This project report summarizes work done by a student team to apply homework calculations to measurements taken from a 2004 Mitsubishi Galant. The team measured vehicle specifications, completed three homework problems involving calculations of engine speed, torque, fuel consumption, effective mass, acceleration, and brake performance. Calculations were shown for static and dynamic load conditions, effects of passengers and a trailer, and comparisons of wheel loads with and without a load-equalizing hitch. Braking forces and times to accelerate to 300 mph in a quarter mile were also estimated.

1. AME 451
Project Report
Submission Date: August 13,2014
Instructor: Mukherjee, Jyoti
Team members: Molani, Mohammad Salem
Jun, Youra
Cao, Zhe

2. Table of Contents
Table of Contents ...............................................................................................................................2
Introduction .......................................................................................................................................2
Results and calculations ......................................................................................................................2
Part 1. Vehicle information and measurements..................................................................................2
Part 2. Homework Problems ............................................................................................................4
HW 1A.......................................................................................................................................4
HW 2..........................................................................................................................................7
HW3 ........................................................................................................................................10
Conclusion.......................................................................................................................................19
Introduction
How would it be if the students get to work and apply the homework problems calculations into
a real car? This project is one of the best opportunities for the group members in order to face the
real engineering duty. The group started from identifying the vehicle information and gets to
know about the vehicle features and how things work. After manually measuring the data needed
from the car to solve the homework problems, the group came up with reasonable results beside
that, there are numbers assumed by the group members for solving the problems.
Results and calculations
Part 1. Vehicle information and measurements
 Vehicle Used : 2004 Mitsubishi Galant 3.8 liter GTS
 Engine ID: 6G75
 Vehicle ID (VIN): 4A3AB76S74E106239
 Tire ID: DOT 93 ENB0621311
 Wheelbase (L): 108.5”
 Tread Width: 8.5”
 Static Loaded Radius(R): 13”

3.  When Car is empty;
 Gross Vehicle Weight : 4541 lbs
 Static Front Wheel Load(Rf) : 2491 lbs
 Static Rare Wheel Load(Rr) : 2050 lbs
 Distance from the ground to the bottom of the vehicle (hi) : 8.75”
 Vertical location of the center of gravity (ycg) : 24” from the ground
 Horizontal location of the center of gravity (xcg) : 49” from the front wheel
o Calculating the horizontal location of the center of gravity
∑ 𝑇𝑓 = 4541𝑙𝑏𝑠 ∙ 𝑥 𝑐𝑔 − 2050𝑙𝑏𝑠 ∙ 108.3𝑖𝑛 = 0
x =
2050𝑙𝑏𝑠 ∙ 108.3𝑖𝑛
4541𝑙𝑏𝑠
= 49𝑖𝑛
 When there are two passengers;
 Weight of two passengers (Wp) : 398.7 lbs
 Distance from the ground to the bottom of the vehicle (hf) : 7.5”
 Vertical location of the center of gravity of the passengers (yp) : 29” from the ground
 Horizontal location of the center of gravity of the passengers(xp) : 19” from the front
wheel.

4.  Spring rate (k):
o Calculating the spring rates
k =
W
2δ
=
W 𝑝
2(ℎ 𝑖−ℎ 𝑓)
=
W 𝑝
2(ℎ 𝑖−ℎ 𝑓)
=
398 .7lbs
2(8.75in−7.5𝑖𝑛)
= 159.48 𝑙𝑏𝑠/𝑖𝑛
 Brake Gain of the front wheel (Gf) : 0.22
N∙m
kPa
per wheel
 Brake Gain of the rear wheel (Gr) : 0.15
N∙m
kPa
per wheel
Part 2. Homework Problems
HW 1A
Usingthe data foundinpart 1, answerthe followingquestions.
a) Find the longitudinal position of the combined CG.
x =
∑(𝑤𝑒𝑖𝑔ℎ𝑡 ∙ 𝑋)
∑ 𝑤𝑒𝑖𝑔ℎ𝑡
=
4541𝑙𝑏𝑠(49")+ 398.7𝑙𝑏𝑠(89.5")
4541𝑙𝑏𝑠 + 398.7𝑙𝑏𝑠
= 52.3"

5. b) Determine the axle loads of the loaded vehicle.
Balancing the moment about front axle.
R 𝑟 ∙ 108.5" = (4541lbs + 398.7lbs)(52.3")
R 𝑟 = 2381.07𝑙𝑏𝑠
Balancing vertical direction forces
𝑅 𝑟 + 𝑅𝑓 = 4541𝑙𝑏𝑠 + 398.7𝑙𝑏
R 𝑓 = 2558.63𝑙𝑏𝑠
c) Find the vertical position of the combined CG.
y =
∑(𝑤𝑒𝑖𝑔ℎ𝑡 ∙ 𝑦)
∑ 𝑤𝑒𝑖𝑔ℎ𝑡
=
4541𝑙𝑏𝑠(24") + 398.7𝑙𝑏𝑠(29")
4541𝑙𝑏𝑠 + 398.7𝑙𝑏𝑠
= 24.4"
d) What are the axle loads when it accelerates at 3m/sec2?
Forward acting force
F = ma = (
4541𝑙𝑏𝑠
32.2𝑓𝑡
𝑠𝑒𝑐2
)(
3𝑚
𝑠𝑒𝑐2
)(
3.28𝑓𝑡
1𝑚
) = 1387.68 𝑙𝑏𝑠
Balancing moments about the front axle
(1387.68lbs)(24.4")+Rr(108.5")=(4541lbs+398.7lbs)(52.3")
𝑅 𝑟 = 2069𝑙𝑏𝑠

6. Balancing moments about the rear axle
(1387.68lbs)(24.4")-Rf(108.5")=(4541lbs+398.7lbs)(108.5"-52.3")
𝑅 𝑟 = 2870.69𝑙𝑏𝑠
e) What are the axle loads when the trailer shown in the figure is connected up? The trailer
hitch on the car is 90mm behind the rear axle.
Balancing moments about hitch
610kg(9.81m/𝑠2)(2.8m) = 𝑅𝑡(3.25𝑚)
𝑅𝑡 = 5155.5𝑁 = 1159.0 𝑙𝑏𝑠
Balancing the force in order to find the force at hitch
𝑅ℎ = 610kg(9.81m/s2
)− 5155.5N = 828.6N = 186.28lbs
Taking the car and balancing moment about front axle
𝑅 𝑟(108.5")=(4541lbs+398.7lbs)(52.3") + 186.28𝑙𝑏𝑠((0.98𝑚)(
39.37"
1m
) + 108.5")
𝑅 𝑟 = 2633.59lbs
Balancing vertical direction forces
𝑅 𝑟 + 𝑅𝑓 = 4541𝑙𝑏𝑠 + 398.7𝑙𝑏 + 186.28𝑙𝑏𝑠
R 𝑓 = 2492.39𝑙𝑏𝑠

7. f) Because of the high rear axle load on the car, the owner installs a “load-equalizing” hitch.
This hitch puts a torque of 1500Nm (=13276.1lbs*in) into the connection point as
illustrated in the figure below. What are the wheel loads (car and trailer) with the load
equalizer?
Balancing the moment about the hitch.
610kg(9.81m/𝑠2)(2.8m) + 1500N ∙ m = 𝑅𝑡(3.25𝑚)
𝑅𝑡 = 5617𝑁 = 1262.75𝑙𝑏𝑠
load at hitch = Rℎ = 610𝑘𝑔(9.81𝑚/𝑠2 )− 5617𝑁 = 367.1N = 82.53𝑙𝑏𝑠
Taking the car and balancing moment about front axle
𝑅 𝑟(108.5") + 13276.1𝑙𝑏𝑠 ∙ 𝑖𝑛
= (4541𝑙𝑏𝑠 + 398.7𝑙𝑏𝑠)(52.3") + 82.53𝑙𝑏𝑠(108.5" + 38.57")
𝑅 𝑟 = 2370.58𝑙𝑏𝑠
HW 2
There are 3 problems in the HW 2. We are going to use all data we measured.
Q1. If the vehicle is running at 80km/hr up a 2% grade. The drag from aerodynamics and
rolling resistance is 572N. It is operating in third gear, which has a ratio of 1.2, and has a
final drive ratio of 3.5.With the given data and the measurements in part 1, answer the
following questions.

8. (a) What is the engine speed (RPM)?
Angular velocity =
𝑣
2∗𝜋∗𝑟
=
80
𝑘𝑚
ℎ𝑟
∗10.936
𝑖𝑛
𝑠
2 ∗𝜋∗13 in
= 422.86
𝑟𝑎𝑑
𝑠
𝑛1
𝑛2
=
𝑡1
𝑡2
,
𝑤2
𝑤1
=
𝑡1
𝑡2
422.86
𝑤1
= 1.21, 𝑤1 = 349.47 𝑟𝑎𝑑/𝑠
(b) How much torque (N-m) does the engine develop to maintain a constant speed if the
driveline efficiency is 91%.
4541 lb = 2059.88 kg = 20207.45 N; 13 in = 0.33 m
Ft = 572 N + 0.02 * 20207.45 N = 976.15 N
Tt = Ft * r = 976.15 N * 0.33 m = 322.13 Nm
T2 =
𝑇𝑡
𝑁𝑟∗𝑁𝑓∗ 𝜂
=
322 .13 𝑁𝑚
1.21∗3.5∗0.91
= 83.59 𝑁𝑚
(c) If the engine has a brake-specific fuel consumption of 0.3 kg/kw-h at these conditions,
what is the fuel consumption (kg/km)? Also give it in liters/100 km, assuming gasoline
weighs 0.78 kg/liter.
Power developed by the engine = T2 * w1 = 83.59 Nm * 349.47 rad/s = 29.21 KW
Fuel consumption = 29.21 KW * 0.3
𝑘𝑔/ℎ𝑟
𝑘𝑤
= 8.76
𝑘𝑔
ℎ𝑟
=
8.76 𝑘𝑔/ℎ𝑟
80 𝑘𝑚/ℎ𝑟
= 0.11
𝑘𝑔
𝑘𝑚
= 0.11
𝑘𝑔
𝑘𝑚
∗
1 𝑙𝑖𝑡𝑒𝑟
0.78 𝑘𝑔
= 0.14
𝑙𝑖𝑡𝑒𝑟
𝑘𝑚
= 14
𝑙𝑖𝑡𝑒𝑟𝑠
100 𝑘𝑚
Q2. The engine of a vehicle has a moment of inertia of 0.277 kg-m2. It is operating in first
gear (ratio of 7.53), has a final drive ratio of 4.11, and the tire radius is 0.508m.
(a) What is its effective mass of the engine?
𝑀 𝑒𝑓𝑓 =
𝐼∗ 𝑁2
𝑟2 =
0.277 𝑘𝑔−𝑚2
∗ (7.53∗4.11)2
(0.33 𝑚)2 = 2436.27𝑘𝑔
(b) What is the “mass factor” for this condition?
Mass Factor =
𝑀+ 𝑀 𝑒𝑓𝑓
𝑀
=
2059 .88 𝑘𝑔+2436 .27 𝑘𝑔
2059.88 𝑘𝑔
= 2.18

9. Q3. The vehicle can accelerate to 300mph in a quarter mile. The performance can be
modeled rather simply by assuming that the engine delivers constant power during
acceleration, and neglecting rolling resistance and aerodynamic drag forces.
(a) Derive and solve the differential equation relating velocity to distance for this case.
𝐹𝑥 = 𝑀 ∗ 𝑎 𝑥 = 𝑀 ∗
𝑑𝑉
𝑑𝑡
P = 𝐹𝑥 ∗ 𝑉; 𝐹𝑥 =
𝑃
𝑉
; 𝑎𝑛𝑑 𝑉 =
𝑑𝑥
𝑑𝑡
𝑃
𝑉
∗ 𝑑𝑡 =
𝑃
𝑉
∗
𝑑𝑥
𝑉
= 𝑀 ∗ 𝑑𝑉
∫ 𝑑𝑥
𝑥
0
=
𝑀
𝑃
∗ ∫ 𝑉2
∗ 𝑑𝑉
𝑉
0
X =
𝑀
𝑃
∗
𝑉3
3
(b) If the dragster weighs 4541 lb, how much power is develop to achieve acceleration to 300
mph in the quarter mile? (Neglect driveline inertia)
X =
5280
4
= 1320 𝑓𝑡
300mph = 440
𝑓𝑡
𝑠𝑒𝑐
P =
𝑀
𝑋
∗
𝑉3
3
=
4541 𝑙𝑏
32.2
𝑓𝑡
𝑠𝑒𝑐2 ∗1320 𝑓𝑡
∗
(440
𝑓𝑡
𝑠𝑒𝑐
)
3
3
= 3,033,601
𝑓𝑡−𝑙𝑏
𝑠𝑒𝑐
= 5515.42 HP
(c) How much time is required to cover the quarter mile?
∫ 𝑑𝑡
𝑇
0
=
𝑀
𝑃
∫ 𝑉 ∗ 𝑑𝑉
𝑉
0

10. T =
𝑀
𝑃
𝑉2
2
=
4541 𝑙𝑏
32.2
𝑓𝑡
𝑠𝑒𝑐2 ∗ 3,033,601
𝑓𝑡 − 𝑙𝑏
𝑠𝑒𝑐
(440
𝑓𝑡
𝑠𝑒𝑐
)2
2
= 4.5 𝑠𝑒𝑐
HW3
Q. You have been asked to help design the brake system on a new light truck. The truck
must meet FMVSS 105 performance requirements for deceleration (take this to mean 20
ft/sec2 lightly loaded, and 19 ft/sec2 fully loaded on an 81 Skid Number surface), as well
as corporate requirements for 0.25 g braking performance on a 30 Skid Number surface.
The important specifications of the vehicle are:
L=108.5” Rtire=13”
Brake gains (per wheel) 0.22 Nm/kPa (front) 0.15Nm/kPa (rear)
lightly-loaded fully-loaded
CG height 24" 24.4"
front axle weight 2491lbs 2558.63lbs
rear axle weight 2050lbs 2381.07lbs
You are to select a pressure proportioning system (break pressure and rise rate) and then
determine the efficiency of the system for the lightly-loaded and fully-loaded conditions. Do this
via the following steps:

11. (a) Calculate and plot the four performance triangles on the front/ rear brake force diagram
for all conditions. (Show your calculations for the intercept points.)
 First step is to find the maximum front and rear brake forces for each condition.
𝐹𝑥𝑓𝑚 =
𝜇 𝑝(𝑊𝑓𝑠 +
ℎ
𝐿
𝐹xr)
1 −
𝜇 𝑝ℎ
𝐿
o Front brake force axis intercepts:
Lightly-loaded & 81 SN, 𝐹𝑥𝑓𝑚 =
0.81(2491𝑙𝑏𝑠 )
1−
0.81(24" )
(108 .5")
= 2458.1lbs
Fully-loaded & 81 SN, 𝐹𝑥𝑓𝑚 =
0.81(2558.63𝑙𝑏𝑠)
1−
0.81(24.4")
(108.5" )
= 2534.09𝑙𝑏𝑠
Lightly-loaded & 30 SN, 𝐹𝑥𝑓𝑚 =
0.3(2491𝑙𝑏𝑠)
1−
0.3(24" )
(108 .5")
= 800.42lbs
Fully- loaded & 30SN, 𝐹𝑥𝑓𝑚 =
0.3(2558 .63𝑙𝑏𝑠 )
1−
0.3(24.4")
(108.5")
= 823.12lbs
𝐹𝑥𝑟𝑚 =
𝜇 𝑝(𝑊𝑓𝑠 −
ℎ
𝐿
𝐹xr)
1 +
𝜇 𝑝ℎ
𝐿
o Rear brake force axis intercepts:
Lightly-loaded & 81 SN, 𝐹𝑥𝑓𝑚 =
0.81(2050𝑙𝑏𝑠)
1+
0.81(24" )
(108.5")
= 1408.19lbs
Fully-loaded & 81 SN, 𝐹𝑥𝑓𝑚 =
0.81(2381.07𝑙𝑏𝑠)
1+
0.81(24.4")
(108.5" )
= 1631.48𝑙𝑏𝑠
Lightly-loaded & 30 SN, 𝐹𝑥𝑓𝑚 =
0.3(2050𝑙𝑏𝑠 )
1+
0.3(24" )
(108 .5")
= 576.73lbs

12. Fully- loaded & 30SN, 𝐹𝑥𝑓𝑚 =
0.3(2381 .07𝑙𝑏𝑠 )
1+
0.3(24.4")
(108.5")
= 669.17lbs
 Slope of front axle
𝑚 =
𝜇 𝑝ℎ
𝐿
1 −
𝜇 𝑝ℎ
𝐿
Lightly-loaded & 81 SN, 𝑚 =
0.81(24" )
108 .5"
1−
0.81(24" )
108.5"
= 0.2183
Fully-loaded & 81 SN, m =
0.81(24.4")
108 .5"
1−
0.81(24.4")
108.5"
= 0.2227
Lightly-loaded & 30 SN, m =
0.3(24" )
108 .5"
1−
0.3(24")
108.5"
= 0.0711
Fully- loaded & 30SN, m =
0.3(24.4")
108.5"
1−
0.3(24.4")
108.5"
= 0.0723
 Slope of rear axle
𝑚 =
−
𝜇 𝑝ℎ
𝐿
1 +
𝜇 𝑝ℎ
𝐿
Lightly-loaded & 81 SN, 𝑚 =
−
0.81(24")
108.5"
1+
0.81(24" )
108 .5"
= −0.1519
Fully-loaded & 81 SN, m =
−
0.81(24.4")
108.5"
1+
0.81(24.4")
108 .5"
= −0.1541
Lightly-loaded & 30 SN, m =
−
0.3(24")
108.5"
1+
0.3(24" )
108.5"
= −0.0622
Fully- loaded & 30SN, m =
−
0.3(24.4")
108 .5"
1+
0.3(24.4")
108 .5"
= −0.0632

13.  Intersection point
F 𝑥𝑓i = 𝜇 𝑝 (𝑊𝑓𝑠 + 𝜇 𝑝 𝑊 (
ℎ
𝐿
))
F 𝑥𝑟i = 𝜇 𝑝 (𝑊𝑟𝑠 − 𝜇 𝑝 𝑊 (
ℎ
𝐿
))
Lightly-loaded & 81 SN, F 𝑥𝑓i = 0.81(2491𝑙𝑏𝑠 + 0.81(4541𝑙𝑏𝑠) (
24"
108.5"
)) = 2676.74lbs
F 𝑥𝑟i = 0.81 (2050𝑙𝑏𝑠 − 0.81(4541lbs) (
24"
108 .5"
)) = 1001.47lbs
Fully-loaded & 81 SN, F 𝑥𝑓i = 0.81(2558.63𝑙𝑏𝑠 + 0.81(4939.7𝑙𝑏𝑠) (
24.4"
108.5"
)) = 2801.33lbs
F 𝑥𝑟i = 0.81 (2381.07𝑙𝑏𝑠 − 0.81(4939.7lbs)(
24 .4"
108.5"
)) = 1199.83lbs
Lightly-loaded & 30 SN, F 𝑥𝑓i = 0.3(2491𝑙𝑏𝑠 + 0.3(4541𝑙𝑏𝑠) (
24"
108 .5"
)) = 837.70lbs
F 𝑥𝑟i = 0.3(2050𝑙𝑏𝑠 − 0.3(4541lbs) (
24"
108.5"
)) = 524.6lbs
Fully- loaded & 30SN, F 𝑥𝑓i = 0.3 (2558.63𝑙𝑏𝑠 + 0.3(4939.7𝑙𝑏𝑠) (
24.4"
108.5"
)) = 867.57lbs
F 𝑥𝑟i = 0.3(2381.07𝑙𝑏𝑠 − 0.3(4939.7lbs) (
24 .4"
108.5"
)) = 614.34lbs
Fxmf (lbs) Fxmr (lbs)
slope for
front
slope for
rear
Fxfi (lbs) Fxri (lbs)
81SN,lightly
loaded
2458.14 1408.19 0.2183 -0.1519 2676.74 1001.47
81SN,fully
loaded
2534.09 1631.48 0.2227 -0.1541 2801.33 1199.83
30SN,lightly
loaded
800.42 576.73 0.0711 -0.0622 837.70 524.60

14.  Deceleration line
Lightly-loaded & 81 SN, Ftot = ma =
(2491lbs+2050lbs)
32.2𝑓𝑡
sec2
×
20ft
sec2 = 2820.5lbs
Fully-loaded & 81 SN, Ftot = ma =
(2558 .63lbs+2381.07lbs)
32.2𝑓𝑡
sec2
×
19ft
sec2 = 2914.73𝑙𝑏𝑠
Lightly-loaded & 30 SN, Ftot = ma =
(2491lbs +2050lbs)
32.2𝑓𝑡
sec2
× 0.25 (
32.2ft
sec2 ) = 1135.25lbs
Fully-loaded& 30S, Ftot = ma =
(2558 .63lbs+2381 .07lbs)
32.2𝑓𝑡
sec2
× 0.25 (
32.2ft
sec2 ) = 1234.93𝑙𝑏𝑠
 Using calculator, find the intersection between the deceleration line and the force line.
Lightly-loaded & 81 SN, (297.43, 2523.07) and (1155.27, 1665.26)
Fully-loaded & 81 SN, (1397.71, 1517.02) and (311.31, 2603.42)
Lightly-loaded & 30 SN, (539.69, 595.56) and (312.60, 822.65)
Fully- loaded & 30SN, (631.00, 603.93) and (384.04, 850.89)
(b) Select the parameters for a proportioning valve (pressure break point and slope) to be
used and show it on the diagram
The diagram below shows the performance triangles for this vehicle. Since the brake
gains were given be proportioning line must have an initial slope of
0.22
0.15
= 1.467. this is
30SN,fully
loaded
823.12 669.17 0.0723 -0.0632 867.57 614.34

15. slope is satisfactory for the lower triangles, but must change a higher slope to meet the
high lower triangles.
It would be reasonable to select the pressure break point for the pressure proportioning
valve equivalent to 4003.4 N (900 lb) of force on the front brakes. The associated
pressure is:
𝑃𝑎 =
4003.4 𝑁∗ 0.33 𝑚
2 ∗ 0.22 𝑁 − 𝑚/𝐾𝑝𝑎
= 3002.55 𝑘𝑃𝑎
At this pressure the rear brake force is:
𝐹𝑟 =
2∗0.15 𝑁−
𝑚
𝐾𝑝𝑎
∗3002.55 𝐾𝑝𝑎
0.33 𝑚
= 2729.6 𝑁 = 613.64 lbf
A reasonable pressure proportioning can be obtained by choosing the second segment of
the line to go to a rear brake force of 6227.51 N (1400 lb) when the front brakes go up to
13345 N (3000 lb). The associated brake pressures at this condition are:
𝑃 𝑎𝑓 =
13345 𝑁 ∗ 0.33 𝑚
2 ∗ 0.22 𝑁 − 𝑚/𝐾𝑝𝑎
= 10008.75 𝐾𝑝𝑎
𝑃𝑎𝑟 =
6227.51 𝑁 ∗ 0.33 𝑚
2 ∗ 0.15 𝑁 − 𝑚/𝐾𝑝𝑎
= 6850.26 𝐾𝑝𝑎
The slope for the proportioning is:
S =
∆𝑃𝑎𝑟
∆𝑃 𝑎𝑓
=
6850.26 − 3002.55
10008.75 − 3002.55
= 0.549
The proportioning valve is denoted as
3002.55
0.549
%Graphing Front Brake Force Vs Rear Brake Force
% For lightly loaded & 81 SN,
fxr1=[1001.47 1408.19];fxf1=[2676.74 0];
plot(fxr1,fxf1);
hold on
fxr2=[0 1001.47];fxf2=[2458.14 2676.74];
plot(fxr2,fxf2);
% For fully loaded & 81 SN,
fxr3=[1199.83 1631.48];fxf3=[2801.33 0];
plot(fxr3,fxf3);
fxr4=[0 1199.83];fxf4=[2534.09 2801.33];
plot(fxr4,fxf4);
% For lightly loaded & 30 SN,

16. fxr5=[524.60 576.73];fxf5=[837.70 0];
plot(fxr5,fxf5);
fxr6=[0 524.60];fxf6=[800.42 837.70];
plot(fxr6,fxf6);
% For fully loaded & 30 SN,
fxr7=[614.34 669.17];fxf7=[867.57 0];
plot(fxr7,fxf7);
fxr8=[0 614.34];fxf8=[823.12 867.57];
plot(fxr8,fxf8);
%Drawing the deceleration line
fxr9=[297.43 1155.27];fxf9=[2523.07 1665.26];
fxr10=[1397.71 311.31];fxf10=[1517.02 2603.42];
fxr11=[539.69 312.60];fxf11=[595.56 822.65];
fxr12=[631.00 384.04];fxf12=[603.93 850.89];
plot(fxr9,fxf9);plot(fxr10,fxf10);plot(fxr11,fxf11);plot(fxr12,fxf12);
%Drawing the proportioning line
fxr13=[0 700]; fxf13=[0 1.467*700];
fxr14=[700 1400]; fxf14=[1.467*700 3000];
plot(fxr13,fxf13); plot(fxr14,fxf14);
xlabel('Rare Brake Force (lbs)'); ylabel('Front Brake Force (lbs)');
title('Brake Force Diagram');
hold off

17. (c) Calculate and plot the braking efficiency for the lightly-loaded and fully-loaded vehicle
as a function of application pressure for the design you selected.
Lightly loaded
Gf=0.22 Gr=0.15 R(tire)=0.33m Pa=3002.55kPa m(prop)=0.549
Wfs=2491lbs=11080.52N Wrs=2050lbs=9118.85N h=0.6096m WB=2.7559m

18. Pf(kPa) Pr(kPa) Fxf(N) Fxr(N) Dx(g) Wf(kg) Wr(kg) µf µr ηf ηr ηb
10 10 13.33 9.09 0.001 1130.02 929.04 0.001 0.001 0.923 1.113 0.923
100 100 133.33 90.91 0.011 1134.57 924.49 0.012 0.010 0.927 1.108 0.927
500 500 666.67 454.55 0.056 1154.79 904.27 0.059 0.051 0.943 1.083 0.943
1000 1000 1333.33 909.09 0.111 1180.08 878.98 0.115 0.105 0.964 1.053 0.964
2000 2000 2666.67 1818.18 0.222 1230.64 828.42 0.221 0.224 1.005 0.992 0.992
3000 3000 4000.00 2727.27 0.333 1281.20 777.86 0.318 0.357 1.046 0.932 0.932
4000 3501.28 5333.33 3182.98 0.422 1321.54 737.52 0.411 0.440 1.025 0.958 0.958
5000 4001.28 6666.67 3637.52 0.510 1361.85 697.21 0.499 0.532 1.022 0.959 0.959
6000 4501.28 8000.00 4092.07 0.599 1402.17 656.89 0.582 0.635 1.029 0.943 0.943
7000 5001.28 9333.33 4546.61 0.687 1442.48 616.58 0.660 0.752 1.042 0.914 0.914
8000 5501.28 10666.67 5001.16 0.776 1482.79 576.26 0.733 0.885 1.058 0.877 0.877
9000 6001.28 12000.00 5455.70 0.864 1523.11 535.95 0.803 1.038 1.076 0.833 0.833
10000 6501.28 13333.33 5910.25 0.953 1563.42 495.64 0.869 1.216 1.096 0.784 0.784
11000 7001.28 14666.67 6364.80 1.041 1603.74 455.32 0.932 1.425 1.117 0.731 0.731
Fully - loaded
Gf=0.22 Gr=0.15 R(tire)=0.33m Pa=3002.55kPa m(prop)=0.549
Wfs=2558.63lbs=11381.35N Wrs=2381.07lbs=10591.53N h=0.620m WB=2.7559m
Pf(kPa) Pr(kPa) Fxf(N) Fxr(N) Dx(g) Wf(kg) Wr(kg) µf µr ηf ηr ηb
10 10 13.33 9.09 0.001 1160.69 1079.15 0.001 0.001 0.872 1.188 0.872
100 100 133.33 90.91 0.010 1165.32 1074.52 0.012 0.009 0.875 1.183 0.875
500 500 666.67 454.55 0.051 1185.89 1053.95 0.057 0.044 0.890 1.161 0.890
1000 1000 1333.33 909.09 0.102 1211.60 1028.24 0.112 0.090 0.910 1.132 0.910
2000 2000 2666.67 1818.18 0.204 1263.03 976.82 0.215 0.190 0.948 1.076 0.948
3000 3000 4000.00 2727.27 0.306 1314.45 925.39 0.310 0.300 0.987 1.019 0.987
4000 3501.28 5333.33 3182.98 0.388 1355.48 884.36 0.401 0.367 0.966 1.056 0.966

19. 5000 4001.28 6666.67 3637.52 0.469 1396.48 843.36 0.487 0.440 0.964 1.067 0.964
6000 4501.28 8000.00 4092.07 0.550 1437.48 802.36 0.567 0.520 0.970 1.059 0.970
7000 5001.28 9333.33 4546.61 0.632 1478.49 761.36 0.644 0.609 0.982 1.038 0.982
8000 5501.28 10666.67 5001.16 0.713 1519.49 720.36 0.716 0.708 0.996 1.008 0.996
9000 6001.28 12000.00 5455.70 0.794 1560.49 679.36 0.784 0.819 1.013 0.970 0.970
10000 6501.28 13333.33 5910.25 0.876 1601.49 638.36 0.849 0.944 1.032 0.928 0.928
11000 7001.28 14666.67 6364.80 0.957 1642.49 597.35 0.910 1.086 1.052 0.881 0.881
Conclusion
The following major assumptions that the team has made so far where included in the homework
problems and applied resulting not actual results:
 Gross Vehicle Weight
 Static Front Wheel Load(Rf)
0.000
0.200
0.400
0.600
0.800
1.000
1.200
0 2000 4000 6000 8000 10000 12000
Efficiency
Pressure (kPa)
Efficiency
Lightly-Loaded
Fully-loaded

20.  Static Rare Wheel Load(Rr)
 The Break gains, CG heights, and Horizontal location of passenger CG.
Last but not least, the project was one of the best ways to learn about real life vehicle. The group was
facing lots of problems. One of the problems that the group faced is identifying the vehicle, the
engine, and the tire ID. The group googled it and get the answer which states that the information
should be inside the hood behind the engine in a silver sticker. Also, measuring the value was one of a
kind experience to the group when trying to measure the lightly loaded and fully loaded.