Error Calculation for Triangular Plate Sides

In summary, the length of the remaining side is 145.7 millimetres and the maximum possible error in this result is +/- 2.6 millimetres.
  • #1
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Homework Statement


Two Sides of a triangular plate are measured as 125mm and 160mm, each to the nearest millimetre. The included angle is quoted as 60+-1. Calculate the length of the remaining side and the maximum possible error in this result

My first question is to the nearest millimetre? What does this mean, i think +-0.5 but when i think about this if it was 0.5 biggest it would be rounded up..

Homework Equations


δu = Fxδx + Fyδy + Fzδz

The Attempt at a Solution


So firstly to find the length of the other side i used the cosine rule
a^2 = b^2 + c^2 - 2bccos(A)
Where b = 125
c = 160
A = pi/3
Which comes out as 145.7 which is correct.
The next stage i guess is to take partial derivatives.
My equation for a is a^2 so i assume i have to square root it so i can find δa I am not sure but i guess i have to??
δa = partial with respect to b * δb + partial with respect to c *δc + partial with respect to A δA

Therefore a = (b^2 + c^2 - 2bcCos(A))^1/2
Okay this is where it gets messy.
Partial with respect to b:
[1/2 * 2b-2cCosA/(b^2+c^2-2bcCos(A))^1/2] *δb

Partial with respect to C
[1/2 * 2c - 2bcos(A)/((b^2+c^2-2bcCos(A))^1/2) ]*δc

Partial with respect to A
[2bcsinA/((b^2+c^2-2bcCos(A))^1/2)] *δA

When i input values of δb and δc = +-0.5 and δA = +-1 i don't seem to get the right answer which is +- 2.6, i really don't see where I'm going wrong, as i did the similar method with a previous exercise.
 
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  • #2
My first question is to the nearest millimetre? What does this mean, i think +-0.5
Right.
but when i think about this if it was 0.5 biggest it would be rounded up..
True, but 0.4999999 (add as many finite 9 as you like) would be rounded down, the smallest upper bound is 0.5.

My equation for a is a^2 so i assume i have to square root it so i can find δa I am not sure but i guess i have to??
You can calculate δa^2 as intermediate result if you like.

I think there are some brackets missing in your derivatives.
Which answer do you get?
 
Last edited:
  • #3
My answers are
Partial b = [((2b-2cCosA)/2*(b^2+c^2-2bcCos(A))^1/2] *δb
= [((250 - 160)/2*((125^2 + 160^2 - 2*160*125*0.5)^0.5))] * 0.5
= (90/291.4) * 0.5 = +-0.154
partial c has the same denominator
[((2c - 2bcos(A)) / 291.4) ]*0.5
= 195/291.4 * 0.5 = +-0.335
Partial A has the same denominator aswell
= (125*250*2*sin(pi/3)) / 291.4
Which is a really large number, not sure what's going wrong there..

The correct answer is 2.6, i don't seem to be close to that at present, i guess there must be a simple mistake I'm making.
 
  • #4
Did you convert ##\delta A## to radians as well?
 
  • #5
Nope completely forgot about that, thanks
Well that leaves me with
[(125*250*2*sin(pi/3) / 291.4 ] *pi/180 = 3.24 which is still off, i can't see where I've made a mistake, I've done problems after this one and been fine, i just can't seem to get this one.
 
  • #6
The 250 in the numerator should be 160, no?

Also, I get a slightly different expression for the partial derivative wrt A than you do.
 
  • #7
δa = [itex]\frac{\partial a}{\partial b}[/itex] δb + [itex]\frac{\partial a}{\partial c}[/itex] δc + [itex]\frac{\partial a}{\partial A}[/itex] *δA

δb = 0.5 δc = 0.5 δA = pi/180

[itex]\frac{\partial a}{\partial b}[/itex] = [itex]\frac{b-cCos(A)}{\sqrt{b^{2}+c^{2} - 2bccos(A)}}[/itex]
[itex]\frac{\partial a}{\partial b}[/itex] = [itex]\frac{125-160*Cos(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{45}{145.688}[/itex]
[itex]\frac{\partial a}{\partial b}[/itex] = 0.309
[itex]\frac{\partial a}{\partial b}[/itex] * δb = 0.1545

[itex]\frac{\partial a}{\partial c}[/itex] = [itex]\frac{c-bCos(A)}{\sqrt{b^{2}+c^{2} - 2bccos(A)}}[/itex]
[itex]\frac{\partial a}{\partial c}[/itex] = [itex]\frac{160-125*Cos(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{97.5}{145.688}[/itex]
[itex]\frac{\partial a}{\partial c}[/itex] = 0.669
[itex]\frac{\partial a}{\partial c}[/itex] * δc = 0.335

[itex]\frac{\partial a}{\partial A}[/itex] = [itex]\frac{cbsin(A)}{\sqrt{b^{2}+c^{2} - 2bccos(A)}}[/itex]
[itex]\frac{\partial a}{\partial A}[/itex] = [itex]\frac{160*125*sin(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{365.52}{145.688}[/itex]
[itex]\frac{\partial a}{\partial A}[/itex] = 2.509
[itex]\frac{\partial a}{\partial A}[/itex] * δA = 0.044

I know for a fact these won't add up to 2.6, so i very confused at which stage I am going wrong.
 
  • #8
160*125*sin(pi/3) is not 365.52. Make sure that you use radians for the sine, not degrees.
 
  • #9
oo what a silly mistake
[itex]\frac{\partial a}{\partial A}[/itex] = [itex]\frac{160*125*sin(\frac{pi}{3})}{\sqrt{125^{2}+160^{2} - 2bccos(\frac{pi}{3})}}[/itex] = [itex]\frac{17320.51}{145.688}[/itex]
[itex]\frac{\partial a}{\partial A}[/itex] = 118.888
[itex]\frac{\partial a}{\partial A}[/itex] * δA = 2.075

2.075 + 0.1545 + 0.335 = δa
δa = 2.56 → 2.6 which is the correct answer yey, thanks very much.
 

What is the formula for calculating the error for a triangular plate side?

The formula for calculating the error for a triangular plate side is: Error = 0.5 * (a + b - c), where a, b, and c are the measured lengths of the three sides of the triangle.

Why is it important to calculate the error for triangular plate sides?

Calculating the error for triangular plate sides is important because it helps to determine the accuracy and precision of the measurements. It also allows for proper adjustments to be made in the design and construction of the plate to ensure its stability and structural integrity.

Can the error for triangular plate sides be negative?

No, the error for triangular plate sides cannot be negative. It represents the difference between the measured lengths and the expected length of the sides, so it will always be a positive value.

What are some common sources of error when calculating the error for triangular plate sides?

Some common sources of error in calculating the error for triangular plate sides include measurement errors, rounding errors, and errors in the construction or assembly of the plate. Environmental factors such as temperature and humidity can also affect the accuracy of the measurements.

How can the error for triangular plate sides be minimized?

The error for triangular plate sides can be minimized by using precise and accurate measuring tools, taking multiple measurements and averaging them, and minimizing any external factors that may affect the measurements. It is also important to follow proper construction and assembly techniques to ensure the most accurate results.

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