I have been playing around with creating a UI in Java's awt class.  I made a drop down menu that animates open and close when hovering.  Right now I have a linear gradient that runs from the top of the drop down to the bottom and I don't like the look of it.  I want the gradient to still go from top to bottom, but each Rectangle2D in the menu should have a solid color of that gradient.  I want to use a for loop to progress throught the menu's rectangles and fill each with a slowly darkening/lightening shade of the first color or maybe slowly progress from one color to another. 
 
Something like this is what I was thinking but the color seem to jump from dark to light and its not a smooth transition.
 
int tempColor = 0x0000ff;
 
for(int i = 0; i < rectList.size(); i++){
        tempColor -= (i * 10);// if I was wanting a red transition I think I might multiply by 10000
        g.setColor(tempColor);
        g.fill(rectList.get(i));//fills with new gradient color
        g.setColor(Color.black);
        g.draw(rectList.get(i));//border
    }
 
On a side note this is the first time I have ever asked for help online, but thought I would give it a try.  Any help would be appreciated.


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#2 Glass_Knife   Moderators   -  Reputation: 4085

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Posted 05 August 2014 - 02:02 PM

It is a common mistake when trying to interpolate a color encoded in a 32bit value to just adjust the value.  But that doesn't work (as you have found out).

 

1.  Use the Color( float, float, float ) constructor that takes values from 0.0 - 1.0, and perform your linear interpolation on each value separately.

2.  Keep a different integer for each red, green, and blue component, clamped between (0, 255), and then OR them together to create the final color.  
    int color = 0xFF << 24 | red << 16 | green << 8 | blue;  // the 0xFF is the alpha value.


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#3 Hankenstien   Members   -  Reputation: 128

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Posted 05 August 2014 - 02:11 PM

Thanks a lot Glass_Knife!  It works great!



#4 Chris_F   Members   -  Reputation: 2226

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Posted 05 August 2014 - 04:55 PM

If banding is noticeable you might want to consider adding a simple random dither to it.



#5 Waterlimon   Crossbones+   -  Reputation: 2458

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Posted 05 August 2014 - 05:13 PM

I did something like that once and faced an additional problem. Something to do with "gamma correction".

 

Basically even if the numbers were linear (as in 0.1, 0.2, 0.3), the actual color produced didnt look like it brightened in linear steps.

 

So what I did, was take the linearly interpolated value (between 0 and 1!) and take the square root of it (actually you should do pow(value, 1.0f/2.2f) but I found square root to be close enough for my purposes) and use that instead

 

so see if it looks any better if you use

 

Color(sqrt®,sqrt(g),sqrt(b)) instead of Color(r,g,b)

 

I dont know too much about this topic though.


o3o


#6 Chris_F   Members   -  Reputation: 2226

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Posted 05 August 2014 - 05:17 PM

I did something like that once and faced an additional problem. Something to do with "gamma correction".

 

Basically even if the numbers were linear (as in 0.1, 0.2, 0.3), the actual color produced didnt look like it brightened in linear steps.

 

So what I did, was take the linearly interpolated value (between 0 and 1!) and take the square root of it (actually you should do pow(value, 1.0f/2.2f) but I found square root to be close enough for my purposes) and use that instead

 

so see if it looks any better if you use

 

Color(sqrt®,sqrt(g),sqrt(b)) instead of Color(r,g,b)

 

I dont know too much about this topic though.

 

Gamma correction isn't necessary unless you really desire a linear gradient. Gradients in Photoshop, for example, are not linear.



#7 Hankenstien   Members   -  Reputation: 128

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Posted 05 August 2014 - 05:40 PM

If banding is noticeable you might want to consider adding a simple random dither to it.

do you mean just a random float that is less than what i subtracted added back into the amount or just subtracting a random float from the value to get the new value?



#8 Chris_F   Members   -  Reputation: 2226

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Posted 05 August 2014 - 06:57 PM

 

If banding is noticeable you might want to consider adding a simple random dither to it.

do you mean just a random float that is less than what i subtracted added back into the amount or just subtracting a random float from the value to get the new value?

 

 

Normally you would use floor(value * 255.0f + 0.5f) to convert your [0, 1] value to a 24-bit image. Instead use floor(value * 255.0f + random), where random is a float in the range [0, 1]. You can see the effect this has here.



#9 Hankenstien   Members   -  Reputation: 128

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Posted 05 August 2014 - 08:05 PM

 

 

If banding is noticeable you might want to consider adding a simple random dither to it.

do you mean just a random float that is less than what i subtracted added back into the amount or just subtracting a random float from the value to get the new value?

 

 

Normally you would use floor(value * 255.0f + 0.5f) to convert your [0, 1] value to a 24-bit image. Instead use floor(value * 255.0f + random), where random is a float in the range [0, 1]. You can see the effect this has here.

 

Thanks Chris this explains a lot with problems I have been having.







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