How Do You Calculate Uncertainty in Physics Equations?

[tmilford]

Homework Statement

I am writing a report where I got a value of distance of let's L= 0.250 +/- 0.005 m. Then I need to use this value to calculate another distance value and an angle. How would I incoporate the uncertainity of this value into my other calculated values?

Homework Equations

The equations I am using are

0.11 = L tan A

d sin A = 600 x 10^-9

Where d is the distance I am trying to incorporate the uncertainity into and A is the angle I am trying to get the uncertainity of.

The Attempt at a Solution

I tried directly using the uncertainity value in the equations but I got a huge uncertainity value for my angle and a really small value for the distance, d, so I don't know that's the correct way of doing it.

 

[HallsofIvy]

If L tan A= 0.11 and L= 0.25+- .005, then
$$tan A= \frac{0.11}{L}$$

L may be as small as 0.25- 0.005= 0.245. In that case,
$$tan A= \frac{0.11}{0.245}= 0.4490$$
Determine A from that.

L may be as large as 0.25+ 0.005= 0.255. In that case,
$$tan A= \frac{0.11}{0.245}= 0.4314$$
Determine A from that.

Those two values of A give the possible range for A.

 

[tmilford]

Okay I understand what you did, but there isn't an exact value for the uncertainity for let's the say the angle. The difference between the max value and actual value is different than the difference between the min value and the actual value.

I am using a table to record these inputs and thought I could simply put a value down for the uncertainity not the range of two numbers, do you think I could average the two out or is there another way to do it?

 

[gneill]

Okay I understand what you did, but there isn't an exact value for the uncertainity for let's the say the angle. The difference between the max value and actual value is different than the difference between the min value and the actual value.

I am using a table to record these inputs and thought I could simply put a value down for the uncertainity not the range of two numbers, do you think I could average the two out or is there another way to do it?

You could use error analysis theory. If some result R depends upon a function f(x,y,z,...) of several variables, each with its own independent, gaussian error (+/- δ value), then

$$\delta R = \sqrt{\left(\frac{\partial f}{\partial x}\delta_x\right)^2 + \left(\frac{\partial f}{\partial y}\delta_y\right)^2 + \left(\frac{\partial f}{\partial z}\delta_z\right)^2 + ...}$$

In your case you have

$$A = arctan\left(\frac{0.11m}{L}\right)$$

So, only one variable with an error term (L +/- δ_L). Do the math!