Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
vuoi
o PayPal
tutte le volte che vuoi
Cars
Tires
Functions
- Carry vehicle load
- Provide a certain stiffness; it is a spring that filters out the load irregularities (hyperbolic)
- Exchange forces with ground
The tire Ft-z characteristic is non-linear
So it can be linearized and modelled as a spring Ft = k t ⇒ S ∝ z ≤ zat
The compliance gives the tire's deformability; it depends on the vehicle type
- Do force vectors given Ft have S ⇒ k ↓ ⇒ large deformation
Longitudinal force
Fx = μx Fz
Braking k = (V - RW) / V
Traction k = (RW - V) / RW
k = longitudinal slip
μ = μ (environmental conditions) "μ = μ (speed)"
If k < klimit; inside the friction cone; in the static case lower than the friction limit, all the wires merge NB
Dry asphalt (μ > 1 ⇒ Fx > Ft)
Wet asphalt
Wet, snow, ice
klim ≈ 1
L = 2πR
V = ωR
KT < L V < ωR
KT = ωR - V/ωR
KB > L V > ωR
KB = V - ωR/V
ice melts under vehicle load
- water film, corner the tire
- broken water film
- tire and ground contact, the forces
αp = built in lateral slip
@ α = ∅ => fu = fp
- effect of pseudo side slip angle
- to go straight α = αp
- ply-steer + conicity
- effect of pseudo camber angle
to go straight you need fu = ∅ , so this is the working point of the vehicle
so ply-steer
no conicity
yet ply-steer
no conicity
balanced
fu ≠ ∅
no ply-steer
yes conicity
proper turning
not balanced
apply a torque
Radial Stiffness of the Tire
fw - pibc + 2S cosϕ ≈ eq vertical direction
tgϕ = S sinϕ / S cosϕ = p{(af - W)/2} = (af - W)/2 / (bc - h)/2
=> S cosϕ = p(bc - h)/2
=> fw - pibc + 2p(bc - h)/2 = fw - pib = Ø
=> fw = pib - in a tire the vertical force is balanced by inner pressure and tire geometry (bc)
Lateral Stiffness of the Tire
Sγs + fd = Sγd eq horizontal direction
Sγs = S cos(90 - ϕ + β) ≈ sinϕ - cosϕ·β (ϕ, β small angles)
Sγd = S cos(90 - ϕ + β) ≈ sinϕ + cosϕ·β
tgβ = d / (af - W) ≈ β
=> fd = 2S cosϕ·β = p{(bc - h)}·d / (af - W) => the lateral force depends on inner pressure and tire geometry
if fw = Ø => W = h = Ø => fd / d = p·bc / af = tire lateral stiffness
so if the α = bf / bc < 1 => fd / d ↑ rageometric (α ≈ %)
for this reason nowadays α≈1 and α ≈ 15% and torus like shapes are not used because they have α=1
Rigid wheel on plastic soil
The pressure acting under the tire due to soil is p = k zn
where n is a function of soil properties and k
assuming the soil compressed only vertically the work per unit surface
L0 = ∫0z0 ρ dz = k z0n+1 / n+1
rolling resistance R · e = L b e → R = b k z0n+1 / n+1 mechanical work diminished by rolling resistance → actually it doesn’t depend on the vertical force
@ This point I need P to find the rolling resistance
P = ∫αφ cosθ · dN, R = ∫αφ γ L θ dN
where cosθ · dN = p b dx is the elementary force in the vertical direction
→ P = ∫αφ p · b dx = b · k √D z0z0 (3-n) / 3
if sand in φ → R = b k √D z0 P = b k √D z0
→ R = b · k ∣ P2 / b2 k2 D ∣ = P2 / bkD
so from a general role, where L = characteristic length
- P = k1 L3
- D = k2 L
- b = k3 L
Rsp = R/P α L/k so to work avoiding a lower rolling resistance it is better to have lower vehicles, high b and D
on the other side the max maximum tractive force is:
T = c · A + P · tg φ
where c φ = f(soil), A = b so it is better to have b↓ = high contact area and the term P tgφ is the frictional force
Discomfort felt by human
Specific level
Stochastic systems
Black box system
- Input: Sξ
- Output:
- Sxz
- SFz
- Sxz-x1
H1 = H1xz → Sxz = |H1|2 Sξ
H2 = HFz → SFz = |H2|2 Sξ
H3 = Hxz-x1 → Sxz-x1 = |H3|2 Sξ
H1 = P.F of sprung mass acceleration due to road irregularities ξ=1
H1 = -ω2Xz
H2 = P.F of tire force Ft due to road irregularities ξ=1
H2 = K1(1-X1)
H3 = P.F of relative displacement due to road irregularities ξ=1
H3 = Xz - X1
vehicle running in a curve in ss @ V = ax -> fx = ∅
[HP] δ remains same.
- torque neglected
f frame non dimensione e tře
characteristics it is possible to
define the vehicle's behavior
remember δ = δ0 + ΔA - Δβ
we can plot the handling
diagram from non dimensione the characteristics,
giving a simple representation of δ needed to
perform different maneuvers
=> we design the commerce vehicles to have underveering,
cos (∂δ/∂β) |v1/2 > ∅, that is because the drive is
more intuitive, turning more the steering wheel
However, more characteristics can be complex and highly
non linear, giving rise to more branches in the
handling diagram.
- for the linearized
models of the vehicle
the vehicle stability @ a certain v2/r value is
evaluated with the Routh-Hurwitz criterion:
given ΦF = d(fyF/fzf)/dα
and Φp = d(fyRfzf)/dα => slope coeffs
to have a stable system it is needed:
Φf : Φp > ∅
Φf : Φp
| ∂δ |v->ax => there is the minimum condition
∂e/lr
can occur in overreering branch
McPherson suspension
- cheap € → one arm less
- less space needed
- independent wheels
- low camber recovery
- slider friction → risk remains
- lateral slip
Longitudinal arm suspension
- simple construction
- cheap €
- no lateral slip
- low space required
→ θwheel = θtrans
Rigid axle body connection
Payload led | Watt linkage
Works well for small displacements, for big ones the body moves laterally.
Coriolis effect
d2rs/dθ2rdθ
momentum impulse
dP = Mot = Rop
momentum
dθ = -dt; Γ = TJw
Rotational given by either up/und
→ D M rw 1/2 = TJw
(C)
The solid axle cannot rotate in the horizontal plane.
Accedi a tutti gli appunti
Tutor AI: studia meglio e in meno tempo
