Finding the Volume of a Horizontal TankDate: 2/13/96 at 10:3:30 From: Gary Silva Subject: Horizontal cylinder What is the formula for finding the volume of a horizontal cylinder? Date: 2/13/96 at 16:16:55 From: Doctor Byron Subject: Re: Horizontal cylinder Hi Gary, I'm assuming that by horizontal cylinder you mean a shape like an ordinary soft drink can. For any rectangular solid (one in which the sides are perpendicular to the base and the top and bottom faces are parallel) the volume is equal to the area of the base times the height. For example, in a box, the area is the length times the width (giving you the area of the base) times the height. A cylinder is only different in the formula you use to find the area of the base. The area of any circle is given by Pi times the radius squared. Therefore, the equation you're looking for looks like this: (Pi)*(radius)^2*(height) Pi is approximately equal to 3.1416. It is the ratio between a circle's diameter and its circumference. This ratio never changes regardless of the size of the circle. -Doctor Byron, The Math Forum Date: 2/13/96 at 20:21:21 From: Gary Silva Subject: Horizontal cylinder > I'm assuming that by horizontal cylinder you mean a shape like an > ordinary soft drink can. But, by horizontal, I mean that the soft drink can would be lying on its side. > (Pi)*(radius)^2*(height) Wouldn't this be for a "vertical" cylinder? My problem is that I have a round tank lying on its side so that the round ends are at the front and back as opposed to the top and bottom. Is there a formula to compute the volume of the cylinder at any level, when it is in this position? In the upright position, it's easy since the height is the liquid level. Date: 2/13/96 at 21:14:3 From: Doctor Byron Subject: Re: Horizontal cylinder Thanks for the clarification. This is certainly a much more interesting problem. The basic idea, however, is still the same. The idea is to find the area of the region where the water hits an end of the tank and multiply it by the length of the tank. So the base times height idea is still in effect. Consider a tank with the water level below the midpoint. Now imagine the sector of the circle that contains the area of water that intersects the end of the tank (a sector essentially a "slice of the pie.") The area in the sector that is not part of the water's area forms an isosceles triangle. By finding the area of the sector (which is simply a fraction of the circle's area) and subtracting the area of the triangle, you will find the area that forms the 'base' of your volume. In the case of a water level above the midpoint, the procedure is similar except that you have a sector and a triangular area that needs to be added to that sector. -Doctor Byron, The Math Forum Date: 2/14/96 at 11:6:6 From: Gary Silva Date: Wed, 14 Feb 1996 10:52:00 -0500 Subject: Horizontal cylinder Thank you for getting back to me. What I have is this: Tank 1: horizontal (on its side) with diameter 95" and length 52". Tank 2: horizontal (on its side) with diameter 95" and length 116" I need to be able to determine the contents of each tank at .125" intervals. I have set up a spreadsheet for this and want to set up a formula in the spreadsheet that will take the level measurement of the contents and compute the volume. l. Is that possible? 2. What is the formula? G. Silva Date: 2/14/96 at 16:58:31 From: Doctor Byron Subject: Re: Horizontal cylinder First of all, you will be glad to hear that what you are asking is indeed possible. I cannot give you specific instructions about how to use your particular spreadsheet, but I hope some basic guidelines and the volume equation will be helpful. Here is an equation for the volume of the water of height h, provided that h is less than r. It was derived using the method described in the previous letter: -1 r - h V = L * [[r^2 * cos (-------)] - [sqrt(2rh - h^2) * (r-h)]] r Where L is length, h is height of the water, r is the radius, sqrt refers to the square root, and cos^(-1) is the inverse cosine. Once you have found the volumes up to this point, you can find the remaining volumes by the symmetry of the tank. That is, you can use the distance from the top of the tank for h, and subtract the result for V from the total volume (this finds the volume of air in the top and subtracts from the total volume to find the amount of water). As for doing this with a spreadsheet, I recommed you create one row or column with a series of the height values which you wish to find. You should then be able to write a formula into the spreadsheet which will allow you to map all the values for height in that column/row to values for volume in some other column/row. This can vary greatly between spreadsheets, however. Best of luck in your efforts. -Doctor Byron, The Math Forum Date: 2/27/96 at 17:8:16 From: Gary Silva Subject: More help I'm back again with that horizontal tank problem. I was able to get the formula that you gave me to work in a spreadsheet. But, it doesn't seem to give accurate answers. For instance. When I use the formula for a vertical tank (r^2h) I get 1595.6 gallons for a full tank (which is correct by my tank literature), and 797.8 gallons at the halfway point. But, when I use the horizontal formula, I get 819 gallons at the halfway point. I'm using 231 to convert from cubic inches to gallons. This 23 gallons wouldn't be of great concern except that when I compare other results to those actually given me on a chart, I find I'm as much as 200 gallons off at some points. Where's the glitch??? G. Silva Date: 2/27/96 at 17:30:46 From: Doctor Byron Subject: Re: More help Hi again Gary, I am going to check over the formula which I sent you earlier. There admittedly may have been some problem with it, for I derived it myself, not having been able to find it worked out ahead of time. I will look over what I did carefully. In the meantime, would you mind sending me the output from the spreadsheet program which you used? -Doctor Byron, The Math Forum Date: 2/28/96 at 9:58:32 From: Gary Silva Subject: Re: More help >I will look over what I did carefully. In the meantime, would you >mind sending me the output from the spreadsheet program which you used? =(L*(r^2*(ACOS((r-h)/r)))-(SQRT(2*(r*h)-2*h)(r-h)))/231 I have two tanks, both 95" in diameter. I'm using r=47.5 for both. Tank 1: 52" long w/max capacity of 1595.6 gallons @h= chart formula 11.875 115.1 393.4 23.750 311.9 551.9 35.625 546.5 688.8 47.5 797.8 819.0 Tank 2: 116" long w/max capacity of 3559.4 gallons I didn't run any figures using this tank's measurements G. Silva Date: 2/28/96 at 10:30:14 From: Doctor Byron Subject: Re: More help I've done some checking on your results. The main problem seems to be that the equation you sent me: (L*(r^2*(ACOS((r-h)/r)))-(SQRT(2*(r*h)-2*h)(r-h)))/231 isn't quite what I had intended. It should actually look something like: (L*((r^2*(ACOS((r-h)/r)))-(SQRT(2*(r*h)-h^2)(r-h))))/231 You'll notice that the 2*h has become an h^2, and the L now multiplies the entire expression. If you're still not sure, looking at it in the form I sent it before may make things a little easier to see: / -1 r - h \ V = L * | [r^2 * cos (-------)] - [sqrt(2rh - h^2) * (r-h)] | \ r / (Actually I changed it slightly to make it clearer that the L multiplies the whole thing.) Good luck. I hope this helps you get some better results. -Doctor Byron, The Math Forum |
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