Experiment I: Technic Pneumatic Cylinder (part no:47224c01 and 2793c01) pushing In this test the cylinder is pushing the lever up on one site. On the other site of the lever the weighing device is pulling down and measuring the force in Kilogram force. At the seem time the NXT is displaying the pressure within the cylinder in mBar.  I did 12 measurements with different pressure levels and wrote down the Kilogram force displayed at the weighing device. In the chart below you can see the difference between the actual measured force and the calculated force. The pressure measurements I have made 262 ,436, 720, 968, 1211, 1460, 1631, 1716, 1978, 2008, 2158 en 2379 in mBar. Using these pressure measurements we can calculated the force performed by the cylinder: [area of the piston = 0,0001368 m²] The force is calculated as followed: F = P (pressure in Pascal) x A (area in m²) F = 100 x pressure in mBar x 0,0001368 = F = 100 x 262 x 0.0001368 = 3.58 Newton F = 100 x 436 x 0.0001368 = 5.96 Newton F = 100 x 720 x 0.0001368 = 9.85 Newton etc... The weight measurements I have made: 0.72, 1.02, 1.62, 2.06, 2.56, 3.06, 3.30, 3.62, 4.04, 4.08, 4.28 and 4.90 KG Using these weight measurements we can calculated the force measured by the weighing device: [gravity = 9.807 meters per second] [lever proportion = 6/10] The force can be calculated as followed: F = weight x 6/10 lever proportion x gravity F = weight x 0.6 x 9.807 = F = 0.72 x 0.6 x 9.807 = 4.24 Newton F = 1.02 x 0.6 x 9.807 = 6.00 Newton F = 1.62 x 0.6 x 9.807 = 9.53 Newton etc.. note: The calculated force should be a straight line and the measured force should be at least nearly a straight line, but I was not able to made samples with a equal interval. Excel can't display the values on the horizontal axle correct. It displays the pressure values on the horizontal axle with a fixed interval, but as I said, my samples ain't! Click here for the Excel sheet with the results of experiment I:  It turn out to be that the calculated force is a bit higher. But I found it remarkable that theory and practise come this close! The difference between theory and practise is probably a result of friction and tolerance in the calculation of the pistion area of the cylinder. Experiment II: Technic Pneumatic Cylinder (part no:47224c01 and 2793c01) pulling In this test the cylinder is pulling the lever up on one site. On the other site of the lever the weighing device is pulling down and measuring the force in Kilogram force. At the seem time the NXT is displaying the pressure within the cylinder in mBar.  I did 12 measurements with different pressure levels and wrote down the Kilogram force displayed at the weighing device. In the chart below you can see the difference between the actual measured force and the calculated force. The pressure measurements I have made: 99, 165, 201, 485, 659, 783, 1068, 1145, 1360, 1454, 1838 en 2426 in mBar. Using these pressure measurements we can calculated the force performed by the cylinder: [area of the piston = 0,0001368 m²] The force is calculated as followed: F = P (pressure in Pascal) x A (area in m²) F = 100 x pressure in mBar x 0,0001368 = F = 100 x 99 x 0.0001368 = 1.16 Newton F = 100 x 165 x 0.0001368 = 1.93 Newton F = 100 x 201 x 0.0001368 = 2.36 Newton etc... The weight measurements I have made: 0.28, 0.42, 0.46, 0.88, 1.18, 1.38, 1.88, 2.02, 2.20, 2.46, 3.12 and 4.16 KG Using these weight measurements we can calculated the force measured by the weighing device: [gravity = 9.807 meters per second] [lever proportion = 6/10] The force can be calculated as followed: F = weight x 6/10 lever proportion x gravity F = weight x 0.6 x 9.807 = F = 0.28 x 0.6 x 9.807 = 1.65 Newton F = 0.42 x 0.6 x 9.807 = 2.47 Newton F = 0.46 x 0.6 x 9.807 = 2.71 Newton etc.. note: The calculated force should be a straight line and the measured force should be at least nearly a straight line, but I was not able to made samples with a equal interval. Excel can't display the values on the horizontal axle correct. It displays the pressure values on the horizontal axle with a fixed interval, but as I said, my samples ain't! Click here for the Excel sheet with the results of experiment II:  Again it turn out to be that the calculated force is a bit higher. The difference between theory and practise is probably a result of friction and tolerance in the calculated piston area. Experiment III: Small technic Pneumatic Cylinder In this test the cylinder is pushing the lever up on one site. On the other site of the lever the weighing device is pulling down and measuring the force in Kilogram force. At the seem time the NXT is displaying the pressure within the cylinder in mBar.  I did 8 measurements with different pressure levels and wrote down the Kilogram force displayed at the weighing device. In the chart below you can see the difference between the actual measured force and the calculated force. The pressure measurements I have made 500, 797, 957, 1106, 1620, 1818, 1990 and 2318 in mBar. Using these pressure measurements we can calculated the force performed by the cylinder: [area of the piston = 0,0000301 m²] The force is calculated as followed: F = P (pressure in Pascal) x A (area in m²) F = 100 x pressure in mBar x 0,0000301 = F = 100 x 500 x 0.0000301 = 1.51 Newton F = 100 x 797 x 0.0000301 = 2.41 Newton F = 100 x 957 x 0.0000301 = 2.89 Newton etc... The weight measurements I have made: 0.20, 0.34, 0.44, 0.50, 0.74, 0.82, 0.88 and 1.08 KG Using these weight measurements we can calculated the force measured by the weighing device: [gravity = 9.807 meters per second] [lever proportion = 6/10] The force can be calculated as followed: F = weight x 6/10 lever proportion x gravity F = weight x 0.6 x 9.807 = F = 0.20 x 0.6 x 9.807 = 1.18 Newton F = 0.34 x 0.6 x 9.807 = 2.00 Newton F = 0.44 x 0.6 x 9.807 = 2.59 Newton etc.. note: The calculated force should be a straight line and the measured force should be at least nearly a straight line, but I was not able to made samples with a equal interval. Excel can't display the values on the horizontal axle correct. It displays the pressure values on the horizontal axle with a fixed interval, but as I said, my samples ain't! Click here for the Excel sheet with the results of experiment III:  I noticed again during the measurements that there is difference between the calculated and the measured force performed by the LEGO pneumatic cylinder. Experiment IV: Small technic Pneumatic Cylinder In this test the cylinder is pulling the lever up on one site. On the other site of the lever the weighing device is pulling down and measuring the force in Kilogram force. At the seem time the NXT is displaying the pressure within the cylinder in mBar.  I did 8 measurements with different pressure levels and wrote down the Kilogram force displayed at the weighing device. In the chart below you can see the difference between the actual measured force and the calculated force. The pressure measurements I have made 500, 797, 957, 1106, 1620, 1818, 1990 and 2318 in mBar. Using these pressure measurements we can calculated the force performed by the cylinder: [area of the piston = 0,0000301 m²] The force is calculated as followed: F = P (pressure in Pascal) x A (area in m²) F = 100 x pressure in mBar x 0,0000301 = F = 100 x 500 x 0.0000301 = 1.51 Newton F = 100 x 797 x 0.0000301 = 2.41 Newton F = 100 x 957 x 0.0000301 = 2.89 Newton etc... The weight measurements I have made: 0.20, 0.34, 0.44, 0.50, 0.74, 0.82, 0.88 and 1.08 KG Using these weight measurements we can calculated the force measured by the weighing device: [gravity = 9.807 meters per second] [lever proportion = 6/10] The force can be calculated as followed: F = weight x 6/10 lever proportion x gravity F = weight x 0.6 x 9.807 = F = 0.20 x 0.6 x 9.807 = 1.18 Newton F = 0.34 x 0.6 x 9.807 = 2.00 Newton F = 0.44 x 0.6 x 9.807 = 2.59 Newton etc.. note: The calculated force should be a straight line and the measured force should be at least nearly a straight line, but I was not able to made samples with a equal interval. Excel can't display the values on the horizontal axle correct. It displays the pressure values on the horizontal axle with a fixed interval, but as I said, my samples ain't! Click here for the Excel sheet with the results of experiment IV:  Again it turn out to be that the calculated force is a bit higher. The difference between theory and practise is probably a result of friction and tolerance in the calculated piston area. I have made an other page about measurements and calculations on LEGO pneumatic cylinders: Pneumatic LEGO & laws of physics I Check the complete Pneumatic LEGO & the laws of physics II high resolution photo gallery at Google Picasa... Copyright © 2010 www.techbricks.nl. All Rights Reserved.
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Pneumatic LEGO & the laws of physics II
Pneumatic LEGO & the laws of physics II
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