braking brake components disk brakes (hsw) drum brakes (hsw) master cylinders (hsw) braking...
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Braking
Brake componentsDisk brakes (HSW)Drum brakes (HSW)Master cylinders (HSW)
Braking performance (Eng. Char., SEP)
Equilibrium forces Weight shift confounds analysis! Sequence of events during braking Brake pedal/cylinder forces Method of calculating
Braking Performance Measures
Deceleration rate, Dx
Stopping time, ts
Stopping distance, SD
Energy (braking)Power (braking)
FBD of vehicle during braking
)loads hitch includes(sin
0sin
WDFFRDg
W
Dg
WDWRFRFRF
aDletMaF
Axfxrhxx
xAxfbfxrbrhxx
xxxx
θ
c Wf
Fbf
Rxf
Wr
Fbr
Rxr
x
z
ha
DA
W/g Dx
Wcosθ
Wsinθ
W
Rhz
Rhxhh
h
L
dh
b
Axle Loads during Braking
sin...cos WLh
LAh
ADLhd
hzRLh
hxRxDLh
gW
WLc
fW h
sin...cos WLh
Lh
ADLLd
hzRLh
hxRxDLh
gW
WLb
rWAhh
xDLh
gW
WfW fs
xDLh
gW
WrsrW
On level road, with no hitch or drag loads…
To find the axle loads we need to know the deceleration … but we can’t calculate the deceleration without the braking force on the tires and the braking forces can’t be calculated until we know the axle loads! Huh? A different approach is required! ASSUME deceleration rate Dx
Stopping time, ts
xtxfxrxx FFFDgW
Ma
forces grade and drag hitch, rolling, small for
MFVV
t
ttt
VVttMF
dVdtMF
dtdV
MF
D
xt
fs
sf
ffxt
xt
xtx
0
0
00
time stopping
and0 let
)(
rearrange and force brake constant Assume
Stopping Distance
x
20
xt
20
0
0
20
2
2DMF
2SD distance stopping
and 0 and 0x ..stopping.. When
)(2
g..rearrangin and rule, chain the Using
VV
V
xxMFVV
dxMF
VdV
VdxdV
dtdx
dxdV
dtdV
MF
D
f
fxtf
xt
xtx
Stopping distance w/drag
b
b
Vb
SD
bxt
xt
FCVF
CM
CVFVdV
Mdx
CVFF
VCF
20
0
2
2
ln2
SD distance stopping
and , factor drag on depends When
0
Energy and power dissipated
stVM
fVM
20
20
20
2tEnergy
Power
)(2
Energy
See examples on pages 57-58
What really happens during braking?
Driver pushes on brake pedal (and brake lever) Brake lever pushes on push rod of master
cylinder Hydraulic fluid becomes pressurized Hydraulic line pressure pushes on caliper piston Caliper piston pushes on brake pads Pads clamp rotor/drum causing friction forces &
torque Brake torque causes braking force at contact
patch Weight is shifted to front axle Tires grab and slow vehicle, OR One or more tires skid (too much braking,
“brakes locked”), also slowing vehicle, not as much though.
Brake Pedal / Master Cylinder Forces
a b
pedalpedF
MCF
pedal andlever brake
MCF
cylinder master
pedMC
pedMC
pedMC
pivot
FPRF
ratio" pedal"PRb
alet
Fb
aF
aFbF
M
)(
0
:
psip
lbsFassume
Fp
sqinAandPRletA
FPR
A
Fp
ApF
F
h
ped
Ph
pedMCh
hMC
x
1200
100
)12(
5.06
)(
0
:
ApF hpiston
Calculating brake force, Fb & brake gain G
psi
lbinG
in
1
inlblb
in in
in
lb
lbin in ArG
in A
in r
2
2
2
2
calpad
2cal
pad
pad
2.19
2.19)3(4.0)8(22
3
4.0
8
32
surfaces) (2)( acalf PANF
tire) (one )2(2
)(2
2
t
a
t
acalpad
t
acalpadb
acalpadbt
fpadbt
fw
rGP
r
PAr
r
PArF
PArFr
FrFr
TT
fF
tr
padr
bF
)(roughlyis
tires2.19for
ab pF
in
Skidding= F(brake)>F(road friction) ME485/585 Vehicle Design S2009
xrsr
xfsf
DgWLhWW
DgWLhWW
)/)(/(
)/)(/(
ractxrfactxf WFWF )( and)(
ractxrbxr
factxfbxf
WFF
WFF
)()(
)()(
Recall actual axle loads during deceleration And available road friction forces: Will brake forces be greater than those the road can provide?
Rear wheel lockup
1. Front wheels turning and steering2. Little braking force on rear wheels
“unstable equilibrium”
“spin out”
Braking forces, coefficients, efficiency (handout)ME485/585 Vehicle Design S2009
W
FFD xrxf
x
r
PGF
r
PGF r
rbxrf
fbxf 2)( and2)(
for )(
for
1121
1
KPKPKKP
KPPP
PP
aar
aar
af
1. Given changeover pressure K1, and pressure proportion K2, determine front and rear
application pressure:
2. Calculate the estimated front and rear braking forces, assuming no skidding:
3. Calculate vehicle deceleration:
K1
Pf, Pr
Pa
Pr
Pf
Braking forces, coefficients, efficiency cont’dME485/585 Vehicle Design S2009
xrsr
xfsf
DgWLhWW
DgWLhWW
)/)(/(
)/)(/(
ractxrfactxf WFWF )( and)(
actxrbxr
actxfbxf
FF
FF
)()(
)()(
),max( rf
xb
r
xrr
f
xff
D
W
F
W
F
4. Calculate actual axle loads
5. Assume friction static road and calculate actual front and rear braking forces
6. Test for skidding, i.e. are estimated brake forces greater than those the road can provide?
7. If no skidding, calculate braking coefficients rf and and braking efficiency b
(see web page)
ME 485/585 Vehicle Design Braking Efficiency RJE 2/8/2009
Variable symbol value unitswheelbase L 108.5 inCG height h 20.5 intire radius rt 12.11 in
front axle wt Wfs 2210 lbsrear axle wt Wrs 1864 lbs
total wt W 4074 lbsBrake gain front Gf 20 in-lb/psiBrake gain rear Gr 14 in-lb/psivalve change P K1 290 psi
rate K2 30% n/askid number sn, μ 0.9
Decel.Pa Pf Pr Fxf Fxr Dx Wf Wr μWf μWr μf μr
pedal front rear front rear front rear front rear front rear Braking E(3-27) E(3-28)100 100 100 330 231 0.138 2316 1758 2084 1582 0.143 0.132 0.966 2444 1386200 200 200 661 462 0.276 2422 1652 2180 1487 0.273 0.280 0.985 2491 1338300 300 293 991 677 0.410 2525 1549 2273 1394 0.392 0.437 0.936 2535 1290400 400 323 1321 747 0.508 2601 1473 2341 1326 0.508 0.507 0.999 2550 1242500 500 353 1652 816 0.606 2676 1398 2409 1258 0.617 0.584 0.982 2564 1194600 600 383 1982 886 0.704 2752 1322 2477 1190 0.720 0.670 0.977 2578 1146700 700 413 2312 955 0.802 2827 1247 2545 1122 0.818 0.766 0.981 2592 1098800 800 443 2642 1024 0.900 2903 1171 2613 1054 slip 0.875 slip 2606 1050900 900 473 2973 1094 0.998 2978 1096 2680 986 slip slip slip 2621 10021000 1000 503 3303 1163 1.096 3054 1020 2748 918 slip slip slip 2635 954
Pressure Brake force Axle loads Braking CoefficientMax Brake force
),max(
)( and)(
)/)(/(
)/)(/(
2)( and2)(
for )(
for
1121
1
rf
xb
r
xrr
f
xff
ractxrfactxf
xrsr
xfsf
xrxfx
rrbxr
ffbxf
aar
aar
af
D
W
F
W
F
WFWF
DgWLhWW
DgWLhWWW
FFD
r
PGF
r
PGF
KPKPKKP
KPPP
PP
Braking forces change due to weight shift
0.000 0.200 0.400 0.600 0.800 1.000 1.2000
500
1000
1500
2000
2500
3000
Front Rear
Deceleration (g's)
Brak
ing
Forc
e (lb
s)
Conclusions
Non-skid deceleration affected by weight shift
If tires skid, use μkinetic/sliding
Skidding deceleration may be unstable