Page 219 - Mechatronics with Experiments
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MECHANISMS FOR MOTION TRANSMISSION 205
is assumed to be 100%. Define the two steady-state curves of the torque converter
(( ) ) 2
w
T (w eng , w turb ) = 1000 − 4000 ⋅ turb − 0.5 (3.345)
p
w
eng
( )
w
N (w eng , w turb ) = 2.5 ⋅ 1 − turb (3.346)
r
w
eng
where w is the rated speed where torque converter performance is given, and N is
p
r
the gear ratio of the planetary gear set at the selected gear = {−1, 0, 1, 2, 3, 4, 5}, N =
p
{−2.5, 0, 2.5, 1.5, 1.0, 0.75, 0.5}.
3. There are two identical differentials: one for the front wheel pair, and one for the rear wheel
pair. Let us assume they are all in locked condition. The final drive gear ratio does not exist.
In other words the output shafts of the differentials drive the wheels directly.
1
J ⋅ ̇ w wheel (t) = T (t) − T brake (t) − ⋅ F traction (t) (3.347)
lp
out
R wheel
4. Friction between the wheel tires and ground is based on a constant friction coefficient and
assumed to be the same for all.
m vehicle ⋅ ̈ x(t) = F traction (t) − F load (t) (3.348)
where
1
F traction = ⋅ ⋅ T (t) (3.349)
r
out
R wheel
F load = F friction + F drag + F gravity (3.350)
The whole vehicle weight, W = m ⋅ g, is equally distributed among the four tires, where
v
v
2
m is the total mass of the vehicle, g = 9.81m∕s , is the traction coefficient. Assume identical
v
t
traction conditions at each tire. The student is asked to specify reasonably realistic parameters and
performance for the components as follows: Given the performance curves and parameters,
T ( , w ) engine map (3.351)
eng throttle eng
w = 2400 rpm (3.352)
r
N (−1, 0, 1, 2, 3, 4, 5) = { −2.5, 0, 2.5, 1.5, 1.0, 0.75, 0.5 } (3.353)
p
J = 1.0kg m 2 (3.354)
lp
R wheel = 0.5 m (3.355)
m vehicle = 1 000 kg (3.356)
= 0.0 for case 1 (3.357)
r
= 0.9 for case 2 (3.358)
F (t) = 0.0 (3.359)
friction
F (t) = 0.0 (3.360)
drag
◦
F gravity (t) = m vehicle ⋅ 9.81 ⋅ sin(15 ) (3.361)
Simulate a condition as follows. The vehicle speed is zero initially at time t = 0.0 s. The engine
runs at a constant speed and at a constant throttle, 100%. The initial direction of motion and steering
◦
is a straight line motion. The slope of the ground is 15 and the vehicle is climbing. The gear is shifted
from neutral to first gear at time t = 1.0 s and the throttle is at 100%. Until that time, the engine
is simply running at a steady-state speed of 2200 rpm. The friction coefficient at each tire–ground
contact is constant and the same for all tires, and the vehicle weight is equally distributed in all tires.
Note: This model can be made more accurate by including dynamics for the engine response
and its controls, dynamics of the clutch and its transient engagement/disengagement, inertia-stiffness
and gear-change transient dynamics of the transmission, inertia of the axles and differential, and a
more accurate model of friction at each tire and ground contact. Furthermore, the steering motion
can be included along with the differential dynamics in unlocked condition.