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How to convert scroll bar position to gamma correction? – Photoshop

I want to build a constant array(1..200) of single numbers to make a ‘lookup table’ to convert the position of the scrollbar which has values 1 – 200 to a value to use for applying gamma correction to an image.

The first array value would have a value somewhere anound 7.0 – 9.9 (photoshop uses 9.9). The middle one, array value 100, would need to be 1.0 exactly. Array value 200 would be about 0.01.

Can anyone tell me which algorithm to use?

I have been attempting to make the array using ‘trial and error’ in some test code based around the function Power(i, 2.2), but got nowhere.

I am using Delphi. I’m not asking the code though, just a steer in the right direction. Any help would be much appreciated.

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2

Answers


  1. 0

    Why not just take the difference and split it up by the number of steps.

    Ex. 9.9-1 / 100 then you know the amount to increment.

    I made a simple example in javascript to show what i mean:

    doWork = function() {

    var result = '';
    var first = 9.9;
    var middle = 1;
    var last = 0.01;

    var firstSteps = (first - middle) / 99;
    var lastSteps = (middle - last) / 100;

    result = result + 'first=' + firstSteps + '<br>';
    result = result + 'last=' + lastSteps + '<br>';

    var value = first;
    for (var i = 1; i < 201; i++) {
        var showValue = Math.round(value * 100) / 100;
        result = result + i + ' => ' + showValue + '<br/>';
        if (i > 99) {
            value = value - lastSteps;
        }
        else if (i == 99) {
            value = middle;
        }
        else {
            value = value - firstSteps;
        }
    }

    document.getElementById('info').innerHTML = result;

}

    Here is the fiddle:
    https://jsfiddle.net/tuv5vfst/

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  2. 0

    This is not really a programming question, but one of math.

    Assume a quadratic function, which is in the form of y=ax2+bx+c.

    Fill in the three value-pairs known:

    9.9  = a + b + c
1    = 10000a + 100b + c
0.01 = 40000a + 200b + c

    Three equations + three unknows = solvable by simple substitution.

    9.9 =   a + b + c
c   =   9.9 - a - b

1   =   10000a + 100b + c
1   =   10000a + 100b + 9.9 - a - b
1   =   9999a + 99b + 9.9
1 - 99b =   9999a + 9.9
-99b    =   9999a + 9.9 - 1
-99b    =   9999a + 8.9
b   =   -101a - 8.9/99

0.01    =   40000a + 200b + c
0.01    =   40000a + 200b + 9.9 - a - b
0.01    =   39999a + 199b + 9.9
0.01    =   39999a + 199(-101a - 8.9/99) + 9.9
0.01    =   39999a - 20099a - 17.89 + 9.9
0.01    =   19900a - 7.99
0.01 - 19900a   =   -7.99
-19900a =   -8
a   =   8/19900
a   =   0.000402

b   =   -101a - 8.9/99
b   =   -808/19900 - 8.9/99
b   =   -0.1305

c   =   9.9 - a - b
c   =   10.0301

    In other words: Gamma = 0.000402 * Pos^2 - 0.1305 * Pos + 10.0301

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