Difference between accuracy and resolution

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Accuracy refers to how close a measurement is to the true value, while resolution indicates the smallest increment that a measuring device can detect. In the example of two tape measures, one with 1/8-inch increments (A) is less precise than the other with 1/32-inch increments (B), which has higher resolution. However, if tape measure B is not used correctly, its accuracy can be compromised due to potential play in its design. Thus, a device can have high resolution but still yield inaccurate measurements if not handled properly. Understanding the distinction between these two concepts is crucial for effective calibration in systems like ventilation and heating.
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I have a few questions regarding measuring devices in systems such as ventilation, heating, cooling etc. When calibrating these I come across the terms "accuracy" and "resolution" (directly translated from norwegian). I a bit confused about the difference of these. Can someone explain this to me...?
 
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A layperson's answer - might not be as formal as someone in the industry:

I have two tape measures A and B, A measures down to 1/8th inch increments, while B measures down to 1/32 inch increments. B has higher resolution.

B has a little hook on the end on a slidey bit with a rivet (it took me a long time to figure out what that was for). Anyway, the slidey bit allows for too much play when I measure; it can slide around by as much as 1/16th of an inch.


Ironically, while tape measure B has higher resolution, if I don't use it correctly, it is actually less accurate.
 
Just to echo what Dave mentioned:

Accuracy: The closeness of a measurement to the actual value being measured.

Resolution: The smallest detectable increment that an instrument will measure/display to.
 
Accuracy: accuracy is the maximum spread in measurements made of slide movements during successive runs at a number of target points
Resolution: Resolution refers to the smallest units of measurement that the system can recognize.
 
I have Mass A being pulled vertically. I have Mass B on an incline that is pulling Mass A. There is a 2:1 pulley between them. The math I'm using is: FA = MA / 2 = ? t-force MB * SIN(of the incline degree) = ? If MB is greater then FA, it pulls FA up as MB moves down the incline. BUT... If I reverse the 2:1 pulley. Then the math changes to... FA = MA * 2 = ? t-force MB * SIN(of the incline degree) = ? If FA is greater then MB, it pulls MB up the incline as FA moves down. It's confusing...

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