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Written by Techbricks.nl
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Thursday, 27 March 2008 21:18 |
I decided to look into the laws of physics and what they can tell about pneumatic LEGO. It could give me answers how to use pumps, tubes, air tanks and cylinders. For example: How many times do I need to move a pump back and forward before a cylinder can push with a certain force? What is the influence of a long tube or one or more air tanks in a pneumatic system? How fast will pressure reduce when a cylinder is filled with air?
Boyle's Law states that the quantity P (pressure) x V (volume) is a constant. (in a situation where the temperature in constant)
note: PSI means pounds per square inch.
1 Bar (1000 mBar) = 14.5038 PSI
1 PSI = 0.068947mBar
(for more info look at http://www.convertworld.com/en/pressure)
So when you have a cylinder connected by a tube to a pump which is pressed for a couple of times then the pressure is rising:
P x V = P x V <=>
P x (cylinder volume + tube volume + pump volume) = initial pressure x (cylinder volume + tube volume + (times the pump is pressed x pump volume)) <=>
P= initial pressure x (cylinder volume + tube volume + (times the pump is pressed x pump volume)) / (cylinder volume + tube volume + pump volume)
note: I use a Mindsensors pneumatic pressure sensor for measuring the air pressure. When it is not connected to a pneumatic system it indicated 0 PSI. Of course that's not true. The air pressure in our environment (around sea level) is about 1000 mBar (14,5 PSI). Sometime it can raise to 1030 mBar (14,9 PSI) (very stable weather), but it can also decline to around 960 mBar (13,9 PSI) (very strong winds). Use a barometer to be sure of the correct pressure level. The pressure that the Mindsensor indicate is relative to what you barometer measures. Pressure = barometer + Mindsensor
These formula tells us that it is important to know what volume size the cylinders, pumps, tubes and tanks have and what the pressure level is. Lets find out!
note: I used a caliper to take the measurements of the LEGO pneumatic parts. The measurements are accurate to a 0.1 millimetre.
note: Most countries in Europe work with metres in stead of inches. Check http://www.convertworld.com/en/length to convert the values to your most favourite system.
1 m (metre) = 100 cm (centimeter) = 1000 mm (milimeter)
1 m² = 10000 cm² = 1.000.000 mm²
1 m³ = 1.000.000 cm³ = 1.000.000.000 mm³ = 1000 litre
1 inch = 2.54 cm
1 cm = 0.3937 inch
1 square inch = 6.4516 cm²
1 cm² = 0.1549 inch
1 cubic inch = 16.38 cm³
1 cm³ = 0.06102 cubic inch
Technic Pneumatic Pump New Complete Assembly (part no: 2797c01)
Diameter of the cylinder = d = 13mm
Area of the piston = a = ∏/4xd² = 3.14159/4x0.013² = 0.0001327 m² = 1.327 cm²
Length of the cylinder = l = 25mm
Volume of the cylinder = l x a = 0.025 x 0.0001327 = 0.000003318 m³ = 3.3183 cm³
Technic Pneumatic Pump Small Complete Assembly (part no: x191c01)
Diameter of the cylinder = d = 6.2mm
Area of the piston = a = ∏/4xd² = 3.14159/4x0.0062² = 0.00003019 m² = 0.3019 cm²
Length of the cylinder = l = 16.5 mm
Volume of the cylinder = l x a = 0.0165 x 0.00003019 = 0,0000004981 m³ = 0.4981 cm³
note: A cylinder does have two in/outlets. The chamber connected to these in/outlets do have a different volume because of the connecting rod. The connection rod, the axle with the pin hole connected to the piston, takes in a significant amount of volume which decreases the volume of that chamber.
Technic Pneumatic Cylinder New Complete Assembly (part no: 2793c01)
Length of the cylinder = l = 27.8 mm
Diameter of the piston = d = 13.2 mm
Right chamber:
Area of the piston surface = a = ∏ / 4 x d² = 3.14159 / 4 x 0.0132² = 0,0001368 m² = 1.368 cm²
Volume of the cylinder = l x a = 0.0278 x 0,0001368 = 0,000003804 m³ = 3.804 cm³
Left chamber:
Diameter connection rod = D = 5 mm
Area of the piston surface = a = ∏ / 4 x (d² - D²) = 3.14159 / 4 x (0.0132² - 0.005²) = 0.0001172 m² = 1.172 cm²
Volume of the cylinder = l x a = 0.0278 x 0.0001172 = 0.000003258 m³ = 3.258 cm³
Technic Pneumatic Cylinder New with Hole Base (Complete Assembly) (part no:47224c01)
Length of the cylinder = l = 27.5 mm
Diameter of the piston = d = 13.2 mm
Right chamber:
Area of the piston surface = a = ∏ / 4 x d² = 3.14159 / 4 x 0.0132² = 0,0001368 m² = 1.368 cm²
Volume of the cylinder = l x a = 0.0275 x 0,0001368 = 0,000003762 m³ = 3.762 cm³
Left chamber:
Diameter connection rod = D = 5 mm
Area of the piston surface = a = ∏ / 4 x (d² - D²) = 3.14159 / 4 x (0.0132² - 0.005²) = 0.0001172 m² = 1.172 cm²
Volume of the cylinder = l x a = 0.0275 x 0.0001172 = 0,000003223 m³ = 3.223 cm³
Technic Pneumatic Cylinder Small Complete Assembly (part no: x189c01)
Length of the cylinder = l = 14.9 mm
Diameter of the piston = d = 6.2 mm
Right chamber:
Area of the piston surface = a = ∏ / 4 x d² = 3.14159 / 4 x 0.0062² = 0.00003019 m² = 0.3019 cm²
Volume of the cylinder = l x a = 0.0149 x 0,0001368 = 0.0000004498 m³ = 0.4498 cm³
Left chamber:
Diameter connection rod = D = 3.2 mm
Area of the piston surface = a = ∏ / 4 x (d² - D²) = 3.14159 / 4 x (0.0062² - 0.0032²) = 0.00002214 m² = 0.2214 cm²
Volume of the cylinder = l x a = 0.0149 x 0.0001172 = 0.00000033 m³ = 0.33 cm³
note: A pump does have a outlet with a one-way valve. When the pump is pressed the air is leaving the pump through the valve. When the pump is pulled air is sucked into the pump from the outside.
Technic Pneumatic Tubing (part no: x188 or 5102cxx)
diameter = 2 mm
volume per metre = l x ∏ / 4 x d² = 1 x 3.14159 / 4 x 0.002² = 0.000003141 m³ = 3.141 cm³ per metre
diameter = 2.5 mm
volume per metre = l x ∏ / 4 x d² = 1 x 3.14159 / 4 x 0.0025² = 0.000004908 m³ = 4.908 cm³ per metre
Technic Pneumatic Airtank (part no: 67c01)
volume air tank = 28 cm³ (filled the tank with water and measured the amount of water in a measuring jug)
Technic Pneumatic Switch (part no: 4694)
Technic Pneumatic Switch - New (part no: 4694b)
note: Measurements done with Mindsensors pressure sensor can be read out by NXT-G, RobotC and NXC/NBC software. The sensor it self is quite accurate: the raw value goes from about 130 to 1023, which should match with 0 to 35 PSI. At www.mindsensors.com Mindsensors offers a pressure NXT-G block that calculate the pressure value in PSI. That's not very accurate because NXT-G calculate in absolute figures without a decimale comma.
I have asked Dr. Nitin Patil (founder and CEO of Mindsensors) to release a NXT-G block that calculate a more accurate value in mBar. He send me a new block that even calculate the raw sensor value (25 times more accurate). Read more about it here...
Overview of the Pneumatic parts with the calculated volumes:
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part description
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part nr at peeron or bricklink
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volume (right) chamber |
volume (left) chamber |
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cubic inch |
cm³ |
cubic inch |
| Technic Pneumatic Pump New Complete Assembly |
2797c01 |
3.31 |
0.201 |
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| Technic Pneumatic Pump Small Complete Assembly |
x191c01 |
0.49 |
0.029 |
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| Technic Pneumatic Cylinder New Complete Assembly |
2793c01 |
3.80 |
0.231 |
2.25 |
0.137 |
| Technic Pneumatic Cylinder New with Hole Base (Complete Assembly) |
47224c01 |
3.76 |
0.229 |
3.28 |
0.200 |
| Technic Pneumatic Cylinder Small Complete Assembly |
x189c01 |
0.44 |
0.026 |
0.33 |
0.020 |
| Technic Pneumatic Tubing 2 mm |
x188 or 5102cxx |
3.14 per metre |
0.191 per metre |
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| Technic Pneumatic Tubing 2 mm (non official) |
x188 |
4.90 per metre |
0.299 per metre |
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| Technic Pneumatic Airtank |
67c01 |
28 |
1.708 |
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Experiment I: a pump and a cylinder.
In this experiment I pressed the pomp 4 times. The pressure is raised from 0, 7, 17, 25 to 33 PSI.
According the calculated volumes of the pump, the tube and the cylinder we can now calculate the pressure after we pressed the pump a couple of times:
[ pump volume = 3.3 cm³ ][ cylinder volume = 3.8 cm³ ][ tube length = 13 cm ][ tube volume = 4.9 cm³/m ][ initial pressure = barometer = 14.3 PSI ]
P= initial pressure x (cylinder volume + tube volume + (times the pump is pressed x pump volume)) / (cylinder volume + tube volume) - initial pressure
p= 14.3 x (3.8 + (0.13 x 4.9) + (1 x 3.3)) / (3.8 + 0.13 x 4.9) - 14.3 = 9.99 PSI
p= 14.3 x (3.8 + (0.13 x 4.9) + (2 x 3.3)) / (3.8 + 0.13 x 4.9) - 14.3 = 19.98 PSI
p= 14.3 x (3.8 + (0.13 x 4.9) + (3 x 3.3)) / (3.8 + 0.13 x 4.9) - 14.3 = 29.97 PSI
etc...
Click here for the Excel sheet with the results of experiment I:
It turn out to be that the calculated pressure 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 air leaks and tolerance in the calculated volumes.
Experiment II: a small pump.
In this experiment I pressed the pump only one time. The pressure raised from 0 to 27 PSI.
According the calculated volumes of the pump and the tube we can now calculate the pressure after we pressed the pump:
[ pump volume = 0.5 cm³ ][ tube length = 4 cm ][ tube volume = 4.9 cm³/m ][ initial pressure = barometer = 14.3 PSI ]
P = initial pressure x (tube volume + pump volume) / (tube volume) - initial pressure
p = (14.3 x (0.045 x 4.9 + 0.5) / 0.04 x 4.9) - 14.3 = 32,4 PSI
The calculated pressure is about 5 PSI lower which is again probably a result of air leaks and tolerance in the calculated volumes.
Experiment III: a big pump.
In this experiment I pressed the pump one time. The pressure level got to high and the tubing disconnected from the pump.
According the calculated volumes of the pump and the tube we can calculate the pressure after we pressed the pump when the tubing wasn't disconnected from the pump:
[ pump volume = 3.3 cm³ ][ tube length = 4 cm ][ tube volume = 4.9 cm³/m ][ initial pressure = barometer = 14.3 PSI ]
P = initial pressure x (tube volume + pump volume) / (tube volume) - initial pressure
p = (14.3 x (0.045 x 4.9 + 3.3) / 0.04 x 4.9) - 14.3 = 214 PSI
After calculating the pressure level it is clear why the tube disconnected!
Experiment IV: a long tube.
In this experiment I took a 1,8 metre long tube and connected one end to the pump and the other to the pressure sensor. I pressed the pump 7 times. The pressure raised from 0 to 3, 9, 14, 19, 24, 28, 32 PSI.
According the calculated volumes of the pump and the tube we can now calculate the pressure level after we pressed the pump 7 times:
[ pump volume = 3.3 cm³ ][ tube length = 1.8 m ][ tube volume = 4.9 cm³/m ][ initial pressure = barometer = 14.3 PSI ]
P= initial pressure x (tube volume + (times the pump is pressed x pump volume)) / tube volume - initial pressure
p= 14.3 x ((1.8 x 4.9) + (1 x 3.3)) / (1.8 x 4.9) - 14.3 = 5.3 PSI
p= 14.3 x ((1.8 x 4.9) + (2 x 3.3)) / (1.8 x 4.9) - 14.3 = 10.7 PSI
p= 14.3 x ((1.8 x 4.9) + (3 x 3.3)) / (1.8 x 4.9) - 14.3 = 16.0 PSI
etc...
Click here for the Excel sheet with the results of experiment IV:
Again it turn out to be that the calculated pressure is a bit higher. The difference between theory and practise is probably a result of air leaks and tolerance in the calculated volumes.
Experiment V: an airtank and a big pump.
In this experiment I connected an pneumatic airtank to a pump and the pressure sensor. I used 12 cm of pneumatic hose. I pressed the pump 26 times. The pressure raised from 0 to 1 ,2 ,4 ,5 ,6 ,8 ,9 ,10 ,12 ,13 ,14 ,15 ,16 ,18 ,19 ,20 ,21 ,22 ,24, 25 ,26 ,27 ,28 , 29, 30, 31 PSI.
According the calculated volumes of the pump, the air tank and the tube we can now calculate the pressure level after we pressed the pump 28 times:
[ pump volume = 3.3 cm³ ][ tube length = 12 cm ][ tube volume = 4.9 cm³/m ][ initial pressure = barometer = 14.3 PSI ][ volume airtank = 28 cm³ ]
P= initial pressure x ((tube + airtank volume) + (times the pump is pressed x pump volume)) / (tube + airtank volume) - initial pressure
p= 14.3 x ((0.12 x 4.9 + 28) + (1 x 3.3)) / (0.12 x 4.9 + 28) - 14.3 = 1.6 PSI
p= 14.3 x ((0.12 x 4.9 + 28) + (2 x 3.3)) / (0.12 x 4.9 + 28) - 14.3 = 3.3 PSI
p= 14.3 x ((0.12 x 4.9 + 28) + (3 x 3.3)) / (0.12 x 4.9 + 28) - 14.3 = 4.9 PSI
etc...
Click here for the Excel sheet with the results of experiment V:
I noticed during the measurements that above 30 PSI the pressure was dropping when I took a break to write down the values. Again we see a difference between the calculated and the measured pressure.
Experiment VI: an airtank and a small pump.
In this experiment I have pushed the piston 185 time forwards and backwards. Every time the pressure level was raised with 1 PSI, I wrote down the times I had pushed the piston: 17, 23, 28, 32, 38, 43, 48, 53, 59, 64, 71, 76, 82, 88, 94, 101, 107, 114, 121, 129, 136, 143, 150, 156, 164, 171, 178, and 185 times.
According the calculated volumes of the pump, the air tank and the tube we can now calculate the pressure level after we pressed the pump 28 times:
[ pump volume = 0,498 cm³ ][ tube length = 12 cm ][ tube volume = 4.9 cm³/m ][ initial pressure = barometer = 14.3 PSI ][ volume airtank = 28 cm³ ]
P= initial pressure x ((tube + airtank volume) + (times the pump is pressed x pump volume)) / (tube + airtank volume) - initial pressure
p= 14.3 x ((0.12 x 4.9 + 28) + (1 x 0.498)) / (0.12 x 4.9 + 28) - 14.3 = 0.23 PSI
p= 14.3 x ((0.12 x 4.9 + 28) + (2 x 0.498)) / (0.12 x 4.9 + 28) - 14.3 = 0.49 PSI
p= 14.3 x ((0.12 x 4.9 + 28) + (3 x 0.498)) / (0.12 x 4.9 + 28) - 14.3 = 0,74 PSI
etc...
note: This graphic drawing may looks unusual to you in comparing to the other experiments. Like I said I wrote down the times I pressed the pump instead of writing down the pressure every time I pressed the pump. The measurements on the horizontal axis of the graphic drawing is not linear! Therefor the measured pressure is draw like a straight line and the calculated pressure isn't!
Click here for the Excel sheet with the results of experiment VI:
With the small pump it takes quite a long time before the pressure is raising. I also noticed that air is leaking away when that pressure is above 27 PSI, probably caused by the one-way valve of the pump.
Experiment VII: a pneumatic switch, pump and cylinder.
This experiment is done with a pump, airtank, cylinder and switch. The experiment starts after I disconnected the cylinder (by the switch) from rest of the pneumatic system. Then I used the pump to raise the pressure to 30 PSI. I also pushed the cylinder piston in, so all the air will push out of the camber.
I measure the pressure while I release air into the cylinder by using the switch. Doing so will push out the cylinder piston. Ones this is completed, I measure the pressure, disconnect the cylinder again and push cylinder piston in, so all air will push out of the chamber again. I have repeated these steps 6 times. The pressure went down from 30 to 24, 20, 17, 14, 12, 10 PSI.
According the calculated volumes of the air tank, the cylinder and the tube we can now calculate the pressure level after we fill the cylinder 6 times:
note: The pump doesn't take part in this experiment. It is only there to reach the starting pressure level of 30 PSI!
[ length tube1 = 17 cm ][ length tube2 = 3 cm ][ tube volume = 4.9 cm³/m ][ out site pressure = barometer = 14.3 PSI ][ volume airtank = 28 cm³ ] [ cylinder volume = 3.76 cm³ ]
P = ((tube1 + airtank) x opening pressure) / (tube1 + airtank + (times filled the cylinder x (tube2 + cylinder)))
p = ((0.17 x 4.9 + 28) x (30 + 14.3)) / (0.17 x 4.9 + 28 + (1 x (0.03 x 4.9 + 3.76))) - 14.3 = 24.71 PSI
P = ((0.17 x 4.9 + 28) x (30 + 14.3)) / (0.17 x 4.9 + 28 + (2 x (0.03 x 4.9 + 3.76))) - 14.3 = 20.55 PSI
P = ((0.17 x 4.9 + 28) x (30 + 14.3)) / (0.17 x 4.9 + 28 + (3 x (0.03 x 4.9 + 3.76))) - 14.3 = 17.20 PSI
etc...
Click here for the Excel sheet with the results of experiment VII:
I am surprised that theory and practise are almost equal. I must have made one or more errors during the measurements or...
I have made an other page about measurements and calculations on LEGO pneumatic cylinders: Pneumatic LEGO & laws of physics II
Check the complete Pneumatic LEGO & the laws of physics high resolution photo gallery at Google Picasa...
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Pneumatic LEGO & the Laws of Physics I
