4 ms·
Am I the only one who thinks the concept of relative error is not meaningful in this context? It gives a disproportionate meaning to 0 without real physical co
by fchu 6y ago
Am I the only one who thinks the concept of relative error is not meaningful in this context?
It gives a disproportionate meaning to 0 without real physical consideration, eg:
- 0.1⁰C ± 0.1 (wow 100% relative error)
- 273.25K ± 0.1 (meh 0.04% relative error)
- jcheng 6y agoThis is talking about the error in the difference between two values with the same units though. For temperature, it wouldn’t matter if you’re using C, K, or F for your starting values, the % error of the difference would be the same (I think).
- eru 6y agoYes. What would make a difference is to change to https://en.wikipedia.org/wiki/Thermodynamic_beta https://en.wikipedia.org/wiki/Thermodynamic_beta (Essentially the same difference as miles per gallons vs litres per 100 km.)
- Archelaos 6y agoI am sometimes joking with friends by suggesting: Let's meet at 12 o'clock ± 5%.
- andrepd 6y agoYes, because there is an arbitrary choice of origin which renders the relative error dependant on units. If you're measuring a length, for instance, or an interval of time, the relative error is independent of which units you choose. If you're measuring e.g. a distance to some point, then again you have an arbitrary choice of origin.
- _Microft 6y agoThe Celsius temperature scale is an interval scale [0] which means that it is possible to calculate differences but not ratios. The Kelvin temperature scale is a ratio scale [0] (it has an "absolute zero") that allows to do that. Beside that if there are uncertainties involved, one should do proper propagation of uncertainty anyways. [1] [0] https://en.wikipedia.org/wiki/Level_of_measurement https://en.wikipedia.org/wiki/Level_of_measurement [1] https://en.wikipedia.org/wiki/Propagation_of_uncertainty https://en.wikipedia.org/wiki/Propagation_of_uncertainty