Vehicle dynamics final project

Vehicle dynamics final project

• 1.
  AME 451 Project Report SubmissionDate: August 13,2014 Instructor: Mukherjee, Jyoti Team members: Molani, Mohammad Salem Jun, Youra Cao, Zhe
• 2.
  Table of Contents Tableof 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 Caris 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 theaxle 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 aboutthe 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 ofthe 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.
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  (a) What isthe 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 vehiclecan 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:
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  (a) Calculate andplot 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 F𝑥𝑓i = 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 satisfactoryfor 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=[0524.60];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 andplot 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 6666.673637.52 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 RareWheel 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.