# Error propagation in calculations

1. Jul 24, 2011

### jakerue

This is an issue I am running into at the beginning of my physics course.

1. The problem statement, all variables and given/known data

Given distance and time in minutes, calculate the time in hours (part of a larger average velocity question) and graph over 170minutes. Include error bars in the graph

2. Relevant equations

So if tm = 10.0 +/- 0.1min and I use min -> hr conversion as 1hr/60min = 0.0167

th = 10.0min * 0.0167 hrs/min = 0.167 min

3. The attempt at a solution

The error in min is +/- 0.1 min. Now if I am using 0.0167hrs/min as an exact constant factor I should use z=k*x and $\Delta$z= k $\Delta$x

So I will use hours = 0.0167 * minutes and my $\Delta$hours = 0.0167 * 0.1min making the $\Delta$hours = 0.00167hours.

By sig fig this value is 0.002 hours correct?

At 10 minutes th = 0.167 +/- 0.002hrs

at 120 minutes th = 2.00 hrs +/- 0.002 hrs

I think I am right but I want to make sure my answer is correct and that this error holds true for all values of minutes from 0-170min.

Thanks for any help, most appreciated.

JakeRue

2. Jul 24, 2011

### Delphi51

Looks good, JakeRue. Certainly 10±0.1 minutes = 0.167±0.00167 hr
and it makes sense to round the .00167 to .002.
Very likely your 2 hours is ±.002 as well, though you haven't given any information from which that can be determined. The ±0.1 minute was given in the question for the 10 minute time. Was it also given in the question for the 2 hours? In reality it would depend on how the time was measured. Always a good idea to post the question exactly as given so we can check your interpretation of it.

3. Jul 25, 2011

### jakerue

The only errors given in the question are with the minutes (+0.1 min) and with the distance (+0.1 km). The question asks to convert minutes to hours and then graph distance over time.

I then use the km/h to calculate average velocity and best fit line.

I am asking about the 2 hours because the minutes is given in the form of 10.0min and then 120.0 min (3 sig fig, then 4 sig fig). If I calculate the error using only the conversion coefficient of 0.00167 * $\Delta$tm then the error would stay the same no matter if the sig fig is 3 digits or 4 digits - correct?

What is tripping me up with the error propagation is that with minutes having a + of 6 seconds, or 10% error, the hour then has +0.00167 which is an error of less than 10% of the hour (I could be wrong here).

I don't understand how can I get less error when I multiply by a unit conversion coefficient (1/60)?

4. Jul 25, 2011

### Delphi51

The error is fixed at 0.1 minute for both 10 minutes and 120 minutes, so certainly the % error will be 12 times less for the 120 minutes. But the size of the error bars will be the same - 0.1 minute for all times.

Changing the units did not change the % error. You had .1 minute per 10 minutes, which is 1%. It converted to .00167 hr per .167 hr, which is also 1%.

How will you calculate the average velocity? You could draw a line of best fit and its slope would be the average velocity. Or just use the last point (total distance and total time).

5. Jul 25, 2011

### jakerue

OK I can see the error being at 1% for both minutes and hours. So converting 120.0min +0.1min using the 0.0167 hrs/min conversion coefficient will give me 2.000 hours +0.002hrs.

I was doing velocity by rise/run from two points (does it matter which two I choose since it won't be the actual value?) and getting y-intercept from that.

When I get my slope I think I must multiply the error of both the time in hours (0.1km) and distance in km (0.002) to get a final average velocity.

(209.0 - 17.5km)/(2.00 - 0.167hrs) = 191.5km/1.833 hrs = 104.5km/hr and the error is

∆average velocity =[velocity] [($\Delta$time/time + $\Delta$distance/distance)]
∆average velocity = 104.5 [(0.0167/1.833hrs) + (0.1/191.5km)] = 104.5(0.00911+0.000522) = 104.5*0.009632

∆average velocity = 1.006544 then sig fig brings it to 1km/hr

So final average velocity = 104.5km/hr +1 km/hr