Estimating the maximum possible percentage error


by benji123
Tags: error, estimating, maximum, percentage
benji123
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#1
Aug10-10, 01:53 PM
P: 4
1. The problem statement, all variables and given/known data
Hi all! Im new to the forum, came across it from googling the problem which I had and someone was having difficuilty with a question like this on here! Here's the question I have...

The coefficient of rigidity, n, of a wire of length L and uniform diameter d is given by

n=AL/d^4

If A is a constant and the parameters L and d are known to within 2% of their correct values, estimate the maximum possible percentage error in the calculated value of n.

2. Relevant equations

n=AL/d^4

3. The attempt at a solution

dn = dn/dL x dL + dn/dA x dA

dn = A/d^4 dL + L/d^4 dA

n = A/d^4 x ΔL + L/d^4 x ΔA

n = ΔL/L + ΔA/A

Where do I go from here? Any help will be greatly appreciated :)
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Mark44
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#2
Aug10-10, 04:32 PM
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Quote Quote by benji123 View Post
1. The problem statement, all variables and given/known data
Hi all! Im new to the forum, came across it from googling the problem which I had and someone was having difficuilty with a question like this on here! Here's the question I have...
Welcome, benji. You have come to a good place for help.
Quote Quote by benji123 View Post

The coefficient of rigidity, n, of a wire of length L and uniform diameter d is given by

n=AL/d^4

If A is a constant and the parameters L and d are known to within 2% of their correct values, estimate the maximum possible percentage error in the calculated value of n.

2. Relevant equations

n=AL/d^4

3. The attempt at a solution

dn = dn/dL x dL + dn/dA x dA
A is a constant, so dn/dA doesn't make any sense. Instead, you want dn/dd. For the record, dn/dL and dn/dd are really the partial derivatives of n. I.e.,
[tex]\frac{\partial n}{\partial L} \text{and} \frac{\partial n}{\partial d}[/tex]
Quote Quote by benji123 View Post

dn = A/d^4 dL + L/d^4 dA

n = A/d^4 x ΔL + L/d^4 x ΔA

n = ΔL/L + ΔA/A

Where do I go from here? Any help will be greatly appreciated :)
For the percent change in n, you want dn/n.
benji123
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#3
Aug10-10, 05:02 PM
P: 4
Quote Quote by Mark44 View Post

For the percent change in n, you want dn/n.
Thanks for the reply Mark :) and correcting me with the dn/dA and dn/dd situation!

I was wondering, how do I get the estimation when I have no values to work from for L and d?

Mark44
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#4
Aug10-10, 05:23 PM
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Estimating the maximum possible percentage error


You don't have L and d, but you have their relative errors: ΔL/L and Δd/d.
benji123
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#5
Aug10-10, 06:04 PM
P: 4
Quote Quote by Mark44 View Post
You don't have L and d, but you have their relative errors: ΔL/L and Δd/d.
What do I do with the relative error? As "Δd" will be 2% what will I use for "d" ?
Mark44
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#6
Aug10-10, 08:02 PM
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No, Δd is not a percent value; it is the error in determining d. The relative error, which is what you are given, is Δd/d, and that is .02. Actually, what is given is that |Δd|/d <= .02. Same with L and ΔL.
benji123
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#7
Aug11-10, 05:31 AM
P: 4
So would it be...

% change in n = % change in L - (4 times % change in d)
% change in n = -2 - (4 times -2)
% change in n = -2 +8
% change in n = + 6%

Therefore, 2% change in L and d = 6% change in n?
Mark44
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#8
Aug11-10, 09:01 AM
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Quote Quote by benji123 View Post
So would it be...

% change in n = % change in L - (4 times % change in d)
% change in n = -2 - (4 times -2)
% change in n = -2 +8
% change in n = + 6%

Therefore, 2% change in L and d = 6% change in n?
No. Where are you getting the -2 numbers?

You started off in the right direction, but didn't fix the errors I pointed out earlier, so let's start from the beginning.

n = AL/d4
[tex]dn = A(\frac{\partial n}{\partial L}~dL + \frac{\partial n}{\partial d}~dd)[/tex]

The percent change in n is Δn/n. Fill in the partial derivatives in the equation above and divide both sides by n.
Dickfore
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#9
Aug11-10, 09:20 AM
P: 3,015
There is a more elementary way.

[tex]
n \propto L
[/tex]

means that n increases with increasing L, while

[tex]
n \propto d^{-4}
[/tex]

means that n decreases with increasing d.

Therefore, the smallest value for n is obtained by taking the smallest value for L and the largest value for d:

[tex]
n_{\mathrm{min}} = A \, \frac{L_{\mathrm{min}}}{d^{4}_{\mathrm{max}}}
[/tex]

The largest value for n, on the other hand, is obtained by taking the largest value for L and the smallest value for d:

[tex]
n_{\mathrm{max}} = A \, \frac{L_{\mathrm{max}}}{d^{4}_{\mathrm{min}}}
[/tex]

In this way, you obtain an interval for the possible values of n:

[tex]
n \in [n_{\mathrm{min}}, n_{\mathrm{max}}]
[/tex]

Instead of the interval notation, one usually uses the "techincal notation":

[tex]
n = \bar{n} \pm \Delta n
[/tex]

which actually means:

[tex]
\left\{\begin{array}{l}
n_{\mathrm{min}} = \bar{n} - \Delta n \\

n_{\mathrm{max}} = \bar{n} + \Delta n
\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}
\bar{n} = \frac{1}{2} \, (n_{\mathrm{min}} + n_{\mathrm{max}}) \\

\bar{n} = \frac{1}{2} \, (n_{\mathrm{max}} - n_{\mathrm{min}}) \\
\end{array}\right.
[/tex]

Then, of course, the relative uncertainty, expressed in percent, is defined as:

[tex]
\delta_{n} \equiv \frac{\Delta n}{\bar{n}} \cdot 100\%
[/tex]

It is up to you to:
1. Find Lmin and Lmax by knowing [itex]\bar{L}[/itex] (the nominal value) and [itex]\Delta L = \delta_{L}/{100 \%} \, \bar{L}[/itex] (the absolute uncertainty);

2. Do the same for dmin and dmax;

3. Find nmin and nmax according to the above formulas;

4. Find [itex]\bar{n}[/itex] (nominal value) and [itex]\Delta n[/itex] (absolute uncertainty) according to the above formulas;

5. Find the relative uncertainty [itex]\delta_{n}[/itex].


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