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For the first iteration, a value of 0 = 20944 rad/s is chosen as an initial guess of the speed at the start of the cycle For j = 1, j 1 = 0 = 20944 rad/s is placed in column 7 As column 7 is filled, column 8 will be formed as the summation 0 + 2 1 + 2 2 + + 2 j 1 required in Eq (726) Bj is then evaluated from Eq (726) using the values of a, Cj , 0, and 2 /J = 2( /5)/0085 = 14784 rad/(lb s2 ft) and placed in column 9 With A j = 0697( /5)(05)/0085 = 2576 rad/s, Eq (724) can be evaluated to give j For j = 1, 1 = 2087 rad/s is thus placed in column 10.

One message can give your location, and another message could be used to describe your station operating conditions The characters of the Morse code are represented by sequences of long and short beeps known as dots and dashes (or dits and dahs) There are one to six beeps or elements in the characters of the standard Morse code The first step in writing a program to send Morse code is to devise a compact way to represent sequences of elements, and an efficient way to play them back The chart shown in Table 12-1 illustrates the encoding scheme used in this program A single byte represents a Morse character The highest 5 bits of the byte represent the actual dots (0s) and dashes (1s), while the lower 3 bits represent the number of elements in the character.

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This value is then brought over to column 7 for j = 2 and the process is repeated until column 10 is filled The last value in column 10, 2159 rad/s, is the speed at the end of the cycle using the initial guess for 0 Then 2159 rad/s is used as the 0 value for the second iteration, producing columns 11 through 14 This process is repeated until the final speed for the cycle equals the initial speed that was assumed In this example this takes three iterations, giving the speed variation given in column 18 The maximum and minimum speeds, from column 18, are 2164 rad/s and 2032 rad/s From Eq (71), Cs is then this range divided by the average: Cs = 2164 2032 max min = = 0063 05( max + min) 05(2164 + 20.

32) (730).

7.3.1 Rim-Type: No Bending If interaction with the spokes is ignored, the stress in a thin-rim flywheel is the uniform hoop stress: = r 2 2 V 2 a = g g (7.31)

An object is created. The object is used. The object is deleted. Something that didn t know the object was deleted calls a method on that object.

Since the stress is a function of velocity, the strength of flywheel materials can be given in terms of maximum rim velocity (see Sec. 7.5.1).

For example, the letter F is a dot, dash, dot, dot, so it is encoded as 0010X100, where X is a don t care bit Since Morse characters can contain up to six elements, we have to handle the exceptions Fortunately, there is some excess capacity in the number-of-elements portion of the byte, which can represent numbers up to 7 Therefore, we can assign a six-element character ending in a dot the number 6, while a six-element character ending in a dash gets the number 7 The program shows how these bytes can be played back to produce Morse code The table of symbols at the beginning of the program listing contains the timing data for the dots and dashes themselves If you need or want to change the program sending speed, you will just enter new values for the dit_length, dah_length, etc.

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

7.14 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

Sometimes your app will crash immediately when this happens. However, sometimes the object will have been deleted, but the memory on the heap hasn t been allocated to something else yet. So, the result is that the variable still points to what looks like an object in memory, and the function call works. In doing so, if that method starts changing member variables, it can now be changing memory on the heap that it doesn t own anymore, resulting in a stomp of other newly allocated objects. For an example of one of these situations, see Figure 4-6.

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