# Calibration error & zero error

1. May 10, 2009

### Shukie

1. The problem statement, all variables and given/known data
I have an equation with a number of variables, of which the values and their uncertainties are known. Two of these variables are measured with a ruler, h and l. The question is, will a zero error in the ruler affect the answer or the uncertainty in the answer? The same question for a possible calibration error.

2. Relevant equations
$$\mu = \left(\frac{\pi*R^4}{8Q}\right)*\left(\frac{p*g*(h1 - h2) - p*g*(h2 - h3)}{l1 - l2}\right)$$

$$R = (400 \pm 4)*10^{-6} m$$
$$Q = (100 \pm 0.5)*10^{-9} \frac{m^{3}}{s}$$
$$l1 = (163.8 \pm 0.2)*10^{-3} m$$
$$l2 = (101 \pm 0.2)*10^{-3} m$$
$$h1 = (223 \pm 1)*10^{-3} m$$
$$h2 = (83 \pm 1)*10^{-3} m$$
$$h3 = (23 \pm 1)*10^{-3} m$$
$$g = 10 \mbox{and} p = 1000$$

3. The attempt at a solution
Evaluating this function I get $$\mu = (1.28 \pm 0.07)*10^{-3}$$. The actual zero error isn't given, so I'll assume an error of 1%. Can I simply multiply all the h and l values by 1.01 and then evaluate $$\mu$$ again to see if it deviates from the above answer? If so, do I multiply the uncertainties in h and l by 1.01 as well, or only their actual values?

As for the calibration error, how is this different from the zero error?

2. May 10, 2009

### LowlyPion

As I would read it, I would think they want you to calculate the error with and without the uncertainties in h and l, which I think means that the nominal result wouldn't change but the error may, if it's weight is significant in determining the original % error.

In the case where you have a systematic error like calibration maybe consider what the effect is on the difference terms in the equation.

3. May 11, 2009

### Shukie

I don't understand. If I have a zero error of 1%, can I simply multiply the uncertainties in h and l by 1.01 as well? If I do that and re-evaluatie the equation, I get $$\mu = (1.28 \pm 0.06)*10^{-3}$$.

4. May 11, 2009

### davieddy

For a zero error of 2 mm, (e.g using the end of the ruler instead of the zero mark)
true value of h = measured value of h - 2 mm

For a calibration error of 1% (e.g. due to thermal expansion of the ruler)
true value = 1.01*(measured value)

5. May 11, 2009

### Shukie

Thanks. Now if I want to calculate the uncertainties again, do I substract 2 mm from the uncertainties in h and l as well or do those stay the same?

6. May 11, 2009

### davieddy

The "uncertainties" represent the limited "precision" of the measurements.
They are called "random" errors as suggested by the +/-. They can be reduced
(and estimated more reliably) by repeating the experiment several times,
but are inescapable and should always be quoted.
BTW when you are quoting the uncertainty, be careful not to introduce
a "rounding error" by not quoting enough significant figures:
L=101 +/- 0.2 is criminal!

Zero, calibration (and rounding!) errors affect the "accuracy" of the result,
and are called "systematic" errors. They have the same sign, and won't
cancel out with repeating the experiment. They should be eliminated or
compensated for.