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Volume of a Cylindrical Tank

Date: 3 Feb 1995 07:34:24 -0500
From: Blockwood
Subject: Volume of partially filled cylinder on its side?

Okay, here's a stumper.

I have to keep an inventory of how much is kept in a farm of tanks 
outside my school. The tanks are cylindrical, which would be no 
problem if they were standing on end, but they're not, they're lying 
on their sides. 

             x      x
          x------------x      Liquid level which is variable,
         x              x           can go up or down
        x                x 
         x              x
            x        x 

Imagine the above is a circle. I know the diameter of the circle, 
and I can measure from the top of the circle down to the liquid level. 
How can I use this information to derive the amount of liquid in the 
circle? I hope to make a formula I can use in a spreadsheet. 

Thanks so much.

Date: 3 Feb 1995 19:51:37 GMT
From: Dr. Math
Subject: Re: Volume of partially filled cylinder on its side?

Hello there!  

Thanks for writing to Dr. Math!  You asked a good question.

Before we start, let's go over the "law of sines"  (we'll be using
it shortly).  Given a triangle ABC where A, B, and C are vertices, 
the following relation holds:

      sin A       sin B     sin C
     --------  = ------- = -------
        A           B         C

where  sin A  means sin of the angle at vertex A, and A (when it 
appears in the denominator) is the length of the side of the 
triangle opposite A.

Okay, now on to the problem. You did a nice job with your drawing 
on the computer, so I'm not going to draw another picture. Instead, 
you draw a picture on paper and label the following:

  Call the distance from the top of the circle to the  liquid level h.
  Call the radius of the circle r.  
  (these are given quantities)

Now, draw in the center of the circle. Draw a line connecting the 
center of the circle to the top of the circle. Then draw two more 
lines from the center of the circle to the points where the liquid 
level lines intersect the circle. Note that these lines are just radii, 
so they have length r. 

Now you have 3 triangles drawn, 2 of which are identical and make up the
third. Look at one of the small triangles. Call the angle at the center
angle z. One of the other angles is a 90 degree angle. Call the third
angle y.  Now, note that the distance between the center of the circle
and the top of the circle is r (that's the definition of radius). We
already said that the distance between the top of the circle and the
liquid level was h, so that means the distance between the center of the
circle and the liquid level must be r-h, right? So, our small triangle
has a side of length r-h opposite the angle y, a side of length r
opposite the 90 degree angle, and a side of length x opposite angle z.
x is an unknown quantity we will need later on in the problem.  

By the law of sines, we have the following relations:

      r        r-h         x
    -----  = --------  = -----
*   sin 90    sin y      sin z

sin 90 = 1, so we have:

  ----- = r
  sin y

Solve this equation for y to get:

y =  arcsin  [(r-h)/r]

Note also that z  = 90  - y.

And, sin z  = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]]  =
(1/r)Sqrt(2rh -  h^2)

(this is by trig identities and stuff...if you have questions, 
write back).

So, then equation * becomes:
           r  =   --------------
                   Sqrt (2rh - h^2)

So, x   = Sqrt (2rh - h^2)

Now the problem is a lot easier. To find the area of the whole
circle, let's first consider the area contributed by the segment 
we've been talking about that is of angle 2z from the center.  

The area of this segment occupied by liquid is just the area of 
the 2 small triangles. We know the base of each triangle is x 
(see above) and the height is r-h. So, the total area contributed 
by this segment is x(r-h).

Now consider the area outside of the segment. That is, we want 
the area of a portion of the circle. If we measure z in degrees we 
want the portion of the circle that takes up 360 - 2z degrees of the 
circle. So, the fraction we are dealing with here is:

                     360 - 2z

The area of the portion we want then is:

                     360 - 2z
                     -------- (Pi)r^2

Now add these two areas together and you are done. Thus, the total 
area is going to be:

   360 - 2(90 - arcsin  [(r-h)/r])
   -------------------------------  (Pi)r^2 +  (r-h)Sqrt (2rh - h^2)

You can simplify this to get something that looks nicer, but I'll 
leave that to you.
Please do not hesitate to write back if anything is confusing or 
doesn't seem right, okay? Hope this helps!

- Sydney, "Dr. Math"
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Trigonometry

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