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BALL – ZOOKA

BALL – ZOOKA. Presented by Steve Taylor Dan Wixted Jonathan Goding Paige Beaumont. Define Problem Safe to use device Ball must not exceed 50 mph Release mechanism will be 6’ from ground and 6” in diameter Distance constraints of 13’ to 20’ Developed probable solutions Crossbow

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BALL – ZOOKA

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  1. BALL – ZOOKA Presented by Steve Taylor Dan Wixted Jonathan Goding Paige Beaumont

  2. Define Problem Safe to use device Ball must not exceed 50 mph Release mechanism will be 6’ from ground and 6” in diameter Distance constraints of 13’ to 20’ Developed probable solutions Crossbow Single Phase Motor Catapult Guided tubular system Identify Problem/Research

  3. Chinese Crossbow

  4. Crossbow History • Archeologists have found crossbows in 2,500-year-old Chinese graves, and some historians believe that they existed in China as early as 2,000 BC. • Two Types • Mediterranean Crossbow • Uses a lever and a rotating nut as a trigger mechanism • Chinese Crossbow • Uses hooks and levers as a trigger

  5. Two factors for figuring out the amount of energy a bow can hold Draw weight The amount of force required to draw the bow Draw length The distance between the bowstring’s position at rest and its position when drawn Simplified A bows overall strength depends on how hard it is to pull the string and how far back you are able to pull it The bow's energy, measured in foot-pounds or joules The arrow's velocity, measured in feet or meters per second Elements for Crossbow Building

  6. Physics of a Crossbow

  7. Hundman Lumber 2”x4”x10’ 3.70 Ace Hardware 4” O.D. X 10’ 18.99 2-1/2” O.D. X 10’ 6.99 2” O.D. X 10’ 4.99 Average Elbow 3.00 Cost

  8. Prove it will work Velocity of projectile measured in feet or meters per second Downward vertical acceleration The bow's energy, measured in foot-pounds or joules Mathematically Solve to find the optimal solution

  9. Finding Velocity:

  10. Sheet1 Spreadsheet's Author: Jonathan GodingLast Updated with Excel 97** This cell's data may not be accurate. Created with xlhtml 0.5

  11. Formula to figure out time it takes for the ball to hit the target: Distance= Initial Velocity * time + ½ *Acceleration*time2 Our initial velocity was 651.36 in/sec which we converted into m/sec to get 16.5554 m/sec Acceleration is 10 m/sec2 Because we knew our distance was 13-20 feet we plugged these values in to get the time it took the ball to hit the target. Mathematically find the time it takes to hit the target:

  12. Formula used to find vertical distance traveled: Dvertical= ½ * Acceleration * Time2 Using the same amount for Acceleration as before and the value for time we just calculated we could plug into this formula and get the distance dropped vertically. Mathematically find the amount that the ball falls vertically:

  13. 13 ft: 3.96m=16.5554*(t)+1/2*10*(t)2 t= .2241 seconds ½*10*(.2241)2=.251227m =9,89837in Thus using the formula angle=opposite/adjacent we get that the angle should be .133963 Degrees. 14 ft: 4.2669m=16.5554*(t)+1/2*10*(t)2 t= .24030 seconds 1/2*10*(.24030)2=.288721m=11.37563in Angle=.1331882 Degrees Our Answers:

  14. 15 ft: 4.571m=16.5554*(t)+1/2*10*(t)2 t= .256309 seconds ½*10*(.256309)2=.32847m =12.94181in angle=.133010 Degrees. 16 ft: 4.87m=16.5554*(t)+1/2*10*(t)2 t= .2721853 seconds 1/2*10*(.2721853)2=.370424m=14.5947203in Angle=.1333058 Degrees Our Answers:

  15. 17 ft: 5.1813m=16.5554*(t)+1/2*10*(t)2 t= .2879 seconds ½*10*(.2879)2=.4152m =16.3322in angle=.1133 Degrees. 18 ft: 5.4861m=16.5554*(t)+1/2*10*(t)2 t= .3035 seconds 1/2*10*(.3035)2=.4607m=18.1522in Angle=.1349 Degrees Our Answers:

  16. 19 ft: 5.790m=16.5554*(t)+1/2*10*(t)2 t= .3190 seconds ½*10*(.3190)2=.5089m =20.05in angle=.1361 Degrees. 20 ft: 6.0957m=16.5554*(t)+1/2*10*(t)2 t= .3344 seconds 1/2*10*(.3344)2=.5591m=22.03in Angle=.1376 Degrees Our Answers:

  17. Resources • http://science.howstuffworks.com/crossbow.htm • http://crossbowhunters.com/_sgt/m3_1.htm • http://www.vintageprojects.com/archery/crossbowfull.pdf • http://et.wcu.edu/ET_students-ARES_PG.htm

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