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June 4th, 2019, 07:06 AM   #1
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Finding approximately the greatest possible percentage error

A swamp interferes with the direct measurement of a surveyline AB. Hence, an auxiliary point C is established and the distance AC is measured with an accuracy of 1 part in 1000. Then the angles BAC and ABC are found to be 45 and 30 degrees respectively, with a possible error of 2 minutes in each. Find approximately the greatest possible percentage error in the computed length of AB.
Ans. +-0.23%

From what I understand, you're supposed to find dAB. So I drew a triangle ABC with the given angles, then used Sine Law to obtain the equation:

$\displaystyle AB/\sin C = AC/\sin B $

And when I take the derivative of the equation, I end up with an equation with unknown values. Need help.

Last edited by skipjack; June 4th, 2019 at 01:07 PM.
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June 7th, 2019, 09:16 PM   #2
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I get $\approx \pm 0.201$ %:

First, assuming $AC=1$, and since $\angle C = 180^\text{o} - \angle A - \angle B$ (neglecting the effects of the curvature of the earth ), we can calculate the nominal value of $AB = 1.93185$ (with nominal values of $\angle A = 45^\text{o}$, $\angle B = 30^\text{o}$, and $\angle C = 105^\text{o}$ ).


Also, $\Delta AC = 0.001$ (since $AC=1$), and $\Delta A$ and $\Delta B$ are $\large \pm \frac{2}{60 \cdot 360} \cdot 2 \pi \approx \pm 0.00058178 \text{ rad}$ max/min.


So, by your relationship:

$\displaystyle AB \cdot \sin B \ = \ AC \cdot \sin C$

We can use the product rule and chain rule to differentiate (I feel more comfortable using the $\Delta$ sign rather than the $\delta$ sign):

$\displaystyle \Delta AB \cdot \sin B + AB \cdot \cos B \cdot \Delta B=\Delta AC \cdot \sin C + AC \cdot \cos C \cdot \Delta C$


Note that the terms immediately following the $+$ sign on each side of the $=$ sign could also be derived with the following identity:

$\displaystyle \sin (x + \Delta x) - \sin(x)=2 \sin \left(\frac{\Delta x}{2} \right) \cos \left (\frac{2x+\Delta x}{2} \right ) \approx \Delta x \cdot \cos x \quad \text{ for small } \Delta x$



Since $\Delta C = - \Delta A - \Delta B$ we have:

$\displaystyle \Delta AB \cdot \sin B + AB \cdot \cos B \cdot \Delta B=\Delta AC \cdot \sin C - AC \cdot \cos C (\Delta A + \Delta B)$


Collecting like terms, and re-arranging, we have:

$\displaystyle \Delta AB \cdot \sin B = \Delta AC \cdot \sin C - \Delta B (AB \cdot \cos B + AC \cdot \cos C) - \Delta A \cdot AC \cdot \cos C$


The same equation, but with approximate nominal and min/max values, with appropriate sign, to find the max $+ \Delta AB$ :

$\displaystyle \Delta AB \cdot \underbrace{\sin B}_{0.5} = \underbrace{\Delta AC}_{0.001} \cdot \overbrace{\sin C}^{0.965926} - \underbrace{\Delta B}_{-0.00058178} \overbrace{(AB \cdot \cos B + AC \cdot \cos C)}^{1.4142136} - \underbrace{\Delta A}_{+0.00058178} \cdot \overbrace{AC}^1 \cdot \underbrace{\cos C}_{-0.258819}$



This gives (for max):

$\displaystyle \frac{\Delta AB}{AB} \approx 0.002008$



We could also note that in order to get a maximum, $AC$ needs to go to its high limit, $\angle B$ needs to go to its low limit, $\angle A$ needs to go to its high limit, and just plug in the values to get the maximum excursions (and go the opposite way for the low limit). Since these directions do not change over the ranges of all variables, I believe that in this case, this gives the true max and min (not true in all cases though).



Over the years, in my work I have had to do some specification min/max calculations, so I wrote a script to look at the derivatives of the variables in an equation and calculate the min/nom/max using the sign of the derivatives. The script also warns if a derivative changes sign at min/max from that at nom.

The output of my script for this problem is (there were no derivative change warnings):

Code:
Function name: distance

Minimum: 1.927977369304497
angle_a        0.784816387
angle_b        0.524180552
ac             0.999



Nominal: 1.9318516510238812
angle_a        0.785398163
angle_b        0.523598776
ac             1.0



Maximum: 1.935734403101186
angle_a        0.785979939
angle_b        0.5230170000000001
ac             1.001

Which gives +0.0020099, -0.0020055 for $\frac{\Delta AB}{AB}$ .
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June 8th, 2019, 05:36 AM   #3
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As a follow up, I think I know why the answer was given as $\approx \pm 0.23$ %:

If we start with one of the equations given above:

$\displaystyle \Delta AB \cdot \sin B + AB \cdot \cos B \cdot \Delta B=\Delta AC \cdot \sin C + AC \cdot \cos C \cdot \Delta C$

and we try to maximize $+\Delta AB$ with approximate nominal and min/max values, with appropriate sign:

$\displaystyle \Delta AB \cdot \underbrace{\sin B}_{0.5} + \overbrace{AB}^{1.93185} \cdot \underbrace{\cos B}_{0.866025} \cdot \hspace -0.3 cm \overbrace{\Delta B}^{-0.00058178}=\underbrace{\Delta AC}_{0.001} \cdot \overbrace{\sin C}^{0.965926} + \underbrace{AC}_1 \cdot \overbrace{\cos C}^{-0.258819} \cdot \hspace -4.75 cm \underbrace{\Delta C}_{\hspace 5cm -(0.00058178+0.00058178 )=-(\Delta A + \Delta B)}$


I hope that this renders correctly on all screens, otherwise, w/o the \hspace(s):

$\displaystyle \Delta AB \cdot \underbrace{\sin B}_{0.5} + \overbrace{AB}^{1.93185} \cdot \underbrace{\cos B}_{0.866025} \cdot \overbrace{\Delta B}^{-0.00058178}=\underbrace{\Delta AC}_{0.001} \cdot \overbrace{\sin C}^{0.965926} + \underbrace{AC}_1 \cdot \overbrace{\cos C}^{-0.258819} \cdot \underbrace{\Delta C}_{-(0.00058178+0.00058178 )=-(\Delta A + \Delta B)}$



This gives $\frac{\Delta AB}{AB} \approx 0.0023$ .

The problem is, of course, $\Delta B$ was taken to be both negative (on the left side) and positive (on the right side) within the same equation, so I think that $\pm 0.20$ % as given above is more accurate.
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