`SOLUTION 1 - The Field of View (FOV)``
``is .2 + (.02x2) = .240 inches.`` This gives a
tolerance of the maximum disk size with .01 inches on top and bottom
for location variation.` `Calculate the magnification using
the row dimension of 129 elements (34.52 mile). This is the dominant
dimension in this case since the diameter of the disk need to be
contained within the field of view of the camera. The column
dimension of 514 elements is 174.016 mils.`

`M = .240 inches / .03452 inches = 7.0``
`

`Resolution in the Row Axis = 7.0 x .26772 mils = 1.87 mils
static resolution. Resolution in the column axis = 7.0 x.33858 mils =
2.37 mils static resolution. However, the 174.016 mils column axis
times 7.0 = 1.218 inches. The FOV window at a magnification of 7.0 in
space is .2416 inches by 1.218 inches. This gives a lot of space for
the disk to move around, yet it can still be accurately
measured.`

`This means that if we project the OpticR~4 array into the
object plane, each row axis pixel will have a pitch of 1~87 mils and
each column axis pixel will have a pitch of 2.37 mils.`

`Using the chart for C-mount lenses (``Figures
F-2``) for a magnification of 7.0, the lens to image
distance for different lenses could be:`

`12.5 mm = 2.5"``16 mm = 3.75"``25 mm = 5.75"``50 - mm = 14"``75 mm = 22"`

`To find the image distance, find 7.0 on the magnification
axis. Follow it until it intersects the lens types and read off the
walues of the working distance on the other axis.`

`The disk height variation of 40 mils creates a change of
dimension (magnification change). The percent of dimensional change
is related to the height variation, divided by the lens-to- object
distance times 100 ((Z/C')*100). If a lens extender is required, the
extender length can be calculated by dividing the lens focal length
by the required magnification. Units are in millimeters. The
resulting image will focus when the lens focus control is set in its
mid-point position. The following lenses can all be used to give a
magnification of 7.0:`

`12.5mm`

`=`

__Lens to Image Distance__

`2.5"`

__%Deviation Versus Z Change__

`1.6%`

__Spacer Length__

`1.8mm`

__Focal Lens Selection__

`16mm`

`=`

__Lens to Image Distance__

`3.75``"`

__%Deviation Versus Z Change__

`1.1%`

__Spacer Length__

`2.3mm`

__Focal Lens Selection__

`25mm`

`=`

__Lens to Image Distance__

`5.75"`

__%Deviation Versus Z Change__

`.67%`

__Spacer Length__

`3.75mm`

__Focal Lens Selection__

`50mm`

`=`

__Lens to Image Distance__

`11"`

__%Deviation Versus Z Change__

`.28%`

__Spacer Length__

`7.1mm`

__Focal Lens Selection__

`75mm`

`=`

__Lens to Image Distance__

`17"`

__%Deviation Versus Z Change__

`.18%`

__Spacer Length__

`10.7mm`

`The 75mm lens will provide the least amount of magnification
distortion. If there is enough physical space, then selecting the
75mm lens with a 10.7mm extender ring places the camera ajd lens 22"
above the disk conveyor.`

`The dynamic property of the system is the smudge. As the part
passes the field of view of the camera, the edge of the part is
smudged across several pixels as the camera integrates the light
entering the camera. Since the part is traveling at 15 ft. per
minute, what must the integration time be so that only one pixel will
be smudged? `

`Converting feet per minute to inches per second =`
`15 ft/min * 12 inches/ft * 1 min/60 sec = 3
inches/sec`

`As calulated before,`` ``1 pixel of
the row dimension = 1.87 mils.`` ``This means
that for each frame scan the part can only move 1.87 mils per scan
and since the part travels at 3 inches/second, then:`

`.00187"/scan * see/3" = .000623 see/scan = 623
microsec/scan`

`This is clearly too fast for the camera, which can operate at
only ``4 scans per second.`` What is the
solution? At each scan, the disk moves:`

`3 inches/sec * .25 seconds/scan = .75
inches/scan`

`The part is only .24 inches in diameter. This means for every
scan, the part can move approximately four times its diameter through
the field of view of the camera. The solution is to place a photo
transistor looking across the conveyor to an LED. As the disk blocks
the LED light to the photo transistor, it triggers a strobe light
that is mounted below the translucent conveyor. Select a strobe light
with a flash of peak energy shorter than 613 microseconds. `

`The setup is shown in ``Figure
F-5``. (GIF 12k)`

`As the strobe light flashes, it also triggers the software
that brings in the camera data. The camera integration time is
directly linked with the part pitch. However, care must be taken so
that the integration time does not exceed where the ambient light or
dark current rises above the camera threshold. If the conveyor stops
or no parts come down the conveyor, this fact must be sent to the
software where it will input data from the camera and throw it away
(dummy` `read) to refresh the pixels to keep the camera in the
alert condition. By having a photo-transistor that precedes the
strobe photo-transistor, the first photo-transistor does a dummy
read. This arms the camera and after the flash the camera will
contain the correct data. A strobe light is an effective tool to
freeze action in dynamic situations. However, in many situations a
strobe light may not be required.`

`SOLUTION 2 -`` This solution shows how to
approach the problem without using a strobe light. Assume that an
incandescent light is used to backlight the part, and the OpticRAM is
operated at ``120 frames per second``, which
translates to 8.33 msec/frame. The part is still moving at
``3 inches/sec,`` as we calculated in the
previous solution.`

`Calculate the distance over which the disk is
smudged:`

`.083 see/scan * 3 inches/sec = .025 inches/scan
smudge`

`From scan to scan, the part moves .025 inches. Therefore, the
field of view needs to be the size of the part (.24") plus 2x the
smudge to allow for the smudge of the leading and trailing edges.
Dividing the FOV (.29") by the row dimension (.03452) we are able to
calculate a magnification constant of 8.4.` `Assuming that a
75mm lens was selected gives a distance of 26" from camera to scene
and a deviation of .15 percent of Z-axis magnification change with a
spacer of 8.9mm. The row axis resolution is determined by the product
of .268 mils * 8.4 giving 2.25 mils. The column axis resolution is
the product of .33858 mils * 8.4 giving 2.84 mlls.`

`This means that each edge has a gradient (in this case) of 12
pixel smudge motion. ``See figure
F-6.`` (GIF 12k) If the threshold is centered to the
midpoint of the light amplitude, the 12 pixels that are smudged will
go to 6 pixels (actual edge) on each side. The actual size can be
realized by either changing the intensity of the lamp via a fixed
threshold or by changing the threshold and holding the intensity of
the lamp constant. However, since size is directly related to light
versus threshold levels, the lamp output needs to be accurately
stablilized.`