MAGNIFICATION USING THE ROW AXIS

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:

Focal Lens Selection

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.