I want to quantify the flow from photos taken from a sensor such as this one: Flowmeter

For a limited number of readings, I can use a plot digitization method. However, I need to do this many times for hundreds of photos to construct the flow as a function of time. The location of the sensor is somewhat challenging, so I can't use a camera stand either (meaning the position of min and max are probably not constant in time due to hand held recording device). Therefore, I am trying to make this process automated. The location of the top of the ball with respect to the scale is what I am looking for. The approach I have in mind follows, but if anyone can suggest a better approach that would be even more appreciated.

I imported the image and called it img1. Then I binarized it:

bin = Binarize[img1, 0.3]

Then I negated the image and cleaned it up slightly:

cleanup = DeleteSmallComponents[ColorNegate@bin, 50]

MMA's text recognition doesn't seem to be working with these numbers, so I gave up trying to use it. However, after many tests I found MorphologicalPerimeter quite useful:

HighlightImage[img1, MorphologicalPerimeter[cleanup]]


Also, I found out these thresholds can isolate the scale:

SelectComponents[cleanup, #Elongation > .4 && #AdjacentBorderCount == 
0 &]

and for the ball:

DeleteSmallComponents[cleanup, 1500]

However, now I am stuck. I don't know how to construct a tangent on top of the ball and determine where it lands on the scale e.g. where it intersects the line connecting the middle of indicators for 0.1 and 1.0.

  • 1
    $\begingroup$ It’s probably easier to annotate manually two of the numbers, e.g. 0.2 and 1, on the axis along which the ball moves. By determining the ball’s position with regards to these two annotated points, the value can then be determined using linear interpolation. Would something like this be acceptable? $\endgroup$
    – C. E.
    Mar 21 at 2:45
  • $\begingroup$ That's a good idea but for a limited number of cases. I need to do this completely automated as a part of a larger code, for many photos created by extracting frames from a video file. Because I'm holding my phone in my hand for recording, the locations of min and max aren't fixed in time. The manual method would probably work with existing plot digitization responses and packages already available from other posts. Thanks for the comment though :) $\endgroup$
    – MathX
    Mar 21 at 3:01
  • 2
    $\begingroup$ Then in that case would it be possible to tape markers onto the tube that the ball moves through? The markers can be made easy to find, and finding them would again provide us with the same kind of reference manual annotation would. One could also use a 2D marker (think identifiable rectangle) to find the perspective transformation, which would increase the precision of the solution. I understand if you don’t want to solve the problem in this way, but just making sure. $\endgroup$
    – C. E.
    Mar 21 at 3:37
  • $\begingroup$ Oh I guess now I understand what you mean. That could actually work. But before making physical changes to the sensor, what would be the next steps? When my hand is shaking the location of those tapes would probably be changing too. Do you have any suggestions on how to detect the tape locations, and can't whatever method is used to do that be used to find the current existing features? $\endgroup$
    – MathX
    Mar 21 at 3:59
  • 1
    $\begingroup$ I think you need to take @C.E.'s comment seriously. Adding three markers to the plywood supporting the sensor will make this a more tractable problem especially since it sounds like you will not use some stabilizing support for your video recording device. $\endgroup$ Mar 21 at 12:48

Here is the start of a solution. It won't be fully automated if the different images have different lighting conditions, such that different thresholds are needed for Binarize to pick up all the scale tick marks.

Starting with the OP's code:

bin = Binarize[img1, 0.3]
cleanup = DeleteSmallComponents[ColorNegate@bin, 50]

enter image description here

Assuming the ball is always the left-most component:

comps = MorphologicalComponents[cleanup];
ball = First[SortBy[ComponentMeasurements[comps, "Centroid"], #[[2, 1]] &]]
 (* 8 -> {308.094, 423.722} *)

HighlightImage[img1, Point[ball[[2]]]]

enter image description here

The top of the ball can be estimated from the bounding box:

ballbounds = 
 ComponentMeasurements[SelectComponents[comps, #Label == ball[[1]] &],"BoundingBox"]
 (* {8 -> {{285., 400.}, {331., 447.}}} *)

balltop = ballbounds[[1, 2, 2, 2]]
 (* 447. *)

enter image description here

To get the scale, I selected components whose left edges were less than 100 pixels away from the right edge of the ball:

ballright = ballbounds[[1, 2, 1, 1]];
scalecomps = 
   SelectComponents[comps, #Label != ball[[1]] &], "BoundingBox"], (#[[2, 1, 1]] - ballright < 100) &];

Then defined scale points as {x,y} = {left, bottom+(top-bottom)/2}

scalepts = {#[[1, 1]], #[[1, 2]] + (#[[2, 2]] - #[[1, 2]])/2} & /@ scalecomps[[All, 2]];
 (* {{377., 662.5}, {375., 611.5}, {373., 567.5}, {371., 526.5}, {369., 475.5}, {366., 427.}, {364., 376.}, {362., 326.}, {360., 271.5}, {356., 208.5}, {359., 163.5}} *)

HighlightImage[img1, Point@scalepts]

enter image description here

Then calibrate the points with the scale values (I couldn't figure out how to get rid of the logo below the scale, so I called it the 0 tick, even though it's not quite right...)

f = Fit[Transpose[{Sort[scalepts[[All, 2]]], Range[0, 1, .1]}], {1, x}, x]
 (* -0.338423 + 0.00199797 x *)

 ListPlot[Transpose[{Sort[scalepts[[All, 2]]], Range[0, 1, .1]}]],
 Plot[f, {x, Min[scalepts[[All, 2]]], Max[scalepts[[All, 2]]]}]]

enter image description here

The measured flow rate, using the top of the ball, is:

f /. x -> balltop
 (* 0.554672 *)
  • $\begingroup$ This is great, thanks! I will get rid of the logo somehow, either by removing the lowest bounding box in the code or by trying to hide it under a light-colored tape in line with the comments. $\endgroup$
    – MathX
    Mar 21 at 13:33
  • $\begingroup$ @MathX - both good ideas. $\endgroup$
    – MelaGo
    Mar 21 at 16:47

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