How to calculate absolute uncertainties in log values

[@icemaths9997]

Great video - very clearly explained. However, I wonder if anyone has any idea why the second example doesn't seem to follow this method for all of the values in the table. Calculating the uncertainties in ln(V) should give 0.05, 0.04, 0.03, 0.03, 0.02, whereas the table has the first and last of these as 0.06 and 0.03. I recognise this question, and can see that it is from a real exam paper - it would be very confusing for students to work out the missing uncertainty by a method which doesn't give the given answers to all of the other rows. Do you think OCR have a different rule? Their practical skills handbook doesn't address this at all as far as I can see.

[@PhysicsHQ]

Hi and thanks. You’re right that the ln(V) values in the second table don’t all follow my method; whereas all log values in table 1 do. This is perplexing. I haven’t yet figured out the method used by OCR to calculate uncertainties in ln(V). I’m sure that my method is correct - all the more so in the absence of explicit instruction from OCR. I should, perhaps, contact them for comment. It is interesting to see the uncertainty values calculated by my method stated to more d.p. : 0.0668 ➝ 0.07 0.0541 ➝ 0.05 0.0400 ➝ 0.04 0.0313 ➝ 0.03 0.0274 ➝ 0.03 0.0247 ➝ 0.02 Now we can see that if someone was working from a table in a spreadsheet in which the values were displayed to 3 d.p. that would lead to the last value being mistakenly rounded up to 0.03. And that would then leave only one inexplicable discrepancy as 0.054 wouldn’t be rounded to 0.06. If it is just one discrepancy between my results and OCR’s I reckon that OCR have used this method but made an error in that one value.

[@icemaths9997]

@@PhysicsHQ Thanks for the very thorough analysis. I agree that someone should contact them and ask what they are doing. Do you want to or shall I? I think one enquiry is enough, don't want them to feel put upon!

[@PhysicsHQ]

@@icemaths9997 😊 I will make contact. Agreed, particularly at this time - they might have a lot on their hands with covid-19 arrangements.

[@PhysicsHQ]

Hi again. I’ve now heard back from OCR. What follows below is the official comment on the matter. In summary, the approach I used is valid but not ideal in some cases with natural logs. OCR will always give credit for this method in examinations. I will nonetheless update my advice to students, particularly the more able students. From OCR: “The method in your video is a common approach and has received credit in recent mark schemes. I believe the discrepancy in the data and your approach for the 2017 H556/03 question is because the logarithms of the extreme values will not be equidistant from the logarithm of the central value so another method that would eliminate this problem is to use the log of the central bound - log of the lower bound as the uncertainty for ln V.”

[@icemaths9997]

@@PhysicsHQ So they are saying that for the second value in the table, for example, we should be using ln(3.7) - ln(3.5) as the uncertaintly. This does give the 0.06 that they have, and the other value we disagreed with them on also works out this way. But how anyone was supposed to know this I have no idea - there is nothing I can see about it in the practical skills handbook, and I don't remember seeing it explained in any text books. This is also going to apply to any non-linear transformation, such as the square or the reciprocal of a measured value, so would seem to be something of a big deal. Thanks so much for chasing this up. I do find uncertainty questions to be more of a challenge than exam boards seem to think, especially compared to the low number of marks these questions seem to earn.

[@kiruthikpranav5047]

Thanks a lot, I was doing a constant % uncertainty giving me insane absolute uncertainties. Great vid, helped a lot.

[@PhysicsHQ]

Great to hear! All the best with your studies.

[@SGayanFernando]

This video was a great help! Thank you very much! 😍

[@PhysicsHQ]

Glad it was helpful! All the best with your studies.

[@stevehu6813]

This really helped, Thank you sir for existing

[@PhysicsHQ]

You are most welcome

[@fabihaareeba9657]

FINALLY GETTING MOTIVATION TO STUDY. THANK YOU SIRRRR

[@PhysicsHQ]

You are welcome! All the best with your studies!

[@alessandromaniscalco1235]

Thank you! Also, to simplify further, (log x )/2 is the same as: log√ x

[@woohyunchoi9572]

Found this in 2021. Thank you so much

[@PhysicsHQ]

You’re welcome. All the best for rest of 2021 and 2022.

[@rayyanxiety]

Thankyou so much for this. I was stuck, not anymore, tho.

[@PhysicsHQ]

You’re more than welcome. All the best for any upcoming exams.

[@prestonchui]

what if you get a negative value after deducting a value of its absolute uncertainty. then you won't be able to log it since you cant log negative values. For example: if your value is 0.45 +/- 2.00.

[@PhysicsHQ]

In the first instance I would suggest trying to retake the measurements at higher precision since this error occurs when the abs uncertainty is larger in magnitude than the measurement. That suggests an experiment redesign is probably necessary. If that isn't possible then you can add a constant to all of the measurements before logging them to ensure that all values will be larger than zero after the uncertainty is subtracted.

[@giovannimariastrampelli8988]

Great video, I have a question probably related though. What if my original values have not symmetrical errors (+0.2/-0.1 for example) and I want that to be reflected in the log error? What is the best approach then?

[@PhysicsHQ]

Hi. Thanks. What do you mean by not symmetrical errors - do you mean one set of measurements are made with one device and have an associated uncertainty and a second set of measurements are made with another device and have a different uncertainty?

[@MA-wv8bt]

for part b i get an uncertainty of 0.028?

[@PhysicsHQ]

I think you used log base 10. But that question uses natural log (ln). [log₁₀(3.2/2.8)/2 = 0.02899…] Try it again with ln(3.2/2.8)/2 and let me know if you still get an error.

[@MA-wv8bt]

@@PhysicsHQ ah sorry!it must be exam nerves haha exam tomorrow!

[@PhysicsHQ]

Hey no problem at all - it’s a stressful time of year. Better to make the mistake today rather than during a live exam.

[@MA-wv8bt]

@@PhysicsHQ haha true! Also thank you so much for all your tiny cards they are amazing you've been a great help! hopefully ocr don't let me down :)

[@PhysicsHQ]

My pleasure. Glad you like the Tinycards; I enjoy making them too.

[@BiologyWK]

Thanks 😊

[@PhysicsHQ]

You’re welcome. All the best with your studies

[@zuhayrroha4392]

What if lg (I/A) value and absolute uncertainty value is given and from there I have to find current (I/A).I know how to get current,10^(lg value) but how to get current uncertainty.

[@PhysicsHQ]

It's a similar starting point as with the log uncertainty - half of the range: ±10^x = ±½ |10^xU - 10^xL| (where xU is the upper value and xL is the lower value). As far as I know this equation can't be simplified further.

[@sippaphashirunyanitiwatna3648]

This method can only be used with logs only right?

[@PhysicsHQ]

Kind of. This is the method of calculating uncertainties applied to log values. But yes you wouldn’t use the calculations as they stand with non-log values.

[@aarv7851]

Hero

[@PhysicsHQ]

Thanks 🙏 All the best