# Rigorous Uncertainty Analysis

012anonymousx
A position of a particle in linear motion is given by:
x = vt + 0.5at2

Calculate x with the error for:
t = 25.3 ± 0.5s
v = 10.1 ± 0.4m/s
a = 2.5 ± 0.3m/s2

So for calculating vt:

q = (10.1) (25.3) = 255.53 (exact)

Δq = (10.1)(25.3) * √ (0.4/10.1)2 (0.5/25.3)2
= 11.31...

Therefore, vt = 255.53 ± 11.3 (the 11.3 is rounded because I heard uncertainties always rounded to precision of last uncertainty)

Now my question is:

How do you round 255.53? And how many sig figs?

Is it 260.0 ± 11.3?
Or 255.5 ± 11.3?
OR something else?
And why?

I thought 260.0

Because when there is just a number to a certain position with no uncertainty given, like 123.4 it is implied that:
123.4 ± 0.1
So the number is rounded to the lowest precision of the uncertainty (I think)
(i.e. if the uncertainty was ± 0.12, the rounding would still occur at the 10th decimal place.
In this case the tenth decimal place.

In the case of 255.53 ± 11.3, it is the 10's.
Therefore 260.0

I'd appreciate if someone would type of a full solution (doesn't have to be pretty).

I guess a similar question is lets say you have the number
1234.5678 +/- 123.45
How should 1234.5678 be rounded?

Using a common sense approach, 255.53 or even 255. is preferable. If you put in the errors, you are choosing between the following: [244.23,266.83] and [248.7,271.3]. Just looking at the second would be quite misleading.

If you look up any measured physics data, you will see that the error term is usually at least two significant figures.

Staff Emeritus
2021 Award
I guess a similar question is lets say you have the number
1234.5678 +/- 123.45
How should 1234.5678 be rounded?

That number makes no sense. 123.45 means you know the uncertainty to about 0.01% of its value, since you didn't write down 123.44 or 123.46. However, 1234.5678 +/- 123.45 implies that you know the central value to about 10% of its value - that is, you know the uncertainty 100x better than what you are measuring.

This almost never happens.

The number you want is most likely 1230 +/- 120

012anonymousx
So what about the original problem then?

255.53 ± 11.3?

Mentor
2021 Award
A position of a particle in linear motion is given by:
x = vt + 0.5at2

Calculate x with the error for:
t = 25.3 ± 0.5s
v = 10.1 ± 0.4m/s
a = 2.5 ± 0.3m/s2
You will want to use the standard propagation of errors formula, explained here: http://www.foothill.edu/psme/daley/tutorials_files/10. Error Propagation.pdf

##\sigma^2_x=\sigma^2_t \left( \frac{\partial x}{\partial t} \right)^2 + \sigma^2_v \left( \frac{\partial x}{\partial v} \right)^2 + \sigma^2_a \left( \frac{\partial x}{\partial a} \right)^2 ##

mrspeedybob
You will want to use the standard propagation of errors formula, explained here: http://www.foothill.edu/psme/daley/tutorials_files/10. Error Propagation.pdf

##\sigma^2_x=\sigma^2_t \left( \frac{\partial x}{\partial t} \right)^2 + \sigma^2_v \left( \frac{\partial x}{\partial v} \right)^2 + \sigma^2_a \left( \frac{\partial x}{\partial a} \right)^2 ##

I don't understand the reasoning behind using such a complicated method.
In the case cited in the OP shouldn't you simply calculate x given the minimum values of v, a, and t. Then calculate x again using the maximum values of v, a, and t. The value and error of x should then be simply ((Xmax+Xmin)/2)±((Xmax-Xmin)/2)

Mentor
2021 Award
I don't understand the reasoning behind using such a complicated method.
In the case cited in the OP shouldn't you simply calculate x given the minimum values of v, a, and t. Then calculate x again using the maximum values of v, a, and t. The value and error of x should then be simply ((Xmax+Xmin)/2)±((Xmax-Xmin)/2)
You could do that, but it would overestimate the error as well as introduce some bias. In the case of the OP your method would get x = 1060 ± 143 whereas the full method would give x = 1056 ± 103.

The reason that your method gives an artifically high estimate of the error is a consequence of the fact that it is highly unlikely that you will get a maximum error in all three inputs at the same time. In fact, errors in one variable can offset errors in another variable.

Last edited: