# Errors in Measurement [of time]

1. Sep 16, 2014

### ChrisVer

I wanted to ask something concerning the errors...
Why sometimes the errors are taken to be $\sqrt{N}$ where N is your measured value (such as the number of counts from a detector) and why is it sometimes given as half of your device precision (eg a common ruler's error is 0.5mm)?
In my case I had an electronic timer which was able to measure the time of a wheeler going left or right, with a precision of 10ms .. since it was automatic I think it gets the measured time from electric signals [as it does for the counted events], so I am not sure whether its error is 5ms or the $\sqrt{t}$ of the measured time t.

2. Sep 16, 2014

### clem

It is not $\sqrt{t}$, since the accuracy of a timer is not dependent on the length of time.

3. Sep 16, 2014

### ChrisVer

even if the timer works in the same way as the counts?
In my device I had a wheel of "target" which moved left and right [trying to use Doppler effect], and the detector was measuring then my photons.
While the wheel was moving left, I had the counts measured for left and the time elapsed [for several left runs] until I stopped the wheel. The same for the right...
So for some speed of the wheel $u$ I had counted 5,000 events left and the wheel was moving for 100ms in total (let's say 7 runs) on some distance d each (so the total distance for left is $s=7 \cdot d$... In this case I want to measure the velocity (something like calibrating the motor with the velocity), and get also the error.
$u_L= \frac{s}{t_L}$
$\delta u_L = u \sqrt{ (\frac{\delta t_L}{t_L})^2 + (\frac{\delta s}{s})^2 }$

So you say the $\delta t_L$ should come from the systematic error of the device? and it wouldn't be $\sqrt{t_{L}}$?
Also shouldn't that depend on the runs ? because it stops (when right-to-left ← motion stops) and then reruns (when left-to-right → stops and right-to-left ← reruns).

http://www.chemicool.com/img1/graphics/mossbau2.gif
my setup, the only difference is that we didn't move the source, but the target.

Also from my script:
Also the electronic circuit is attached:

#### Attached Files:

• ###### Screenshot_1.png
File size:
25.1 KB
Views:
105
Last edited: Sep 16, 2014
4. Sep 17, 2014

### Meir Achuz

I was thinking of a simpler set up. Anyway, if the timer has of plus or minus 5ms error, then I think the total error for 7 runs should be 5 times the square root of 7. Maybe someone does Mossbauer experiments can give a better answer.

5. Sep 17, 2014

### Khashishi

In this case, the error of your clock is half the precision, since the error simply comes from the resolution of the device.

You use $\sqrt{N}$ when you are measuring something with a Poisson distribution, which arises frequently when you are counting independent random events. In this case, you are asking a totally different question. Your measurement is presumed to be exact, since it is an integer count, but you have an uncertainty in the population. That is, you could measure exactly 55 photons hitting a screen, but if you measure it again, you might expect to measure $55 \pm \sqrt{N}$ photons to hit the screen this time.

6. Sep 17, 2014

### Khashishi

You didn't say how you measure the velocity. Do you take the counts and divide by time? In this case you would have to take the uncertainty in the counts (sqrt(N)) and combine it with the uncertainty of the time (+-5ms). You combine errors in the usual way.

7. Sep 17, 2014

### ChrisVer

the velocity?
I am measuring the time and I know the distance my wheel is covering in that time...
For example if (to get 5,000 events for Left to Right) I needed some time Left to Right=1000ms and 10 runs (each run is 25 mm) then the velocity is given by:
$u = \frac{10 ~runs \times 25 mm}{1000 ms}$
Dividing the counts by time I'm getting the count rate (which I do) and yes, I can understand the propagation of error. I just could not understand why you needed the sqrt error or when you get the half of precision.