powerpoint motor presentation

F.I.R.S.T. Motors &F.I.R.S.T. Motors &Drive System FundamentalsDrive System Fundamentals

December 7, 2002Team Ford FIRST

Paul CopioliUtica Community Schools & Ford Motor Company

The ThunderChickens (Team #217)

AgendaAgenda

1. Introduction (why we are here)

2. Intro to Things Mechanical

3. First Motor Characteristics

4. Robot Drive Systems - Design Objectives

5. Questions & Answers

IntroductionIntroductionWho 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

Intro to Things MechanicalIntro to Things Mechanical

Force - units are Pounds (Lbf) & Newtons (N)Force - units are Pounds (Lbf) & Newtons (N) Velocity - meters/sec, ft/sec, MPHVelocity - meters/sec, ft/sec, MPH Acceleration - m/sAcceleration - m/s22, ft/s, ft/s22, g = 9.81 m/s, g = 9.81 m/s2 2

Angular Velocity - RPM, rad/sec, deg/secAngular Velocity - RPM, rad/sec, deg/sec Torque - N*m, ft*LbfTorque - N*m, ft*Lbf Torque = Force * Lever Arm (Wheel Radius)Torque = Force * Lever Arm (Wheel Radius) Velocity = Ang. Velocity * Wheel RadiusVelocity = Ang. Velocity * Wheel Radius

Formulas & UnitsFormulas & Units

Unit conversions of interestUnit conversions of interest 1lbs = 4.45 N1lbs = 4.45 N 1 inch = 0.0254 meters1 inch = 0.0254 meters 1 in-lbs = 0.11 N-m 1 in-lbs = 0.11 N-m 1 RPM = 60 Rev / Hour = 0.105 Rad / Sec 1 RPM = 60 Rev / Hour = 0.105 Rad / Sec 1 mile = 5280 X 12 inches = 63,000 inches1 mile = 5280 X 12 inches = 63,000 inches

Power = Force (N) X Velocity (m/s)Power = Force (N) X Velocity (m/s) Power = Torque (N-m) X Angular Velocity (Rad/Sec)Power = Torque (N-m) X Angular Velocity (Rad/Sec) Electrical Power = Voltage X CurrentElectrical Power = Voltage X Current

FIRST MotorsFIRST Motors

1. Motor Characteristics (Motor Curve)

2. Max Power vs. Power at 30 Amps

3. Motor Comparisons

4. Combining Motors

Motor CharacteristicsMotor Characteristics

Torque v Speed CurvesTorque v Speed Curves Stall Torque (T0)Stall Torque (T0) Stall Current (A0)Stall Current (A0) Free Speed (Wf)Free Speed (Wf) Free Current (Af)Free Current (Af)

SpeedT

orqu

e, C

urre

nt

T0

Wf

Af

A0

K (slope)

Slope-Intercept (Y=mX + b)Slope-Intercept (Y=mX + b)

Y=Motor TorqueY=Motor Torque m=K (discuss later)m=K (discuss later) X=Motor SpeedX=Motor Speed b=Stall Torque (T0)b=Stall Torque (T0)

Speed

Tor

que,

Cur

rent

T0

Wf

Af

A0

K (slope)

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

(Y=mX + b) Continued ...(Y=mX + b) Continued ...

Y=Motor TorqueY=Motor Torque m=K = -T0/Wfm=K = -T0/Wf X=Motor SpeedX=Motor Speed b=Stall Torque = T0b=Stall Torque = T0

Speed

Tor

que,

Cur

rent

T0 (b)

Wf

Af

A0

K (-T0/Wf)

Equation for a motor:

Torque = (-T0/Wf) * Speed + T0

Current (Amps) and FIRSTCurrent (Amps) and FIRST

What are cutoff Amps?What are cutoff Amps? Max useable ampsMax useable amps Limited by breakersLimited by breakers Need to make assumptionsNeed to make assumptions

Speed

Tor

que,

Cur

rent

T0

Wf

Af

A0

Cutoff Amps

Can our Motors operate above 30 amps?

- Absolutely, but not continuous.

When designing, you want to be able to perform continuously; so finding motor info at 30 amps could prove to be useful.

Torque at Amp LimitTorque at Amp Limit

T30 = Torque at 30 AmpsT30 = Torque at 30 Amps W30 = Speed at 30 AmpsW30 = Speed at 30 Amps

Speed

Tor

que,

Cur

rent

T0

Wf

Af

A0

Cutoff AmpsCurrent Equation:

Current = (Af-A0)/Wf * Speed + A0

Motor Equation:

Torque = (-T0/Wf) * Speed + T0

S @ 30A (W30) = (30 - A0) * Wf / (Af-A0)

T @ 30A (T30) = (-T0/Wf) * W30 + T0

Power - Max vs. 30 AmpsPower - Max vs. 30 Amps

Speed

Tor

que,

Cur

rent

T0

Wf

Af

A0

PowerPower = Torque * Speed

Must give up torque for speed

Max Power occurs when:

T = T0/2 & W=Wf/2

What if max power occurs at a current higher than 30A?

Power is Absolute - It determines the Torque - Speed tradeoff!

Paul’s Tip #1: Design drive motor max power for 30A!

Motor ComparisonsMotor Comparisons

• Chiaphua Motor

• Drill Motor

• Johnson Electric Fisher-Price Motor

Let’s Look at Some FIRST Motors

We will compare T0, Wf, A0, Af, T30, W30, max power (Pmax), amps @ max power (Apmax), and power at 30 amps (P30).

We will be using Dr. Joe’s motor spreadsheet updated to handle the new motors.

Motor ComparisonsMotor Comparisons

T0 Wf A0 Af Pmax T30 W30 P30N-m RPM Amps Amps Watts N-m RPM Watts

Chiaphua 2.2 5,500 107 2.3 321 0.58 4045 245.7Johnson F-P 0.51 20,000 109 1.84 273 0.14 15000 219.9Bosch Drill 0.65 20,000 117 2.5 340 0.15 15333 240.9

Motor

Motor Equations:

1. Fisher-Price: T = (-0.51/20,000) * W + 0.51

2. Bosch Drill: T = (-0.65/20,000) * W + 0.65

3. Chiaphua: T = (-2.2/5,500) * W + 2.2

Combining MotorsCombining MotorsUsing 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!!

FP and drill will match 1:1

Wf FP(drill) / Wf Chiaphua = 20000/5500 = 40/11

Gear ratio to match Chip & FP(drill) is 40/11.

We will use an efficiency of 95% for the match gear.

More to come on Gear Ratio & Efficiency in the Second Half!

Combined Motor DataCombined Motor DataT0 Wf Pmax T30 W30 P30

N-m RPM Watts N-m RPM WattsF-P & Drill 1.16 20,000 607 0.29 15,000 456F-P & Chip 3.96 5,500 570 1.09 4,045 462Drill & Chip 4.45 5,500 641 1.13 4,045 479F-P, Drill, & Chip 6.21 5,500 894 1.63 4,045 690

Motor

Motor Equations:

1. F-P & Drill: T = (-1.16/20,000) * W + 1.16

2. F-P & Chip: T = (-3.96/5,500) * W + 3.96

3. Drill & Chip: T = (-4.45/5,500) * W + 4.45

4. F-P, Drill, & Chip: T = (-6.21/5,500) * W + 6.21

Motor Q & AMotor Q & A

Robot Drive SystemsRobot Drive Systems

1. Drive System Terms

2. Types of Mechanisms

3. Traction Basics

4. Gearing Basics

5. Design Condition

Drive System TermsDrive System Terms1. Gear Ratio: Can be described many ways

- Motor Speed / Output Speed- When GR > 1 Torque Increaser

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

Types of Drive MechanismsTypes of Drive Mechanisms

2. Spur GearsEfficiency ~ 95% - 98%GR = N2/N1

N1 N2

1. Chain & BeltEfficiency ~ 95% - 98%GR = N2/N1

N2N1

Types of Drive MechanismsTypes of Drive Mechanisms

4. Other Types- Worm Gears Efficiency ~ 40% - 70%

- Planetary Gears Efficiency ~ 80% - 90%

3. Bevel GearsEfficiency ~ 90% - 95%GR = N2/N1

N2

N1

Traction BasicsTraction BasicsTerminologyTerminology

The friction coefficient for any given contact with the floor, multiplied by the normal force, equals the maximum tractive force can be applied at the contact area.

Tractive force is important! It’s what moves the robot.

normalforce

tractiveforce

torqueturning the

wheel

maximumtractiveforce

normalforce

frictioncoefficient= x

weight

Friction coefficient = Mu

Traction FundamentalsTraction Fundamentals“Normal Force”“Normal Force”

weightfront

The normal force is the force that the wheels exert on the floor, and is equal and opposite to the force the floor exerts on the wheels. In the simplest case, this is dependent on the weight of the robot. The normal force is divided among the robot features in contact with the ground.

normalforce(rear)

normalforce(front)

Traction FundamentalsTraction Fundamentals“Weight Distribution”“Weight Distribution”

more weight in backdue to battery andmotors

front

The weight of the robot is not equally distributed among all the contacts with the floor. Weight distribution is dependent on where the parts are in the robot. This affects the normal force at each wheel.

morenormalforce

lessnormalforce

less weight in frontdue to fewer partsin this areaEXAMPLE

EXAMPLEONLYONLY

Traction FundamentalsTraction FundamentalsWeight Distribution is Not ConstantWeight Distribution is Not Constant

arm position inrear makes the weightshift to the rear

front

arm position in frontmakes the weightshift to the front

EXAMPLEEXAMPLEONLYONLY

normalforce(rear)

normalforce(front)

Traction FundamentalsTraction FundamentalsWeight Distribution is Not ConstantWeight Distribution is Not Constant

Where the weight is in the robot is only part of the story!

When the robot accelerates (changes speed), inertial forces tend to change the weight distribution.

(Example of inertial force: the force exerted by the seat on your back in a Z06 Corvette as it accelerates.)

So, it is important to consider how the weight distribution changes when the robot changes speed.

Traction FundamentalsTraction Fundamentals“Weight Transfer”“Weight Transfer”

robot acceleratingfrom 0 mph to6 mph

inertial forcesexerted bycomponentson the robot

EXAMPLEEXAMPLEONLYONLY

more normal force is exertedon the rear wheels becauseinertial forces tend to rotatethe robot toward the rear

less normal force is exertedon the front wheels becauseinertial forces tend to rotatethe robot away from the front

In an extreme case (with rear wheel drive), you pull a wheelieIn a really extreme case (with rear wheel drive), you tip over!

Gearing BasicsGearing Basics• Consecutive gear stages multiply:

N1

N2

N3

N4

• Gear Ratio is (N2/N1) * (N4/N3)• Efficiency is .95 *.95 = .90

Gearing Basics - Wheel AttachmentGearing Basics - Wheel Attachment

N1

N2

N3

N4

• 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

Motor Shaft

Wheel Diameter - Dw

Dw = Rw * 2

Fpush

Design ConditionDesign 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 {on a flat surface}

Now T = F * Rw ----> F = Tout / Rw

Design Condition ContinuedDesign Condition Continued• Tout = T30 * GR * eff {@ each wheel}

The above gives you the best combination of gear ratio and wheel diameter for maximum pushing force!

Ffriction = Tout / Rw: Mu*(W + Wt)/n = T30 * GR * eff / Rw

Mu*(W + Wt) GR/Rw = ---------------------------

n*T30*eff

Design Condition ContinuedDesign Condition Continued

O.K. So what is my top speed?

0.85 * Wfree * π * 2 * Rw Vmax [m/sec] = ------------------------------

60 * GR

Where Wfree is in RPM, Dw is in meters.

The 0.85 accounts for drive friction slowing the robot down.

Design Condition ContinuedDesign Condition Continued

0.85 * Wfree * π * 2 * Rw 0.85 * Wfree * π * 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

Drive System FundamentalsDrive System Fundamentals

QUESTIONS?