Basler slA1000-30fm, User Manual, Page: 145 / 166

Using Multiple Cameras on a Single Bus and Managing Bandwidth
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You may be asking why we multiply the percentage for camera 1 by 4096 and the percentage for
camera 2 by 8192. The reason is:
During the part of the bus cycle when the packet for camera 1 is transmitted, the bus will operate
at S400 speed. At S400, the maximum number of bytes that can be transmitted in a bus cycle
is 4096.
During the part of the bus cycle when the packet for camera 2 is transmitted, the bus will operate
at S800 speed. At S800, the maximum number of bytes that can be transmitted in a bus cycle
is 8192.
Example 2: Assume that you have three cameras on the bus and that you want these camera to
capture and transmit images simultaneously. Camera one is operating at S800 speed and is set for
a packet size of 4200 bytes. Camera two is operating at S800 speed and is set for a packet size of
1800 bytes. Camera 3 is operating at S400 speed and is set for a packet size of 1000 bytes. How
much of the available bandwidth would each camera use?
For camera 1, the calculation would be:
4200 / 8192 = 51.3%
For camera 2, the calculation would be:
1800 / 8192 = 22.0%
For camera 3, the calculation would be:
1000 / 4096 = 24.4%
If you add these three results together, you find that 97.7% of the available bandwidth is being used.
Keep in mind that if the sum was greater than 100%, you would need to lower the packet size setting
for one or more of the cameras.