Error calculation

  • Thread starter steve2510
  • Start date
  • #1
36
0

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 im 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.
 

Answers and Replies

  • #2
35,261
11,513
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 im 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
36
0
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 whats 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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,881
1,456
Did you convert ##\delta A## to radians as well?
 
  • #5
36
0
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 cant 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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,881
1,456
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
36
0
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 im going wrong.
 
  • #8
35,261
11,513
160*125*sin(pi/3) is not 365.52. Make sure that you use radians for the sine, not degrees.
 
  • #9
36
0
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.
 

Related Threads on Error calculation

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
0
Views
1K
Replies
2
Views
5K
Replies
6
Views
7K
Replies
1
Views
3K
Replies
1
Views
9K
Replies
7
Views
3K
Replies
22
Views
3K
Top