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# Braking performance 4

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It is obvious that vehicle weight has a linear relationship
with the energy to be dissipated (stored) and the change
in velocity required has a exponential relationship.
• Deceleration times and stopping distances vary
somewhat for all vehicles on a given road surface.
• It should then be obvious that sizing the brake system
components has critical importance with respect to the
potential vehicle velocity and the mass of the vehicle.
• Note that heavy trucks generally have greater stopping
distances as compared to typical passenger cars.

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### Braking performance 4

1. 1. Braking Performance 3/27/2012 ١ME 467 Vehicle Dynamics
2. 2. Stopping Distance • It is the distance between the moment when a hazard is recognized and the time when the vehicle comes to a complete stop. • It is the sum of the distance traveled during the reaction time at given speed and active braking time. 3/27/2012 2 ME 467 Vehicle Dynamics
3. 3. Reaction Time • The reaction time is the period which elapses between the recognition of the object, the decision to brake, and the time it takes for the foot to hit the brake pedal. 3/27/2012 3 ME 467 Vehicle Dynamics
4. 4. Reaction Time • The reaction time is not a fixed value: it ranges from 0.3 to 1.7 s, dependingdepending upon the driver and on external factors. 3/27/2012 4 ME 467 Vehicle Dynamics
5. 5. Stopping Sight Distance (SSD) • Worst-case conditions – Poor driver skills – Low braking efficiency – Wet pavement • Perception-reaction time = 2.5 seconds • Equation: 3/27/2012 5 ME 467 Vehicle Dynamics rtV a V SSD 1 2 1 2 +=
6. 6. Speed and Stopping Distance • Look carefully at stopping distance for each vehicle speed !!! 3/27/2012 6 ME 467 Vehicle Dynamics
7. 7. Vehicle Braking • It is obvious that vehicle weight has a linear relationship with the energy to be dissipated (stored) and the change in velocity required has a exponential relationship. • Deceleration times and stopping distances vary somewhat for all vehicles on a given road surface.somewhat for all vehicles on a given road surface. • It should then be obvious that sizing the brake system components has critical importance with respect to the potential vehicle velocity and the mass of the vehicle. • Note that heavy trucks generally have greater stopping distances as compared to typical passenger cars. 3/27/2012 7 ME 467 Vehicle Dynamics
8. 8. Braking and Deceleration 3/27/2012 8 ME 467 Vehicle Dynamics
9. 9. Retarding Forces • The retarding forces on a vehicle are generally of the type shown in the following equation, where q is uphill grade: – Front and rear brake forces – Aerodynamics drag– Aerodynamics drag – Rolling resistance – Climbing resistance and downgrade force 3/27/2012 9 ME 467 Vehicle Dynamics sinθWFFFFD g W aM RollingcAerodynamirBrake,fBrake,xx -----=÷÷ ø ö çç è æ -= FBf , FBr include: 1) brake torque 2) bearing friction 3) drive line drag.
10. 10. Rolling Resistance • It is the product of deformation processes which occur at the contact patch between tire and road surface. • Frr = f Mg Depends on: 3/27/2012 10 ME 467 Vehicle Dynamics Depends on: – bearing friction, – deformation of tires, – road surface. Rxf + Rxr = fr (Wf + Wr) = frW fr : rolling resistance coefficient. Rolling resistance: depends on the distribution of loads on axles. : it equivalent to about 0.01 g (0.3 ft /sec2).
11. 11. Rolling Resistance • It is directly proportional to the level of deformation and inversely proportional to the tire radius. • It increase with greater loads, higher speeds, and lower pressure. 3/27/2012 11 ME 467 Vehicle Dynamics
12. 12. Aerodynamic Drag • It is empirical value and depends on vehicle shape and speed. Fa = CdV2 Depends upon: 3/27/2012 12 ME 467 Vehicle Dynamics Depends upon: – frontal area, – relative vehicle speed to air speed DA = C . A . V2 DA: can be neglect at low speed, DA » 0.03 g at high speed (1 ft /sec2). A: frontal area V: relative speed C: aerodynamic drag coefficient.
13. 13. Drive line drag (inertia) • Inertia (adds to the effective mass of vehicle). • Drag arises from internal friction in gears and bearings and from engine braking. • Engine braking arising from internal friction and air pumping losses.and air pumping losses. • Engine braking multiplies by the gear ratio selected. • If vehicle decelerates faster than drive line, slowing down from drive line drag will have lower contribution in braking effort. 3/27/2012 ١٣ME 467 Vehicle Dynamics
14. 14. Grade Rg = W sinθ Grade increase the braking effort of vehicle in uphill road and decrease it in down hill road.road and decrease it in down hill road. Rg = W sinθ = W θ , θ in (radians) ≈ grade for small angles. Grade of 4% (0.04) will be equivalent of deceleration (! 0.04g) →1.3 ft /sec2 3/27/2012 ١٤ME 467 Vehicle Dynamics
15. 15. Active Braking and Kinetic Energy • Brake systems convert the kinetic energy of the vehicle at velocity to some other form of energy. • Current design brake systems in conventional automobiles convert the kinetic energy of the vehicle into thermal energy (heat) which is subsequently dissipated into the atmosphere.into the atmosphere. • Hybrid vehicles attempt to recover the kinetic energy as stored battery or capacitor electrical energy. • Kinetic energy (KE) is mathematically represented as ½ the product of the vehicle mass and the velocity of the vehicle squared. 3/27/2012 15 ME 467 Vehicle Dynamics 2 2 1 vmKE =
16. 16. Energy Conversion during Braking Conversions that might be possible would include: Kinetic Energy to Thermal Energy • Typically a friction system generating heat between a rotating wheel part and a vehicle fixed friction component which can be modulated by the driver.be modulated by the driver. Kinetic Energy to Stored Electrical Energy • Regenerative system where electric motors can be used as electric generators under braking and subsequently turn some portion of the kinetic energy into stored electrical energy. • The stored electrical energy can be later used to supplement performance and/or range. 3/27/2012 16 ME 467 Vehicle Dynamics
17. 17. Energy Conversion during Braking Kinetic Energy to Stored Mechanical Energy • Regenerative system where mechanical systems can be engaged under braking to impart kinetic energy to a flywheel subsequently turning some portion of the kinetic energy into stored mechanical energy. • The stored energy can be released as required to augment acceleration and/or extend the range of the vehicle.acceleration and/or extend the range of the vehicle. Kinetic Energy to Stored Hydraulic Pressure Energy • Regenerative system where hydraulic motors can be used as hydraulic pumps under braking and subsequently turn some portion of the kinetic energy into stored hydraulic energy. • Storage of the fluid under pressure can later be released to the pump, driving it as a hydraulic motor which can be used to augment acceleration and/or extend the range of the vehicle. 3/27/2012 17 ME 467 Vehicle Dynamics
18. 18. Constant Deceleration • Assuming the forces on the vehicle are constant: • This equation can be integrated because F is constant dt dV M F =D xtotal x = • This equation can be integrated because Fx is constant from initial velocity Vo to final velocity Vf: 3/27/2012 18 ME 467 Vehicle Dynamics òò = sf 0 t 0 V V xtotal dt M F dV
19. 19. Constant Deceleration • The fundamental relationship equations that govern braking are: òò =-=>= t s xtotal f0 V xtotal t F VVdt F dV f Where ts is the time for the velocity change 3/27/2012 19 ME 467 Vehicle Dynamics òò =-=>= 0 sf0 V t M VVdt M dV o
20. 20. Stopping Distance (SD) • can substitute for dt in equation (1) and integrate to obtain the relationship between velocity and distance: x M F 2 VV xtotal, 2 f 2 0 = - Where x = distance traveled during deceleration • In the case where Vf = 0, then x = stopping distance 3/27/2012 20 ME 467 Vehicle Dynamics M2 x 2 o D2 V x =
21. 21. Braking Energy • The time to stop is then: • Thus, all things being equal, the time to stop is x 0 xtotal 0 s D V M F V =t = • Thus, all things being equal, the time to stop is proportional to the velocity, where as the distance to stop is proportional to the velocity squared. • The energy absorbed by the brake system is the kinetic energy of motion: 3/27/2012 21 ME 467 Vehicle Dynamics )V(V 2 M =Energy 2 f 2 0 -
22. 22. Deceleration with Wind Resistance • Aerodynamic drag effects: 3/27/2012 22 ME 467 Vehicle Dynamics
23. 23. Types of braking systems • Desk and Drum types 3/27/2012 23 ME 467 Vehicle Dynamics
24. 24. Physics of braking •d 3/27/2012 24 ME 467 Vehicle Dynamics
25. 25. 3/27/2012 25 ME 467 Vehicle Dynamics
26. 26. Brakes Drum brakes Advantage - High brake factor (low actuation effort). - Easy to integrate with park brake. Disadvantage - not consistent in torque performance.- not consistent in torque performance. Disc brakes Advantage - more consistent torque Disadvantage - low brake factor. - high actuation effort. 3/27/2012 ٢٦ME 467 Vehicle Dynamics
27. 27. Brake factor Is a mechanical advantage in drum brakes to minimize required actuation effort? Moment equation about pivot of shoe A åMp = e Pa + n m NA – m NA = 0 Pa : application of an actuation force. NA : normal force e, n, m: distances as shown in (fig). A e, n, m: distances as shown in (fig). NA = Pa · e , NB = Pa · e m - m n m + m n Friction forces on brake shoes: FA = m NA , FB = m NB , FA = Pa · m · e , FB = Pa · m · e m - m n m + m n FA = m e , FB = m e Pa m - m n Pa m + m n ٢٧
28. 28. FA = m e , FB = m e Pa m - m n Pa m + m n TL = FA · r TT = FB · r TL : brake torque of leading shoe, TT : brake torque of trailing shoe. Braking torque developed by leading shoe greater than brake torque developed by trailing shoe. The moment produced by the friction force on the leading shoe acts to rotateThe moment produced by the friction force on the leading shoe acts to rotate against the drum and increase the friction force developed (self servo). For this reason two leading shoe type of brake usually using in front brakes. In leading shoe If µ gets too large ≈ 1 brake factor goes to infinity ( m ≈ n) and brake will lock. Consequences: 1- sensitivity to the lining coefficient of friction 2- high noise ٢٨ME 467 Vehicle Dynamics
29. 29. Leading show A :- Moment produced by the friction force on the show acts to rotate it against the drum and increase the friction force developed (self-servo which characterized by brake factor). Trailing show B :- friction force acts to reduce the application force, lower brake factor and higher braking forces required to achieve desired braking torque. 3/27/2012 ٢٩ME 467 Vehicle Dynamics
30. 30. Braking torque during stopping processBraking torque during stopping process Tb = f ( Pa , velocity , temperature ) •Torque normally increase linearly with Pa •Torque increases as velocity increase. •Torque decreases as temperature increase. Disk brake shows less torque variation during stopping. So, in drum brakes it is difficult to maintain proper balance between front and rear braking effort during max braking. 3/27/2012 ٣٠ME 467 Vehicle Dynamics
31. 31. Braking force at the ground Fb = Tb – Iw aw r That if wheels still rotating during braking Iw : rotational inertia of wheels aw: rotational deceleration of wheels.aw: rotational deceleration of wheels. If wheels lock up: aw related to acceleration of vehicle aw = ax , r Iw: part of vehicle mass Fb = Tb r 3/27/2012 ٣١ME 467 Vehicle Dynamics
32. 32. road friction)-Traction (Tire Friction force can be increase to the limit of frictional coupling between the tire and road Mechanisms of friction: • Adhesion, it arises from intermolecular bonds between the rubber and road surface. Higher on dry road, reduced with wet road surface (wheel rotating). • Hysterises, Represents energy loss in the rubber as it deforms when sliding over road surface not affected by water, thus it has better wet traction. ٣٢
33. 33. - Both Mechanisms of friction depend on small amount of slip occurring in the contact patch. - As braking force develops, additional slip is observedadditional slip is observed as a result of deformation of the rubber elements. - Deformation increases from front to back and the force developed increases proportionality. 3/27/2012 ٣٣ME 467 Vehicle Dynamics
34. 34. Both mechanisms depends on two factors:- (brake force coefficient and slip - which are coexistent) 1) slip of tires slip = v – w . r , % v v : forward velocityv : forward velocity w: tire rotational speed w¯….slip- , w=0….slip=100% 2) brake force coefficient (µ)= Fx Fz Fx: braking force, Fz: axle load Brake coefficient µ increases with slip to about 10-20% slip (see fig.) µp – peak coefficient (it establishes the max brake force that can be obtained from particular tire and road friction). Max braking coefficient develops at µp . Then if Slip- … µ ¯ until µs . ms … at full wheel locking. in ABS ms return to mp in a repeated cycles. 3/27/2012 ٣٤ME 467 Vehicle Dynamics
35. 35. Friction in braking depends on: 1. tire design 2. road surface type 3. velocity (both peak and slide friction decrease with velocity increase). 4. Inflation pressure. Wet road: -P ® -mp ,ms Dry road: P less affect on mDry road: P less affect on m 5.vertical load: (Brake force coefficient=Fx /Fz ), increasing of vertical load reduce brake force coefficient levels, if load increase mP ,ms don’t increase proportionally. 3/27/2012 ٣٥ME 467 Vehicle Dynamics
36. 36. Brake proportioning: describes the relation between front and rear brake forces, determined by the pressure applied to each brake wheel cylinder and the gain of each.and the gain of each. Ideal design is to bring both axles up to lock up simultaneously 3/27/2012 ٣٦ME 467 Vehicle Dynamics
37. 37. Brake proportioning Dynamic load during braking W , W : static loadsWfs , Wrs : static loads Wd : dynamic load transfer. Maximum brake forces on an axle: mp- peak coefficient of friction ٣٧
38. 38. Fxm = f(m , Dx) (see the fig) Substituting Dx in previous eq. of Fxmf, Fxmr (shown again below) yields the equations in the square 3/27/2012 ٣٨ME 467 Vehicle Dynamics
39. 39. The intersection point can be determined by manipulating equations of Fxmf, Fxmr. 3/27/2012 ٣٩ME 467 Vehicle Dynamics
40. 40. Brake gain and Brake proportioning The gain of brakes on front and rear wheels is an important factor to determine brake proportioning. Brake force on each wheel Fb =Tb =G Pa r rr r G: brake gain (in–lb/psi) Pa: application pressure. 3/27/2012 ٤٠ME 467 Vehicle Dynamics
41. 41. Hydraulic proportioning valve Hydraulic proportioning valve: provide equal pressure to both front and rear brakes up to certain pressure level and then reduce pressure to rear brake. So, it adjusts the brake torque output on front and rear wheels in accordance to the peak traction forces possible or to loads. Example:- Hydraulic proportioning valve identification number:(500 / 0.3) that means:- For Pa < 500 psi Pf = Pr = Pa = application pressure. For Pa > 500 psi Pf = Pa , Pr = 500 + 0.3 (Pa – 500) If Pa = 700 psi Pf = Pa = 700 psi Pr = 500 +0.3(700 – 500) = 570 psi. 3/27/2012 ٤١ME 467 Vehicle Dynamics