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# Troublshooting your pc for dummies - Bourg D.M.

Bourg D.M. Troublshooting your pc for dummies - Wiley publishing , 2002. - 350 p.
ISBN 0-596-00006-5 Previous << 1 .. 4 5 6 7 8 9 < 10 > 11 12 13 14 15 16 .. 128 >> Next Let’s look at a simple 2D example demonstrating how to apply the formulas discussed in this section. Suppose you’re working on a top-down view auto racing game in which you want to simulate the automobile sprite based on 2D rigid bo<-ly dynamics. At the start of the game the player’s car is at the starting line, full of fuel arid ready to go. Before starting the simulation, you need to calculate the mass properties of the car, driver, and fuel load at this initial state. In this case, the body is made up of three components: the car, driver, and full load of fuel. Later on during the game, however, the mass of this body will change as fuel burns off and the driver gets thrown after a crash. For now, let’s focus on the initial condition as illustrated in Figure 1-8.
Car
Driver /
NOT TO SCALE
Figure 1-8. Example Body Consisting of Car, Driver, and Fuel
The properties of each component in this example are given in Table 1-2. Note that length is measured along the x-axis, width is measured along the y-axis, and height would be coming out of the paper. Also note that the coordinates, in the form (x, y), to the centroid of each component are referenced to the global origin.
Iable 1-2. Example Properties
Car
Length = 15.5 ft Width = 6.0 ft Height = 4.1 ft Weight = 3913.0 lb Centroid = (100,100) ft
Driver (seated)
Length = 3,0 ft Width = 1.5 ft Height = 3.5 ft Weight = 190.01b Centroid = {103,105) ft
Fuel -
length = 1.5 ft
Width = 3.0 ft
Height = 1.0 ft
Density of Fuel = 1.45 slug/ft1
Centroid = (93,100) ft
The first mass property we want to calculate is the mass of the body. This is a simple calculation, since we are already given the weight of the car and the driver. The only other component of weight we need is that of the fuel. Since we are given the mass density of the fuel and the geometry of the tank, we can calculate the volume of the tank and multiply by the density and the acceleration due to gravity to get the weight
12 [ Chapter 1: Basic Concepts
of the fuel in the tank. This yields 210 lb of fuel as shown here:
Wfuel = pvg = (1.45 slug/ft3) (1.5 ft) (3 ft) (1 ft) (32.174 ft/ s2) = 2101b -
Now, the total weight of the body is
l^tocal = l^car + ^driver + ^fuel
Wmtll = 3913 lb + 190 lb + 210 lb = 4317 lb
To get the mass of the body, you simply divide the weight by the acceleration due to gravity:
Mtota, = W»«i/g = 4317 lb/(32.174 ft/s2) = 134.2 slugs
A slug is a strange-sounding unit that you might not feel comfortable using, so converting to SI units for mass, we get 1958.2 kg, nearly 2 metric tons.
The next mass property we want is the location of the center of gravity of the body In this example we will calculate the centroid relative to the global origin and will apply the first moment formula twice, once for the x-coordinate and again for the y-coordinate, as shown here:
?^cgbody — {(Xcg car X^car) “1“ (^cgdriver)(^driver) H" C^cgfuel) (^uel)}/ ^tocal
Xcgbody = {(100 ft) (3913 lb) 4- (103 ft) (1901b) + (10 0 ft) (210 lb)}/4317 tb
Xcgbody = 99.7 ft
X:gbody — {(.yeg car )(l^car) + (itg driver) (Waiver) 4“ (ycgfuelX^tuel)}/ WrotaS
Ycgbody = ((100 ft) (3913 lb) + (105 ft) (190 lb) + (100 ft) (210 lb)}/4317 lb
Ycgbody = 100.1 ft
Notice that we used weight in these equations instead of mass. Remember that we can do this because the acceleration due to gravity built into the weight value is constant and appears in both the numerator and denominator, thus canceling out.
Now it’s time to calculate the mass moment of inertia of the body. This is easy enough in this 2D example, since we have only one rotational axis, coming out of the paper, and therefore need to perform the calculation only once. The first step is to calculate the local moment of inertia of each component about its own neutral axis. Given the limited information we have on the geometry and mass distribution of each component, we will make a simplifying approximation by assuming that each component can be represented by a rectangular cylinder and will therefore use the corresponding formula for moment of inertia. In the equations that follow, I’ll use a lowercase
Mass, Center of Mass, and Moment of Inertia | 13
iv co represent width so as to not confuse it with weight, for which I’ve been using a capital W.
r
Iocar = (m/U)(w2 + L2)
Iocar = ((3913 lb/32.174 ft/s2)/12)((6.0 ft)2 + (15.5 ft)2} = 2800lb-s2-ft
/„dri™r = (m/12)(w2 + t1) lodriver = ((150 lb/32.174ft/s2)/12)((1.5 ft)2 + (3.0 ft)2) = 5.5lb-s2-ft
lofuel = (m/12)(iv2+L2)
Iofud = ((210 lb/32.174 ft/s2}/12)((3.0 ft)2 + (1.5ft)2) - 6.11b-s2-ft
Since these are the moments of inertia of each component about its own neutral axis, we now need to use the parallel axis theorem to transfer these moments to the neutral axis of the body, which is located at the body center of gravity that we recently calculated.
To do this, the distance from the body center of gravity to each component’s center of
gravity must be found. The distances squared from each component to the body center of gravity are Previous << 1 .. 4 5 6 7 8 9 < 10 > 11 12 13 14 15 16 .. 128 >> Next 