Camera order

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rprueckl
Posts: 4
Joined: Wed Jan 15, 2020 3:28 am

Camera order

Post by rprueckl » Wed Jan 15, 2020 3:31 am

Hi,

in Motive there is the option to sort the cameras according to their serial number instead of the 'arbitrary' ordering.
Is something like this also possible via the Camera SDK? I have found the function

Code: Select all

SuggestCameraIDOrder
which could bethe right direction, but I do not know how to use it.

many thanks for your advice,
robert

beckdo
NaturalPoint Employee
NaturalPoint Employee
Posts: 520
Joined: Tue Jan 02, 2007 2:02 pm

Re: Camera order

Post by beckdo » Wed Jan 15, 2020 12:47 pm

Hi Robert,

You are correct. The Camera ID of a camera is primarily a human readable number that can be displayed on the camera and/or in your software and UI. For USB cameras, their Camera ID is fixed based on it's physical connection. Now for Ethernet cameras, the Camera ID is 'arbitrarily' assigned as you suggest.

If you want to have your Ethernet cameras report Camera ID's in a certain order, create a list of camera serial numbers and pass it to SuggestCameraIDOrder. Following the call to SuggestCameraIDOrder, you should be able to call CameraID() on your cameras and those Camera IDs will reflect the order you suggested in your call to SuggestCameraIDOrder.

Thanks,
Doug

rprueckl
Posts: 4
Joined: Wed Jan 15, 2020 3:28 am

Re: Camera order

Post by rprueckl » Mon Feb 10, 2020 8:17 am

Hi Doug,

many thanks for the advice. I will give that a try as soon as I get back to the cameras and report back here.

best,
Robert

rprueckl
Posts: 4
Joined: Wed Jan 15, 2020 3:28 am

Re: Camera order

Post by rprueckl » Wed Apr 01, 2020 11:44 pm

Hi again,

I tried your suggestion and it worked so far. However, it did not influence the order in which

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CameraLibrary::CameraList list
delivers the cameras or the order of frames coming in from the cameras.

Due to that, I decided to manage a separate list of serial numbers in the desired order to manage the cameras and influence the frames of the cameras.

Thanks,
Robert

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