The volume of a cylinder (with a radius of r and a length L ) in the horizontal position filled to a depth (d) can be calculated with the following formula:
L((r2)*(arcos((r-d)/r)) - (r-d)*sqrt(2rd-d2))
Note: Calculator must be set to work in radians as opposed to degrees
For a cylinder, Volume = Area x Depth. Rule: Do not mix increments of measure ( eg. if measuring inches, convert all measurements to inches).
FootnoteThe above is true for the volume of any right cylinder. But if the cylinder is horizontal and the depth is considered to be parallel to a radius, then the answer would be different.
The mathematics underlying the volume-depth relationship for a horizontal cylinder (with plane or domed ends), plus several other shapes, is given in
http://pkukmweb.ukm.my/~jsm/pdf_files/SM-PDF-39-1-2010/16.pdf
The computer program used to produce the illustrations in this article may be downloaded from
http://pyweb.swan.ac.uk/~evans/software.htm
Take the diameter of the base, (D) multiply it by pi, (3.14) and then multiply that by the height of the cylinder (H). Assuming a cylinder of uniform cross-sectional area, the formula is cross-sectional area x vertical height of liquid column If a circular cylinder, then (pi) x (square of radius) x (vertical height of liquid column).
The volume of a cylinder (with a radius of r and a length l ) in the horizontal position filled to a depth (d) can be calculated with the following formula:
L((r2)*(arcos((r-d)/r)) - (r-d)*sqrt(2rd-d2))
Note: Calculator must be set to work in radians as opposed to degrees
The volume of a cylinder (with a radius of r and a length l ) in the horizontal position filled to a depth (d) can be calculated with the following formula:
L(r2*(arcos((r-d)/r)) - (r-d)*sqrt(2rd-d2))
Note: Calculator must be set to work in radians as opposed to degrees
no, because there will be no change in volume
The method you would use is to take a graduated cylinder, face the markings towards you, fill the cylinder with a liquid that won't react with said object (usually water is fine) to an appropriate measure (around half way depending on the size of the object), record the volume before placing the object in the fluid, place the object in the fluid and record the volume after, subtract the initial from the total (liquid +object's displacement) and you will have the objects volume. However you must be aware that certain objects will float or at least have some amount of buoyancy making determining it's volume harder to ascertain (if at all possible) with this method.
Use the Archimedes' method, fill a container with water (right to the top) with a spill tray underneath to catch the water over flow, drop the rock in and measure the amount of water displaced. This method will be about as accurate as you can get.
Density = mass / volume. An object will float if it has less density than the fluid in which it is placed. The buoyant force is equal to the volume (this may be the submerged part of the volume) times the density of the displaced fluid.
The buoyancy force is equal to the WEIGHT of the volume of fluid displaced by the object.
You need more information. It all depends on the size of the cylinder
certain fluid at 10 bar is contained in cylinder behind a piston ,the initial volume being 0.05 m3 calculate the work done by the fluid when it expands reversibly i) according to a law P=(A/ V2 ) -(B/ V) ,to a final volume of 0.1 m3 and a pressure of 1bar,where A and B are constants.Answer(19200 j)
Level with the bottom of the fluid's meniscus
A graduated cylinder is simply a beaker with parallel sides and equally spaced volume markings along the side. As the sides are parallel the volume increases proportionately to the level of fluid in the beaker. Equally spaced markings ("graduations") are marked on the side of the cylinder to indicate the volume of fluid to that point.If you are using a graduated cylinder you will notice that the level of fluid (eg water) will seem to cling to the sides of the glass near the edge in a small radius due to the surface tension of the fluid. This radius is called the miniscus. Always read the volume of fluid from the marking at the bottom of the miniscus.
A graduated cylinder is simply a beaker with parallel sides and equally spaced volume markings along the side. As the sides are parallel the volume increases proportionately to the level of fluid in the beaker. Equally spaced markings ("graduations") are marked on the side of the cylinder to indicate the volume of fluid to that point.If you are using a graduated cylinder you will notice that the level of fluid (eg water) will seem to cling to the sides of the glass near the edge in a small radius due to the surface tension of the fluid. This radius is called the miniscus. Always read the volume of fluid from the marking at the bottom of the miniscus.
Check out "horizontal cylindric segment" in Wolfram Alpha Online. That is the correct term for the solid you are looking for. Wolfram Alpha - "The solid cut from a horizontal cylinder of length L and radius R by a single plane oriented parallel to the cylinder's axis of symmetry (i.e., a portion of a horizontal cylindrical tank which is partially filled with fluid) is called a horizontal cylindrical segment."
It depends on what information you have. If the liquid is stored in a container of which the dimensions are known, then you must calculate the volume of the container. You can simply search google for the formulae for the volume of a cube, cylinder, sphere etc. If the dimensions are not known, but the weight and density of the fluid is, then the volume can be calculated as: volume = weight (divided by) density
Capacity of the container = (pi) x (radius of the round end)2 x (height of the cylinder). That's the capacity of the container. If the volume of the fluid in it is really what you want, then you can use the same formula, but instead of the full height of the container, use only the height of the fluid column, i.e. what we professionals would technically refer to as the "depth".
Check out "horizontal cylindric segment" in Wolfram Alpha Online. That is the correct term for the solid you are looking for. Wolfram Alpha - "The solid cut from a horizontal cylinder of length L and radius R by a single plane oriented parallel to the cylinder's axis of symmetry (i.e., a portion of a horizontal cylindrical tank which is partially filled with fluid) is called a horizontal cylindrical segment."
The solution is easy for a vertical 'cylinder' (i.e. a cylinder with its faces at right angles to the direction of gravity, like barrel standing on a horizontal surface. In this case the volume is given by: V=h*(1/2)*pi*(r^2) Where V=Volume, h is the height to which the cylinder is filled, pi is the number Pi (3.142) and r is the radius of the cylinders faces. If the cylinder is vertical (i.e. a barral lying on the gound so that it could easily be rolled away) it gets a bit more tricky: V=l*(pi*r^2/2-r^2*arcsin(1-h/r)-(r-h)*sqrt(h*(2r-h)) Where l is the length of the cylinder. Note: h is always measured from the lowest point of the fluid contained in the cylinder to the the fluid's surface arcsin is the inverse of sin sqrt denotes the squareroot of the following bracket Good luck! felixmschubert@yahoo.de
It is 41.2 millilitres!
You're going to have to be more specific.