Area Measurement: Planimeters & Green's Theorem

Introduction

    Determining the areas of irregular objects is a common and important problem. How large is the lake? How large is the drainage basin? What is the surface area of the leaf? What is the area under the peak obtained from a gas chromatograph? How much sealer do I need for the deck? How much sod do I need for my yard?

    Perhaps the most basic examples come from surveying: How do you determine the size of a parcel of land? Many fields are bordered by creeks, rivers, and lakes, thus  irregular boundaries occur frequently. The best practical information on measuring area is not found in math books, but in surveying books.

    If you had a map of a lake and wanted to determine its surface area, what would you do? You could xerox the map onto a piece of graph paper and count squares. You could xerox the map and cut out the lake and weigh the paper! What else? If you had to do this type of thing very often, you would probably invest in an electronic or mechanical planimeter. What are these devices, and how do they work?    

Electronic planimetry

    Consider the problem of measuring the area of a lake, given a good map. One could be a little more sophisticated by dividing the lake into various geometric objects of known area, not just squares. For example, one might divide the lake into a combination of squares, rectangles, trapezoids, triangles, etc., and add up the area of all these. Generalizing a little bit more, the boundary can be approximated by a finite number of straight line segments. These form a polygon.

    The area of an any polygon can be found exactly by dividing it into trapezoids.

[Graphics:HTMLFiles/index_1.gif]

The area of polygon ABCDEF is the area between ABCD and the x axis minus the area between DEFA and the x axis. Note that the area between ABCD and the x axis is the sum of the three trapezoids ABB'A', BCC'B', and CDD'C'. Similarly, the area between DEFA and the x axis is the sum of three trapezoids. The area of a trapezoid is the average height times the base, thus the area of ABB'A' is (Overscript[AA ', _]+Overscript[BB ', _])/2 times Overscript[B ' A ', _]. If we use the subscript one for the x and y coordinates of point A, and the subscript 2 for the coordinates of point B, then the area of this trapezoid in terms of its coordinates is

(y_1 + y_2)/2 (x_1 - x_2)

which is positive. If we start and end at the same point (say A), and go around the polygon in a counter-clockwise direction, the signed area of the polygons underneath the closed curve will automatically be negative. For example, the polygon FAA'F' will have signed area determined by

(y_6 + y_1)/2 (x_6 - x_1)

in this example. Hence if we add up the trapezoids all the way around and make the ending point the same as our starting point we will get the area of the enclosed polygon. You might think the area is

Underoverscript[∑, i = 1, arg3] (y_i + y_ (i + 1))/2 (x_i - x_ (i + 1))

1/2 (x_1 - x_2) (y_1 + y_2) + 1/2 (x_2 - x_3) (y_2 + y_3) + 1/2 (x_3 - x_4) (y_3 + y_4) + 1/2 (x_4 - x_5) (y_4 + y_5) + 1/2 (x_5 - x_6) (y_5 + y_6) + 1/2 (x_6 - x_7) (y_6 + y_7)

but this isn't quite right because we need to set

{x_7, y_7} = {x_1, y_1} ;

then we get the correct result:

Underoverscript[∑, i = 1, arg3] (y_i + y_ (i + 1))/2 (x_i - x_ (i + 1))

1/2 (x_1 - x_2) (y_1 + y_2) + 1/2 (x_2 - x_3) (y_2 + y_3) + 1/2 (x_3 - x_4) (y_3 + y_4) + 1/2 (x_4 - x_5) (y_4 + y_5) + 1/2 (-x_1 + x_6) (y_1 + y_6) + 1/2 (x_5 - x_6) (y_5 + y_6)

This simplifies to

1/2Underoverscript[∑, i = 1, arg3] (x_i y_ (i + 1) - x_ (i + 1) y_i)

1/2 (-x_2 y_1 + x_6 y_1 + x_1 y_2 - x_3 y_2 + x_2 y_3 - x_4 y_3 + x_3 y_4 - x_5 y_4 + x_4 y_5 - x_6 y_5 - x_1 y_6 + x_5 y_6)

This algorithm is the one used by electronic planimeters. Edward Donley in Finding Areas with the Gauss-Green Formula gives an excellent example using Mathematica.

    In summary, the area of a polygon with n vertices can be found from

1/2Underoverscript[∑, i = 1, arg3] (x_i y_ (i + 1) - x_ (i + 1) y_i)

but we have to be careful to "close the path" by replacing x_ (n + 1) by x_1, and y_ (n + 1) by y_1. Also note that we need a simple closed curve which is a curve that doesn't intersect itself. Otherwise parts of the curve might not be traversed in a counter-clockwise direction which would give the wrong signed area. As an exercise these formulas should be compared to those for numerical integration using the Trapezoidal Rule. How and why do they differ?

    A planimeter is an instrument for determining the area of a plane figure. Pictures and descriptions of electronic planimeters are available from manufacturers such as Lasico. See Appendix A for a Mathematica program for the coordinate area of a polygon, and Appendix B for a description of how to use Mathematica itself as a digitizer (your own electronic planimeter!).

Green's Theorem

    Green's Theorem is a higher dimensional analogue of the Fundamental Theorem of Calculus. It relates the double integral over a closed region to a line integral over its boundary:    

∫∫ (∂Q/∂x - ∂P/∂y) A = ∮ F r

Applications include converting line integrals to double integrals or vice versa, and calculating areas. Green first published the theorem in 1828, but it did not become well known until 1846 when it was republished by Lord Kelvin.

    In the last section we derived the coordinate formula for the area of a polygon using only geometry:

1/2Underoverscript[∑, i = 1, arg3] (x_i y_ (i + 1) - x_ (i + 1) y_i)

Is it possible to derive this formula from Green's Theorem? The objective is to pick Q and P such that

∂Q/∂x - ∂P/∂y = 1

then the integral on the left is the area A which may be obtained from the line integral on the right. One possible choice for P and Q is

P[x_, y_] := -y/2 ;

Q[x_, y_] := x/2 ;

F[{x_, y_}] := {P[x, y], Q[x, y]} ;

First just consider the line integral from {x_i,y_i} to {x_ (i + 1),y_ (i + 1)}. This line can be represented parametrically by

r[t_] := {x_i, y_i} + t {x_ (i + 1) - x_i, y_ (i + 1) - y_i}

as t goes from 0 to 1. Then

F[r[t]]

{1/2 (-y_i - t (-y_i + y_ (1 + i))), 1/2 (x_i + t (-x_i + x_ (1 + i)))}

from which it follows that

∫_0^1F[r[t]] . r '[t] t

1/2 (-x_ (1 + i) y_i + x_i y_ (1 + i))

Our closed path is the sum of all the straight line segments making up the polygon or

1/2Underoverscript[∑, i = 1, arg3] (x_i y_ (i + 1) - x_ (i + 1) y_i)

and we have derived the coordinate formula for the area of a polygon using Green's Theorem.

Mechanical planimeters

     These ingenious mechanical devices, invented by the Swiss mathematician Jacob Amsler in 1854, are a source of fascination to many. Although there isn't a great deal of readily available information on mechanical planimeters in books or journals, there is on the web. For general information on mechanical planimeters see Larry's Planimeter Platter by Larry Leinweber and planimeters by Robert Foote.

    An excellent geometric explanation of these devices has been given by Tanya Leise. These devices can also be explained using Green's Theorem, although it is a bit more difficult than it was for the electronic planimeter (see Tanya Leise's planimeter site). A simpler explanation using Green's Theorem, but taking a slightly different approach is in Calculus: Multivariable by Brian Blank and Steven Krantz (Key College Publishing, 2005). Another reference containing both approaches (using Green's Theorem) is Calculus of One & Several Variable by Robert Seeley, (Foresman & Co., 1973).

Conclusion

    One of the fascinating things about mathematics is seeing and finding relationships between things that at first glance may appear unrelated. We have considered electronic planimeters in detail where we found they could be easily understood from the geometric formula for the area of a polygon. The interesting part was that they could also be explained using Green's Theorem from multivariable calculus. A similar situation exists for mechanical planimeters – they can be explained geometrically or using Green's theorem. Please pursue the links and references to see for yourself! (Work through the Connected Curriculum Project on Green's Theorem and the Planimeter.)

    Note that Amsler invented the polar planimeter in 1854 and Green's Theorem became widely known in 1846. Did Amsler know about Green's Theorem? If so, what role did it play in his invention of the planimeter?    

Appendix A: Mathematica program for coordinate area of a polygon

     If you are interested in programming, the following (clever) Mathematica program by Stan Wagon (Mathematica in Action, Second Edition,1999) is another implementation:

area[poly_] := Plus @@ Flatten[poly ({1, -1} Reverse[#] &) /@RotateLeft[poly]]/2

This program is much more difficult to understand than Donley's. It is a good programming exercise to figure out how this "one-liner" works. The only variable here is poly which should be the list of the coordinates of the vertices of the polygon. In the above example:

ABCDEF = Table[{x_i, y_i}, {i, 1, 6}]

{{x_1, y_1}, {x_2, y_2}, {x_3, y_3}, {x_4, y_4}, {x_5, y_5}, {x_6, y_6}}

area[ABCDEF]

1/2 (-x_2 y_1 + x_6 y_1 + x_1 y_2 - x_3 y_2 + x_2 y_3 - x_4 y_3 + x_3 y_4 - x_5 y_4 + x_4 y_5 - x_6 y_5 - x_1 y_6 + x_5 y_6)

Appendix B: Mathematica as a Digitizer

    It is possible to use Mathematica as a digitizing tablet to digitize a small drawing or map. First make a plot similar to the following:

xrange = {a, b} = {0, 2} ;

yrange = {c, d} = {0, 3} ;

Plot[d , {x, a, b}, PlotRange-> {xrange, yrange}, AspectRatio->Automatic, GridLines->Automatic] ;

Use values for a,b,c, and d appropriate for your drawing. (For example, if you want to digitize the boundary of a lake located on a 30 mile by 30 mile section of the map, and the scale of the map is 10 miles per inch, you would set a=0, b=30, c=0, d=30.) Select the Mathematica graphic and resize it so the scale of the graphic matches the scale of your drawing. Do this by holding the drawing on top of your computer screen so that you can see through it to resize the graphic. If your paper is too opaque to see through use a copy machine to make a transparency, or try increasing the brightness setting on your monitor.

    Next, use Post-it Notes or some drafting tape (to minimize the risk of damaging the coating on your computer screen) to attach your drawing to the screen with the origin of the drawing matching the origin of the Mathematica graphic.

    Select the graphic; while holding down the Command key (Macintosh), click on the points of your graphic you wish to digitize. Copy (use the Edit menu) the points to the clipboard, and then they can be pasted into Mathematica as a list.

    That's all there is to it! Now you can use Mathematica to do whatever you want with your list of data. Look at  it with ListPlot (use the option PlotJoined->True), calculate area, center of mass, moment of inertia, transform it, etc.


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