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!