Error analysis: propagation of error

[Claric1]

Homework Statement

The following may appear to be a rather strange question but I have no access to a lab to conduct experiments so I have to rely on my imagination, and questions from textbooks, to contrive questions that will hopefully improve my understanding of a topic.

I have recently began Taylor's, 'An Introduction to Error Analysis' (1982) and am currently working through the third chapter. In conjunction with this text, I am studying O level mathematics texts from the 1950's which was prior to the advent of the calculator. I came across a question in one of these texts and thought I could apply some error analysis to my solution to the question to see firstly if I could correctly identify how any uncertainties were propagated and secondly to see if my estimated uncertainties were anywhere near correct. I apologise in advance if what follows is not correct, I have only recently began to self-teach error analysis and am finding it a fascinating yet tricky subject.

The question from an O level maths text is quite simple and involves scale drawing to obtain a solution. It is written as follows:

An aeroplane flies 10 miles SE, 15 miles E 25 deg N then 10 miles N 56 deg W. Construct a scale drawing and find the distance and direction in which the aircraft must now fly to return to its starting-point.

Before I completed the drawing I completed a quick freehand sketch, resolved the vectors in a north and east orientation and calculated the resultant vector and compass bearing, see attachment 253.

My results gave the following:
distance 13.3 miles
bearing S 68.6 deg W

I then decided on the following scale: 1 inch represents 5 mile (my ruler is graduated in inches and tenths). I then thought that the uncertainty in the length of any line I drew was +/- 0.1 inches (this seemed rather a lot to me but I had previously heard that one should measure to the nearest small graduation and half that value, 0.05 inches, would be your error at that end of the ruler; however, as there are two ends to consider the total error per line drawn would be 0.1 inches.)

I then looked at my sketch and the thought occurred to me that I couldn't just sum these errors as they are in different directions, see attachment 252. My thought was to resolve the errors, in the same way I had the vectors to obtain t

wo perpendicular errors then us Pythagoras' Theorem to give the resultant error, see attachment 254.

My scale drawing and calculated errors gave distance of 13 +/- 1.5 miles and a bearing of S 69 deg W +/-3 deg. The range of both agree with calculated answer. Furthermore, solution in text (also probably found by scale drawing) gives 13 miles and S 68 deg W.

Homework Equations

My questions are as follows:

Is my general method of calculating error propagation correct? I have simply added 1 degree error three times for the three angles drawn. However, I have not included my error in reading the resultant length or angle, Should I have done this to get a better (but larger) uncertainty?

I have also noticed that if I halve my uncertainty throughout to 0.05 inches and 0.5 degrees per angle, I get approximately half the uncertainty in my final solution, i.e. +/- 0.8 miles. This still gives an adequate range which contains the 'exact' answer. Therefore, is my original estimated uncertainty of 0.1 inches too large?

I fully realize that this may be a rather convoluted way of looking at error analysis but, as mentioned above, I have no lab and attend no college so have to rely on myself to think of suitable ways to test the knowledge I think I have acquired.

Many, many thanks in advance for any advice anyone can offer.

 

[haruspex]

as there are two ends to consider the total error per line drawn would be 0.1 inches.)

If you are trying to measure an existing distance then you will align one end with a gradation. The error in the measurement at that end depends only on how precisely you do that (parallax etc.); the error at the other end will be ± half a gradation.
But if you are trying to draw a new length which should be a whole number of gradations then you only have your own precision to worry about at each end.

couldn't just sum these errors as they are in different directions,

Quite so, and you also need to think about errors in the angles. Each point you construct will have an area of error. A long distance amplifies the angle error, so the area might be a lot wider in the transverse direction than in the longitudinal direction.

 

[Claric1]

Thank you very much for getting back to me. I think perhaps I need to reformulate the question in order to make it applicable to error analysis. I will try and prepare something today.

With regards to your comment on half a gradation, I would completely agree with you as my intuition tells me that should be so; the Taylor book also agrees with you (so far). However, I have consulted some A level texts and they seem to use +/- 1 grad in their error analysis. Could it be that this is rather subjective point that has not been universally agreed upon. The following links show two videos with the same opinion as the A level text (the latter by Walter Lewin) which both seem to point towards 1 grad as the error.





As stated above, I will try and prepare something today that reformulates the original question so that I can find out if my original method of finding total error is correct.

Many thanks once again.

 

[Claric1]

Okay, I have rewritten the question as best I can using Latex and have attached a pdf of this below. I did not add the LaTeX code as the document contains a diagram and I am unsure if MathJax can cope with diagrams.

As I am still unsure with regards to which uncertainty to take as correct (see comments by myself and haruspex above) and have therefore incorporated them both into the question. I hope, worded in this form, my original question can be answered using error analysis, if so I respectfully request approval/disapproval of my original method of finding resultant error.

If possible, I really need to know if my original method for finding resultant error, albeit reworked into this question, is the right approach to take in finding such an error. I also take haruspex's point in that there will be an area of uncertainty with this solution and I assume this could be represented by a locus of points but am not sure and would appreciate some guidance.

As previously stated, I am very new to this area of error analysis so would really appreciate as much detail as anyone can give.

My apologies once again for labouring over the same question but I really want to understand this. So far I've learned that if I want an answer, I need to formulate the correct question.

Sincerest thanks in advance for any help anyone can offer and thank you once again to haruspex for their reply to my original post.

 

[haruspex]

some A level texts and they seem to use +/- 1 grad in their error analysis.

What the video refers to as Zero Error is what I mentioned as "parallax etc."
But I find it quite bizarre to treat it always as half a gradation. The might be reasonable where the gradations are very fine, such as 1mm, but gets ridiculous for coarser grades. In reality, it depends on how good your eyes are, how thick the ruler is, and how wide the gradation mark is (the black line).
Anyway, as I mentioned, if constructing a line of given length then you have Zero Error at each end, not measurement error.

 

[haruspex]

there will be an area of uncertainty with this solution

Create a variable to represent each of the angles and distances that you have to measure out in drawing the graphical solution. Write the final answer as an algebraic expression in terms of those variables. By taking derivatives, you can find the error in the answer resulting from small errors in the individual variables.

 

[Claric1]

Thank you very much for your reply. I had read in the error text by Taylor that I’m reading about derivatives but I wasn’t sure if that was applicable as the angles are known so it appeared to me as each length was just getting multiplied by a scalar. I transferred this logic to my error solution thinking I could multiply each individual component by the absolute value of the scalar quantity and purely sum the result, then finally use Pythagoras’ Theorem to obtain a complete error. Thank you letting me know that the calculus approach is the way to go. It will be interesting to see how this differs from my result.
Did you get chance to look at the LaTeX pdf I’ve written. I would appreciate your input if you have time but please don’t worry if you’re busy.
Many thanks for all your help and advice once again.

 

[haruspex]

I wasn’t sure if that was applicable as the angles are known

The angles are no more "known" than the distances are. Both are subject to error.
You do need to ensure your variables correspond to your construction method. Suppose you want to go distance a from the origin at angle α above the horizontal then distance b from there at angle β above the horizontal. If you plot it that way, i.e. with angle β measured from a known horizontal grid line, then you would write that the x coordinate is now x+Δx=(a+Δa)cos(α+Δα)+(b+Δb)cos(β+Δβ).
But you might choose to measure angle β as angle θ=β-α above the slope of the line length a. In this case x+Δx=(a+Δa)cos(α+Δα)+(b+Δb)cos(α+θ+Δα+Δθ).

 

[Claric1]

Thank you for the explanation, this problem seems to have a more difficult solution than I first thought. I’m assuming partial differentiation is the way forward.
I will try the calculation tomorrow and let you know how it goes.
Thank you for putting so much and effort into your replies, it really is very much appreciated.

 

[haruspex]

partial differentiation is the way forward.

Yes.

 

[Claric1]

This is Taylor's rule from his text

 

[Claric1]

Just wanted let you know that I have been trying to crack this today but am not there yet. It's been a long time since I've done any calculus and this partial differentiation is very new to me. That being said, I have had a go but from slightly different point of view. I read a rule in the Taylor text regarding functions of several variables (summarised in the rule 5 at the end of the following link):
http://web.mit.edu/fluids-modules/www/exper_techniques/2.Propagation_of_Uncertaint.pdf

I'm afraid my maths skills aren't as good as I first thought: I am unable to partially differentiate anything of the form you mentioned yesterday, i.e. x+Δx=(a+Δa)cos(α+Δα)+(b+Δb)cos(β+Δβ). Are the deltas on the RHS to be given a specific value?

However I can partially differentiate if I use the form from rule 5 in the link above, i.e .Rx +Δx= R_1 cos alpha + R_2cos beta +R_3cos gamma + Δx(in the form of rule 5 of the above link).

I had no idea such a simple question (initially) could be so perplexing.

I am so sorry to keep asking you about this but unfortunately, you are the only one who has replied to my post.

If you think I should leave this as it's too difficult for me then I completely understand. I would just like to see this problem through to completion as I've been thinking about it for a few days now.

Gratitude in advance for both your time and understanding.

 

[haruspex]

Just wanted let you know that I have been trying to crack this today but am not there yet. It's been a long time since I've done any calculus and this partial differentiation is very new to me. That being said, I have had a go but from slightly different point of view. I read a rule in the Taylor text regarding functions of several variables (summarised in the rule 5 at the end of the following link):
http://web.mit.edu/fluids-modules/www/exper_techniques/2.Propagation_of_Uncertaint.pdf

I'm afraid my maths skills aren't as good as I first thought: I am unable to partially differentiate anything of the form you mentioned yesterday, i.e. x+Δx=(a+Δa)cos(α+Δα)+(b+Δb)cos(β+Δβ). Are the deltas on the RHS to be given a specific value?

However I can partially differentiate if I use the form from rule 5 in the link above, i.e .Rx +Δx= R_1 cos alpha + R_2cos beta +R_3cos gamma + Δx(in the form of rule 5 of the above link).

I had no idea such a simple question (initially) could be so perplexing.

I am so sorry to keep asking you about this but unfortunately, you are the only one who has replied to my post.

If you think I should leave this as it's too difficult for me then I completely understand. I would just like to see this problem through to completion as I've been thinking about it for a few days now.

Gratitude in advance for both your time and understanding.

As regards the error resulting from the individual errors, it's the product rule. If z=f(x)g(y) then Δz=f'(x)Δx g(y)+f(x)g'(y)Δy.

So in the simple example of your problem, x+Δx=(a+Δa)cos(α+Δα)+(b+Δb)cos(β+Δβ)
≈a cos(α)+Δa cos(α)-a sin(α)Δα+b cos(β)+Δb cos(β)-b sin(β)Δβ
Δx≈Δa cos(α)-a sin(α)Δα+Δb cos(β)-b sin(β)Δβ
Then you just fill in the error values for Δa etc.

But that is not quite the same as the uncertainty. That is a statistical matter, which allows for the possibility that some errors partly cancel others.
For that you use the root-sum-square formula you quote in post #11.

 

[Claric1]

As always thanks again, your advice is very much appreciated. I will try again tomorrow and hopefully succeed.
I think I’ve become confused due, in part, to the wording early on in the text. Initially the author states that, to begin with the two words ‘error’ an ‘uncertainty’ will be interchangeable. However, he also states that some authors introduce additional definitions of errors which the book discusses later in the text but I’ve not got that far yet.
I think I’ve also got a bit confused with your notation though the fault is mine and not yours. I assumed your capital delta had the same meaning as the lower case delta from the text where the author uses it for uncertainty/error but I think you use it to denote the derivative. I’m sorry I didn’t understand.
Many thanks once again.

 

[haruspex]

I assumed your capital delta had the same meaning as the lower case delta from the text where the author uses it for uncertainty/error

Yes.

 

[Claric1]

Now I think understand, that’s very slick! I wish applied maths was more my thing. Not seen the product rule of two variables expressed like that before.