# Measurement and Uncertainty - an example

I previously talked about measurements (some) when I looked at the uncertainty in the distance to the Sun. One of the simple ways of determining the uncertainty of a calculated quantity is to use the uncertainty of the measured variables and find the max and min that calculated quantity can be. The example I used was in calculating the uncertainty for the area of a rectangle. So, the maximum and min areas would be:

And then the uncertainty in the area can be described as:

Yes, I know this is not as sophisticated as the normal procedure for error propagation, but it works. This is the appropriate level for algebra-based lab courses or high school. Uncertainty is important in labs. It is important to realize that everything that is measured is not exact. How do you compare two quantities without uncertainty? Ok. On with the example. In this post, I want to measure the dimensions of a metal block, find it's mass and determine the density.

What do I need to measure?

• Lenght (in cm)- of course you could use different units.
• Width (in cm)
• Height (in cm)
• Mass (in grams)

So, I made a spreadsheet with all the calculations in it. You can change the values and the uncertainties and it gives the density with uncertainty.

First, the important thing to note here is the max and min density. Since density is mass/volume, to get the max density you would need to use the max mass and the MINIMUM volume (since you are dividing by volume). Next, how would I report this density? It should be as:

Here the uncertainty in the density starts at the 1/10 of a g/cm3 decimal place. So, the density should be reported to this same significant figure place.

If you can't remember how to find the uncertainty for measured values, in this case I used "half of the smallest division - or greater". The "or greater" means that the measurer can use his/her judgment to make it bigger (imagine measuring a tennis ball with a ruler).

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Excellent post. This is also a nice example of error propagation because the "usual" formula with the derivatives and adding errors in quadrature and whatnot works very well here, too.

A nice variant might be a cuboid with length >> (height, width), to show that the error in your final answer depends more strongly on some of the length errors than on others.

By Anonymous Coward (not verified) on 20 Mar 2009 #permalink