How to use Unit CamS3



  • Hello,
    I've been trying to use the CamS3 for a simple project (capture photos on demand as response to a trigger) and I have a few issues:
    -when configuring the camera by connecting to it on IP 192.168.4.1, there is a "stream" button leading to address 192.168.4.1/api/v1/stream. The only issue is that stream quality is 0.3 MP. I'd like to use the full capability of 2MP. The API also supports "capture" command on address 192.168.4.1/api/v1/capture. Does anyone know how I could capture images on 2MP? Does API v1 support this? Any chance of API v1 docs to figure it out?
    -if I configure the camera to save to SD card, quality is set to 640x480 = 0.3 MP. How can I change that to 2MP?
    -if I configure the camera to post images to EzData, only 1 picture (the first ever) is posted, and then nothing else, no matter what I try. Has anyone had luck with that? It would be ideal for my needs (again, in full resolution)
    -any plans to have other software installed? Maybe possibility to use it with UiFlow?

    Thank you.



  • To partially answer my own question, I have found useful information about the cam webserver API here:
    https://github.com/easytarget/esp32-cam-webserver/blob/master/API.md
    In short, there are a few other useful endpoints, like /status and /control. In order to change resolution, the uri to access would be http://<IP-ADDRESS>/control?var=framesize&val=8
    Not all the commands in the link work, unfortunately (and, quite important, the 'save_prefs' does not) but it can be used to programatically access and save frames at the specific resolution one wants, which is all I need.
    It appears that the cam webserver is the standard ESP32 CameraWebServer, but I could not find an official API doc just yet. Some more info here: https://github.com/espressif/arduino-esp32/blob/master/libraries/ESP32/examples/Camera/CameraWebServer/app_httpd.cpp
    I find it odd that a simple API documentation of the CameraWebServer is not offered on the product page, it would be most useful.