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Changing Angle of a Tank

Date: 06/12/2003 at 23:21:09
From: Daniel
Subject: Geometry



Dimensions a,b,r are known, as is the linear distance from point A to 
point B. I need a formula to determine the value of h.

Points A and B represent pressure sensors in fixed positions on the 
base of a round tank. The chord through CD represents the water level 
in the tank. Lines a and b are the heights of water registered by each 
sensor.

Thanks.


Date: 06/13/2003 at 08:56:23
From: Doctor Peterson
Subject: Re: Geometry

Hi, Daniel.

Your picture looks something like this:

                      *********
                ******         ******
             ***                     ***
           **                           **
         **                               **
        *                                   *
       *                                     *
      *                                       *
     *                                         *
     *                                         *
    *          r                                *
    *---------------------O---------------------*
    *                     |                     *
     *                    |                    *
     *                    |                    *
      *                   |                   *
       *                  |                  *
        *-----------C-----+----------D------*
         **         |     |         b|    **
           **       |a    |h         |  **
             ***    |     |          B**
                ****A*    |    ******
                      *********

It doesn't show the chord AB, whose length I assume is what you mean 
by "the linear distance from A to B." It would be easier if you meant 
the distance CD, but we can get the latter from the former.

If the sensors are in fixed positions, then you should be able to 
measure more than just the distance between A and B in order to locate 
them. I would just think in terms of a coordinate system with the 
y-axis through the center, and the x-axis through the bottom of the 
circle. Then you can simply measure the coordinates of A and B; and 
the depth of water h is just the height a or b plus the height (y 
coordinate) of A or B respectively. If you can't measure those 
directly, then we could make some calculations based on whatever you 
can measure conveniently.

The important thing is that we don't have to calculate h directly 
from a, b, and r each time, as if A and B could move around. Rather, 
we can find the positions of A and B once, and then get h with little 
calculation. The downside is that now either a or b alone should be 
enough to determine the depth of the water; what you have is two 
redundant sources of information that you would want to average in 
some way to eliminate error. I suppose the error could come either 
from inaccuracies in the location of the sensors, or in the sensors 
themselves. I would definitely prefer to determine their locations 
without reference to their own outputs, in order to control the source 
of error. This is like saying I would not want to use a plane's 
altimeter to determine the altitude of an airport, but to use 
surveying to do that, and then calibrate the altimeter by the known 
number!

Does it make sense to determine the locations of A and B before using 
their data? What measurements can you make other than the distance 
between them?

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 06/15/2003 at 01:03:21
From: Daniel
Subject: Geometry

Sorry, what I didn't make clear in the previous response is that the 
angle of the tank changes over time and thus points A and B are not 
at set locations from the line h, but at set points on the outside of 
the tank. The position of the line h changes as the angle of the tank 
changes, as the line h represents the deepest point in the tank. I 
hope this clears things up a little

Thanks


Date: 06/16/2003 at 12:13:19
From: Doctor Peterson
Subject: Re: Geometry

Hi, Daniel.

Here is my picture again:

                      *********
                ******         ******
             ***                     ***
           **                           **
         **                               **
        *                                   *
       *                                     *
      *                                       *
     *                                         *
     *                                         *
    *          r                                *
    *---------------------O---------------------*
    *                     |                     *
     *                    |                    *
     *                    |r-h                 *
      *                   |                   *
       *                  |                  *
        *-----------C-----+----------D------*
         **         |     |         b|    **
           **       |a    |h         |  **
             ***    |     |          B**
                ****A*    |    ******
                      ****E****

We know r, a, b, and d = AB.

I'd like to start by using the angle t = AOB to define the "distance" 
between the sensors. Then their actual location can be defined by 
another angle u = AOE; and angle BOE = t-u. Later we can find t in 
terms of d.

This gives us these equations:

  cos(u) = (r-h+a)/r
  cos(t-u) = (r-h+b)/r

or

  r cos(u) = r - h + a
  r cos(t-u) = r - h + b

I'd like to solve first for u. We can eliminate h by equating two 
expressions for r-h:

  r cos(u) - a = r cos(t-u) - b
               = r[cos(t)cos(u) + sin(t)sin(u)] - b

  cos(t)cos(u) + sin(t)sin(u) - cos(u) = (b-a)/r

  sin(t)sin(u) - (1 - cos(t))cos(u) = (b-a)/r            [1]

  (1 - cos(t))cos(u) - sin(t)sin(u) = (a-b)/r            [1]

Now,

  (sin(t))^2 + (1 - cos(t))^2 = 2 - 2 cos(t) = 2(1 - cos(t)

If we define an angle v for which

  sin(v) = sin(t)/sqrt(2(1 - cos(t)))
  cos(v) = (1 - cos(t))/sqrt(2(1 - cos(t))) = sqrt((1 - cos(t))/2)

then equation [1] becomes

  cos(v)cos(u) - sin(v)sin(u) = (a-b)/[r sqrt(2(1 - cos(t)))]

  cos(v + u) = (a-b)/[r sqrt(2(1 - cos(t)))]

  v + u = arccos[(a-b)/[r sqrt(2(1 - cos(t)))]]

  u = arccos[(a-b)/[r sqrt(2(1 - cos(t)))]] -
      arccos[sqrt((1 - cos(t))/2)]

Now,

  h = a + r(1 - cos(u))

This is sufficient to find h. I'd like to try simplifying it, but I 
won't take the time to do so just now.

We still have to find t in terms of d, assuming the latter is easier 
for you to measure:

  sin(t/2) = (d/2)/r
  cos(t/2) = sqrt(1 - d^2/(2r)^2)

  sin(t) = 2 sin(t/2) cos(t/2)
         = 2 * d/(2r) * sqrt(4r^2 - d^2)/(2r)
         = d/(2r^2) sqrt(4r^2 - d^2)
  cos(t) = 1 - 2 sin^2(t/2)
         = 1 - 2 d^2/(2r)^2
         = (2r^2 - d^2)/(2r^2)

Let me know if that does what you need.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 06/17/2003 at 09:17:15
From: Doctor Peterson
Subject: Re: Geometry

Hi, Daniel.

I've been thinking about this problem off and on, trying to come up 
with a simnpler solution. I finally realized that I had been 
forgetting a basic trick for simplifying a problem: recast it in a 
form that is symmetrical.

So rather than solve immediately for h, I'd like to first find the 
midpoint of the chord AB

                      *********
                ******         ******
             ***                     ***
           **                           **
         **                               **
        *                                   *
       *                                     *
      *                                       *
     *                                         *
     *                                         *
    *                         x                 *
    *                C----O------+------------D *
    *                |   / \ ----+   r        | *
     *               |   /  \    |-----       |*
     *               |  /    \  y|     -----  |*
      *              | r/     \  F------------B
       *             | /       \ |c      ---- *
        *            | /        \|  ----d   *
         **          E-----------M--      **
           **       c|/     ----        **
             ***     |  ----d        ***
                *****A--       ******
                      *********

Note that I am using d this time for HALF the distance AB.

Rather than work directly with a and b, we'll just use

  c = (a-b)/2

This is the vertical distance between either A or B and M, the 
midpoint of the chord. Once we find y, the vertical distance from M 
to the center of the circle O, we can use the fact that the distance 
from the center down to the water line can be expressed in two ways:

  r - h = (y + c) - a

to solve for h:

  h = (r + a) - (y + c)
    = r + (a+c) - y
    = r + (a+b)/2 - y

From right triangles AEM, OCE, and ODB, taking EM = e, we have three 
equations:

  e^2 = d^2 - c^2                           [1]
  (e-x)^2 + (y+c)^2 = r^2                   [2]
  (e+x)^2 + (y-c)^2 = r^2                   [3]

Adding together the last two equations (after expanding them), we have

  2e^2 + 2x^2 + 2y^2 + 2c^2 = 2r^2

or (dividing by 2 and replacing e^2 with d^2 - c^2)

  d^2 - c^2 + x^2 + y^2 + c^2 = r^2

  x^2 + y^2 = r^2 - d^2                     [4]

Subtracting equation [2] from [3],

  4ex - 4cy = 0

which yields

  ex = cy

  y^2 = e^2x^2/c^2 = x^2 (d^2 - c^2)/c^2    [5]

Plugging this into [4], we get

  x^2 + x^2(d^2/c^2 - 1) = r^2 - d^2

  x^2 d^2/c^2 = r^2 - d^2

  x^2 = c^2/d^2 (r^2 - d^2)

We want y, so from [5]

  y^2 = c^2/d^2 (r^2 - d^2) * (d^2 - c^2)/c^2

      = (r^2 - d^2)(d^2 - c^2)/d^2

  y = sqrt[(r^2 - d^2)(d^2 - c^2)]/d

    = sqrt[(r^2 - d^2)(d^2 - (a-b)^2/4)]/d

Finally,

  h = r + (a+b)/2 - y
    = r + (a+b)/2 - sqrt[(r^2 - d^2)(d^2 - (a-b)^2/4)]/d

Try using that, and see if it works out right. As a check, we can take 
some special cases:

  if a=b,

    h = r + a - sqrt[(r^2 - d^2)d^2]/d
      = r + a - sqrt(r^2 - d^2)

  which works out right if you draw a picture.

  if d=r (so that AB has to be a diameter),

    h = r + (a+b)/2 - sqrt(0) = r + (a+b)/2

  which again works out right.

So there's at least a good chance I got it right. Using the right 
variables kept the work simple enough that I didn't make the mistakes 
I was making with more direct methods.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Calculus
College Conic Sections/Circles
College Higher-Dimensional Geometry
High School Calculus
High School Conic Sections/Circles
High School Higher-Dimensional Geometry

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