CEE320 Vehicle Dynamics Course Outline, Concepts, Resistance, Tractive Effort, Acceleration, Braking, Stopping Distance
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This document contains lecture slides for a CEE320 Vehicle Dynamics course. It covers topics such as resistance forces that impede vehicle motion like aerodynamic, rolling, and grade resistance. It also discusses tractive effort, vehicle acceleration, braking forces, and stopping sight distance. Formulas and examples are provided for calculating forces, acceleration, braking distance, and stopping sight distance. Key concepts are defined including resistance, tractive effort, and the relationships between engine torque, gear ratios, and vehicle speed.
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CEE320 Vehicle Dynamics Course Outline, Concepts, Resistance, Tractive Effort, Acceleration, Braking, Stopping Distance
- 2. CEE320 Winter2006 Outline 1. Resistance a. Aerodynamic b. Rolling c. Grade 2. Tractive Effort 3. Acceleration 4. Braking Force 5. Stopping Sight Distance (SSD)
- 3. CEE320 Winter2006 Main Concepts • Resistance • Tractive effort • Vehicle acceleration • Braking • Stopping distance grla RRRmaF +++=
- 4. CEE320 Winter2006 Resistance Resistance is defined as the force impeding vehicle motion 1. What is this force? 2. Aerodynamic resistance 3. Rolling resistance 4. Grade resistance grla RRRmaF +++=
- 5. CEE320 Winter2006 Aerodynamic Resistance Ra Composed of: 1. Turbulent air flow around vehicle body (85%) 2. Friction of air over vehicle body (12%) 3. Vehicle component resistance, from radiators and air vents (3%) 2 2 VACR fDa ρ = 3 2 VACP fDRa ρ = sec 5501 lbft hp ⋅ = from National Research Council Canada
- 6. CEE320 Winter2006 Rolling Resistance Rrl Composed primarily of 1. Resistance from tire deformation (∼90%) 2. Tire penetration and surface compression (∼ 4%) 3. Tire slippage and air circulation around wheel (∼ 6%) 4. Wide range of factors affect total rolling resistance 5. Simplifying approximation: WfR rlrl = += 147 101.0 V frlWVfP rlrlR = sec 5501 lbft hp ⋅ =
- 7. CEE320 Winter2006 Grade Resistance Rg Composed of – Gravitational force acting on the vehicle gg WR θsin= gg θθ tansin ≈ gg WR θtan= Gg =θtan WGRg = For small angles, θg W θg Rg
- 8. CEE320 Winter2006 Available Tractive Effort The minimum of: 1. Force generated by the engine, Fe 2. Maximum value that is a function of the vehicle’s weight distribution and road-tire interaction, Fmax ( )max,minefforttractiveAvailable FFe=
- 10. CEE320 Winter2006 Engine-Generated Tractive Effort • Force • Power r M F de e ηε0 = ( ) π2 min sec 60 rpmengine 550 lbfttorque sec lbft 550hp × × ⋅ = ⋅ Fe = Engine generated tractive effort reaching wheels (lb) Me = Engine torque (ft-lb) ε0 = Gear reduction ratio ηd = Driveline efficiency r = Wheel radius (ft)
- 11. CEE320 Winter2006 Vehicle Speed vs. Engine Speed ( ) 0 12 ε π irn V e − = V = velocity (ft/s) r = wheel radius (ft) ne = crankshaft rps i = driveline slippage ε0 = gear reduction ratio
- 13. CEE320 Winter2006 Maximum Tractive Effort • Front Wheel Drive Vehicle • Rear Wheel Drive Vehicle • What about 4WD? ( ) L h L hfl W F rlf µ µ − − = 1 max ( ) L h L hfl W F rlr µ µ + + = 1 max
- 15. CEE320 Winter2006 Vehicle Acceleration • Governing Equation • Mass Factor (accounts for inertia of vehicle’s rotating parts) maRF mγ=− ∑ 2 00025.004.1 εγ +=m
- 16. CEE320 Winter2006 Example A 1989 Ford 5.0L Mustang Convertible starts on a flat grade from a dead stop as fast as possible. What’s the maximum acceleration it can achieve before spinning its wheels? μ = 0.40 (wet, bad pavement) 1989 Ford 5.0L Mustang Convertible Torque 300 @ 3200 rpm Curb Weight 3640 Weight Distribution Front 57% Rear 43% Wheelbase 100.5 in Tire Size P225/60R15 Gear Reduction Ratio 3.8 Driveline efficiency 90% Center of Gravity 20 inches high
- 17. CEE320 Winter2006 Braking Force • Front axle • Rear axle ( )[ ] L fhlW F rlr bf ++ = µµ max ( )[ ] L fhlW F rlf br +− = µµ max
- 18. CEE320 Winter2006 Braking Force • Ratio • Efficiency ( ) ( ) rear front fhl fhl BFR rlf rlr = +− ++ = µ µ µ η maxg b =
- 19. CEE320 Winter2006 Braking Distance • Theoretical – ignoring air resistance • Practical • Perception • Total ( ) ( )grlb b fg VV S θµη γ sin2 2 2 2 1 ±+ − = ± − = G g a g VV d 2 2 2 2 1 pp tVd 1= ps ddd += a VV d 2 2 2 2 1 − = For grade = 0
- 20. CEE320 Winter2006 Stopping Sight Distance (SSD) • Worst-case conditions – Poor driver skills – Low braking efficiency – Wet pavement • Perception-reaction time = 2.5 seconds • Equation rtV G g a g V SSD 1 2 1 2 + ± =
- 21. CEE320 Winter2006 Stopping Sight Distance (SSD) from ASSHTO A Policy on Geometric Design of Highways and Streets, 2001 Note: this table assumes level grade (G = 0)
- 22. CEE320 Winter2006 SSD – Quick and Dirty ( ) ( ) ( ) ( ) a VV V V Ggag VV d 22 2 22 1 2 2 2 1 075.1 2.11 075.1 2.11 1 2 47.1 02.322.112.322 047.1 2 ==××= +× −× = ± − = 1. Acceleration due to gravity, g = 32.2 ft/sec2 2. There are 1.47 ft/sec per mph 3. Assume G = 0 (flat grade) ppp VttVd 47.147.1 1 =××= V = V1 in mph a = deceleration, 11.2 ft/s2 in US customary units tp = Conservative perception / reaction time = 2.5 seconds ps Vt a V d 47.1075.1 2 +=
- 24. CEE320 Winter2006 Primary References • Mannering, F.L.; Kilareski, W.P. and Washburn, S.S. (2005). Principles of Highway Engineering and Traffic Analysis, Third Edition). Chapter 2 • American Association of State Highway and Transportation Officals (AASHTO). (2001). A Policy on Geometric Design of Highways and Streets, Fourth Edition. Washington, D.C.
Editor's Notes
- Power is in ft-lb/sec
- Rolling resistance = 2 components Hysteresis = energy loss due to deformation of the tire Adhesion = bonding between tire and roadway
- Low profile tires reduce r and increase tractive effort
- Torque and HP always cross at 5252 RPM. Why? Look at the equation for HP
- For 4WD Fmax = μW (if your 4WD distributes power to ensure wheels don’t slip, which is common)
- For a front wheel drive car, sum moments about the rear tire contact point: -Rah – Wsinθh + Wcosθlr + mah - WfL = 0 cosθ = about 1 for small angles encountered -Rah – Wsinθh + Wlr + mah - WfL = 0 WfL = -Rah – Wsinθh + Wlr + mah WfL = + Wlr – Wsinθh – Rah + mah Wf = (lr/L)W + (h/L)(-Wsinθ – Ra + ma) But… Wsinθ = Rg Substituting: Wf = (lr/L)W + (h/L)(-Rg – Ra + ma) We know that… F = ma + Ra + Rrl + Rg Therefore, -F + Rrl = -ma – Ra– Rg Wf = (lr/L)W + (h/L)(-F + Rrl) Now, Fmax = μWf and Rrl = frlW Substituting: Fmax = μ((lr/L)W + (h/L)(-Fmax + frlW)) Simplifying: Fmax + (μh/L)Fmax = μ((lr/L)W + (h/L)(frlW)) Fmax(1 + μh/L) =( μW/L)((lr + hfrl)
- Tire size P = passenger car 1st number = tire section width (sidewall to sidewall) in mm 2nd number = aspect ratio (sidewall height to width) in tenths (e.g. 60 = 0.60) 3rd number = wheel diameter
- Practical comes from V22 = V12 + 2ad (basic physics equation or rectilinear motion) a = 11.2 ft/sec2 is the assumption This is conservative and used by AASHTO Is equal to 0.35 g’s of deceleration (11.2/32.2) Is equal to braking efficiency x coefficient of road adhesion γb = 1.04 usually