apple pi robotics gaining a mechanical advantage

Apple Pi Robotics

Gaining a Mechanical Advantage

Mechanics 101 –Gaining an Advantage

Levers

Belts/Pulleys – Chain/Sprockets

Gears

Let's talk Forces!

Forces –

F4

F1

F2

Force has Magnitude & Direction

If F1= F2 & F3 = F4 ?

F3 M

Forces –

F4

F1

F2

If F1= F2 & F3 < F4 ??

F3 M

Forces –

F4

F1

F2

If F1< F2 & F3 < F4 ??

F3 M

Forces –

F4

F1

F2

If F1< F2 & F3 < F4

F3 M

Practical Example: Static Model

F = mass x a (gravity) = Weight

F/2 F/2

Practical Example: Dynamic Model

F1

f2

(F1-f2) = mass x acceleration

or

(F1-f2)/mass = acceleration

Provided there is enough reaction force (F2) to support F1

F2 = u * wt/2 (approx)

F1 is all about Torque at Wheels

(HP is the product of speed and torque)

f2 is friction due to windage

F2 is reaction force (tire friction)

F2

wt

Practical Example:

F=ma (mass of body x gravitational acceleration)

Windage is proportional to speed2

ma = uwV2

We have reached terminal velocity

uwV2

Types of Forces

F(external)= mass * acceleration

F(friction)=u * Wt

F(weight) = Wt = mass* acceleration(gravity)

F(wind)= windage const. * velocity2

F(impulse) = mass * velocity change

Coefficients of Friction(examples)

Aluminum Steel .61Copper Steel .53Brass Steel .51Cast Iron Copper 1.05Concrete (wet) Rubber .30Concrete (dry) Rubber 1.0Polyethylene Steel .20

Materials Coefficient

Mechanical Advantage – The Lever

L1 L2

F1

F2

F1 x L1 = F2 x L2

Mechanical Advantage – The Lever

4 ft 2 ft

10 lbs

F2

F2 = ??

Mechanical Advantage – The Lever

4 ft 2 ft

10 lbs

F2

10 lbs x 4 ft = F2 x 2 ft

F2 = 10 lbs x 4 ft / 2 ft = 20 lbs

Mechanical Advantage – The Lever

D2

F1

F2

Trade Off - Force vs DistanceD1/D2 is Inversely Proportional to F1/F2

D1

2 ft

4 ft

Mechanical Advantage – The Lever

D2

F1

F2

So ……….. If D1 is 1ft, then D2 = ??If velocity of the left most side of lever is 6 inches/sec then velocity of the right most side is = ??Examples of real devices??

D1

2 ft

4 ft

Mechanical Advantage – The Lever

D2

F1

F2

-- See Saw -- Catapult-- Oars-- Pry Bar

D1

L2

L1

Mechanical Advantage – The LeverFirst Order

Another Configuration

Examples?

F1

F2

L2

L1

Mechanical Advantage – The LeverFirst Order

Crow/Pry BarClaw Hammer

F1

F2

L2

L1

Balancing Forces – Arm/Boom

L1

F1

F2

Fulcrum

M

If F1 = F2 Equilibrium/No Motion - Right?

Balancing Forces/Equilibrium –

L1

F1

F2

This is the REAL mechanical system! So F2 must be greater to balance out the effect of boom weight

m=Wt/32.2M

Balancing Forces/Equilibrium –

L1

F1

F2

If (F1 x L1)+ (Wt x L1/2) = F2 x L1, then equilibrium

Wt = ma = Fw

L1 2

Center of Gravity (CG)

M

Balancing Forces/Equilibrium –

L1

F1Wt ?

L2/2

L2

How heavy is counterbalance?

F2

10 lbs

2 lbs

Balancing Forces/Equilibrium –

L1

F1

2 lbs

10 lbs

44 lbs

L2/2

L2

(2 * L2) + (10*L2*2) = Wt *L2/2

Wt = (2*L2)+(10*L2*2)/(L2/2)

Wt = 22*L2/(L2/2) = 44 lbs

Let's Take a Look atMechanism on “Little Foot”

F1

F2(F2=F1)

3” 21”

F2 x u

mass on end of arm

Frictional Force

Bungee Force

Velocity = v

Let's Take a Look atMechanism on “Little Foot”

F1

F2

3” 21”

Weight

F2 x u

Frictional Force = F2 * u = 15lbs * .70 = 10.5 lbsEstimated Velocity of 4 ft/secWeight = mass * 32.2 = 6 oz. Approx.

Frictional Force

Bungee Force

Velocity = v

Gain lever advantage of 7:1 (21” / 3”)

Assume 4 ft/sec velocity at end of arm

We have accelerated a 6 oz. weight and will be decelerating “very quickly” (.02 secs?)

m = 1/3 lb / 32.2 ft/sec2

Kinetic energy stored = mv

– Pulse Energy = m (v1 -v2) = F(t1-t2)

For our example this calculates to approx. 15 lbs force

Approx. 50% greater than our calculated frictional force

Inertial SwitchMechanism on “Little Foot”

BUT...What happens if for some unknown reason

it doesn't work?

F1

F2

3” 21”

Wt = 6 oz.

F2 x u

What can we do to effect the outcome?

Frictional Force

Bungee Force

Resolution of Forces -

L2

F2

F1

All forces applied to the arm can be “resolved” into perpendicular (rotational) and parallel (compression/extension) force components

L1

Resolution of Force

Y

X

F

Fx

Fy

Resolution of Force How? A Little Math

Using SIN, COS & TAN Functions -If any two are known you can solve for any others

F

Fx

O

Fy

Resolution of Force How? A Little MathY

X

F

Fx

Fy

O

SOH-CAH-TOA ?

Resolution of Force How? A Little MathY

X

F

Fx

Fy

O

SOH-CAH-TOA

Sin (O) = Opposite/HypotenusCos(O) = Adjacent/HypotenusTan(O) = Opposite/Adjacent

Resolution of Force How? A Little MathY

X

F

Fx

Fy

O

Examples: For angles of 0 – 90 degrees, Sin & Cos vary between 0 – 1, Tan varies between 0 and infinitySin(30) = .5Sin(45) = .707Cos(30)= .866

Resolution of Force How? A Little MathY

X

F

Fx

Fy

O

Examples: Sin(30) = .5Sin(45) = .707Cos(30)= .866

If (O) = 30 degrees and F = 5 lbs, thenSin (30) = Fx/5 lbs or Fx = Sin(30) x 5 lbs = 2.5 lbs

Resolution of Forces -

L2

F2

f2xf2y

F1

f1xf1y

All forces applied to the arm can be “resolved” into perpendicular and parallel force componentsMOVIE CLIP – 2008 Robot

L1

Apple Pi 2008 Robot -

L1F1

F2

If f1x x L1 = f2x x L2 Equilibrium

PneumaticCylinder

f1xf1y

f2yf2x

L2

Belts-Sheaves and

Chain-Sprockets

Belts, Sheaves & Sprockets

Belt Types V - belt

Does slip Depends on tension

Flat belt Does slip Depends on tension

Cogged generally solidly “coupled” to sprocket Teeth molded into belt Mesh with slots in pulley Flat or V groove No slip/highly efficient

Pulleys/Sprockets-

Configuration of pulleys with belt to gain mechanical advantage

Dia. = 1”

Dia. = 4”

Each pulley has a circumference of (Pi) x DSo pulley 1 has a circumference of 3.14” while pulley 2 has a circumference of 12.56”. It will take 4 revolutions of P1 to “feed out” enough belt to allow for 12.56” of the circumference of P2 to be “conveyed”. (assuming no slippage)

Pulley #2 will go at ¼ the speed (RPM's) as P #1 and generate 4 x's the torque

P#1

P#2

Pulleys/Sprockets-

Another configuration of pulleys with belt Different diameter pulleys

Dia. = 1”RPM x1

Dia. = 4”RPM x ¼

Dia. = 2”RPM x ½

Pulleys/Sprockets-

What are speeds if “N” is 10 revs/sec.?

Dia. = 1”

Dia. = 4”

Dia. = 2”

N = 10 revs/sec

Belts/Pulley vs. Chain/SprocketWhats the Difference?

* Belts generally used for lower torque requirements* Belts can slip (could be good?)* Chain can be separated (maintenance)* Belts do not need lubrication

Gears-

Gears behave like “closely coupled” pulleys Differe

Dia. = 4”

Each gear has a circumference of (Pi) x DSo gear #1 has a circumference of 3.14” while gear #2 has a circumference of 12.56”. It will take 4 revolutions of G1 to “circum navigate” the perimeter of G2 to allow for 12.56” of the circumference of G2 (assuming no slippage)

Gear #2 will go at ¼ the speed (RPM's) as Gear #1

G#2

G#1Dia. = 1”

Gears behave like “closely coupled” pulleys Differences – G#2 turns opposite direction

& no chance for slippage

Compound Gears-

G#2 Dia.4”

Previous example showed that G#2 will go at ¼ the speed of G#1. G#3 coupled directly to G#2 so it goes at ¼ the speed also. Then G#4 will go at 1/3 the speed of G#3 since is has 3 times the circumference (or diameter).

Total mechanical advantage is 4 x 3 or 12.If G#1 is going 120 RPM's, then G#4 is ?

This compound gear set provides 12:1 ratio (1/12 the speed) with 12 x's the torque.

G#2G#1

G#1 & 3 Dia. = 1”

G#3

G#4

G#4 = Dia. 3”

“MANTIS” Drive Train

Design Speed to be 12 fps CIM Motors have 5000 rpm top speed Will use 2 speed transmission

– High range has 9.4:1 gear ratio

– Chain sprockets available are 12, 15, 22, 26, and 30 tooth

Drive wheel is 8” diameter Select drive components after transmission

“MANTIS” Drive Train For 12 fps Wheel circumference of 8” x 3.14 = 25” (approx)

25 in/rev/12in/ft = 2.08 ft/rev

X revs/sec = 12 ft/sec/2.08 ft/rev = 5.77 revs/sec 5000 rpm = 83 rev/sec (5000rev/min/60secs/min)

Total ratio is 83 rev/sec/5.77 rev/sec = 14.38:1 Ratio of transmission is 9.4:1 so we need

additional 14.38/9.4 (1.53:1) ratio Looking at available sprockets (12, 15, 22, 24

and 30) we picked 22 and 15 (1.46:1 ratio)

That's All Folks!