`We have talked so far about what happens to the middle of the
part, now we need to talk about what happens at the left or right
edge of the part in a dynamic situation. (Refer back to
``Figure F-4`` GIF
20k)`

`Assuming the right most edge or left most edge covers a pixel,
the question is, for what duration is the pixel covered? Assume from
scan to scan that the disk moves ``.025
inches``. Using the formula for a chord of a circle
``(Figure F-7),``
``(GIF 16k) we need to determine the error at point A and
point B.`

`From earlier calculations, ``the pixel width in the
column axis of 2.84 mils, with .1 inch radius is:`

`Co = SQRT(4(2*.00284*.1 - .00284~)) = .0476`

`.0476" * scan/.025" = 190%``of the time the A and B pixels are dark suggesting that the error at points A and B is neglible. (190% of the time is an awkward way of saying that the disk travels only about half the distance between points A and B in one scan period. then the percentage exceeds 60%, we can say for certain that the left/right edge pixel represents the part. Motion is always a problem even in static situations because between the camera and the scene there is vibration which may require careful attention to detail.`

`Once data is captured either by a strobe lamp or by back
lighting (shadowing) and stored in the computer memory, statistical
averaging is then done in order to improve the data. `

`EXAMPLE:`` Using the formula to find how many
row pixels should come dark at the same time at the entry and exit,
``The row resolution is .00225 inches per pixel. Using the
formula for a chord of a circle:`

`.042 mils * pixel/.00225 mils = 18
pixels`

`This indicates that if 18 pixels are averaged at the max/min
points then the resolution and accuracy can be increased by a value
of:`

`SQRT(number of pixels) / 2`

`Find the midpoint of the circle, then average the 10 pixels on
either side (20 pixels)`

`SQRT(20)/2 = 4.5/2 = approx 2`

`This suggests a half a digit increase in accuracy.`

`1.84 mile/.2 diem * 100 = .92% + .23% for Z axis Motion =
1.15%`

`1.15% / 2 = .575% resolution (after
calibration).`

`From disk to disk, one should be able to resolve each disk to
about .6% The design goal was 1%. If it is desired, an out of round
figure of merit can also be calculated:`

`area = pi * R*R`

`circumference = 2 * pi * R`

`area/circumference = R/2`

`Adding the area pixels and dividing by the edge pixels, should
give a number close to half the radius pixel as a ratio. The ratio
should hold. If it does not, this is an indication of out of
roundness. One can also sort parts for rough out-of-round
tolerances.`