# Tag Info

1

It looks like Pythagoras, but it is only remotely related. The important concept, as presented in SteveB's answer, is that the variables are considered to be independent, i.e. one does not affect the other. In mathematics, independent parameters are said to be orthogonal , and can thus be assigned to separate axes in Cartesian N-space. It just so happens ...

1

The general formula for error propagation is: $$\Delta f(x_1,x_2,\ldots)=\sqrt{(\frac{\partial f}{\partial x_1}\Delta x_1)^2 + (\frac{\partial f}{\partial x_2}\Delta x_2)^2 + \cdots}$$ Where does this come from? We assume that the errors are relatively small (ignore $\Delta x_i \Delta x_j$ terms etc.), and that the errors are independent (in the ...

1

The formula $$\frac{\Delta{A}}{A} \approx \frac{\Delta{X}}{X} + \frac{\Delta{Y}}{Y}$$ is an approximation because you are ignoring $\Delta X$$\Delta Y A better approximation would be$$\Delta A=\frac{\partial A}{\partial X}\Delta X+\frac{\partial A}{\partial Y}\Delta Y$$Since errors always add we take the absolute magnitude of \frac{\partial ... 0 The square root is there as a better estimator of the error than just adding the errors together. If you add the errors together you are finding the maximum possible error which will happen when both quantities are a maximum(or minimum) together. This is an unlikely event compared with all the other domination of errors. The square root formula you you ... 1 Your method is perfectly correct for estimating the error, especially at a high school or early university level. As Farcher correctly points out, it's unlikely for both errors to be at their extreme concurrently, so your error is more like a maximum bound on the error than a typical representative error. The further you go in physics, the more you'll want ... 1 Formula for value -$$v_i=\sqrt{{g}\over{2}}RH^{-\frac{1}{2}}$$General formula for error -$$ \delta v_i = \sqrt{(\frac{\partial v_i}{\partial R}\delta R)^2+(\frac{\partial v_i}{\partial H}\delta H)^2} $$First Calculate the derivatives -$$ \frac{\partial v_i}{\partial R} = \sqrt{{g}\over{2}}H^{-\frac{1}{2}}  \frac{\partial v_i}{\partial H} = ... 1 Your method assumes that when one measurement is at one extreme the other is also at an extreme. This is not likely to happen very often and therefore you have overestimated your error. If you have two quantities$A\pm a\%$and$B\pm b\%$a better estimate of the percentage error in$A \times B$or$\frac A B$is$\sqrt {a^2+b^2}$. The percentage error in ... 0 It appears that none of Physical Review Style and Notation Guide, the AIP Style Manual, the IAU Style Manual, or The ACS Style Guide: A Manual for Authors and Editors, weigh in at all on this matter, so I would say it is to some extent up to personal taste. Some recent examples from the literature:$6356 ± 8 \:\mathrm{keV}75.3a_0^3 ± 0.4a_0^3$... 3 I assume that$t$is the step size. If$t$is the integration step, you will make a certain error in calculating the integral. As you make the step$t\$ smaller, the maximum error you make will reduce - and if the integrand is reasonably well behaved (in the sense that word was used when somebody calculated the scaling of the error) then you can expect that ...

3

You're making two measurements, each with an error, and then subtracting. But, here's the key: Assume... it has a constant length, thus the two variables are not independent. This isn't true: the two variables are completely independent. That you made an error in one direction at one end of the object indicates nothing about your error at the other end ...

Top 50 recent answers are included