Power = Pressure * Pipe Size Area

Pneumatics Pushing Force

Force = Surface Area * Pounds Per Square Inch

= Pi*R2 * 60 lbs/inch2

= 3.14*12 inch2 * 60 lbs

= 188 lbs

60 PSI

1”

Pneumatics Pulling Force

Force = Surface Area * Pounds Per Square Inch

= Pi*R2 - Pi*r2 * 60 lbs/inch2

= 3.14*12 - 3.14*(.625/2)2 * 60 lbs

= 170 lbs

60 PSI

R=1”

r = ¼”

Pneumatic Forces

Piston Rod Force Force

Diameter Diameter Push(pi*R2) Pull (pi*(R2-r2))

2” .625” 188 170

1.5” .44” 106 97

1.0” 3/16 47 45.5

0.75” .25” 26.49 23.55

Pneumatic Forces at Different Pressures

Switches, Solenoid, Remote Control

Computer

Interface Controller

12 v

Switches, Solenoid, Remote Control

Computer

Interface Controller

12 v

Pneumatic Mounting: Open PositionLength = co = 18 inches

Co = 90 degrees

c2 = a2 + b2 - 2a*b*Cos(Co)

182 = a2 + b2

a = sqrt(c2 - b2) b = sqrt(c2 – a2)

a

b

c

C

0

2

4

6

8

10

12

14

16

18

20

0 2 4 6 8 10 12 14 16 18 20

b

a

Pneumatic Mounting: Closed PositionLength = cc = 10 inches

Cc = 40

c2 = a2 + b2 - 2a*b*Cos(Cc)

102 = a2 + b2 - 2a*b*Cos(40)

100 = a2 + b2 – 1.532a*b

0 = a2 – 1.532a*b + (b2 - c2)

a = ½ *(2bCos(Cc) +/- sqrt((-2bCos(Cc))2 – 4 (b2 - c2))

a

b

100 = (324-b2) + b2 -1.532*sqrt(324-b2)b ;from previous page1.532*sqrt(344-b2)b=224Sqrt(324-b2)b=146.2141(324-b2)b2=21378.562 ;square both sides0 = b4-324b2 + 21378.562 b2= (324 +/- sqrt(3242 – 4*21378.562))/2 ; quadratic eqb2= 92.247 or b2 = 231.7527b = 9.6 or b = 15.223a = sqrt(324 - b2)A = 15.223 or a = 9.6

-15

-10

-5

0

5

10

15

20

0 5 10 15 20

b

a

open closed (b-) closed (b+)

c

C

Pneumatic Mounting: Two PositionsClosed:Length = Cc = 10 inchesCc = 10 c2 = a2 + b2 - 2a*b*Cos(Cc) 102 = a2 + b2 - 2a*b*Cos(10) 100 = a2 + b2 – 1.9696a*b100 = (324-b2) + b2 -1.9696*sqrt(324-b2)b1.9696*sqrt(344-b2)b=224Sqrt(324-b2)b=113.72778(324-b2)b2=12934.0080 = b4-324b2 + 12934.008 b2= (324 +/- sqrt(3242 – 4*12934.008))/2b2= 46.631 or b2 = 277.369b = 6.8287 or b = 16.6544a = sqrt(324 - b2)

A = 16.6544 or a = 6.8287

a

b

c

Pneumatic Mounting: Two Positions

a

b

co

co2 = a2 + b2 -2a*b*Cos(Co)

cc2 = a2 + b2 -2a*b*Cos(Cc)

co2 - cc

2 = 2a*b*(Cos(Cc)-Cos(Co))

(co2 - cc

2 )/(Cos(Cc)-Cos(Co))=2ab

ab = k; where k= ½ (co2 - cc

2 )/(Cos(Cc)-Cos(Co))

a = k/b; then plug into equation(1)

co2 = (k/b)2 + b2 -2k*Cos(Co)

0 = b2 +(- 2k*Cos(Co)- co2)+k2/b2

0 = b4 +jb2+k2 where (- 2k*Cos(Co)- co2)

b2 = (-j +/- sqrt(j2 - 4k2))/2

b = sqrt(b2); a = k/b;

Co

a

b

cc

Cc

Pneumatic Mounting: Two Positions

a

b’

co

co2 = a2 + b2 -2a*b*Cos(Co)

cc2 = a2 + b2 -2a*b*Cos(Cc)

co2 - cc

2 = 2a*b*(Cos(Cc)-Cos(Co))

(co2 - cc

2 )/(Cos(Cc)-Cos(Co))=2ab

ab = k; where k= ½ (co2 - cc

2 )/(Cos(Cc)-Cos(Co))

co2 = (k/b)2 + b2 -2(k/b)*b*Cos(Co)

co2 = (k/b)2 + b2 -2k*Cos(Co)

0 = b4 +(- 2k*Cos(Co)- co2)b2+k2

0 = b4 +jb2+k2 where (- 2k*Cos(Co)- co2)

b2 = (-j +/- sqrt(j2 - 4k2))/2

b = sqrt(b2); a = k/b;

Co

a’

b’

cc

Cc

CT=C+asin(b/Hb)+asin(a/Ha) b’=sqrt(b2-Hb

2)a’=sqrt(a2-Ha

2)

a’

b

‘

Speed?Too Fast!!!• Adjustable inlets are available for adjustable speed

to slow down the pistons.

Too Slow???• Air volume through plastic tubing limiting factor.• Larger pistons are more powerful, but need more

air, so are generally slower.• If need speed, can use parallel air tanks,

Team 39 used 3 parallel tanks and

valves with large brass 4 way + connector in 2007 to catapult large ball.

Adjustable Positions• Magnetic reed switches come with the ordered

pistons.

• Can place at desired position and then turn off both input and output valves. This is rumored to work.– I believe it takes two separate festo valves– But may be able to be done with one

Adjustable Force• Use two different regulators to get two

different pressures.

• Add them in the T using two different valves