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JWST499-c07
JWST499-Cetinkunt
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 579
the line pressure which is determined by the load condition and motor displacement
as shown below.
The motor determines how this power is converted to mechanical power to meet the

load demands. In other words, depending on the displacement of the motor, this power
is converted to mechanical power in various combinations of torque and speed,
Power (t) = (p (t) − p (t)) ⋅ D (t) ⋅ w (t) (7.717)
m
r
m
m
m
where p (t) is the pressure on the high side of the motor port, p (t) is the pressure on
m
r
low side (return line), D (t) is motor displacement, w (t) motor output shaft angular
m
m
speed.
For sizing calculations, we can use rated speed of the engine.
Power (t) = Power (t) = Power load (t) (7.718)
p
m
Pump displacement for each one of the two identical circuits is determined by the
load power requirement,
Power (t) = (F l,max ∕2) ⋅ V max (7.719)
p
= p relief ⋅ D (t) ⋅ w eng (t) (7.720)
p
Sizing should be done for worst case load conditions conditions,
(F l,max ∕2) ⋅ V max
D (t) ≥ (7.721)
p
p ⋅ w (t)
relief eng
Let us assume that pressure at the pump output and motor input is same, neglecting the
pressure drop (pressure loss) in the hydraulic line between pump and motor.
p (t) = p (t) (7.722)
p m
In reality, the return line pressure is equal to the charge pressure, which is about 2 MPa in
a typical hydrostatic drive,
p charge (t) = p (t) = 2 MPa (7.723)
r
However, we will neglect the return line pressure

p (t) = 0 (7.724)
r
since the pressure on the return line is much lower than the pressure on the powered line.
For instance, p (t) is around 40–50 MPa range.
p
In motor sizing and control, the main consideration is the load force. The load force
is not under our control and is determined by operating conditions such as the ground
traction conditions and the load the vehicle faces. The objective in motor control is to scale
the available hydraulic power to mechanical power conversion so that the output torque is
equal or greater than the load torque.
Power (t) = p (t) ⋅ D (t) ⋅ w (t) (7.725)
m
m
m
p
= w (t) ⋅ T (t) (7.726)
m
m
= w (t) ⋅ N ⋅ R ⋅ (F (t)∕2) (7.727)
g
t
m
l
where R is the radius of the track output sprocket connected to the motor driven final gear
t
reducer, N is the gear ratio of the final gear reducer between the motor and track sprocket
g
drive, F is the load force the two tracks must overcome. It is clear that
l
T (t) = p (t) ⋅ D (t) = N ⋅ R ⋅ (F (t)∕2) (7.728)
l
p
m
t
m
g