(1+T/273) m/s
Example 1
The example below is taken from a typical first semester lab in which the value of the gravitational constant, "g", is determined. A torpedo falls freely along a strip of waxed paper, leaving a small dot on the strip every 1/60th of a second, the time for which is determined by the standard frequency of AC power. The data set consists of the position and the time for several points. Here's a typical set of data in which the position is measured to the nearest cm and time is known to be multiples of 1/60th of a second:
| Position (cm) | Time (sec) |
| 2.4 | 0/60 = 0.000 |
| 3.1 | 1/60 = 0.0167 |
| 4.1 | 2/60 = 0.0333 |
| 5.1 | 3/60 = 0.0500 |
| 6.2 | 4/60 = 0.0667 |
| 7.6 | 5/60 = 0.0833 |
| ... etc | ... etc |
The average velocity between each interval is just the change in position divided by the change in time. The error represents the uncertainty in your calculated value that arises from the uncertaintly in your measurements. Let's assume that you are confident of your measurements of the dots to ±0.5mm (i.e. you recorded to the nearest mm). And you assumed that the power company kept a steady beat and the error in time was negligible. For the first interval, the calculation of the velocity and its error, you would write
v1 = (3.1±0.05 - 2.4±0.05) cm / (.0167 - 0) sec = (.7±0.1) / .0167 = 42 ±6 cm/s
Note that when we add or subtact numbers, the actual errors add. The ±6 comes from ±0.1 / 0.0167. That's right, a whopping ±6 cm/s ! You will have a similar error for all the velocities.
Important Note: We have expressed the time to 3 significant digits, one more than we have for the position. (Using your calculator, you could have simply multiplied the change in position by 60.) But when we report the velocity, we only show only two significant figures (your calculator may give many more), the number of digits of the least certain number in the calculation. Convince yourself that the next interval yields an average velocity of 60±6 cm/s.
The average acceleration is the time rate of change in the velocity. So, we divide the change in the average velocity by the time taken for this change in velocity. It is a very similar calculation to the previous one. For example, the average acceleration (and its error) during the time the velocity changed from 42 cm/s to 60 cm/s would be
a = (60±6 - 42 ±6 ) cm/s / 0.0167 s = 18 ±12 cm/s / 0.0167 s = 1080±720 cm/s2
The error is almost the size of your value! A typical set of acceleration values might be 1080, 720, 720, 1080, 1440 cm/s2 ...etc. The error analysis easily explains this. The inherent 0.1 cm variation in the measurement of the change of position, results in a large error in the acceleration. Nevertheless, the average of all the acceleration values might still yield a very reasonable result (i.e around 980 cm/s2), since the errors should "average out". Of course, this works very well when you have lots of data and perhaps not so well when you have only a few data points.
Error vs Difference: Finding the difference between your value of g and the value in a text book does not represent the error in your value. It is a difference. Sometimes it is appropriate to compare your calculation to those of others, but it is just a difference, not the error in your calculations.
Example 2
Now consider another typical lab exercise in which the acceleration of a human is measured. Cones are placed every 15 ft and individuals with stopwatches record the time it takes for the runner to reach each cone. The runner repeats the sprint several times, trying to be consistent. The data consists of the times at which the runner passes each cone. There will be a set of data for each trial. For three trials, a typical data set might be:
| Position (ft) | Time #1(sec) | Time #2(sec) | Time (sec) #3 |
| 0 | 0 | 0 | 0 |
| 15.0 | 0.61 | 0.59 | 0.66 |
| 30.0 | 1.10 | 0.95 | 1.12 |
| 45.0 | 1.41 | 1.56 | 1.46 |
| 60.0 | 1.71 | 1.69 | 1.76 |
| 75.0 | 2.03 | 1.99 | 2.10 |
Assume that the errors in the distances are small and may be ignored. (The values are recorded as 15.0 ft , etc, implying an accuracy to within 1/10 ft) Obviously, the real source of error is in the time values. This error arises both from the reaction time of the timers and from the variations between the trials. There are two approaches for handling the error.
Method A
First, you could simply look at the range of average velocity values you got for all the trials. The highest value for the first 15 ft was 15.0 / .59 = 25.4 ft/s and the lowest was 15.0 / .66 = 22.7 ft/s. (Note that we are keeping three significant places for the average velocity, one more than can actually be justified. We can drop the extra digit after the final calculation.) The average of all the runs, using the range for the error, would be
vave = (25.4 + 24.6 + 22.7) / 3 = 24.2 +1.2/-1.5 ft/s
where the +/- indicates exactly the range above and below the average. This method of reporting the error is appropriate when the data set is small or the data is not uniformly distributed about the average. If the average is near the center of the data, it is appropriate to simply report ± half the range. For this example, that would be ±(25.4-22.7)/2 and we would write:
24.2 ±1.4 ft/s
where the error has been rounded from ±1.35 to ±1.4 . A similar procedure would be followed for the error in acceleration. If you have a large number of values, then you may wish to use the Standard Deviation (SD) formulation. Most calculators will provide the mean and SD for any group of numbers, but you can review this formulation at the SCI 301Statistical Analysis page.
Method B
Let's treat each trial separately and make a reasonable assumption about the error in time, based upon the reflexes of the timers. A very typical value often cited in text books for human reaction time is 0.1 sec. So let's assume a liberally sized error of ±0.1 sec. (It would be better to actually checked the reaction times of the individual timers and use those values. But that would require another lab session!) So how do we use the 0.1 sec error in the velocity calculation? When dividing or multiplying numbers, we must add the percentage (or fractional) errors. We are dividing distance by time, so we must add the % errors of both distance and time. For example, 0.1/0.61 = 0.17 or 17% for our time, and we have assumed 0% for the distance, so
v1 = 15±0 ft / (.61±0.1s) = 15±0% / (.61 ±17%+0%) = 24.6 ±17% ft/s = 24.6 ±4.2 ft/s
where 4.2 is 17% of 24.6. (If we had an error of 0.2 ft in the distance, then the % error for the distance would be 0.2/15 = 1.3 % and the total error would be 18.3%, more appropriately rounded to 18%.)
Calculation of acceleration would introduce about another 17% error (since you would be dividing by time again) and the % errors would add.
Method C - the "high/low" method
There is an alternative to the % error method, which we will call the "high/low" method. Simply add the error to the value and calculate the result, then subtract the error and repeat the calculation. This will provide the range about the average or middle value. For example,
vhigh = (15-0) ft / (0.61-0.1s) = 15 / 0.51 = 29.4 ft/s
vlow = (15-0) ft / (0.61+0.1s) = 15 / 0.71 = 21.1 ft/s
for a range of 29.4 - 21.1 = 8.3 and using ±half the range, you express the error as 24.6±4.2 ft/s. Although this method might seem a bit crude, it is straightforward and easy to understand.
This method can actually be a faster approach for problems in which there are many multiplications and divisions (and you do not already have the % errors). But you must take care that you add or subtract errors such that they give maximum or minimum value for the entire calculation. For example, consider the following calculation:
A = (14.0 ±1.2)(12.1 ±2.4) / (4.5 ±0.5)(3.0 ±1.0)
For the high value of A you must add the errors in the numerator but subtract them in the denominator. That is:
Ahigh = (14.0 +1.2)(12.1 +2.4) / (4.5 -0.5)(3.0 -1.0)
And you would do just the reverse for the low value of A.
Example 3
In this example, we are going to compare a theoretical value to an experimental value, both of which have errors. In a experiment with sound, a 500 Hz tuning fork is held above a resonance tube. The length of the tube is varied by altering the position of the bottom. As the bottom position is altered, the experimentor listens for the loudness of the sound coming from the tube to peak, indicating a resonance. Two positions are recorded with an error estimated using techniques similar to those outlined in Example 2. The wavelength produced by the tuning fork is determined from twice the difference in these two positions. Let's assume the positions (with estimated error) are 27.5 ±0.4 cm and 62.0 ±0.4 cm. The calculated wavelength and error would be
l = 2 x (62.0 ±0.4 - 27.5 ±0.4) cm = 69.0 ±1.6 cm
where the errors are added and then multiplied by 2. The velocity of sound in the tube is found by multiplying the frequency of the tuning fork (500 Hz) by the wavelength. The uncertainty in the frequency embossed on the tuning fork can probably be ignored, but for demonstation purposes, let's assume it is ±5 Hz. When multiplying or dividing numbers, we must add the % errors. Our respective % errors are (±1.6/69.0 )x100 = ±2.3% and (±5/500)x100 = ±1%. The experimentally determined velocity is
v = (0.690 ±2.3%) x ( 500 ±1%) = 345 ±3.3% m/s = 345 ± 11 m/s
The theoretical value of the velocity of sound is determine from a text book formula that includes the temperature of the air. The temperature of the air must be measured with a thermometer. Hence, an error is introduced into the theoretical value. A reasonable assumption for the error in the temperature would be the precision of thermometer. Let's assume that precision is ±1oC. The theoretical formula is given by
(1+T/273) m/swhere T is measured in oC. Now assume the temperature in the resonance tube was measured to be 22 ±1oC. What is the error in the theoretical value? Since the quantity with error is part of a sum and embedded within a square root, the safest approach is the high/low method.
vhigh = 331 (1+23/273) = 345 m/s
vlow = 331 (1+21/273) = 343 m/s
resulting in an average value with error of 344 ±1 m/s. We are now in a position to compare the two values.
vexp = 345 ± 11 m/s
vtheo = 344 ± 1 m/s
The conclusion is clear. Within error, the two values are in agreement.
Example 4 - Functions
The error in a function calculation involving numbers with errors is more complicated. Consider the case in which you have measured an angle with a certain error (e). What is the error in the value of the cosine or sine of that angle? There are formulas for calculating function errors and those for a few common functions are listed below.
But you must use these formulas with caution! The formulas are useful only for modest errors. In addition, they are not reliable for all ranges of values. (For example, if your measured angle is very near zero, the formula for error in cosine of the angle will give you zero error!)
A safer method that gives a reasonable error estimate for functions is the "high/low" method illustrated as Method C of Example 2. For example, assume you have measured an angle with error given by a = 12.5 ± 1.5 degrees. What is the error in the value of cosa? The value is cos(12.5) = 0.976, and the range in values is:
the low value is: cos(12.5-1.5) = 0.982
and high value is: cos(12.5+1.5) = 0.970
So we can report a value of 0.976± 0.006. This is consistent with the formula given above, ± 0.026 x sin(12.5) = ± 0.0057. (Note that error in the angle was converted to radians.)