Nt1310 Unit 3 Lab Report
MESEGENA YSIENI
ECE102
WEEK 3, Monday, Sep-15-2014-session
LAB 1
Due date: Sep-22, 2014
1. Problem Statement: How many ping pong balls will be able to fit in T1-112 class room?
2. Assumptions:
The ping pong balls are not going to stack on top of one another in a perfect manner.
The furniture and computers in the room are insignificant.
The ping pong balls stack on top of one another with no space between them.
The ping pong balls stack on top of one another in a three dimensional mode.
3. Heuristics:
The volume of the room is: V=H*W*L.
Measurements are in inches.
Volume of the ping pong ball is V=PI*R^3.
The room measures height of 10ft, width of 23ft, and a length of 35ft.
Diameter of a ping pong ball 40mm = 1.57inches. …show more content…
4. Analysis:
This model is built to fit as many ping pong balls as possible in T1-112 class room.
There are various variables to consider in this process. The lab room is contains in it desks, computers, chairs, two cabinets, and a small desk with a printer on top of it. I will be ignoring the tables and the computers on top of them because their thickness is relatively immaterial in measurement for this task because they don’t possess much of the floor space. The very first thing that needs to be identified is the volume of the room. For this purpose I will be taking the measurement of the room, cabinet, column and a ping pong ball as …show more content…
follows:
Room Width =27fr=324 inches Cabinet W =3.5ft=42inc Column W 3ft=36 inch
Room Length =41ft=492 inches Cabinet H = 6ft = 72inch Column Side 2ft = 24 inch
Room Height=10ft= 120 inches Cabinet depth = 2.5ft=30inc Column H = 10ft = 120 inch
In stacking the ping pong balls it’s important to note that these balls will not stack on top of one another in a perfect manner because they are of a circular shape. Two possible modeling scenarios are put in test and the lower bound is modified and used to determine how many ping pong balls are going to fit in the class room.
Lower Bound Model:
Number of ping pong balls/room = Volume of room/Volume of PP ball (as cube)
The first scenario is assuming a three dimensional pattern.
This setup will use a large amount of space for relatively less amount of balls distributed through the room due to the fact that there will be too much space left in between the balls that is wasted. As a result this would be the lower bound model, the result is too small.
Figure 1.1
Upper Bound Model:
Number of ping pong balls/room = Volume of room/Volume of PP ball (as cube)
This model is under the assumptions that the ping pong balls stack on top of one another with no space left in between them. The issue with this model is, it is impossible to stack them with this manner considering the balls are circular and they do not have perfect edges for them to stack up in this manner. Due to this factor this model overestimates the amount of balls the room can accommodate. It’s displayed in the figure 1.2 below:
Figure 1.2
Modified Lower Bound Model: To calculate the number of ping pong balls in the class room this model considers stacking the balls as displayed in the figure 1.3. The formula for the calculation is as listed
below:
Number of PP balls = [volume of the room-volume of the column-volume of cabinet]/volume of PP ball.
Volume of room = 324*492*120 = 19,128,960 (cube inches)
Volume of Cabinets = (42*72*30)*2, calculating the volume of two cabinets in the room. =181,440(cube inches)
Volume of Column = 36*24*120 = 103,680 (cube inches)
Volume of PP balls = 0.649(vol cube). Number of PP balls = [ (19,128,960-181,440-103,680) / 0.649 ] = 29,035,193
5. Conclusion: In the class room T1-112 it is possible to fit 29,035,193 ping pong balls.
Figure 1.3
ECE102
WEEK 3, Monday, Sep-15-2014-session
LAB 1
Due date: Sep-22, 2014
1. Problem Statement: How many ping pong balls will be able to fit in T1-112 class room?
2. Assumptions:
The ping pong balls are not going to stack on top of one another in a perfect manner.
The furniture and computers in the room are insignificant.
The ping pong balls stack on top of one another with no space between them.
The ping pong balls stack on top of one another in a three dimensional mode.
3. Heuristics:
The volume of the room is: V=H*W*L.
Measurements are in inches.
Volume of the ping pong ball is V=PI*R^3.
The room measures height of 10ft, width of 23ft, and a length of 35ft.
Diameter of a ping pong ball 40mm = 1.57inches. …show more content…
4. Analysis:
This model is built to fit as many ping pong balls as possible in T1-112 class room.
There are various variables to consider in this process. The lab room is contains in it desks, computers, chairs, two cabinets, and a small desk with a printer on top of it. I will be ignoring the tables and the computers on top of them because their thickness is relatively immaterial in measurement for this task because they don’t possess much of the floor space. The very first thing that needs to be identified is the volume of the room. For this purpose I will be taking the measurement of the room, cabinet, column and a ping pong ball as …show more content…
follows:
Room Width =27fr=324 inches Cabinet W =3.5ft=42inc Column W 3ft=36 inch
Room Length =41ft=492 inches Cabinet H = 6ft = 72inch Column Side 2ft = 24 inch
Room Height=10ft= 120 inches Cabinet depth = 2.5ft=30inc Column H = 10ft = 120 inch
In stacking the ping pong balls it’s important to note that these balls will not stack on top of one another in a perfect manner because they are of a circular shape. Two possible modeling scenarios are put in test and the lower bound is modified and used to determine how many ping pong balls are going to fit in the class room.
Lower Bound Model:
Number of ping pong balls/room = Volume of room/Volume of PP ball (as cube)
The first scenario is assuming a three dimensional pattern.
This setup will use a large amount of space for relatively less amount of balls distributed through the room due to the fact that there will be too much space left in between the balls that is wasted. As a result this would be the lower bound model, the result is too small.
Figure 1.1
Upper Bound Model:
Number of ping pong balls/room = Volume of room/Volume of PP ball (as cube)
This model is under the assumptions that the ping pong balls stack on top of one another with no space left in between them. The issue with this model is, it is impossible to stack them with this manner considering the balls are circular and they do not have perfect edges for them to stack up in this manner. Due to this factor this model overestimates the amount of balls the room can accommodate. It’s displayed in the figure 1.2 below:
Figure 1.2
Modified Lower Bound Model: To calculate the number of ping pong balls in the class room this model considers stacking the balls as displayed in the figure 1.3. The formula for the calculation is as listed
below:
Number of PP balls = [volume of the room-volume of the column-volume of cabinet]/volume of PP ball.
Volume of room = 324*492*120 = 19,128,960 (cube inches)
Volume of Cabinets = (42*72*30)*2, calculating the volume of two cabinets in the room. =181,440(cube inches)
Volume of Column = 36*24*120 = 103,680 (cube inches)
Volume of PP balls = 0.649(vol cube). Number of PP balls = [ (19,128,960-181,440-103,680) / 0.649 ] = 29,035,193
5. Conclusion: In the class room T1-112 it is possible to fit 29,035,193 ping pong balls.
Figure 1.3