Measurement Lab

Most of you have probably had to make various kinds of measurements; perhaps measuring ingredients while baking or determining how much wrapping paper is necessary to cover a gift. You might have also confronted the possibility of arriving at a measurement that is not entirely representative of what you set out to measure. This can happen due a faulty measuring device – say a poorly calibrated or damaged ruler, difficulty performing the measurement, or measuring an inherently ambiguous quantity. For example, what is your exact height?

A moment’s thought should convince you that it is impossible to determine the measurement of a physical quantity (e.g. your height) exactly. The term “accuracy pertains to the extent one can estimate a physical quantity’s actual value.

Prior to making any measurements is often possible to estimate a lower limit on one’s measuring accuracy. Using the odometer of your car, it would be impossible to determine the length of a city block to 0.01mile! In fact, using your car’s odometer you might realistically expect to make claims of a block’s length only to 0.05mi.

1. For the following quantities make a rough guess of the accuracy that might be associated with each measurement.  For example, you might think that if you use your stride as a unit of measurement to guess the room's size, you might have an accuracy of 1 meter, but if you used a tape measure, you may think the accuracy is better. It's important to state what type of device you would use to make a measurement of a given accuracy.   

  Accuracy of measurement Measuring Instrument to be Used
a) your Height    
b) the girth of your wrist    
c) the dimensions of this room    
d) your weight    
e) the distance from SMC to SM Pier    

2. Suppose in part (a) above you had access to a measuring device (perhaps a laser) that could determine length measurements to 1/1000 inch. Would this change your answer to (a)? Would your accuracy in such case be 1/1000inch? Explain.

 

 

 

At the front of the room obtain a meter stick and roughly 30 cm (1ft) of string. You will notice that the meter stick has four sides – each with a different scale. In the following take care to use the scale called for in the instructions –do not look at or use a more precise side of the stick for that will defeat the purpose of this exercise!

3. Choose one member of your group that will be measured.
a) Using the side of the stick with 10cm increments measure, to the best of your ability, the height of the person. You will have to estimate, but that’s alright. Also record the number of digits in your answer.

  
             Height = 
 
            # of digits = 

b) Using the side of the stick with 1cm increments measure to the best of your ability, the girth of the person’s wrist. Also record the number of digits in your answer.

  
             Wrist = 
 
            # of digits = 

c) What would you estimate to be your accuracy in parts (a) and (b)? Is this in any way correlated with the number of digits on your answers?

  


             

          

4. Now, using the most precise (1mm) scale of the stick have each of the other people in your group take height and wrist measurements of the designated person in your group. It is important that you do not share your measurement results with one another until all of them have been made. Make sure you have at least three independent measurements for the person’s height and wrist. 

  Average Height   Average Wrist
 Height 1 =     Wrist 1 =   
 Height 2 =   Wrist 2 =
 Height 3 =   Wrist 3 = 

b) Compare each of your three measurements with the average and determine roughly how much they differ from the average. The difference in the measurements comes from the limitations of making an accurate measurement and is referred to as the uncertainty of the measurement. List what factors you believe limited the accuracy in measuring the person’s wrist and height. 

 
 Factors in Height Uncertainty

 

 

 

 

 
 Factors in Wrist Uncertainty

 

 

              
 
  

5. In the case of the height measurements made in Q4, would you regard the average of all the measurements as more representative of the person’s actual height than any of the individual measurements? Explain your reasoning.

 

 

 

 

Generally speaking it makes no sense to quote a measurement with more digits than is consistent with its uncertainty. For example, if one’s uncertainty on a height measurement is 1cm it would make no sense to quote the person’s average height as 185.23cm. Since there is an uncertainty of 1cm, there is no way that you can reasonably state a measurement to 1/100 of a cm! In the case of a 1cm uncertainty the number should be rounded to the nearest cm giving 185cm.

6. a) What are your person’s average height and wrist measurements appropriately rounded to reflect your uncertainties?

 Height =   
 Wrist =           
  

b) Consider a mountaineering company attempting to market a particular type of very light yet strong type of rope.  Why might it be to their advantage to state an accurate limit for the strength of their rope? Why might it be to their disadvantage to state a figure that reflects an accuracy beyond that which they are capable? 

 

 

 

 

There is an aspect of measurement that is not entirely captured by just stating an uncertainty. For example, suppose someone tells you they made a measurement and they have an uncertainty of 1cm, depending on whether they are measuring their finger or the dimensions of a building they could either be very sloppy or extremely careful.

The notion of a percentage offers a useful measure of the above considerations. The percentage error of a measurement is defined by:

Where “average value” refers to the average of all the independent measurements and “error” refers to the amount that the measurements differ from the average. (We'll use "error = 1/2(max. value - min. value).)

7. Determine the percentage error for both the wrist and height measurement for your group.

  
  % error (height) = 
 
  
  % error (wrist) = 
 

8. Given a number representing the length of some quantity is there any relation in the number of digits to the accuracy of the measurement? For example, is there any difference in the percentage errors associated with the numbers 75in and 75.00in? How about .0075in and 75in? Explain your reasoning.

 

 

 

 

If your measurements of some quantity are in error and this measurement is used in a calculation, then the results of that calculation must also then have an associated error. This effect is called propagation of error and is illustrated by the following question.

9. Suppose you measure a square piece of carpet to have an edge of length 10.5in. Assume that your measurement has a 5% error. Then, for all you know the edge could be as long as 11in or as short as 10in. What do you predict the percentage error in the area will be?

 

 

Recalling that the edge could be as long as 11in or as short as 10in, what are the maximum and minimum values for the carpet’s area.  

  
  maximum area  = 
 
  
  minimum area = 
 

What is the percentage error in the carpet’s area?

 

  percentage error in area  =     --------------------------------------   = 

 

  How is this related to the percentage error of the edge length?

  

 

10. Now measure the mass and volume of an unknown material to calculate its density. For the cylinder, measure its height and diameter using Vernier calipers.

  
  m ±
Dm = 
 
  
  h ±
Dh =
 
 D ± DD = 

So now the density is r = m/Volume = m/(hpD2/4). But what is the uncertainty? The mass, height and diameter all have uncertainties in them. In fact, the uncertainty is given by:

The uncertainty in any quantity calculated by multiplying or dividing quantities that have their own errors is always found this way: the sum of the fractional errors of the measured quantities (each added in as many times as it appears in the calculation) gives the fractional error in the calculated quantity. 

So for the density, the uncertainty is given as follows: (Note that the fractional error in the diameter is multiplied by 2 since it came into the original formula twice since it was squared.

So:

 
            r
± Dr