Transcript E-Waste Curriculum

Apple Pi Robotics
Gaining a Mechanical
Advantage
Mechanics 101 –
Gaining an Advantage
Levers
Belts/Pulleys – Chain/Sprockets
Gears
Let's talk Forces!
Forces –
F2
F3
M
F4
F1
Force has Magnitude & Direction
If
F1= F2
&
?
F3 = F4
Forces –
F2
F3
F4
M
F1
If
F1= F2
&
??
F3 < F4
Forces –
F2
F3
F4
M
F1
If
F1< F2
&
??
F3 < F4
Forces –
F2
F3
F4
M
F1
If
F1< F2
&
F3 < F4
Practical Example: Static Model
F/2 F/2
F = mass x a (gravity) = Weight
Practical Example: Dynamic Model
F1 is all about Torque at Wheels
(HP is the product of speed and torque)
wt
f2
f2 is friction due to windage
F2 is reaction force (tire friction)
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)
Practical Example:
Windage is proportional to speed2
uwV2
ma = uwV2
We have reached terminal velocity
F=ma (mass of body x gravitational acceleration)
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)
Materials
Aluminum
Copper
Brass
Cast Iron
Concrete (wet)
Concrete (dry)
Polyethylene
Steel
Steel
Steel
Copper
Rubber
Rubber
Steel
Coefficient
.61
.53
.51
1.05
.30
1.0
.20
Mechanical Advantage – The Lever
F2
L1
F1
F1 x L1 = F2 x L2
L2
Mechanical Advantage – The Lever
F2
4 ft
10
lbs
F2 = ??
2 ft
Mechanical Advantage – The Lever
F2
4 ft
2 ft
10
lbs
10 lbs x 4 ft = F2 x 2 ft
F2 = 10 lbs x 4 ft / 2 ft = 20 lbs
Mechanical Advantage – The Lever
F2
2 ft
D1
4 ft
F1
Trade Off - Force vs Distance
D1/D2 is Inversely Proportional to F1/F2
D2
Mechanical Advantage – The Lever
F2
2 ft
D2
4 ft
D1
F1
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??
Mechanical Advantage – The Lever
F2
L2
D2
L1
D1
F1
-- See Saw
-- Catapult
-- Oars
-- Pry Bar
Mechanical Advantage – The Lever
First Order
L2
F2
L1
F1
Another Configuration
Examples?
Mechanical Advantage – The Lever
First Order
L2
F2
L1
F1
Crow/Pry Bar
Claw Hammer
Balancing Forces – Arm/Boom
F2
Fulcrum
L1
M
F1
If F1 = F2 Equilibrium/No Motion Right?
Balancing Forces/Equilibrium –
F2
L1
M
m=Wt/32.2
F1
This is the REAL mechanical system!
So F2 must be greater to balance out the
effect of boom weight
Balancing Forces/Equilibrium –
F2
L1
L1
2
M
Wt = ma = Fw
F1
Center of Gravity
(CG)
If (F1 x L1)+ (Wt x L1/2) = F2 x L1, then equilibrium
Balancing Forces/Equilibrium –
F2
L1
L2/2
L2
10 lbs
2 lbs
F1
How heavy is counterbalance?
Wt ?
Balancing Forces/Equilibrium –
L1
L2/2
L2
10 lbs
2 lbs
F1
(2 * L2) + (10*L2*2) = Wt *L2/2
Wt = (2*L2)+(10*L2*2)/(L2/2)
Wt = 22*L2/(L2/2) = 44 lbs
44 lbs
Let's Take a Look at
Mechanism on “Little Foot”
Velocity = v
Frictional Force
F2 x u
F1
Bungee Force
F2
(F2=F1)
3”
21”
mass on end of arm
Let's Take a Look at
Mechanism on “Little Foot”
Velocity = v
Frictional Force
F2 x u
F1
Bungee Force
Weight
F2
3”
21”
Frictional Force = F2 * u = 15lbs * .70 = 10.5 lbs
Estimated Velocity of 4 ft/sec
Weight = mass * 32.2 = 6 oz. Approx.
Inertial Switch
Mechanism on “Little Foot”
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
BUT...
What happens if for some unknown reason
it doesn't work?
Frictional Force
F2 x u
F1
Bungee Force
F2
3”
21”
Wt = 6 oz.
What can we do to effect the outcome?
Resolution of Forces -
L1
F1
L2
F2
All forces applied to the arm can be “resolved”
into perpendicular (rotational) and parallel
(compression/extension) force components
Resolution of Force
Y
Fy
F
Fx
X
Resolution of Force
How? A Little Math
F
Fy
O
Fx
Using SIN, COS & TAN Functions If any two are known you can solve for
any others
YF
Resolution of Force
How? A Little Math
F
y
O
SOH-CAH-TOA
?
Fx
X
YF
Resolution of Force
How? A Little Math
F
y
O
Fx
SOH-CAH-TOA
Sin (O) = Opposite/Hypotenus
Cos(O) = Adjacent/Hypotenus
Tan(O) = Opposite/Adjacent
X
YF
Resolution of Force
How? A Little Math
F
y
O
Fx
X
Examples:
For angles of 0 – 90 degrees, Sin & Cos vary
between 0 – 1, Tan varies between 0 and infinity
Sin(30) = .5
Sin(45) = .707
Cos(30)= .866
YF
Resolution of Force
How? A Little Math
F
y
O
Examples:
Sin(30) = .5
Sin(45) = .707
Cos(30)= .866
Fx
X
If (O) = 30 degrees and F = 5 lbs, then
Sin (30) = Fx/5 lbs or Fx = Sin(30) x 5 lbs = 2.5 lbs
Resolution of Forces -
f2y
f1x
f1y
f2x
L1
F1
L2
F2
All forces applied to the arm can be “resolved”
into perpendicular and parallel force components
MOVIE CLIP – 2008 Robot
Apple Pi 2008 Robot -
f2y
f2x
f1y
f1x
F1
L1
L2
F2
If f1x x L1 = f2x x L2
Equilibrium
Pneumatic
Cylinder
Belts-Sheaves
and
Chain-Sprockets
Belts, Sheaves & Sprockets

Belt Types

V - belt



Flat belt



Does slip
Depends on tension
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”
P#1
Each pulley has a circumference of (Pi) x D
So 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)
P#2
Pulley #2 will go at ¼ the
speed (RPM's) as P #1
and generate 4 x's the
torque
Dia. = 4”
Pulleys/Sprockets
Another configuration of pulleys with belt

Different diameter pulleys
Dia. = 1”
RPM x1
Dia. = 2”
RPM x ½
Dia. = 4”
RPM x ¼
Pulleys/Sprockets
What are speeds if “N” is 10 revs/sec.?
N = 10 revs/sec
Dia. = 1”
Dia. = 2”
Dia. = 4”
Belts/Pulley vs. Chain/Sprocket
Whats 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

Differences – G#2 turns opposite direction
& no chance for slippage
Each gear has a circumference of (Pi) x D
So 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)
Dia. = 1”
G#1
G#
2
Gear #2 will go at ¼ the
speed (RPM's) as Gear
#1
Dia. = 4”
Compound GearsPrevious 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 ?
G#4 = Dia. 3”
G#1
G#2
G#1 & 3
Dia. = 1”
G#4
G#3
G#2
Dia.4”
This compound gear set
provides 12:1 ratio (1/12
the speed) with 12 x's the
torque.
“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!