Volume of water in a cylinder
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Volume of water in a cylinder
OK this isn't strictly a computer/internet issue but as this is the cleverest forum here 
I have a cylindrical water tank (on its side). Its 1210 mm in diameter and 3000mm long. Contains aprox 3000l of water. I need to try to calculate the volume of water in the tank at various depths. Obivously at 605mm its half full so thats an easy 1500 litres.
I've googled a few sites including one that came up with
A = pi*a^2/2 - a^2*arcsin(1-h/a) - (a-h)*sqrt(h(2a-h))
(a = radius, h= height of water)
But my head is starting to hurt and i'm not sure why.
Is there a simpler way?
I have a cylindrical water tank (on its side). Its 1210 mm in diameter and 3000mm long. Contains aprox 3000l of water. I need to try to calculate the volume of water in the tank at various depths. Obivously at 605mm its half full so thats an easy 1500 litres.
I've googled a few sites including one that came up with
A = pi*a^2/2 - a^2*arcsin(1-h/a) - (a-h)*sqrt(h(2a-h))
(a = radius, h= height of water)
But my head is starting to hurt and i'm not sure why.
Is there a simpler way?
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Wikipedia is your friend.
http://it.wikipedia.org/wiki/Cilindro_(geometria)
the formula is the one at the bottom.
http://it.wikipedia.org/wiki/Cilindro_(geometria)
the formula is the one at the bottom.
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want another clever little measurment? Ok say you have 2 odd size round pipes carrying a fluid be it gas or other....... and you want to replace it with one to carry the same amount............ measure the diameter across the pipes internally, draw a right angled triangle using the dimensions of the pipes on both 90 degrees angles of the right angled triangle.............. the long diagonal side of the triangle is the internal diameter of the pipe you need to replace the two 
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I think he gave up maths and went to work as a plumber. Certainly pays better!
Ecce Homo! Loquitur...
The Indian mathematician Baudhayana proved the theorem around 800 BC, 300 years before Pythagoras. Johnnie come lately them Greeks.
You got a maths problem, just call the Indian restaurant and ask them to send the answer round.
Mind you, the Chinese take-away will probably claim to be able to get it to you within 30 minutes.....
You got a maths problem, just call the Indian restaurant and ask them to send the answer round.
Mind you, the Chinese take-away will probably claim to be able to get it to you within 30 minutes.....
Spoon PPRuNerist & Mad Inistrator
Ok maybe I'm barking up the wrong tree but the formulae I would use would be -
Pi X Radius squared X height
Pi X Radius squared X height
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I asked for No.6 at my chinese and they sent me a pack of cigarettes 
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
... on its side ? - That'll be "pi r squared" times length then
... Then again you could always replace the tank with a square section one
Coconutty
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
... on its side ? - That'll be "pi r squared" times length then
... Then again you could always replace the tank with a square section one
Coconutty
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Don't need the Greeks - there's a way around everything
If the cylinder is not sealed - pour the water into a measuring jug.
If the cylinder is sealed (which I guess it must be if it's lying on its side and still has the same volume of water in it)....what difference does it make how much is in it? Why do we care (we can't do anything about it)?
If the cylinder is not sealed - pour the water into a measuring jug.
If the cylinder is sealed (which I guess it must be if it's lying on its side and still has the same volume of water in it)....what difference does it make how much is in it? Why do we care (we can't do anything about it)?
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If the cylinder is sealed (which I guess it must be if it's lying on its side and still has the same volume of water in it)....what difference does it make how much is in it? Why do we care (we can't do anything about it)?
Psychophysiological entity
So, is the first calculation in IMP or US? Galls 
Oh, and is there a link for cubes of fluid?
Oh, and is there a link for cubes of fluid?
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Never mind the Greeks - What about the Geeks ?
OK, here we go : I will try and make this sound far more complicated than it really is
The information you need to work out the volume of liquid in your tank is :
1) The Radius of the cylinder ( Half the diameter )
2) The Width of the cylinder ( or Length if you prefer )
3) The depth of the water in it
Assume the tank is less than half full :
Looking end on the shape formed by the water is a "Segment" as represented by the green area :
The depth of water is the height of the segment represented by Line E-D ( also known as the "Sagitta" ) so you need to make up a dipstick or other method of measuring the depth of water left in the tank.
Go to this web site : Daysleeper's Tank Calculator
Check the option for knowing the Radius and Segment Height, fill in the details and click "Calculate" All you need to do then is multiply the answer by the Width ( or Length ) of the tank and you have the volume of water in it.
If the tank is more than half full ( a bit like my Glass
) then work out the total volume of the tank using "PI x R squared x Width ( or Length )" and halve the answer. ( To get the volume of the bottom half of the tank.
Then perform the segment calculation for the top half of the tank - remembering this time that when you dip the tank, the dry bit will be the segment height ! You need to calculate the dry segment area and subtract it from the volume of a half full tank, then add it it to the bottom half of the tank and multiply it by the Width ( or Length ) to get the total volume.
( Or better still wait until the tank is less than half full
)
After a few calculations at various depths you can mark your dipstick ( If you'll excuse the phrase ) and you're done
Coconutty
Edited to sort the typo's and to add :
- I think I might know him 
The information you need to work out the volume of liquid in your tank is :
1) The Radius of the cylinder ( Half the diameter
)2) The Width of the cylinder ( or Length if you prefer )
3) The depth of the water in it
Assume the tank is less than half full :
Looking end on the shape formed by the water is a "Segment" as represented by the green area :
The depth of water is the height of the segment represented by Line E-D ( also known as the "Sagitta" ) so you need to make up a dipstick or other method of measuring the depth of water left in the tank.
Go to this web site : Daysleeper's Tank Calculator
Check the option for knowing the Radius and Segment Height, fill in the details and click "Calculate" All you need to do then is multiply the answer by the Width ( or Length ) of the tank and you have the volume of water in it.
If the tank is more than half full ( a bit like my Glass
) then work out the total volume of the tank using "PI x R squared x Width ( or Length )" and halve the answer. ( To get the volume of the bottom half of the tank.Then perform the segment calculation for the top half of the tank - remembering this time that when you dip the tank, the dry bit will be the segment height ! You need to calculate the dry segment area and subtract it from the volume of a half full tank, then add it it to the bottom half of the tank and multiply it by the Width ( or Length ) to get the total volume.
( Or better still wait until the tank is less than half full
)After a few calculations at various depths you can mark your dipstick ( If you'll excuse the phrase ) and you're done
Coconutty
Edited to sort the typo's and to add :
a huge floater
Last edited by Coconutty; 19th Jul 2006 at 14:10.
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The reason we care is that the cylinder contains the rainwater that is used to flush my loos and wash my clothes. Thus I would like to have a rough idea at what depth of water I should set my system to add mains water so that I dont end up leaving a huge floater in the loo.