Sizing Electric Motors for Mobile Robotics
Short Description
Download Sizing Electric Motors for Mobile Robotics...
Description
Sizing Electric Motors for Mobile Robotics
May 21, 2006
The Basics
May 21, 2006
Unit Conversions 2π
rad sec
1Watt
=
May 21, 2006
=
1
rev sec
1Volt
⋅
1Watt
Ampere
=
N m
=
1
1Volt
⋅
sec
Coulomb sec
Basics The FORCE applied by a wheel is always tangent to the wheel.
Force is measured in units of weight w eight (lb, oz, N) May 21, 2006
Basics The required TORQUE to move a mobile robot is the force times the radius of the wheel.
May 21, 2006
Torque is measured in units of weight w eight x length (lb·ft, oz ·in, N·m)
Procedure for Sizing DC Motors
May 21, 2006
Information Needed • Estimated Weight • Number of wheels and motors • Maximum incline • Desired maximum velocity at worst case • Push/Pull forces
May 21, 2006
Procedure • Step One: Determine total applied force at worst case
May 21, 2006
Friction • Static Friction – Used to determine traction failure
• Rolling Friction – Used to determine motor requirements
• Kinetic Friction
May 21, 2006
Rolling Friction F R
= µ ⋅ N R
Ro lling friction ∀ µR Is the coefficient of Rolling – Using the coefficient of Static friction ( µS) will typically be to high
• To determine µR: – Roll a wheel at a initial velocity, v elocity, v, and measure the time, t, in which it takes to v stop
µ R
May 21, 2006
=
t g ⋅
Rolling Friction • Some typical values for µR – Steel on steel: 0.001 – Rubber on pavement: 0.015
May 21, 2006
Other Forces • Gravity F I
= W ⋅ sin θ
• External θ
May 21, 2006
Total Force • Calculate worst case – Up hill with rolling friction F W ( µ R cos θ
= ⋅
⋅
+ sin θ )
– Up hill with rolling friction, pushing F W ( µ R cos θ sin θ ) – Level ground with rolling friction F µ R W
= ⋅ =
⋅
+
+ F
EX
⋅
– Level ground with rolling friction, pushing F May 21, 2006
= µ ⋅W + F R
EX
Other Cases • Tracks – Set µr =0 – Use a spring scale to determine the force required to pull the chassis in neutral and add that to the t he worst case force
• Gear Trains – Bulky gear trains may significantly affect the outcome – If this is a concern, it may be best to test in the same way as tracks May 21, 2006
Procedure • Step One: Determine total applied force at worst case • Step Two: Calculate power requirement
May 21, 2006
Power Requirement • Determine velocity, v, requirement under maximum load (worst case force) • Using the worst case force and velocity, calculate the power requirement P
= F ⋅ v
• This is the total power, divide by the number of motors if more than one motor is used RULE OF THUMB: 3 TIMES MARGIN May 21, 2006
Procedure • Step One: Determine total applied force at worst case • Step Two: Calculate power requirement • Step Three: Calculate torque and speed requirement
May 21, 2006
Speed/Torque Requirements • Using the velocity requirement, v, and the radius of the wheel, r
ω =
v
r
Speed requirement is in rad/sec
• Using the speed from above and the power per motor T May 21, 2006
=
P
ω
Procedure • Step One: Determine total applied force at worst case • Step Two: Calculate power requirement • Step Three: Calculate torque and speed requirement • Step Four: Find a motor that meets these requirements May 21, 2006
Spec Sheet
May 21, 2006
Spec Sheet
May 21, 2006
Procedure • Step One: Determine total applied force at worst case • Step Two: Calculate power requirement • Step Three: Calculate torque and speed requirement • Step Four: Find a motor that meets these requirements • Step Five: Plot motor characteristics May 21, 2006
Torque vs. Speed Curve T
= T − PK
T PK S NL
• Where T = Torque • TPK = Stall Torque • SNL = No Load Speed ∀ ω = Speed May 21, 2006
⋅ ω
Torque vs. Speed Curve Torque vs. Speed 7.00E-02
From this plot, maximum speed can be determined for a given load.
6.00E-02
5.00E-02
m 4.00E-02 N , e u q r o 3.00E-02 T
2.00E-02
1.00E-02
0.00E+00 0
1000
2000
3000
4000
Speed, rpm
May 21, 2006
5000
6000
7000
8000
Power T
= T − PK
T PK S NL
⋅ ω
ω = (T PK − T )
P P (ω )
P (T ) May 21, 2006
=− =−
= T ⋅ ω
T PK S NL S NL T PK
⋅ ω + T ⋅ ω 2
PK
⋅ T + S ⋅ T 2
NL
S NL T PK
Power Power vs. Speed 1.20E+01
1.00E+01
8.00E+00
s t t a w , 6.00E+00 r e w o P
P (ω )
4.00E+00
2.00E+00
=−
T PK S NL
⋅ ω + T ⋅ ω 2
PK
0.00E+00 0
1000
2000
3000
4000
Speed, rpm
May 21, 2006
5000
6000
7000
Power Power v s. Torque Torque 1.20E+01
1.00E+01
8.00E+00
s t t a w , 6.00E+00 r e w o P
P (T )
4.00E+00
=−
2.00E+00
S NL T PK
⋅ T + S ⋅ T 2
NL
0.00E+00 0
0.01
0.02
0.03
Torque, Nm
May 21, 2006
0.04
0.05
0.06
Power
Power vs. Speed 1.20E+01
1.00E+01
8.00E+00
Power vs. Torque
s t t a w , 6.00E+00 r e w o P
1.20E+01
4.00E+00
1.00E+01
2.00E+00
8.00E+00
0.00E+00
s t t a w , 6.00E+00 r e 6000 w o P
0
1000
2000
3000
4000
Speed, rpm
ω
1 =
2
5000
7000
4.00E+00
ω max
2.00E+00
0.00E+00 0
0.01
0.02
Peak power is obtained at half of maximum torque and speed May 21, 2006
0. 03
0.04
0.05
0.06
Torque, Nm
T
1 =
2
T max
Procedure • Step One: Determine total applied force at worst case • Step Two: Calculate power requirement • Step Three: Calculate torque and speed requirement • Step Four: Find a motor that meets these requirements • Step Five: Plot motor characteristics May 21, 2006
A Few Extra Points
May 21, 2006
Simple DC Motor Model V
= I ⋅ R + e
T
e
= k ⋅ I t
η max = 1 − May 21, 2006
I NL I P
2
= k e ⋅ ω
V
= I ⋅ R + k ⋅ ω e
Motor Inductance • The windings of a DC motor creates an Inductance, L • Change in current through an di V L inductance creates a voltage dt • Switching current to a motor causes di/dt to spike (Flyback) =
May 21, 2006
Flyback voltages can be very high and damage electronics, that is why a flyback diode in the switching circuit is required.
Winches • Similar to drive motors
May 21, 2006
Common Mistakes • Using static or kinetic friction instead of rolling friction – If a wheel is rolling without slipping, the only energy loss is due to deformations in the wheel/surface (rolling friction)
• Using PWM to control a motor reduces the available torque – The average power, speed and torque are reduced, however, effective torque is not significantly effected May 21, 2006
Questions?
May 21, 2006
View more...
Comments