DanielDaniels
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I measured the length of a pencil using a ruler lets say 5 times (N=5). Then I calculated the mean length L (cm).
Now I am utterly confused about calculating the absolute error here. Shall I use this: (largest value) - (smallest value)/2 OR use the standard deviation from mean method? Which is the right one and why?
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Stonebridge
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(Original post by Daniel Atieh)
I measured the length of a pencil using a ruler lets say 5 times (N=5). Then I calculated the mean length L (cm).
Now I am utterly confused about calculating the absolute error here. Shall I use this: (largest value) - (smallest value)/2 OR use the standard deviation from mean method? Which is the right one and why?
They are both acceptable values for the uncertainty. (There are other ways.) What does your school/teacher/tutor say? For AS and A2 physics, the exam boards, as far as I know, say you do not need to use standard deviation.
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DanielDaniels
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(Original post by Stonebridge)
They are both acceptable values for the uncertainty. (There are other ways.) What does your school/teacher/tutor say? For AS and A2 physics, the exam boards, as far as I know, say you do not need to use standard deviation.
I think I don't need the standard deviation. But which one is better, and if possible to briefly explain why?

Furthermore, for any measuring device, I can find the uncertainty of one reading by taking the smallest division/2 right? Are all vernier calipers (for example) having the same uncertainty?
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Stonebridge
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(Original post by Daniel Atieh)
I think I don't need the standard deviation. But which one is better, and if possible to briefly explain why?

Furthermore, for any measuring device, I can find the uncertainty of one reading by taking the smallest division/2 right? Are all vernier calipers (for example) having the same uncertainty?
It's not a case of one being better. They are just different ways of expressing the confidence level you have in a result or reading.
You can have, for example
-possible error
-maximum possible error
-probable error

This web site is a good resource for this.
It will tell you all you need to know but is at a level slightly above A2
http://www.rit.edu/cos/uphysics/unce...rt1.html#range

No. Not all vernier callipers have the same precision.
The page I linked to above explains this.
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DanielDaniels
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(Original post by Stonebridge)
It's not a case of one being better. They are just different ways of expressing the confidence level you have in a result or reading.
You can have, for example
-possible error
-maximum possible error
-probable error

This web site is a good resource for this.
It will tell you all you need to know but is at a level slightly above A2
http://www.rit.edu/cos/uphysics/unce...rt1.html#range

No. Not all vernier callipers have the same precision.
The page I linked to above explains this.
Thanks a bunch, really. Quite helpful.

One last thing, how is this the answer here?

The answer is 0.0025, but am I supposed to look at the smallest division from such figure? :/
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Stonebridge
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(Original post by Daniel Atieh)
Thanks a bunch, really. Quite helpful.

One last thing, how is this the answer here?

The answer is 0.0025, but am I supposed to look at the smallest division from such figure? :/
It depends on the smallest division you can measure with the vernier.
I'm finding it impossible on my screen to see clearly what you have in that diagram to be able to give an answer.

However, take a look at this
Name:  vernier.JPG
Views: 417
Size:  199.4 KB

it may answer your question.
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DanielDaniels
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(Original post by Stonebridge)
It depends on the smallest division you can measure with the vernier.
I'm finding it impossible on my screen to see clearly what you have in that diagram to be able to give an answer.

However, take a look at this


it may answer your question.
Please accept my apologies for the delayed response. Now this takes us to how I can find the smallest division of a vernier caliper.

An example is suitable here:




But before that, I want to make sure I can read correctly. So for the top image, the main scale is in cm right? so I can see it is 21 mm. For the vernie scale, the whole scale represents 1 mm, I believe (thought I don't know why it is 10 in the image?). So from the vernier scale, it is 0.3 mm. Now adding them up, we get 21.30 mm.

Well, my second question is how I can find the smallest division in such situation? I need that smallest division in order to find the error in one reading.

By the way, is this method when we are giving just one reading the only working way?
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Stonebridge
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For the first diagram there is nothing there which tells me what units it is measuring.
All I can say is that the main scale is reading between 2.1 and 2.2 and the vernier aligns at 3 meaning it's measuring 2.13

The 2nd scale is reading between 7 and 8mm with the vernier aligning at 5, giving a reading of 7.5mm

The smallest division on the second scale is 0.1mm so you would be correct to give the reading as 7.5 ± 0.1 mm

This gives the precision of the single reading.
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DanielDaniels
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(Original post by Stonebridge)
For the first diagram there is nothing there which tells me what units it is measuring.
All I can say is that the main scale is reading between 2.1 and 2.2 and the vernier aligns at 3 meaning it's measuring 2.13

The 2nd scale is reading between 7 and 8mm with the vernier aligning at 5, giving a reading of 7.5mm

The smallest division on the second scale is 0.1mm so you would be correct to give the reading as 7.5 ± 0.1 mm

This gives the precision of the single reading.
Thanks a bunch!

Shouldn't I divide the smallest division by 2?

Also, for the second image, if the vernier scale corresponds to 1 mm, and that each division is 0.1 mm, then why they show 1 mm, ... up to 10 mm? Or this means that the 1 mm is divided into 10 divisions.


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Stonebridge
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The plus/minus value you give for a reading is YOUR confidence level in your measurement. There is no right or wrong answer. If you feel you can read the vernier scale to half of the smallest division then that is up to you. Both are acceptable. plus/minus 0.1 is the maximum uncertainty in the single reading. You could say you can read it to less than that if you wish. It's your decision.

I don't understand what you are asking in your second question.
The main scale clearly reads in mm (from 0 to 25mm) and the vernier (sliding) scale reads in 1/10 of these divisions which is 0.1mm
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DanielDaniels
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(Original post by Stonebridge)
The plus/minus value you give for a reading is YOUR confidence level in your measurement. There is no right or wrong answer. If you feel you can read the vernier scale to half of the smallest division then that is up to you. Both are acceptable. plus/minus 0.1 is the maximum uncertainty in the single reading. You could say you can read it to less than that if you wish. It's your decision.

I don't understand what you are asking in your second question.
The main scale clearly reads in mm (from 0 to 25mm) and the vernier (sliding) scale reads in 1/10 of these divisions which is 0.1mm
Very clear. Thanks once again!

One more thing regarding the standard deviation equation:

I found three equations for it; one is having the N replaced by (N-1) and other one where the N is replaced by N(N-1). Which one shall I use and what is the difference? The N and N-1 are probably quite close if we repeated the experiment many times.
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Stonebridge
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There are various ways of writing the formula.
The difference between n and n-1 arises because one refers to the whole population and the other to a sample.
See here

http://www.bbc.co.uk/bitesize/standa...on/revision/2/

or

https://www.mathsisfun.com/data/stan...-formulas.html

You don't need this for A Level physics.
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DanielDaniels
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(Original post by Stonebridge)
There are various ways of writing the formula.
The difference between n and n-1 arises because one refers to the whole population and the other to a sample.
See here

http://www.bbc.co.uk/bitesize/standa...on/revision/2/

or

https://www.mathsisfun.com/data/stan...-formulas.html

You don't need this for A Level physics.
Many thanks really!
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