QUOTE (Originally by Plumb)
I did a little experiment, made a table 1000 units wide, had a ball roll to the flat bottom, copied to photoshop, and duplicated the balls from edge to edge along the bottom, and I came up with, based on the known fact that pinballs are 1 1/16" in diameter...
1000
VP units = 21.25" or 539.75 mm
therefore...
1 vp unit = .53975 mm
47
VP units = 1 inch
therefore...
In Dimension Manager a standard 20,25" x 42" playfield is 952x1974.
This bothered me a bit, since 1 inch is not 47, but 47,05882352941176 VP Units.
Which means a standard table is 952,9411764705882 by 1976,470588235294. Rounded 953 x 1976, not 952x1974.
Still bothered, I also did a little experiment. This time in VPX 10.0.8 beta5.
Made an empty table, 1000 units wide, and dropped pinballs into it. 20 pinballs fit fine.
Then i made the table 997 units wide. The 20 pinballs didn't fit correctly.
And then 998 units wide. And lo and behold, the 20 pinballs fit!
Therefore...
998 VP Units = 20 pinballs
1 pinball = 1.0625 inches
1 pinball = 49,9 VP units
1 inch = 46,96470588235294 VP Units
And a standard 20.25" x 42" table is 951,035 by 1972,517647, rounded 951 x 1973.
So by rounding down from 47.06 to 47, we were closer to the "real" value of 46.96 by chance. But multiplied by 42 or more, still off by a couple of pixels.
What do you think?
Edited by hmueck, 26 September 2023 - 06:10 PM.