# Can you help me understand how to determine uncertainty of measurements and calculations?

Asked by

Mariah (

25883)
November 11th, 2011

I had a lot of trouble in my physics lab in college. One of my problems was understanding how to arrive at the uncertainty of my measurements and calculations. We didn’t really have to do this in high school, and they didn’t teach us in college, just kinda threw us in the lab and said “go at it.”

So say I have a length measurement that I’m finding with a meter stick. The meter stick’s smallest increment is millimeters. I measure something at 25 millimeters. So does that mean that the uncertainty of my measurement is +/- .5 millimeters, because anything between 24.5 and 25.4 millimeters will round to 25 millimeters? Or…what?

Then let’s say I have a calculation to make involving that measurement. For instance, say I need to find the slope of a ramp in which the previous measurement is the hypotenuse. So I also measure the height of the ramp and then use arcsin to find the angle. But what is the uncertainty of that angle? How do I incorporate the uncertainty of my measurements into my calculations?

It’d help me a lot to understand this better. Thanks.

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## 13 Answers

Maybe this video will help you.

@ragingloli thank you so much! I know what I’m doing this winter! MIT, here I come.

My husband deals with this all the time. He works with dynamic positioning systems on boats/ ships/ oil rigs and the data is completely reliant on the information supplied by the Surveyors and THEY don’t seem to understand this concept, so trying to get the data within the few centimetre parameters they work with is IMPOSSIBLE, if the surveyors don’t first supply the measurements correctly. He has had to go in and ‘fire’ the surveyors on jobs, essentially, or get them fired, so that a better company is brought in to fix the errors.

Imagine trying to measure, pitch, roll, and yaw if the measurements from the data collection points are wrong or not within the acceptable parameters for error. Impossible. There’s a real life example for you.

My physics professor always told us that you measured as many digits as you possibly could, and then the very last digit was an estimate. So, if your meter stick has every millimeter marked, you could measure something down to 3.48cm. The 3.4cm would be exactly right since you can see the .4 mark on the meter stick. Then the 0.08 would be your estimated value where you give the best guess you can since it’s between the marks on your instrument.

In your example, if you think it is exactly 25 millimeters, you would record your measurement as 25.0mm, with the zero after the decimal place being your estimate.

In your other example, the uncertainty of the angle depends on how certain your initial measurement was. That’s where significant digits come in. If you measured 25.0 millimeters, then you have three significant digits. So, that’s how many significant digits you keep from the arcsin calculation. Once again, the first two digits would be exact, and the last digit would be considered an estimate. The degree of certainty is determined by how many significant digits you give.

@ragingloli Cool video. It seemed like he was just picking fairly arbitrary measures of uncertainty though. Is that customary?

And when comparing two measurements he maximized the uncertainty, e.g. when both were +/- .003, the maximum difference is when one is – .003 and the other is +.003 so the total difference is .006, and +/- .006 became his new uncertainty. Is that how you handle any kind of operation (adding two measurements vs. multiplying them or using another operation) or does it vary depending on the operation? Gahh :\

@cazzie, God, sounds like a nightmare!

@bobbinhood Okay, I can understand that. The .08 is an estimate. But you have to include at +/- to reflect that that .08 might not be exactly right. That’s what I’m confused about, how do you determine what +/- you need to include?

@Mariah I’ve never heard of that, but I also only took a couple physics classes in college. Perhaps you could go talk to one of the physics professors at your university? Most professors are significantly more helpful in their offices than they are in class. They usually like answering questions. I wish I could be more help, but we never did it that way.

Wait…didn’t you once say that you were doing all of your classes online for now? If so, maybe you could email a professor? Also, if you are unable to get your answer here, Stack Exchange is a really good resource for technical questions. I used their LaTeX forum before and found it to be fantastic. Since they have a physics forum, I’m sure you could get your question answered.

@ratboy I understand significant figures, that’s not the issue. It’s uncertainty; how do you determine that a measurement is accurate “to within 1 millimeter,” for example.

@bobbinhood I’m not enrolled in classes right now; I’m just wanting to learn this for when I go back to school. Thanks for the suggestions!

@ratboy That looks great, I’ll give it a read. Thanks.

For a particular project here, we use the uncertainty of superimposed Gaussian noise on top of the signal to extract statistically significant digits far beyond the granularity of the measurement capability. For example, we will take 100 readings and average them to be 95% confident that our results are within 1/10 of the reading resolution. If we need to go to even finer resolution, e.g. 1/100 of the equipment resolution, we take 10,000 readings and average.

Perhaps I’m oversimplifying in giving this answer, but it seems that the margin of error in your example would be dependent on the smallest division of the measuring instrument.

E.g., if the smallest measurement on a meter stick is a millimeter, then the MOE would be +/- 1 millimeter.

However, if one was using a micrometer, then the MOE would be =/- 0.001 millimeter.

In another case, if you’re measuring a fluctuating quantity, then the MOE could be an estimation of the deviations from the median measurement.

E.g., even with a fixed number of processes running on my computer, my RAM usage is fluctuating from 496 MB to 500 MB. In that case, I could say that my RAM usage is 498 MB, +/- 2 MB.

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