ME 364 – Fall ‘09 Test #2 11/6/09 Closed book, closed notes. This first part of the test is worth 30 points. Turn this part of the test in to receive the second part. 1. Define all of the terms shown in the figure below. Also define longitudinal slip (or slip ratio) while braking. RU V RE RL 2. On the graph below sketch a “typical” lateral force vs. slip angle curve with numbers for a passenger car tire at 500 lb vertical load, a passenger car tire at 1000 lb vertical load, and Lateral force, lb 0 Slip Angle, degrees 20 3. Give a physical description of the concepts of neutral steer, understeer and oversteer. Do not use equations or formulas. Other problems on back → 4. Define the following terminology related to vehicle dynamics: Ackerman steer angle roll center sideslip angle tire scrub Other problems on back → ME 364 – Fall ‘09 Test #2 11/6/09 Closed book, closed notes, one 8.5 x 11 inch page of handwritten formulas allowed. Show all work. 5. Weight and geometric data 700 Vertical Load for a vehicle is given below. 2000 lb Experimental data for the 1800 lb tires used on this vehicle is 600 1600 lb shown in the plot. Determine: 500 1400 lb a) Cornering stiffness of a front tire 1200 lb 400 b) Cornering stiffness of a 1000 lb rear tire 800 lb 300 c) Understeer gradient 600 lb 200 d) Steer angle required for . Fy Force, (lb) Lateral this vehicle to travel a circle of radius 100 R=240 feet at a speed of 60 mph 0 0 0.5 1 1.5 2 Slip Angle, a (deg) Curb Weight: 4370 lb Inches Feet Wheelbase: 116.0 9.67 CG distance behind front axle: 51.04 4.25 CG distance in front of rear axle: 64.96 5.42 CG distance from ground: 23.28 1.94 6. Determine the location for the roll center for the vehicle suspension shown below. Other problems on back → 7. In the suspension shown below, a vertical displacement of 0.50 inches at the wheel causes the spring to shorten by 0.29 inches. a) What is the motion ratio for this suspension? b) If the desired stiffness at the wheel is KW = 150 lb/in, what is the appropriate spring stiffness, KS? 8. A 2-D version of a double-wishbone / double A-arm suspension is shown below. The links are NOT drawn to scale. Use position kinematics to analytically determine c) angle the upright (R2) makes with the horizontal X axis, d) angle the upper A-arm (R3) makes with the horizontal X axis, and e) the length of the shock (RS). Mag Angle mm deg R0 390 90 RS R 370 -10 1 RF R2 440 R3 290 R3 RF 510 80 R1b 210 5 R0 R2b 1 15 R1b R Angle from R1 to R1b is 1=15 deg 2 R1 Other problems on back → .