logajan@ns.network.com (John Logajan) (04/05/91)
Howard Chu suggests alternating frames (flicker palettes) to increase
the apparent number of available intensities per pixel R G or B.
He also (cleverly) suggests an unbalanced duty cycle -- two frames
of palette X and one frame of palette Y.
Dave Baggett mentions that some combinations of flicker intensities
are unsatisfactory, especially flickering between two extremes.
So I have made some tables of the possibilites of such an unbalanced
duty cycle flicker method. In the first table you can see the
redundancy of many of the intensity values. For instance many of
the intensities can be derived four different ways.
Y Palette value (1/3 duty cycle)
7 6 5 4 3 2 1 0
| | | | | | | |
7 - 21 20 19 18 17 16 15 14 . . . . . . . . . . . . . .
6 - . . 19 18 17 16 15 14 13 12 . . . . . . . . . . . .
5 - . . . . 17 16 15 14 13 12 11 10 . . . . . . . . . .
4 - . . . . . . 15 14 13 12 11 10 9 8 . . . . . . . .
3 - . . . . . . . . 13 12 11 10 9 8 7 6 . . . . . .
2 - . . . . . . . . . . 11 10 9 8 7 6 5 4 . . . .
1 - . . . . . . . . . . . . 9 8 7 6 5 4 3 2 . .
0 - . . . . . . . . . . . . . . 7 6 5 4 3 2 1 0
| | | | | | | |
^ 7 6 5 4 3 2 1 0
|
|----- X Palette value (2/3 duty cycle)
In case you can't understand the above table, the one below puts the
Y palette value in better alignment and throws out selected redundancies.
Y Palette value (1/3 duty cycle)
7 6 5 4 3 2 1 0
| | | | | | | |
7 - 21 20 . . . . . .
6 - 19 18 17 . . . . .
5 - . 16 15 14 . . . .
4 - . . 13 12 11 . . .
3 - . . . 10 9 8 . .
2 - . . . . 7 6 5 .
1 - . . . . . 4 3 2
0 - . . . . . . 1 0
^
|
|----- X Palette value (2/3 duty cycle)
Finally, I have a table of average values of intensity.
Note that the difference between X and Y values is never
greater that +or- 1. This should minimize the AC ability
of the eye to detect flicker. If our eyes were DC only,
then any of the other redundant combinations would have
worked as well. :-)
X X Y # Intensity (average)
----- -- ---------
7+7+7 = 21 = 7
7+7+6 = 20 = 6.666
6+6+7 = 19 = 6.333
6+6+6 = 18 = 6
6+6+5 = 17 = 5.666
5+5+6 = 16 = 5.333
5+5+5 = 15 = 5
5+5+4 = 14 = 4.666
4+4+5 = 13 = 4.333
4+4+4 = 12 = 4
4+4+3 = 11 = 3.666
3+3+4 = 10 = 3.333
3+3+3 = 9 = 3
3+3+2 = 8 = 2.666
2+2+3 = 7 = 2.333
2+2+2 = 6 = 2
2+2+1 = 5 = 1.666
1+1+2 = 4 = 1.333
1+1+1 = 3 = 1
1+1+0 = 2 = 0.666
0+0+1 = 1 = 0.333
0+0+0 = 0 = 0
--
- John Logajan @ Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428
- logajan@ns.network.com, 612-424-4888, Fax 612-424-2853