Transcript FIRST Motors - Chief Delphi

F.I.R.S.T. Motors &
Drive System Design
January 4th, 2003
FIRST Novi Kickoff
Paul Copioli
Utica Community Schools & Ford Motor Company
The ThunderChickens (Team #217)
page 1
Agenda
1. Introduction (why we are here)
2. Intro to Things Mechanical
3. First Motor Characteristics
4. Robot Drive Systems - Design Objectives
5. Questions & Answers
page 2
Introduction
Who am I?
•Paul Copioli
•Bachelors of Science - Aeronautical Engineering
•U.S. Air Force Academy
•M.S.E. - Aerospace & Mechanical Engineering
•University of Michigan
•FANUC Robotics North America
•Senior Product Development Engineer
•4th Season with FIRST
page 3
Intro to Things Mechanical
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•
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•
•
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page 4
Force - units are Pounds (Lbf) & Newtons (N)
Velocity - meters/sec, ft/sec, MPH
Acceleration - m/s2, ft/s2, g = 9.81 m/s2
Angular Velocity - RPM, rad/sec, deg/sec
Torque - N*m, ft*Lbf
Torque = Force * Lever Arm (Wheel Radius)
Velocity = Ang. Velocity * Wheel Radius
Formulas & Units
• Unit conversions of interest
–
–
–
–
–
1lbs = 4.45 N
1 inch = 0.0254 meters
1 in-lbs = 0.11 N-m
1 RPM = 60 Rev / Hour = 0.105 Rad / Sec
1 mile = 5280 X 12 inches = 63,000 inches
• Power = Force (N) X Velocity (m/s)
• Power = Torque (N-m) X Angular Velocity (Rad/Sec)
• Electrical Power = Voltage X Current
page 5
FIRST Motors
1. Motor Characteristics (Motor Curve)
2. Max Power vs. Power at 30 Amps
3. Motor Comparisons
4. Combining Motors
page 6
Motor Characteristics
• Torque v Speed Curves
–
–
–
–
Stall Torque (T0)
Stall Current (A0)
Free Speed (Wf)
Free Current (Af)
T0
A0
Af
Speed
page 7
K (slope)
Wf
Slope-Intercept (Y=mX + b)
•
•
•
•
Y=Motor Torque
m=K (discuss later)
X=Motor Speed
b=Stall Torque (T0)
T0
K (slope)
A0
Af
Speed
Wf
What is K? … It is the slope of the line.
Slope = change in Y / change in X = (0 - T0)/(Wf-0) = -T0/Wf
K = Slope = -T0/Wf
page 8
(Y=mX + b) Continued ...
•
•
•
•
Y=Motor Torque
m=K = -T0/Wf
X=Motor Speed
b=Stall Torque = T0
T0 (b)
K (-T0/Wf)
A0
Af
Speed
Equation for a motor:
Torque = (-T0/Wf) * Speed + T0
page 9
Wf
Current (Amps) and FIRST
• What are cutoff Amps?
– Max useable amps
– Limited by breakers
– Need to make assumptions
T0
A0
Cutoff
Amps
Af
Speed
Can our Motors operate above 30(40) amps?
Wf
- Absolutely, but not continuous.
When designing, you want to be able to perform continuously;
so finding motor info at 30 (40) amps could prove to be useful.
page 10
Torque at Amp Limit
• T30 = Torque at 30(40) Amps
• W30 = Speed at 30(40) Amps
Current Equation:
T0
A0
Cutoff
Amps
Current = (Af-A0)/Wf * Speed + A0
Af
Motor Equation:
Torque = (-T0/Wf) * Speed + T0
S @ 30A (W30) = (30 - A0) * Wf / (Af-A0)
T @ 30A (T30) = (-T0/Wf) * W30 + T0
page 11
Speed
Wf
Use 40 Amps
for 2003 Drill &
Chiaphua
Power - Max vs. 30(40) Amps
Power = Torque * Speed
Must give up torque for speed
Power
T0
Max Power occurs when:
T = T0/2 & W=Wf/2
What if max power occurs at
a current higher than 30A
(40A)?
A0
Af
Speed
Wf
Paul’s Tip #1: Design drive motor max power for 30A(40A)!
Power is Absolute - It determines the Torque Speed tradeoff!
page 12
Motor Comparisons
Let’s Look at Some FIRST Motors
• Chiaphua Motor
• Drill Motor
• Johnson Electric Fisher-Price Motor
We will compare T0, Wf, A0, Af, T30, W30, max
power (Pmax), amps @ max power (Apmax), and
power at 30(40) amps (P30).
We will be using Dr. Joe’s motor spreadsheet updated
to handle the new motors.
page 13
Motor Comparisons
Motor
Chiaphua
T0
Wf
A0
Af
Pmax
T40
N-m
RPM
Amps
Amps
Watts
N-m
2.2
5,500
107
2.3
316.8
0.80
Johnson F-P
0.38
15,000
57
1.1
149.2
0.20
Bosch Drill
0.87
19,670
127
4.5
448.0
0.25
Motor Equations:
1. 2003 Fisher-Price: T = (-0.38/15,000) * W + 0.38
2. 2003 Bosch Drill:
T = (-0.87/19,670) * W + 0.87
3. 2002-03 Chiaphua: T = (-2.2/5,500) * W + 2.2
page 14
Combining Motors
Using multiple motors is common for drive trains. We will look
at matching the big 3 motors.
I try to match at free speed, but you can match at any speed
you like!!
Wf Drill/Wf FP 19670/15000 ~ 17/13 = Gear Ratio
Wf drill / Wf Chiaphua = 19670/5500 ~ 18/5 = Gear Ratio
Wf FP / Wf Chiaphua = 15000/5500 ~ 30/11 = Gear Ratio
We will use an efficiency of 95% for the match gears.
More to come on Gear Ratio & Efficiency in the Second Half!
page 15
Combined Motor Data
T0
Wf
Pmax
T40
W40
P40
N-m
RPM
Watts
N-m
RPM
Watts
F-P & Drill
1.46
15,000
573
0.42
10,683
470
F-P & Chip
3.19
5,500
459
1.15
3,510
423
Drill & Chip
5.18
5,479
743
1.86
3,510
684
F-P, Drill, & Chip
6.16
5,483
884
2.22
3,510
816
Motor
Motor Equations:
1. F-P & Drill:
T = (-1.46/15,000) * W + 1.46
2. F-P & Chip:
T = (-3.19/5,500) * W + 3.19
3. Drill & Chip:
T = (-5.18/5,479) * W + 5.18
4. F-P, Drill, & Chip: T = (-6.16/5,483) * W + 6.16
page 16
Motor Q & A
page 17
Robot Drive Systems
1. Drive System Terms
2. Types of Mechanisms
3. Traction Basics
4. Gearing Basics
5. Design Condition
page 18
Drive System Terms
1. Gear Ratio: Can be described many ways
- Motor Speed / Output Speed
2. Efficiency - Work lost due to drive losses
- Friction, heat, misalignment
3. Friction Force - Tractive (pushing) force generated
between floor and wheel.
4. W is rotational speed & V is linear Speed (velocity)
5. N1 is # of teeth on input gear/sprocket
6. N2 is # of teeth on output gear/sprocket
page 19
Types of Drive Mechanisms
1. Chain & Belt
Efficiency ~ 95% - 98%
GR = N2/N1
N2
N1
2. Spur Gears
Efficiency ~ 95% - 98%
GR = N2/N1
N1
page 20
N2
Types of Drive Mechanisms
3. Bevel Gears
Efficiency ~ 90% - 95%
GR = N2/N1
N1
N2
page 21
Types of Drive Mechanisms
4. Worm Gears
Efficiency ~ 40% - 70%
# Teeth on Worm Gear
GR = ------------------------------# of Threads on worm
Worm
Worm gear
page 22
Types of Drive Mechanisms
5. Planetary Gears
Efficiency ~ 80% - 90%
RING GEAR
(FIXED)
SUN GEAR
(INPUT)
CARRIER
(OUTPUT)
PLANET GEAR
Nring
GR = ------- + 1
Nsun
page 23
Traction Basics
• Remember Ken Patton’s Key Points
• Ffriction = m * Fnormal
• Experimentally determine m:
• Fnormal = Weight * cos(q)
• Fparallel = Weight * sin(q)
When Ff = Fp, no slip
Ff = m*Weight * cos(q)
Fp = Weight * sin(q) = m*Weight * cos(q)
m = sin(q) / cos(q)
page 24
m = tan(q)
q
Gearing Basics
• Consecutive gear stages multiply:
N2
N4
N1
N3
• Gear Ratio is (N2/N1) * (N4/N3)
• Efficiency is .95 *.95 = .90
page 25
Gearing Basics - Wheel Attachment
N2
N1
N4
Wheel Diameter - Dw
Dw = Rw * 2
Motor Shaft
N3
Fpush
• Gear 4 is attached to the wheel
• Remember that T = F * Rw
• Also, V = W * Rw
• T4 = T1 * N2/N1 * N4/N3 * .95 * .95
• W4 = W1 * N1/N2 * N3/N4
• F = T4 / Rw
• V = W4 * Rw
page 26
Design Condition
• Assumptions
•Each of the 4 wheels have their own motor.
• Weight is evenly distributed.
• Using all spur gears.
• Terms
• W = Weight of robot
• Wt = Weight transferred to robot from goals
• n = # of wheels on the ground (4)
• p = # driving wheels per transmission (1)
• q = # of transmissions (4)
• Tout = wheel output Torque
• Find the gear ratio & wheel diameter to maximize
push force.
The maximum force at each wheel we can attain is ???
Fmax = Ffriction = Mu*(W + Wt)/n
Now T = F * Rw ----> F = Tout / Rw
page 27
{on a flat surface}
Design Condition Continued
• Tout = T30(40) * GR * eff
{@ each wheel}
Ffriction = Tout / Rw: Mu*(W + Wt)/n = T30(40) * GR * eff / Rw
Mu*(W + Wt)
GR/Rw = --------------------------n*T30(40)*eff
The above gives you the best combination of gear ratio and
wheel diameter for maximum pushing force!
page 28
Design Condition Continued
O.K. So what is my top speed?
Vmax [m/sec] =
0.9 * Wfree * p * 2 * Rw
-----------------------------60 * GR
Where Wfree is in RPM, Rw is in meters.
The 0.9 accounts for drive friction slowing the robot down.
page 29
Design Condition Continued
0.9 * Wfree * p * 2 * Rw 0.9 * Wfree * p * 2 * n * T30 * eff
Vmax = --------------------------------- = -------------------------------------------60 * GR
60 * Mu * (W + Wt)
T30 * GR * eff
Fmax = -------------------- = Mu * (W + Wt)
Rw
Max force and max velocity are fighting each other
page 30
Drive System Fundamentals
QUESTIONS?
page 31