Can't View Original & Processed Image

[OWilliamson]

On our dashboard the original camera image is showing perfectly, but when I add another indicator for our processed image it only shows the processed image and not the original. I put the two images on separate tabs but I'm still getting this issue.

Any ideas on how to make them both show up?

[Joe Ross]

Can you post your code? There are a few things with the way that IMAQ handles images that could cause an issue.

[mesamb1]

two things right off the bat to check. Did you make sure there is two declared imaqs? one to display unprocessed ,and the second to display processed? Secondly, is your process display set to display binary image(assuming you are using a variation of the processing code FIRST sent out). Both are easy to forget, and with out both it is easy to accidentally process your imaq image meant as the original followed by neither showing because they are both binary images, which isnt labview's default output

[OWilliamson]

This is the first part of the project. I only have 1 IMAQ and it is for the host image. I kept the original image indicator but copied in the indicator for the processed image from the example project and put that in later in the code. Do I need to have 2 IMAQs to view them both?

[Greg McKaskle]

Yes. The IMAQ image is unlike other LV datatypes. Due to the large amount of data in an image, it acts as a refnum. Your diagram defines the allocation and destruction of images, and branching the wire doesn't create a new one for you.

To correct your diagram, create another image on the outside left of the loop and wire that into the destination of the threshold. That will preserve the original image so that it can be displayed.

Greg McKaskle

[OWilliamson]

So would I do it like this or do I need to have another Read MJPG with the original not connected at all? Also, do I leave the IMAQ as the RGB U32 like the original image?

[Greg McKaskle]

Good questions.

If you right click on the Image Threshold VI and choose Help, it will take you to the reference manual on this VI. It clarifies ...

Image Dst is a reference to the destination image. If Image Dst is connected, it must be an unsigned 8-bit image.

So, you don't need to get another image. Your diagram is very close, but make the dst be the correct type of image. If you run it the way you have it, you will receive runtime errors at that node.

Greg McKaskle

[OWilliamson]

I updated the dst to be U8, is that all I have to do? I can't test right now because I don't have access to the camera.

[Joe Ross]

You also need to make sure that both images have a unique name.

[OWilliamson]

I changed the name, is there anything else I need to change?

[Greg McKaskle]

Good catch. I missed the name conflict. That should be it.

Greg McKaskle

[OWilliamson]

Thanks for all the help, good luck to you guys going to competition!