Reluctance Sensors (VRS) applications are in pump RPM, plant floor machinery, electric
motor speed, wheel speed sensing, and engine transmission.
c)
Most Suitable Sensing System
For the rubber wheels of trolley running on a steel surface, there are certain sensors that
applicable. For detecting steel surface, the two magnetic technologies applicable are hall-
effect sensors and variable reluctance sensors (Corp 2019, 1). Based on the design parameters
and advantages of each sensor, it is possible to narrow down to the most suitable sensing
system.
A combination of different techniques can also be applied to come with the best
sensor. The best sensor for this task is Light Detection and Ranging system. The LIDAR
sensor measures the speed of trolley on the steel track accurately as well as the curvature and
distance. It has numerous advantages and its range is 0.5 to 100 mph (
Ahmadian, Craft, and
Stuart 2014, 1
). This sensor eliminates need for frequent calibration and mechanical failures.
Task 2
a)
Actuator Requirements
Based on the given parameters, the actuator can be designed as shown below.
Fig. 1: Design
For components to move, a force is required to overcome the force created by the spring
force and the components. The maximum forces created by the actuators are calculated
below. The maximum sinusoidal paths are calculated using the formulas below:
Frequency f = 1/T,
Location on Y-axis y = A sin (2πft),

Sensors and Actuators Design
Velocity v = 2πfA cos (2πft),
Acceleration a = - (2πf)
2
[A sin(2πft) = -4π
2
f
2
y (Webassign.net 2019, 1)
For f = 0.15 Hz and A = 0.5M , we have get maximums as;
Distance = 0.5 sin (90) = 0.5 m
Velocity = 2π × 0.15 × 0.5 cos 0= 0. 471 m/s
Acceleration = - (2π × 0.15)
2
0.5 sin 90 = -0.444 m
2
/s.
For f = 0.15 M and A = -0.5 Hz, we get:
Distance = -0.5 sin (90) = -0.5 m
Velocity = 2π × 0.15 ×- 0.5 cos0= -0. 471 m/s
Acceleration = - (2π × 0.15)
2
(-0.5) sin 90 = 0.444 m
2
/s
For f = 1 Hz and A = 0.05 M, the values obtained are;
Distance = 0.05 sin (90) = 0.05 m
Velocity = 2π × 1 × 0.05 cos0= 0. 314 m/s
Acceleration = - (2π × 1)
2
0.05 sin 90 = -1.973 m
2
/s
For f = 1 Hz and A = -0.05 M, the values obtained are;
Distance = -0.05 sin (90) = -0.05 m
Velocity = 2π × 1 × (-0.05) cos0= -0. 314 m/s
Acceleration = - (2π × 1)
2
(-0.05) sin 90 = 1.973 m
2
/s
From above results, the value of acceleration are a = 1.973
m
2
/s and distance is 0.5 m.
From Newton’s second law f = ma= -ky
Total force = ma + (-ky) = (500 ×1.973) + (2000 × 0.5) = 1986.5 N
Velocity = 0.471 m/s
b)
Outline Design and Parameters
Assumption made:
210 bar supplied and 25% excess flow allowed.

Sensors and Actuators Design
Area , A
=
1.25
Famx
0.67
Ps
=
1.25 □ 1986.5
0.67 □ 210□ 10S
= 0.00017 m
2
Maximum Flow From the valve = Q = AV
max
= 0.00017 × 0.471 = 0.00008 m
3
Peak hydraulic power = 210 × 10
5
× 0.00008 = 1680W =1.68 kW
c)
Block Diagram Model
The block diagram model with mass, actuator, and system is as shown and it relates to
actuator position and voltage:
Fig. 2 Block Diagram Model

Sensors and Actuators Design
From the model above, the input is current and voltage which are both electrical which goes
to the hydraulic power supply. The hydraulic instigates flow rate and constant amount of
supply pressure which are then controlled by the servo valve that produces flow rate and
pressure which are fed to the actuator. The actuator outputs force at a specific mass and some
velocity.

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- Fall '19
- Miles per hour, Actuators Design