In every measurement and estimate we make, there are always errors. Why is error analysis important in physics? Should we be concerned of the result instead of calculating for errors? It is true that we must be more concerned about the measurement's result than its uncertainty but not calculating for error uncertainty is sometimes a great undoing.
Here is a story that gives an example of this mistake of not calculating the tolerance or error:
A thrifty driver observed that his van could run 6 kilometers for every liter of gasoline. Now he will be going onto an interstate journey with a distance of 500 kilometers. Considering his fuel tank has 50 liters capacity, he made a full tank refueling on a nearby gas station. For exact computation 50 liters x 6 km per liter = 300 kilometers. Therefore, he thought that he could run 300 km with his current fuel. Looking at his km reading, he set forth on his journey. After a 250 km travel he refueled at another station and calculated that his fuel could last for another 50 km so he needed fuel to run the other 200km. Thus, 200km/6km per liter = 33.3 liters. Taking a little allowance, he loaded 35 liters. With all the hope of the exact computation, he set off for the next half of his journey. In a great dismay, his van/car stopped in the middle of the street. Why is it? It is because of the error component on the observation or measurement. The 6 km per liter measurement can be 6 km ± 1km depending on traffic and road conditions. It also depends on what gears you are using. Higher gears (used on high speed) use lesser fuel per distance traveled.
Let us compute for the driver's mistake. Using the data 6 km ± 1km per liter, the minimum distance his car can run per liter of gas is 5 km. Therefore, his fuel 50 liters + 35 liters = 85 liters of gasoline has the possibility to run only 85 liters x 5 km/liter = 425 km! It is a long way to go before he can reach his destination!
This is only a simple example and with lesser complication. Imagine what will happen if the same miscalculation is applied for a space shuttle going to mars!
There are two kinds of errors: random and systematic errors.
Random errors are caused by the inefficiency in the method of measurements and other interferences.
Systematic errors are harder to deal with, since the instrument used causes it. It is especially hard if you are using multiple instruments and have no idea which instrument malfunctioned.
In representing error measurement in uncertainty analysis, we use two forms: the absolute form and the relative form.
The absolute error (error represented in absolute form) tells the reader the exact number of units the measurement is uncertain about or the measurement's precision. For example we may say that the weight of a two-year old cow is 200kg ± 10kg. With that statement we see clearly that a two-year old cow is about 190kg to 210kg.
On the other hand, the relative error tells us the how many part of the measured value is uncertain. It is an error relative to measured value. We can compute for the relative error using the formula RELATIVE ERROR = ABSOLUTE ERROR / MEASURED VALUE.
Using our previous example, the relative error is 10kg/200kg, which is equal to .05 or 5%. The relative error is now represented as 200kg ± 5%.
Although the absolute error is sometimes easier to understand, relative error is easier to use in more complicated computation. It is also the only way to compare errors of different units. For example, which is more accurate measuring a cow's weight, which is 200kg ± 10kg or its height, which is 150 cm ±5 cm? We cannot compare 10kg to 10cm since they have the same value. So to compute for accuracy we have to convert this absolute form to relative form. 200kg ± 10kg becomes 200kg ± 5% and 150cm ± 5cm becomes 150 cm ± 3.3%. It is now clear that the weight value has greater error so the height's measurement is more accurate.
CALCULATING FOR ERROR:
Addition and subtraction
In adding or subtracting absolute errors, we add the error's absolute values. For example:
(800m ± 10m) + (500m ± 5m) = 1300m ± 15m
The same is true when subtracting:
(800m ± 10m) - (500m ± 5m) = 300m ± 15m
We cannot add or subtract relative errors with different values. For example, we cannot add 800m ± 1.25% and 500m ± 1%. We can only add if they have same relative errors. Say:
(800m ± 2%) + (500m ± 2%) = 1300 ± 2%. Yes, we just add the measured value and maintain the relative error.
Multiplication and division
Multiplying absolute errors is quite messy. Multiplying it with a constant is just for warm-up. For example you measured a certain distance with a meter stick with length 1m ± 2cm or 1m ± .02m. You finished with the measurement of 50 meters. Now, what is the actual accuracy/error representation?
Accuracy = 50x (1m±.02m) = 50m±1m. We distribute 50 to both the measured value and the absolute error.
Unlike in relative error (given that counting up to 50 to measure 50m with a meter stick is 0% error), we multiply the constants and add the relative error:
Accuracy = (50 ± 0%) x (1m ± 2%) = 50m ± 2%.
Let's have this next example: how much distance can a car with speed 85km/hr ± 10km/hr travel in 6 hrs ± 24min?
First we should convert 24 min to hour, which is 0.4 hour. The formula for Distance is
Distance = Speed x Time, so:
Distance = (85km/hr ± 10km/hr) x (6hr ± 0.4 hr), computing for average distance:
Average Distance = (85 km/hr) x 6 hr = 510 km. Then farthest possible distance:
Farthest Distance = fastest speed x longest travel time, difference is +98km
Farthest Distance = 95km/hr x 6.4 hr = 608 km. The least possible distance traveled is:
Shortest Distance = slowest speed x shortest time
Shortest Distance = 75 km/hr x 5.6 hr = 420km, difference from average is -90km so we can say
Distance = 510km ± 98km or take the mean to be exact:
Mean = (farthest + shortest)/2. Then we got better result of 514km±94 km.
Quite complicated computation using absolute value as you can see. But if we use relative value:
Distance = (85km/hr ± 11.76%) x (6 hr ± 6.67%), we multiply the actual measurement and add the relative error. We get:
Distance = 510km ± 18.43%
To check: 510kmx1.1843 = 603.99km and 510kmx.0.8157=416km
In this part we totally use the relative error. The rule is simple multiply the relative error to the exponent. For example compute the volume of a cube with side 4m±5%.
The formula for Volume of cube is V= s3
V = (4m±5%)^3
V = 64m3 ± 15%
This ends the simple error analysis guide for physics. Everyone can use these formulas to compute for error. Computation of error is essential not only for giving enough tolerance but also minimizing the resources up to the highest possible result only. Furthermore, error analysis isn't isolated for physics; it is also applied in chemistry, mathematics, mechanics, astronomy, etc.
For complete and detailed discussion of error analysis look for the book "An Introduction to Error Analysis" by John R. Taylor.