1. Power = Pressure * Pipe Size Area

2. Pneumatics Pushing Force Force = Surface Area * Pounds Per Square Inch =Pi*R 2 * 60 lbs/inch 2 = 3.14*1 2 inch 2 * 60 lbs = 188 lbs 60 PSI 1”

3. Pneumatics Pulling Force Force = Surface Area * Pounds Per Square Inch = Pi*R 2 - Pi*r 2 * 60 lbs/inch 2 = 3.14*1 2 - 3.14*(.625/2) 2 * 60 lbs = 170 lbs 60 PSI R=1” r = ¼”

4. Pneumatic Forces PistonRodForceForce DiameterDiameterPush( pi*R 2 )Pull (pi*(R 2 -r 2 )) 2”.625” 188170 1.5”.44” 10697 1.0” 3 / 16 4745.5 0.75”.25 ” 26.4923.55

5. Pneumatic Forces at Different Pressures

6. Switches, Solenoid, Remote Control Computer Interface Controller 12 v

7. Switches, Solenoid, Remote Control Computer Interface Controller 12 v

8. Pneumatic Mounting: Open Position Length = c o = 18 inches C o = 90 degrees c 2 = a 2 + b 2 - 2a*b*Cos(C o ) 18 2 = a 2 + b 2 a = sqrt(c 2 - b 2 ) b = sqrt(c 2 – a 2 ) a b c C

9. Pneumatic Mounting: Closed Position Length = c c = 10 inches C c = 40 c 2 = a 2 + b 2 - 2a*b*Cos(C c ) 10 2 = a 2 + b 2 - 2a*b*Cos(40) 100 = a 2 + b 2 – 1.532a*b 0 = a 2 – 1.532a*b + (b 2 - c 2 ) a = ½ *(2bCos(Cc) +/- sqrt((-2bCos(Cc)) 2 – 4 (b 2 - c 2 )) a b 100 = (324-b 2 ) + b 2 -1.532*sqrt(324-b 2 )b ;from previous page 1.532*sqrt(344-b 2 )b=224 Sqrt(324-b 2 )b=146.2141 (324-b 2 )b 2 =21378.562 ;square both sides 0 = b 4 -324b 2 + 21378.562 b 2 = (324 +/- sqrt(3242 – 4*21378.562))/2 ; quadratic eq b 2 = 92.247 or b2 = 231.7527 b = 9.6 or b = 15.223 a = sqrt(324 - b2) A = 15.223 or a = 9.6 c C

10. Pneumatic Mounting: Two Positions Closed: Length = C c = 10 inches C c = 10 c 2 = a 2 + b 2 - 2a*b*Cos(C c ) 10 2 = a 2 + b 2 - 2a*b*Cos(10) 100 = a 2 + b 2 – 1.9696a*b 100 = (324-b 2 ) + b 2 -1.9696*sqrt(324-b 2 )b 1.9696*sqrt(344-b 2 )b=224 Sqrt(324-b 2 )b=113.72778 (324-b 2 )b 2 =12934.008 0 = b 4 -324b 2 + 12934.008 b 2 = (324 +/- sqrt(324 2 – 4*12934.008))/2 b 2 = 46.631 or b 2 = 277.369 b = 6.8287 or b = 16.6544 a = sqrt(324 - b 2) A = 16.6544 or a = 6.8287 a b c

11. Pneumatic Mounting: Two Positions a b coco c o 2 = a 2 + b 2 -2a*b*Cos(C o ) c c 2 = a 2 + b 2 -2a*b*Cos(C c ) c o 2 - c c 2 = 2a*b*(Cos(C c )-Cos(C o )) (c o 2 - c c 2 )/(Cos(C c )-Cos(C o ))=2ab ab = k; where k= ½ (c o 2 - c c 2 )/(Cos(C c )-Cos(C o )) a = k/b; then plug into equation(1) c o 2 = (k/b) 2 + b 2 -2k*Cos(C o ) 0 = b 2 +(- 2k*Cos(C o )- c o 2 )+k 2 /b 2 0 = b 4 +jb 2 +k 2 where (- 2k*Cos(C o )- c o 2 ) b 2 = (-j +/- sqrt(j 2 - 4k 2 ))/2 b = sqrt(b 2 ); a = k/b; CoCo a b c CcCc

12. Pneumatic Mounting: Two Positions a b’ coco c o 2 = a 2 + b 2 -2a*b*Cos(C o ) c c 2 = a 2 + b 2 -2a*b*Cos(C c ) c o 2 - c c 2 = 2a*b*(Cos(C c )-Cos(C o )) (c o 2 - c c 2 )/(Cos(C c )-Cos(C o ))=2ab ab = k; where k= ½ (c o 2 - c c 2 )/(Cos(C c )-Cos(C o )) c o 2 = (k/b) 2 + b 2 -2(k/b)*b*Cos(C o ) c o 2 = (k/b) 2 + b 2 -2k*Cos(C o ) 0 = b 4 +(- 2k*Cos(C o )- c o 2 )b 2 +k 2 0 = b 4 +jb 2 +k 2 where (- 2k*Cos(C o )- c o 2 ) b 2 = (-j +/- sqrt(j 2 - 4k 2 ))/2 b = sqrt(b 2 ); a = k/b; CoCo a’ b’ c CcCc C T =C+asin(b/H b )+asin(a/H a ) b’=sqrt(b 2 -H b 2 ) a’=sqrt(a 2 -H a 2 ) a’ b

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14. 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.

15. 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

16. Adjustable Force Use two different regulators to get two different pressures. Add them in the T using two different valves