Page 154 - Mechatronics with Experiments

140 MECHATRONICS
is basically a precision threaded screw and nut. In a screw and nut pair, typically we turn
the nut and it advances linearly on the stationary screw. In the lead-screw, used as a motion
conversion mechanism, the nut is not allowed to make rotation around the screw, but is
supported by linear (translational) bearings to move. The screw is rotated (i.e., by an electric
motor). Since the screw is not allowed to travel, but only rotate, the nut moves translation-
ally along the screw. The tool is connected to the nut. Hence, the rotary motion of the motor
connected to the screw is converted to the translational motion of the nut and the tool.
Ball-screw design uses precision ground spherical balls in the groove between the
screw and nut threads to reduce backlash and friction in the motion transmission mechanism.
Any lead-screw has a finite backlash typically in the order of micrometer range. By using
preloaded springs, a set of spherical bearing-type balls are used to reduce the backlash.
Hence, they are called ball-screws. Most XYZ table-type positioning devices use ball-
screws instead of lead-screws. However, the load carrying capacity of ball-screws is less
than that of the lead-screws since the contact between the moving parts (lead and nut) is
provided by point contacts of balls. On the other hand, a ball-screw has less friction than
lead-screw.
The kinematic motion conversion factor, or effective gear ratio, of a lead-screw is
characterized by its pitch, p rev∕in or rev∕mm. Therefore, for a lead-screw having a pitch
of p, which is the inverse of the distance traveled for one turn of the thread called lead,
l = 1∕p in∕rev or rev∕mm, the rotational displacement (Δ in units of rad) at the lead shaft
and the translational displacement (Δx) of the nut is related by
Δ = 2 ⋅ p ⋅ Δx (3.49)
1
Δx = ⋅ Δ (3.50)
2 ⋅ p
The effective gear ratio may be stated as
N = 2 ⋅ p (3.51)
ls
Let us determine the inertia and torque seen by the input end of the lead-screw due to a load
mass and load force on the nut. We follow the same method as before and use energy–work
relations. The kinetic energy of the mass m at a certain speed ̇ x is
l
1 2
KE = m ⋅ ̇ x (3.52)
l
2
Noting the above motion conversion relationship,
1
̇ x = ⋅ ̇ (3.53)
2 ⋅ p
Then
1 1
KE = m ⋅ ⋅ ̇ 2 (3.54)
l
2 (2 ⋅ p) 2
1
= J eff ⋅ ̇ 2 (3.55)
2
Then, the effective rotary inertia seen at the input shaft (J ) due to a translational mass on
eff
the nut (m )is
l
1
J = ⋅ m (3.56)
eff 2 l
(2 ⋅ p)
1
= ⋅ m l (3.57)
N 2
ls