CODEWITHSUNDEEP
javac++objectopencvdetection
hannibal
hannibal

Reputation: 21

Object detection with a generic webcam

Here’s my task which I want to solve with as little effort as possible (preferrably with QT & C++ or Java): I want to use webcam video input to detect if there’s a (or more) crate(s) in front of the camera lens or not. The scene can change from "clear" to "there is a crate in front of the lens" and back while the cam feeds its video signal to my application. For prototype testing/ learning I have 2-3 images of the “empty” scene, and 2-3 images with one or more crates.

Do you know straightforward idea how to tackle this task? I found OpenCV, but isn't this framework too bulky for this simple task? I'm new to the field of computer vision. Is this generally a hard task or is it simple and robust to detect if there's an obstacle in front of the cam in live feeds? Your expert opinion is deeply appreciated!

Upvotes: 0

Views: 2639

Answers (5)

Browny Lin
Browny Lin

Reputation: 2507

There are 2 parts involved in object detection. One is feature extraction, the other is similarity calculation. Some obvious features of the crate are geometry, edge, texture, etc...

So you can find some algorithms to extract these features from your crate image. Then comparing these features with your training sample images.

Upvotes: 0

brainwash
brainwash

Reputation: 689

This is not something you should start in Java. When I had this kind of problems I would start with Matlab (OpenCV library) or something similar, see if the solution would work there and then port it to Java.

To answer your question I did something similar by XOR-ing the 'reference' image (no crate in your case) with the current image then either work on the histogram (clustered pixels at right means large difference) or just sum the visible pixels and compare them with a threshold. XOR is not really precise but it is fast.

My point is, it took me 2hrs to install Scilab and the toolkits and write a proof of concept. It would have taken me two days in Java and if the first solution didn't work each additional algorithm (already done in Mat-/Scilab) another few hours. IMHO you are approaching the problem from the wrong angle.

If really Java/C++ are just some simple tools that don't matter then drop them and use Scilab or some other Matlab clone - prototyping and fine tuning would be much faster.

Upvotes: 0

Adamski
Adamski

Reputation: 54725

Here's an approach I've heard of, which may yield some success:

  • Perform edge detection on your image to translate it into a black and white image, whereby edges are shown as black pixels.
  • Now create a histogram to record the frequency of black pixels in each vertical column of pixels in the image. The theory here is that a high frequency value in the histogram in or around one bucket is indicative of a vertical edge, which could be the edge of a crate.

You could also consider a second histogram to measure pixels on each row of the image.

Obviously this is a fairly simple approach and is highly dependent on "simple" input; i.e. plain boxes with "hard" edges against a blank background (preferable a background that contrasts heavily with the box).

Upvotes: 2

Brad Greenway
Brad Greenway

Reputation: 1

Lots of variables here including any possible changes in ambient lighting and any other activity in the field of view. Look at implementing a Canny edge detector (which OpenCV has and also Intel Performance Primitives have as well) to look for the outline of the shape of interest. If you then kinda know where the box will be, you can perhaps sum pixels in the region of interest. If the box can appear anywhere in the field of view, this is more challenging.

Upvotes: 0

Bernd Elkemann
Bernd Elkemann

Reputation: 23560

You dont need a full-blown computer-vision library to detect if there is a crate or no crate in front of the camera. You can just take a snapshot and make a color-histogram (simple). To capture the snapshot take a look here:

http://msdn.microsoft.com/en-us/library/dd742882%28VS.85%29.aspx

Upvotes: 0

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