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Power = Pressure * Pipe Size Area

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

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Power = Pressure * Pipe Size Area

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  1. Power = Pressure * Pipe Size Area

  2. Pneumatics Pushing Force 60 PSI 1” Force = Surface Area * Pounds Per Square Inch = Pi*R2 * 60 lbs/inch2 = 3.14*12 inch2 * 60 lbs = 188 lbs

  3. Pneumatics Pulling Force 60 PSI R=1” 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 r = ¼”

  4. 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

  5. Pneumatic Forces at Different Pressures

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

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

  8. Pneumatic Mounting: Open Position b C Length = 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 c

  9. Pneumatic Mounting: Closed Position b C a Length = 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)) c 100 = (324-b2) + b2 -1.532*sqrt(324-b2)b ;from previous page 1.532*sqrt(344-b2)b=224 Sqrt(324-b2)b=146.2141 (324-b2)b2=21378.562 ;square both sides 0 = b4-324b2 + 21378.562 b2= (324 +/- sqrt(3242 – 4*21378.562))/2 ; quadratic eq b2= 92.247 or b2 = 231.7527 b = 9.6 or b = 15.223 a = sqrt(324 - b2) A = 15.223 or a = 9.6

  10. Pneumatic Mounting: Two Positions Closed: Length = Cc = 10 inches Cc = 10 c2 = a2 + b2 - 2a*b*Cos(Cc) 102 = a2 + b2 - 2a*b*Cos(10) 100 = a2 + b2 – 1.9696a*b 100 = (324-b2) + b2 -1.9696*sqrt(324-b2)b 1.9696*sqrt(344-b2)b=224 Sqrt(324-b2)b=113.72778 (324-b2)b2=12934.008 0 = b4-324b2 + 12934.008 b2= (324 +/- sqrt(3242 – 4*12934.008))/2 b2= 46.631 or b2 = 277.369 b = 6.8287 or b = 16.6544 a = sqrt(324 - b2) A = 16.6544 or a = 6.8287 b a c

  11. co2 = a2 + b2 -2a*b*Cos(Co) cc2 = a2 + b2 -2a*b*Cos(Cc) co2 -cc2 = 2a*b*(Cos(Cc)-Cos(Co)) (co2 -cc2 )/(Cos(Cc)-Cos(Co))=2ab ab = k; where k= ½ (co2 -cc2 )/(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+k2where (- 2k*Cos(Co)- co2) b2 = (-j +/- sqrt(j2 - 4k2))/2 b = sqrt(b2); a = k/b; Pneumatic Mounting: Two Positions b Co b Cc a a co cc

  12. co2 = a2 + b2 -2a*b*Cos(Co) cc2 = a2 + b2 -2a*b*Cos(Cc) co2 -cc2 = 2a*b*(Cos(Cc)-Cos(Co)) (co2 -cc2 )/(Cos(Cc)-Cos(Co))=2ab ab = k; where k= ½ (co2 -cc2 )/(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+k2where (- 2k*Cos(Co)- co2) b2 = (-j +/- sqrt(j2 - 4k2))/2 b = sqrt(b2); a = k/b; Pneumatic Mounting: Two Positions b’ b b’ Co Cc a a’ a’ co cc CT=C+asin(b/Hb)+asin(a/Ha) b’=sqrt(b2-Hb2) a’=sqrt(a2-Ha2)

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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

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