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Referring to Fig 722, C1 is the cross section of the camshaft; C2 is a circular hole with its center located on the x-axis, so that the center of mass of the cam is shifted only in the x-direction Moreover, a and c are the radii of C1 and C2, respectively, and x2 is the distance between the center of C2 and the origin O, ie, the center of the camshaft cross section Now we recall that our reason for removing material from the cam plate is to shift the centroid of the cam plate to the axis of rotation, namely, the axis perpendicular to the x and y axes through point O Hence, we compute the centroid of the new cam plate The centroid of a planar object composed of n given shapes of known centroids can be obtained using xc =

You Try It: Sketch the locus y 2 = x 3 + x + 1 on a set of axes The reader unfamiliar with cartesian geometry and the theory of loci would do well to consult [SCH2]

Part II:

return chrs[idx]; } } // Advance to the next object public bool MoveNext() { if(idx == chrsLength-1) { Reset(); // reset enumerator at the end return false; } idx++; return true; } // Reset the enumerator to the start public void Reset() { idx = -1; } } class EnumeratorImplDemo { static void Main() { MyClass mc = new MyClass(); // Display the contents of mc foreach(char ch in mc) ConsoleWrite(ch + " "); ConsoleWriteLine(); // Display the contents of mc, again foreach(char ch in mc) ConsoleWrite(ch + " "); ConsoleWriteLine(); } }

(787)

Here is the output:

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Here we give a whirlwind review of basic ideas of trigonometry The reader who needs a more extensive review should consult [SCH1] When we rst learn trigonometry, we do so by studying right triangles and measuring angles in degrees Look at Fig 125 In calculus, however, it is convenient to study trigonometry in a more general setting, and to measure angles in radians

In the program, first examine MyClass It encapsulates a small char array that contains the characters A through D An index into this array is stored in idx, which is initialized to 1 MyClass then implements both IEnumerator and IEnumerable GetEnumerator( ) returns a reference to the enumerator, which in this case is the current object The Current property returns the next character in the array, which is the object at idx The MoveNext( ) method advances idx to the next location It returns false if the end of the collection has been reached and true otherwise Reset( ) sets idx to 1 Recall that an enumerator is undefined until after the first call to MoveNext( ) Thus, in a foreach loop, MoveNext( ) is automatically called before Current This is why idx must initially be 1; it is advanced to zero when the foreach loop begins A generic implementation would work in a similar fashion Inside Main( ), an object of type MyClass called mc is created and the contents of the object are twice displayed by use of a foreach loop

where xc = center of mass of the overall object xi = center of mass of the ith part Ai = area of the ith part for i = 1, 2, , n From Fig 722, the centroid location of the cam with removed material is found to be xc = with the notation Acam xcam - A2 x2 Ac - A2 (788)

24:

As the preceding example shows, it is not difficult to implement IEnumerator and IEnumerable However, it can be made even easier through the use of an iterator An iterator is a method, operator, or accessor that returns the members of a set of objects, one member at a time, from start to finish For example, assuming some array that has five elements, then an iterator for that array will return those five elements, one at a time By implementing an iterator, you make it possible for an object of a class to be used in a foreach loop Let s begin with an example of a simple iterator The following program is a modified version of the preceding program that uses an iterator rather than explicitly implementing IEnumerator and IEnumerable

Acam = total area enclosed by the cam pro le A2 = area of circle C2; A2 = pc2 xcam = centroid abscissa of the cam plate x2 = abscissa of the center of C2 To have the centroid coincide with the axis of rotation, we set xc = 0 in Eq (788) and solve for x2, thereby obtaining x2 = Acam xcam pc 2 (789)

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