# Reporting length with error limits

• B
Hello all,

I am in a total state of confusion with regard to this. Here is question.
The length of a rod is [11.05 (+/-) 0.2]cm. What is the length of two rods?

I want to know should the answer be reported like

[22.1(+/-) 0.2] or [22.1(+/-) 0.4].

Regards:)

Related Other Physics Topics News on Phys.org
Staff Emeritus
2019 Award
Yes, it should be reported like either of them. But what I think you are asking "which one is right"?

Yes, it should be reported like either of them. But what I think you are asking "which one is right"?
Yes I want to know which is the correct way out of the two?

Staff Emeritus
2019 Award
What do you think? What have you tried?

What do you think? What have you tried?
I think I need to focus on what they are asking, when they say what is the length of two rods; and I think what they are asking is "what would you report when you place two rods end-to-end to measure it's length with the scale?"

And for this I think the maximum error should still remain the same(that is 0.2cm) as the instrument is the same.

But I am not sure of everything I said and concluded.

Regards

mfb
Mentor
And for this I think the maximum error should still remain the same(that is 0.2cm) as the instrument is the same.
If you put 10000 of them in a row, you can measure 1 km with an uncertainty of 2 mm? If you put 3 billion next to each other, you can measure the distance to the Moon with the same uncertainty?
Do you think this is realistic?

CWatters
Homework Helper
Gold Member
And for this I think the maximum error should still remain the same(that is 0.2cm) as the instrument is the same.
No you can forget how the rods are measured. They might have been made in totally different factories. You can't assume they are being measured with same instrument.

Just consider what the minimum and maximum length of two rods could be. Calculate the nominal length and tolerance from that.

If you put 10000 of them in a row, you can measure 1 km with an uncertainty of 2 mm? If you put 3 billion next to each other, you can measure the distance to the Moon with the same uncertainty?

Yeah I am thinking that, if you put 3 billion next to each other, you can measure the distance to the Moon with the same uncertainty(though I am not challenging that I am thinking correctly). I am considering that error is not linked with my calibrations(faulty calibrations or something) so it's only the two end points of rod(or two end points of many joint rods) from where uncertainty comes and they should be independent of the number of rods I have kept in between.

Do you think this is realistic?
I am unsure of all my claims but at the same time I couldn't feel the other idea that error should increase to be a realistic one.

Regards

No you can forget how the rods are measured. They might have been made in totally different factories.
This is giving me some thought but what if the rods are made not only in the same factories but also in the same mould under same conditions. I mean isn't it possible that they are identical?

You can't assume they are being measured with same instrument.
Why?
Just consider what the minimum and maximum length of two rods could be. Calculate the nominal length and tolerance from that.
Can you give me an example as it will be more helpful.

I am very happy to see you all helping me.

Regards

Staff Emeritus
2019 Award
Now you're getting to the core of the matter - is the deviation something that varies rod to rod (called a statistical deviation) or is it something in common for every rod (called a systematic deviation)?

hilbert2
Gold Member
Hello all,

I am in a total state of confusion with regard to this. Here is question.
The length of a rod is [11.05 (+/-) 0.2]cm. What is the length of two rods?

I want to know should the answer be reported like

[22.1(+/-) 0.2] or [22.1(+/-) 0.4].

Regards:)
If you find a way to generate Gaussian distributed random numbers with Wolfram Mathematica or something, you could try making a dozen pairs of those numbers with mean set to 11.05 and standard error to 0.2 - then calculate the standard error for the set of the sums of each pair. That way you can experiment with how statistical, versus systematic error works.

...is the deviation something that varies rod to rod (called a statistical deviation) or is it something in common for every rod (called a systematic deviation)?
Can you please elaborate I am not good at these statistics so that I can figure out the differences of details of statistical deviation and systematic deviation.

I really want to know it from the core. :)

Regards

If you find a way to generate Gaussian distributed random numbers with Wolfram Mathematica or something, you could try making a dozen pairs of those numbers with mean set to 11.05 and standard error to 0.2 - then calculate the standard error for the set of the sums of each pair. That way you can experiment with how statistical, versus systematic error works.
I know what is standard deviation, but what is standard error and Wolfram Mathematica things I don't know.

I would like if you could take me to the target point as easy as possible but if that's not possible unless I study special statistical methods than thing is different. In that case I just like to know out of the two options for my original question, the more correct answer?

Regards:)

hilbert2
Gold Member
I know what is standard deviation, but what is standard error and Wolfram Mathematica things I don't know.

I would like if you could take me to the target point as easy as possible but if that's not possible unless I study special statistical methods than thing is different. In that case I just like to know out of the two options for my original question, the more correct answer?

Regards:)
Two times variable ##x \pm \delta x## is ##2x \pm 2\delta x##. If you have two different uncertain variables ##x \pm \delta x## and ##y \pm \delta y##, where there's no correlation between the signs of their errors, then their sum is ##x + y \pm \sqrt{(\delta x )^2 + (\delta y )^2}##. Note how the errors add in a "Pythagorean law" like way.

The error of ##2\times (x + \delta x ) + 3 \times (y + \delta y )## would be

##\sqrt{2^2 (\delta x)^2 + 3^2 (\delta y)^2}##.

The error laws of more complex functions of two variables are calculated by differentiation.

This kind of statistical laws can be demonstrated by making a lot of gaussian distributed random variables and looking at the distributions of the values of functions built from them.

Two times variable ##x \pm \delta x## is ##2x \pm 2\delta x##. If you have two different uncertain variables ##x \pm \delta x## and ##y \pm \delta y##, where there's no correlation between the signs of their errors, then their sum is ##x + y \pm \sqrt{(\delta x )^2 + (\delta y )^2}##. Note how the errors add in a "Pythagorean law" like way.

The error of ##2\times (x + \delta x ) + 3 \times (y + \delta y )## would be

##\sqrt{2^2 (\delta x)^2 + 3^2 (\delta y)^2}##.

The error laws of more complex functions of two variables are calculated by differentiation.

This kind of statistical laws can be demonstrated by making a lot of gaussian distributed random variables and looking at the distributions of the values of functions built from them.
Ok thats good. Can you suggest me some(or better only one best) book(s) to me for learning more about error analysis in measurement?

I have one another function( no so complex) where I need to write expression for error. Can I ask?

If it asks for the range of possible lengths of two rods you have to assume they could both be of minimum or maximum length.

The questions straight as it is, are shown in two pictures below.

#### Attachments

• 15.4 KB Views: 337
• 27.4 KB Views: 331
CWatters
Homework Helper
Gold Member
Then I believe it's 22.1(+/-) 0.4. It's not a statistical exercise.

Just for interest it's possible for tolerances to be asymmetrical. The diameter here is specified as D inches +0,000 - 0,002 meaning it won't be bigger (to three decimal places) than the nominal diameter D but might be smaller.

http://www.digicanmc.com/HTML/Sec_6/images/6-23.gif

Ok got it, thank you all.

Regards:)