# Combining uncertainties - percentage and absolute. Brief summary.

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#1
In view of the fact that this question is being asked again and again on this forum, and to save me time posting the same answer again and again, this is a summary of how you combine uncertainties at A Level.*

You have two values, each with an absolute ± uncertainty.

1. If you add or subtract the two (or more) values to get a final value
The absolute uncertainty in the final value is the sum of the uncertainties.
eg.
5.0 ± 0.1 mm + 2.0 ± 0.1 mm = 7.0 ± 0.2 mm
5.0 ± 0.1 mm - 2.0 ± 0.1 mm = 3.0 ± 0.2 mm

2. If you multiply one value with absolute uncertainty by a constant
The absolute uncertainty is also multiplied by the same constant.
eg.
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The constant can be any number. eg Pi

3. If you multiply or divide two (or more) values, each with an uncertainty
You add the % uncertainties in the two values to get the % uncertainty in the final value.
eg
5.0 ± 0.1 mm x 2.0 ± 0.1 mm

This is
5.0 ± 2% x 2.0 ± 5%

Result
10.0 ± 7%

This is 10.0 ± 0.7 mm2
(0.7 is 7% of 10.0)

4. If you square a value
You multiply the % uncertainty by 2
If you cube a value you multiply the % uncertainty by 3
etc
If you need the square root of a value, you divide the % uncertainty by 2.
This is because square root in index form is to the power ½
√x = x½

The general rule is
Multiply the % uncertainty by the index.

What happens to % uncertainty when I multiply by a constant?

The % uncertainty doesn't change. The absolute uncertainty is multiplied by the constant. (see 2 above)
In the example given above we multiplied 5.0 ± 0.1 by a constant, 2
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The absolute uncertainty is multiplied by 2.
The original % uncertainty was 5.0 ± 2%
In the final value of 10.0 ± 0.2 mm
the % uncertainty is still 2%

Note: This is consistent with 3. above.
When you multiply a value by a constant, it is assumed the constant has no uncertainty. We do not associate an uncertainty with the value of Pi or the number 2, for example. So they have a % uncertainty of zero.
So when you multiply the value by the constant and add the % uncertainties, there is only the % uncertainty in the value itself and zero in the constant. Result: no change in % uncertainty.

What if the formula I use to calculate my final value has both adding and multiplication/division?

Let's take an example. Assume you have all the uncertainties in the values in the formula and we want the uncertainty in s

s = ut + ½at²

Step 1.
Add the % uncertainties in u and t to find the % uncertainty in ut
Step 2.
Multiply the % uncertainty in t by 2 (Rule 4 above) and add it to the % uncertainty in a to find the % uncertainty in ½at² (The constant ½ has no uncertainty)
Step 3.
Convert those % uncertainties to absolute uncertainties in ut and in ½at²
Step 4.
Add the absolute ± uncertainties in ut and ½at² found in 3. above to get the absolute uncertainty in the final value of s

*If you would like a more advanced treatment of this topic I recommend the following.
http://www.rit.edu/cos/uphysics/unce...tml#systematic.
54
8 years ago
#2
(Original post by Stonebridge)
In view of the fact that this question is being asked again and again on this forum, and to save me time posting the same answer again and again, this is a summary of how you combine uncertainties at A Level.

You have two values, each with an absolute ± uncertainty.

1. If you add or subtract the two (or more) values to get a final value
The absolute uncertainty in the final value is the sum of the uncertainties.
eg.
5.0 ± 0.1 mm + 2.0 ± 0.1 mm = 7.0 ± 0.2 mm
5.0 ± 0.1 mm - 2.0 ± 0.1 mm = 3.0 ± 0.2 mm

2. If you multiply one value with absolute uncertainty by a constant
The absolute uncertainty is also multiplied by the same constant.
eg.
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The constant can be any number. eg Pi

3. If you multiply or divide two (or more) values, each with an uncertainty
You add the % uncertainties in the two values to get the % uncertainty in the final value.
eg
5.0 ± 0.1 mm x 2.0 ± 0.1 mm
This is
5.0 ± 2% x 2.0 ± 5%
Result
10.0 ± 7%
This is 10.0± 0.7
(0.7 is 7% of 10.0)

4. If you square a value
You multiply the % uncertainty by 2
If you cube a value you multiply the % uncertainty by 3
etc
The general rule is
Multiply the % uncertainty by the index.

What happens to % uncertainty when I multiply by a constant?

The % uncertainty doesn't change. The absolute uncertainty is multiplied by the constant. (see 2 above)
In the example given above we multiplied 5.0 ± 0.1 by a constant, 2
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The absolute uncertainty is multiplied by 2.
The original % uncertainty was 5.0 ± 2%
In the final value of 10.0 ± 0.2 mm
the % uncertainty is still 2%

Note: This is consistent with 3. above.
When you multiply a value by a constant, it is assumed the constant has no uncertainty. We do not associate an uncertainty with the value of Pi or the number 2, for example. So they have a % uncertainty of zero.
So when you multiply the value by the constant and add the % uncertainties, there is only the % uncertainty in the value itself and zero in the constant. Result: no change in % uncertainty.
You might want to add square rooting. 0
#3
(Original post by Zenarthra)
You might want to add square rooting. No need. Work it out yourself.
Rule 4 above.
0
8 years ago
#4
(Original post by Stonebridge)
No need. Work it out yourself.
Rule 4 above.
xD
0
8 years ago
#5
Hopefully should clear up alot of issues people are having 0
8 years ago
#6
(Original post by Stonebridge)
3. If you multiply or divide two (or more) values, each with an uncertainty
You add the % uncertainties in the two values to get the % uncertainty in the final value.
eg
5.0 ± 0.1 mm x 2.0 ± 0.1 mm

This is
5.0 ± 2% x 2.0 ± 5%

Result
10.0 ± 7%

This is 10.0 ± 0.7 mm2
(0.7 is 7% of 10.0)
Thanks for that!

I have a problem with this uncertainty for about a year now.

Take this example

10 ± 2 mm x 11 ± 3 mm

Consider that its actually 12 x 14 - then it will = 168 = 170 (to 2 sf)

But as according to this rule above the answer should be represented by 110 ± 52. That is the upper bound is 162. So this value lies outside this range and so this rule is false.

0
#7
(Original post by RoyalBlue7)
Thanks for that!

I have a problem with this uncertainty for about a year now.

Take this example

10 ± 2 mm x 11 ± 3 mm

Consider that its actually 12 x 14 - then it will = 168 = 170 (to 2 sf)

But as according to this rule above the answer should be represented by 110 ± 52. That is the upper bound is 162. So this value lies outside this range and so this rule is false.

There's no problem.

The rule for adding percentages (or fractional uncertainties) is derived (from calculus) assuming the uncertainty is small.
In this example the uncertainties are rather large (20% and 27% giving a total of 47%) which means this method will produce a different result from the one where you take those upper and lower bounds. As the % uncertainty gets smaller, the results will agree more closely with your other method. You really need to keep % uncertainties below about 10% to keep it valid. 47% is way too high.
The % uncertainty formula is a quick way of finding your total uncertainty and is a lot quicker than using upper and lower bounds, especially when there are many terms to include in the total error.
The only limitation is that you need the uncertainties to be relatively small. Even so, it still gives a good idea of what uncertainties you have in your experiment.
BTW. If you have an experiment with an uncertainty of 47% the fact that the % error formula gives a different result is the least of your worries.
0
8 years ago
#8
(Original post by Stonebridge)
There's no problem.

The rule for adding percentages (or fractional uncertainties) is derived (from calculus) assuming the uncertainty is small.
In this example the uncertainties are rather large (20% and 27% giving a total of 47%) which means this method will produce a different result from the one where you take those upper and lower bounds. As the % uncertainty gets smaller, the results will agree more closely with your other method. You really need to keep % uncertainties below about 10% to keep it valid. 47% is way too high.
The % uncertainty formula is a quick way of finding your total uncertainty and is a lot quicker than using upper and lower bounds, especially when there are many terms to include in the total error.
The only limitation is that you need the uncertainties to be relatively small. Even so, it still gives a good idea of what uncertainties you have in your experiment.
BTW. If you have an experiment with an uncertainty of 47% the fact that the % error formula gives a different result is the least of your worries.
Thanks...guess that clears it now 0
#9
(Original post by RoyalBlue7)
Thanks...guess that clears it now Good question; and a point that needed further explanation.
0
8 years ago
#10
(Original post by Stonebridge)
In view of the fact that this question is being asked again and again on this forum, and to save me time posting the same answer again and again, this is a summary of how you combine uncertainties at A Level.

You have two values, each with an absolute ± uncertainty.

1. If you add or subtract the two (or more) values to get a final value
The absolute uncertainty in the final value is the sum of the uncertainties.
eg.
5.0 ± 0.1 mm + 2.0 ± 0.1 mm = 7.0 ± 0.2 mm
5.0 ± 0.1 mm - 2.0 ± 0.1 mm = 3.0 ± 0.2 mm

2. If you multiply one value with absolute uncertainty by a constant
The absolute uncertainty is also multiplied by the same constant.
eg.
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The constant can be any number. eg Pi

3. If you multiply or divide two (or more) values, each with an uncertainty
You add the % uncertainties in the two values to get the % uncertainty in the final value.
eg
5.0 ± 0.1 mm x 2.0 ± 0.1 mm

This is
5.0 ± 2% x 2.0 ± 5%

Result
10.0 ± 7%

This is 10.0 ± 0.7 mm2
(0.7 is 7% of 10.0)

4. If you square a value
You multiply the % uncertainty by 2
If you cube a value you multiply the % uncertainty by 3
etc

The general rule is
Multiply the % uncertainty by the index.

What happens to % uncertainty when I multiply by a constant?

The % uncertainty doesn't change. The absolute uncertainty is multiplied by the constant. (see 2 above)
In the example given above we multiplied 5.0 ± 0.1 by a constant, 2
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The absolute uncertainty is multiplied by 2.
The original % uncertainty was 5.0 ± 2%
In the final value of 10.0 ± 0.2 mm
the % uncertainty is still 2%

Note: This is consistent with 3. above.
When you multiply a value by a constant, it is assumed the constant has no uncertainty. We do not associate an uncertainty with the value of Pi or the number 2, for example. So they have a % uncertainty of zero.
So when you multiply the value by the constant and add the % uncertainties, there is only the % uncertainty in the value itself and zero in the constant. Result: no change in % uncertainty.

What if the formula I use to calculate my final value has both adding and multiplication/division.

Let's take an example. Assume you have all the uncertainties in the values in the formula and we want the uncertainty in s

s = ut + ½at²

Step 1.
Add the % uncertainties in u and t to find the % uncertainty in ut
Step 2.
Multiply the % uncertainty in t by 2 (Rule 4 above) and add it to the % uncertainty in a to find the % uncertainty in ½at²
Step 3.
Convert those % uncertainties to absolute uncertainties in ut and in ½at²
Step 4.
Add the absolute ± uncertainties in ut and ½at² found in 3. above to get the absolute uncertainty in the final value of s
Hello there

I didn't understand aiii
Can you actually explain it to me by linking to the rules please. Many thanks!
0
8 years ago
#11
So if i had to work out the %uncertainty for 2D, would i multiply the % uncertainty of D by two
0
#12
(Original post by Daniel Atieh)
Hello there

I didn't understand aiii
Can you actually explain it to me by linking to the rules please. Many thanks!
It's rule 2.
you are multiplying the value by a constant. The constant here is 1/10

That's equivalent to dividing by 10
if you divide by a constant you also divide the absolute uncertainty by that constant.
0
#13
(Original post by azar316)
So if i had to work out the %uncertainty for 2D, would i multiply the % uncertainty of D by two
No use rule 2.
you are multiplying the value by a constant.
0
8 years ago
#14
(Original post by Stonebridge)
It's rule 2.
you are multiplying the value by a constant. The constant here is 1/10

That's equivalent to dividing by 10
if you divide by a constant you also divide the absolute uncertainty by that constant.
Awesome! Really got it, thank you so much!!
0
8 years ago
#15
Thanks i understand now. I never read it properly
0
8 years ago
#16
I know logarithms are used in A level physics (although I can't remember if uncertainty calculations are required for them), so it may be wise to put a section in on them, as most 'standard' methods don't work (like as adding absolute or fractional uncertainties in quadrature).

The way I was taught to do them easily and with reasonable accuracy is using the minimum and maximum values.
So say you have an equation and you measure x to be then you may be asked for value and uncertainty of y.
The value is straightforward: , but the uncertainty isn't immediately obvious, as gives highly nonsensical results.

Instead consider the range of values "7" could really be - it may vary from 6.9 to 7.1, so the minimum that can be is and the maximum is . These should give you a range of values of around 0.01 rather than 2 (which is what you get if you take the logarithm of the uncertainty).

You can write this more generally as Or you may choose to have the plus and the minus separate as halving the difference between the extremes is not quite right, but is a good easy approximation.

That's how I do most of my uncertainties these days, but I know that A levels can be very rigid and only accept the awarding body's chosen method so it may not be considered "correct".
0
8 years ago
#17
(Original post by Manitude)
...
I haven't seen functions of x such as sin(x), log(x) and e^(x) to do with any uncertainty calculations in AL ( Edexcel) but I guess they could be easily done using the range method or calculus (finding the derivative/rate of change and multiplying by the uncertainty)?

Posted from TSR Mobile
0
8 years ago
#18
(Original post by Stonebridge)
It's rule 2.
you are multiplying the value by a constant. The constant here is 1/10

That's equivalent to dividing by 10
if you divide by a constant you also divide the absolute uncertainty by that constant.
Hello

Can you please look at this thread and help me with my question: http://www.thestudentroom.co.uk/show....php?t=2674025
Will be really appreciated!
0
8 years ago
#19
(Original post by Stonebridge)
In view of the fact that this question is being asked again and again on this forum, and to save me time posting the same answer again and again, this is a summary of how you combine uncertainties at A Level.

You have two values, each with an absolute ± uncertainty.

1. If you add or subtract the two (or more) values to get a final value
The absolute uncertainty in the final value is the sum of the uncertainties.
eg.
5.0 ± 0.1 mm + 2.0 ± 0.1 mm = 7.0 ± 0.2 mm
5.0 ± 0.1 mm - 2.0 ± 0.1 mm = 3.0 ± 0.2 mm

2. If you multiply one value with absolute uncertainty by a constant
The absolute uncertainty is also multiplied by the same constant.
eg.
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The constant can be any number. eg Pi

3. If you multiply or divide two (or more) values, each with an uncertainty
You add the % uncertainties in the two values to get the % uncertainty in the final value.
eg
5.0 ± 0.1 mm x 2.0 ± 0.1 mm

This is
5.0 ± 2% x 2.0 ± 5%

Result
10.0 ± 7%

This is 10.0 ± 0.7 mm2
(0.7 is 7% of 10.0)

4. If you square a value
You multiply the % uncertainty by 2
If you cube a value you multiply the % uncertainty by 3
etc

The general rule is
Multiply the % uncertainty by the index.

What happens to % uncertainty when I multiply by a constant?

The % uncertainty doesn't change. The absolute uncertainty is multiplied by the constant. (see 2 above)
In the example given above we multiplied 5.0 ± 0.1 by a constant, 2
2 x (5.0 ± 0.1 mm ) = 10.0 ± 0.2 mm
The absolute uncertainty is multiplied by 2.
The original % uncertainty was 5.0 ± 2%
In the final value of 10.0 ± 0.2 mm
the % uncertainty is still 2%

Note: This is consistent with 3. above.
When you multiply a value by a constant, it is assumed the constant has no uncertainty. We do not associate an uncertainty with the value of Pi or the number 2, for example. So they have a % uncertainty of zero.
So when you multiply the value by the constant and add the % uncertainties, there is only the % uncertainty in the value itself and zero in the constant. Result: no change in % uncertainty.

What if the formula I use to calculate my final value has both adding and multiplication/division.

Let's take an example. Assume you have all the uncertainties in the values in the formula and we want the uncertainty in s

s = ut + ½at²

Step 1.
Add the % uncertainties in u and t to find the % uncertainty in ut
Step 2.
Multiply the % uncertainty in t by 2 (Rule 4 above) and add it to the % uncertainty in a to find the % uncertainty in ½at²
Step 3.
Convert those % uncertainties to absolute uncertainties in ut and in ½at²
Step 4.
Add the absolute ± uncertainties in ut and ½at² found in 3. above to get the absolute uncertainty in the final value of s

In step 3 of finding the absolute uncertainty in 1/2 a t^2, you mean multiplying the whole"1/2at^2" by the percentage uncertainty or just multiply the percentage uncertainty by at^2? why ? sorry for asking this . It seems silly but i really want to clear my concept. Thank you!

also for percentage uncertainty , does significant figures matter?

http://www.thestudentroom.co.uk/show....php?t=2672144
0
#20
(Original post by Lamalam)
In step 3 of finding the absolute uncertainty in 1/2 a t^2, you mean multiplying the whole"1/2at^2" by the percentage uncertainty or just multiply the percentage uncertainty by at^2? why ? sorry for asking this . It seems silly but i really want to clear my concept. Thank you!

also for percentage uncertainty , does significant figures matter?

http://www.thestudentroom.co.uk/show....php?t=2672144
To find the absolute uncertainty in ½at2
1. find the % uncertainty as I explained.
2. Then multiply the actual value you have calculated for ½at2 by (the % uncertainty / 100). This is how you find actual uncertainty from % uncertainty for any value.

Normally % uncertainties are expressed to the nearest %. Never 4.8% for example. Call it 5%
0
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