Could uncertainty be that big?

[peripatein]

Another question I have concerns a table's surface area.
If L=122.14±0.14[cm] and W=24.30±0.57[cm], I got that S=2968.00±69.70[〖cm〗^2], using ∆S=√((∂S/∂L)^2 〖∆L〗^2+(∂S/∂W)^2 〖∆W〗^2 ).
Would you kindly confirm this result? Is it plausible that ∆S would be nearly 70 cm^2??

 

[digfarenough]

I do notice that if W had an error of 0.5 cm, then the error in the surface area would be around 0.5 cm * L = (0.5 cm)(122 cm) = 61 cm ^2 which is certainly close to 70 cm^2.

Others will surely have more insightful input!

[Edit. Just to be clear, though my argument is basically just half of the calculation you did, I'm trying to point out it seems pretty reasonable to me! A small error on the width times a long length can produce a sizable change in surface area.]

 

[haruspex]

What exactly do you mean by a ± error? In engineering terms, this usually means the actual limits of error and does not imply any particular distribution beyond that fact. In that model, the range for the area is min length * min width to max length * max width.
Your sum-of-squares approach effectively interprets the ± in the source data as meaning some (unstated) number of standard deviations.
In the numbers you quote, it happens that the (much) larger error range goes with the smaller dimension. As a result, your sum-of-squares calculation produces pretty much the same answer as above; the combination of width * error in length makes hardly any contribution.
If the ± in the source data represents hard limits but for the area you're more interested in standard deviation, you'll need to make some assumption about the source distributions.

 

[peripatein]

Thank you very much for your replies!

 

[blue_raver22]

I understand that uncertainty is a fundamental aspect of any measurement or calculation. It is not uncommon for uncertainty to play a significant role, especially in scientific research and experiments. In fact, some level of uncertainty is expected and even necessary in order to accurately represent the complexity and variability of the natural world.

In regards to the table's surface area calculation, it is plausible that the uncertainty (∆S) could be nearly 70 cm^2. This is because the uncertainty is calculated using the uncertainties of both length (∆L) and width (∆W), and these values are relatively large compared to the actual values of L and W. Additionally, the formula used to calculate ∆S takes into account the partial derivatives of S with respect to L and W, which can also contribute to the overall uncertainty.

As for confirming the result, I would suggest checking the calculations and ensuring that the units are consistent. It may also be helpful to double-check the uncertainties (∆L and ∆W) to ensure they are accurate. If all the calculations are correct, then it is likely that the result is accurate and the uncertainty of ∆S is indeed nearly 70 cm^2.

In conclusion, uncertainty is a natural and important aspect of scientific work, and it is not uncommon for it to be a significant factor in calculations and measurements. It is always important to carefully consider and account for uncertainty in order to accurately represent and interpret scientific data.