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connecting ship and sonar is wrapped around the drum, whose maximum angular
velocity
ω
is calculated according to this formula described by [ 14 ]:
v
r b ¼
2
33 1
s
x ¼
15 ¼
13
:
ð 5 Þ
:
0
x—
radius of the cable drum, m. The rpm
of the drum is calculated according to this formula:
the cable drum angular velocity, 1/s, r b
30 x
p
n
¼
¼
127
:
36 aps
= min
ð 6 Þ
The moment of the inertia of the load:
¼ 90
¼ 2 : 02 kgm 2
J m ¼ m v 2
2 2
13
max
x
:
ð 7 Þ
2
:
33 2
The coef
cient of reduction for cable drum is described ( 8 ):
n mo n
i
¼
20
ð 8 Þ
i
the reduction coef
cient, n mot
the cable drum motor rpm. The torque that is
needed to lift the weight ( 9 ):
M s ¼
mgr
¼
90
9
:
81
0
:
15
¼
132
:
44Nm
ð 9 Þ
M s
the system torque, Nm. The angular velocity of the engine rotor:
n mot p
30 ¼
2400 p
30 ¼
2 1
s
x mot ¼
251
:
ð 10 Þ
x mot
angular velocity of cable drum motor, 1/s. The engine reach the maximum
speed during the time period D t
5 s. We assumed that the moment of the engine
inertia is equal to the reduced moment of the inertia of the load is described as:
¼
0
:
J i 2 ¼ 0 : 01 kgm 2
J ¼ J mot þ
ð 11 Þ
the motor inertia moment, kgm 2 , The general moment of inertia of the motor
rotor is ( 12 ):
J mot
J i 2 ¼
01 kgm 2
J
¼
J mot þ
0
:
ð 12 Þ
Resistance to the axis of the engine torque presented as ( 12 ):
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