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Vehicle dynamics
- 1. CEE320 Spring2008 Vehicle Dynamics CEE 320 Anne Goodchild
- 2. CEE320 Spring2008 Outline 1. Resistance a. Aerodynamic b. Rolling c. Grade 2. Tractive Effort 3. Acceleration 4. Braking Force 5. Stopping Sight Distance (SSD)
- 3. CEE320 Spring2008 Main Concepts • Resistance • Tractive effort • Vehicle acceleration • Braking • Stopping distance grla RRRmaF +++=
- 4. CEE320 Spring2008 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 Spring2008 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 Spring2008 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 Spring2008 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 Spring2008 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 Spring2008 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 Spring2008 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 Spring2008 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 Spring2008 Vehicle Acceleration • Governing Equation • Mass Factor (accounts for inertia of vehicle’s rotating parts) maRF mγ=− ∑ 2 00025.004.1 εγ +=m
- 16. CEE320 Spring2008 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
- 18. CEE320 Spring2008 Braking Force • Front axle • Rear axle ( )[ ] L fhlW F rlr bf ++ = µµ max ( )[ ] L fhlW F rlf br +− = µµ max
- 19. CEE320 Spring2008 Braking Force • Ratio • Efficiency ( ) ( ) rear front fhl fhl BFR rlf rlr = +− ++ = µ µ µ η maxg b =
- 20. CEE320 Spring2008 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
- 21. CEE320 Spring2008 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 + ± =
- 22. CEE320 Spring2008 Stopping Sight Distance (SSD) from ASSHTO A Policy on Geometric Design of Highways and Streets, 2004 Note: this table assumes level grade (G = 0)
- 23. CEE320 Spring2008 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 +=
- 25. CEE320 Spring2008 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 Torque determines acceleration – a car will accelerate at its maximum when torque to the drive wheels is maximum (this is lower than engine output torque due to losses within the system). Max acceleration at max torque. However, as you accelerate you need to up-shift and change the gearing ratio. When you shift, you change the RPM to the drive axle(s) to a lower value. This, in turn, lowers torque. Therefore, as you accelerate, the longer you can stay in a higher gear the less loss in torque overall. One way to express this is through horsepower because horsepower takes into account engine speed. A higher horsepower car can stay in a lower gear longer and thus suffer less torque loss. It is better to make torque at high RPM than low RPM because you can take advantage of gearing. In theory, different gear ratios - most commonly four or five in cars' gearboxes - should mask different torque characteristics by altering engine speed to suit but the reality is that engines which produce high torque figures at low revolutions respond much more readily in give and take driving.The practical advantages come in the form of reduced gear changing, lower engine revs and wear and, invariably, lower fuel consumption in all conditions other than constant speed driving.
- 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