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Hovercraft

Hovercraft. By: Devin Zenner & Grace Overend. Picture of our Hovercraft!. Actual Plans. Pressure. Pressure P=F/A Pressure under hovercraft has to be more than 14.7 psi. PSI Pounds per square inch of pressure “normal” psi is 14.7, what we’re used to everyday

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Hovercraft

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  1. Hovercraft By: Devin Zenner & Grace Overend

  2. Picture of our Hovercraft!

  3. Actual Plans

  4. Pressure • Pressure • P=F/A • Pressure under hovercraft has to be more than 14.7 psi. • PSI • Pounds per square inch of pressure • “normal” psi is 14.7, what we’re used to everyday • As you increase by 10,000 ft. you subtract 4 psi from normal…pressure decreases by altitude. • Estimate the amount of psi that we will need • Take weight that we want to lift divided by the surface area (of hovercraft).

  5. Bernoulli’s Principle • P+1/2P2+Pgh=C • As the Velocity of a fluid increases the Pressure exerted by the fluid decreases. • Applies to leafblower and holes in plastic… • http://library.thinkquest.org/27948/bernoulli.html

  6. Fluid Mechanics • Examples in everyday life • Airplane • Wings • Lift/takeoff • Landings • Army Hovercrafts • HUGE • Holds 12 Hummers • Water AND Land

  7. Fluid Mechanics/Lift • Lift • Molecules provide the necessary lift. • The more molecules that are present the more pressure is built up providing more lift.

  8. Pascal’s Principle • Pascal’s Principle • Fluid (air) in a closed container is transmitted equally to every point of the fluid and to the walls of the container. • Plastic skirt that contains the air provides the necessary lift for the hovercraft. • Air is equally pushed out of all the holes in the bottom • Lifts the hovercraft off of the ground.

  9. Pascal’s Principle Continued • Necessary for the pressure to be equal throughout the skirt • Prevent uneven lift. • Pressure is equal to the initial force divided by the initial area • Also equal to the final force divided by the final area. • P =F1/A1=F2/A2

  10. Example of Pascal’s Principle • The small piston of a hydraulic lift has an area of 0.20m2. A car weighing 1.20x104N sits on a rack mounted on the large piston. The large piston has an area of 0.90m2. How large a force must be applied to the small piston to support the car? • Given: • A1=0.20m2 • A2=0.90m2 • F2=1.20x104 N • Unknown: • F1=? • Use the equation for pressure and apply Pascal’s principle.

  11. Pascal’s Answer • F1=(A1/A2)F2 • =(0.20m2/0.90m2)(1.20x104 N) • F1=2.7x103 N

  12. Pictures! Weeeeee!

  13. Diffraction and the Dangers of Baldness

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