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sekdiy avatar sekdiy commented on August 26, 2024

Let's try the Step-by-step procedure from the Calibration page in order to get on the same page.

  1. Software: I'm assuming you're using the Calibration example.
  2. Setup: What capacity and kFactor did you use as initial values, just so that I can comprehend.
  3. Bucket: What size are you using?

Maybe you can disclose the sensor you'r using, so that I can follow your calculations.

How do we decide which interval value should correspond to the M factor value ?

The reference will always be your sensor's capacity.
This 100% reference is then divided into ten intervals.

PS: Please note that it will not be possible to change the range of your sensor to 40% of it's original capacity. Calibration can increase the measurement precision within your sensor's range, not change the range.

from flowmeter.

sekdiy avatar sekdiy commented on August 26, 2024

Closed due to inactivity.

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sekdiy avatar sekdiy commented on August 26, 2024

Someone asked me:

I want to make sure I understand the M factors setting correctly, can you tell me if the following statements are correct?

So I thought I'll just answer here quickly:

  1. You use only one parameter for upper boundary (capacity) so the lower boundary is always 0

Yes. Capacity really is a scalar property. The usable range is usually smaller, the linear range even more so.
I somehow made the assumption that flow could stop – and that this should be detectable.

  1. If my sensor has a capacity of 5 l/min but I use it only up to 3 l/min I can set the capacity to 3 and be more accurate with the M factors

Yes. Should your flow rate surpass the 3 l/min, the library will just assume the upper boundary:

this->_currentCorrection = this->_properties.kFactor / this->_properties.mFactor[min(decile, ceiling)]; //!< combine constant k-factor and m-factor for decile

  1. If I use capacity of 3 then the M factors cover these ranges: 0-0.3, 0.3-0.6, 0.6-0.9 .... So I actually need to measure the flowrate at 0.15, 0.45, 0.75 to be the most accurate

Yes. I would assume that the linear approximation used should work that way, although I haven't double-checked if that's what the code really does. :D

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