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Guide 1-3d. Graphical Analysis for a Non-Linear Relationship

Cross-sectional area vs. diameter of cylinders

In a previous guide, you saw how a linear relationship between the circumference and diameter of cylinders was determined.

Let's take this a step further and experimentally determine the relationship between the cross-sectional area and the diameter of the 5 cylinders. Our method of measuring the diameter is to place a base of the cylinder onto a sheet of graph paper that is ruled in millimeters and then to trace around the circular base. We then remove the cylinder and count the number of enclosed squares, each with an area of 1.0 mm². We also count the partial squares as best we can and add them in to the total. The process is illustrated to the right. For this diagram, we estimate the area of the circle to be 9.0 mm².

Perhaps you're asking yourself why we don't find the area simply by squaring the diameter and multiplying by π/4. If we did that, we would be assuming what we're trying to show by experiment. We measure the area without reference to the formula for the area of a circle. After that, we'll plot a graph of area vs. diameter and determine the relationship experimentally.

Example data and a graph are shown below. Note that the area is given in units of m² in order to be consistent with the units of diameter.

Cylinder ID Diameter
(m)
Area of cross section
(m²)
1 0.0120 0.00013
2 0.0163 0.00021
3 0.0195 0.00030
4 0.0253 0.00049
5 0.0390 0.00118

We can see that the relationship between area and diameter is non-linear. So we'll need to re-express a variable in order to create a linear relationship. This process is called linearizing the data. Here's where we can use what we know about the theoretical relationship between area and diameter of circles. Since we expect that the area should be proportional to the square of the diameter, we should square the values of diameter and plot area vs. diameter². Of course, this also corresponds with the fact that the curve above has the appearance of a parabola.

The table of data below includes the values of diameter². Below that is the graph of area vs. diameter².  A best fit line has been drawn and the slope determined.

Cylinder ID

Diameter
(m)

Diameter²
(m²)

Area of cross section
(m²)

1 0.0120 0.000144 0.00013
2 0.0163 0.000266 0.00021
3 0.0195 0.000380 0.00030
4 0.0253 0.000640 0.00049
5 0.0390 0.001521 0.00118

slope = (A2 - A1)/(D22 - D12)

         = (0.00111 m2 - 0.00007 m2)/(0.00144 m2 - 0.00009 m2)

         = (0.00104 m2)/(0.00135 m2)

         = 0.770 (units divide out)

Now we prepare a matching table.

Math maps to Physics Value
(graph)
Value
(expected)
Units
y --> A    
m --> pi/4 0.770 0.785 none
x --> D²    
b --> none 0.00 0

The equation of the line is :  A = (0.770)D² + 0.00

Now we compare the experimental result to the theory. We expect from theory that A = π/4)D². Thus, we expect that the slope will be π/4 = 0.785. The experimental value is within 2% of the expected value.

You may find it strange that we went to all this trouble to determine relationships that we already knew. Of course, we were just illustrating the process by which a relationship between two physical variables is determined experimentally and then compared to theory. This is called a verification experiment. The method of determining relationships experimentally can be used to support or refute an existing theory or to provide a predictive relationship when a theory is not known. When the experimental relationship is used in conjunction with an accepted theory, characteristics of a physical system can be determined.  



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