Delta Tau GEO PMAC Manual Do Utilizador
Geo PMAC Drive User Manual
28
System Wiring
In standard metric (SI) units, the kinetic energy of a linearly moving mass is:
2
mv
2
1
K
E
=
where:
E
E
K
is the kinetic energy in joules (J)
m is the mass in kilograms (kg)
v is the linear velocity of the mass in meters/second (m/s)
Here also, to get energy in Joules from English mechanical units, additional conversion factors are
required. To calculate the kinetic energy of a mass having a weight of W pounds, the following equation
can be used:
v is the linear velocity of the mass in meters/second (m/s)
Here also, to get energy in Joules from English mechanical units, additional conversion factors are
required. To calculate the kinetic energy of a mass having a weight of W pounds, the following equation
can be used:
2
Wv
0211
.
0
2
v
g
W
678
.
0
K
E
=
=
where:
E
E
K
is the kinetic energy in joules (J)
W is the weight of the moving mass in pounds (lb)
g is the acceleration of gravity (32.2 ft/sec
g is the acceleration of gravity (32.2 ft/sec
2
)
v is the linear velocity of the mass in feet per second (ft/sec)
Energy Lost in Transformation
Some energy will be lost in the transformation from mechanical kinetic energy to electrical energy. The
losses will be both mechanical, due to friction, and electrical, due to resistance. In most cases, these
losses comprise a small percentage of the transformed energy and can be ignored safely because this leads
to a conservative design. However, if the losses are significant and the system should not be over-
designed, calculate these losses.
In metric (SI) units, the mechanical energy lost due to Coulomb (dry) friction in a constant deceleration to
stop of a rotary system can be expressed as:
losses will be both mechanical, due to friction, and electrical, due to resistance. In most cases, these
losses comprise a small percentage of the transformed energy and can be ignored safely because this leads
to a conservative design. However, if the losses are significant and the system should not be over-
designed, calculate these losses.
In metric (SI) units, the mechanical energy lost due to Coulomb (dry) friction in a constant deceleration to
stop of a rotary system can be expressed as:
d
t
f
T
2
1
LM
E
ω
=
where:
E
E
LM
is the lost energy in joules (J)
T
f
is the resistive torque due to Coulomb friction in newton-meters (N-m)
ω is the starting angular velocity of the inertia in radians per second (1/s)
t
t
d
is the deceleration time in seconds (s)
If the frictional torque is expressed in the common English unit of pound-feet (lb-ft), the comparable
expression is:
expression is:
d
t
f
T
678
.
0
LM
E
ω
=
In metric (SI) units, the mechanical energy lost due to Coulomb (dry) friction in a constant deceleration to
stop of a linear system can be expressed as:
stop of a linear system can be expressed as:
d
vt
f
F
2
1
LM
E
=
where:
E
E
LM
is the lost energy in joules (J)
T
f
is the resistive force due to Coulomb friction in newtons (N)
v is the starting linear velocity in meters/second (m/s)
t
t
d
is the deceleration time in seconds (s)
If the frictional force is expressed in the English unit of pounds (lb) and the velocity in feet per second
(ft/sec), the comparable expression is:
(ft/sec), the comparable expression is:
d
vt
f
F
678
.
0
LM
E
=