Propagation of Uncertainty with Angles

[TheJuke]

Homework Statement

I conducted an experiment which involves measuring two distances (Y and L) and have used tan to determine the angle, then finally calculated the sine of the angles for use in my analysis.

I have uncertainties in both length measurements and am unsure how to propagate the uncertainties the way through.

Homework Equations

Unsure of what equation to use here.

The Attempt at a Solution

Unsure where to start really.

Thanks.

 

[frogjg2003]

What level of mathematics have you taken? If you've taken multivariable calculus, it will be really simple to explain.

 

[TheJuke]

I am currently taking multivariable calculus so that would be great

 

[frogjg2003]

If δf is the uncertainty of a function of f(x₁,x₂,...,x_n) each with error δx_i, then the most general equation for the error is $(\delta f)^2=\sum_{i=1}^n(\delta x_i \frac{\partial f}{\partial x})^2$. If you want a derivation, I would recommend reading Introduction to Error Analysis by Taylor.

 

[TheJuke]

Thanks so much, I think I have it.

Would you mind having a look over my calculations? I am 100% sure of them as I would expect a far greater error.



Example Calculations:
θ=arctan(6/28)
θ=0.21 radians
sinθ=0.21

Propagation of Uncertainty:
Let r= y/L
∆r= (√((∆y/y)^2+ (∆L/L)^2 ))r
θ=arctan(r)
∆θ=(d(arctan(r))/dr) ∙ ∆r
∆θ(1/d)= 1/(r^2+1) (√((∆y/y)^2+ (∆L/L)^2 ))r

At smallest value:
∆θ(0.5)= 1/((6/28)^2+1) (√((0.25/6)^2+ (0.1/28)^2 ))(6/28)
∆θ=0.00857
∆θ=0.01
At maximum value:
∆θ(0.5)= 1/((18/28)^2+1) (√((0.25/18)^2+ (0.1/28)^2 ))(18/28)
∆θ=0.00652
∆θ=0.01
Therefore the uncertainty in all calculations is 0.01 as this is the limit of the precision in the original measurements.

 

[frogjg2003]

I would have used 0.009 and 0.007 for the two accuracies. Propagation of uncertainty is one of the few cases where I would recommend plugging in numbers at every step. That is because
$$\frac{\delta r}{\sqrt{r^2+1}}$$
is easier to calculate than
$$\frac{\sqrt{\left(\frac{\delta y}{L}\right)^2+\left(\frac{y\times\delta L}{L^2}\right)^2}}{\sqrt{\left(\frac{y}{L}\right)^2+1}}$$