Convert Precise Readings to Accurate Values Without Calibration

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Discussion Overview

The discussion revolves around the challenge of converting precise readings from measuring instruments to accurate values without calibration. Participants explore the relationship between precision and accuracy, and the implications of measurement systems in this context.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • Some participants question whether it is possible to convert precise readings to accurate values without calibration, suggesting that the readings are inherently limited by the instrument's accuracy.
  • One participant emphasizes the distinction between accuracy and precision, noting that a measurement system can exhibit various combinations of both qualities.
  • Another participant argues that the original question is nonsensical, as it implies a method of calibration without actually calibrating.
  • A participant shares a personal context, mentioning an assignment related to measuring cylinder accuracy, and expresses uncertainty about achieving accuracy from imprecise scales.
  • Some suggest that applying a correction factor could potentially account for biases in measurements, but this would require knowledge of the necessary correction, which typically involves using a more accurate device.

Areas of Agreement / Disagreement

Participants generally disagree on the feasibility of achieving accurate values from precise readings without calibration. Multiple competing views remain regarding the definitions and implications of accuracy and precision.

Contextual Notes

Limitations include the dependence on the definitions of accuracy and precision, as well as the unresolved nature of how to determine necessary corrections without additional accurate instruments.

Cashlover123
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How can anybody convert precise readings from any measuring instruments to the accurate value without calibration?
 
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Cashlover123 said:
How can anybody convert precise readings from any measuring instruments to the accurate value without calibration?

Beats me. Is it a trick question?
 
Surely you can't. Your readings are only as good as the instrument. I don't see how you could convert that to the actual reading.

If my desk is 40.5 inches and I measure it at 40 +-1 then there's no way to get to the real value. You can only improve the accuracy of your reading by using more precise instruments.
 
Accuracy and precision are two different things. A measurement system can be accurate but not precise, precise but not accurate, neither, or both.

See Accuracy versus precision; the target analogy here:

http://en.wikipedia.org/wiki/Accuracy_and_precision
 
Right, and the implication that has for the OP's question is that you can turn a precise set of readings into an accurate and precise set of readings if you know the offset...which is, of course, the definition of "calibration". So the question begs the answer it rejects. It's like asking "how can you calibrate without calibrating?" It's nonsensical.
 
Maybe i was not clear enough, i got this assignment on how to have accurate readings from a measuring cylinder. Now when you are taking readings, the scales in the cylinder are not accurate. Of course it gives precise measurements but accuracy is important. That's why i was wondering if there is anyway to have accuracy from any measuring device.
BTW, i just wanted to try my luck to see from previous question if there was just any conversion equations, which sounds stupid to me now. Thanks.
 
If it has a fixed scale, you won't get more accurate than is already there.

To improve it, you would need remove the old scale and put a new, more precisely calibrated one on.
 
You can try to account for this bias, i.e. apply a correction. But of course you would need to figure out what this correction needs to be.

Take any measurement, you could think of a ruler as an example. Maybe the measurements you make need to be corrected by an offset. Or maybe you need to multiply the measurements by some factor. Or maybe you need to do both of the previous steps, say measurement corrected=measurement*a+b. Or it might be more complicated than this. So really this correction depends and can only be figured out using another device that is more accurate than the one you have.
 

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