1/4 Tank Dipstick Problem (from Car Talk)

Date: 12/04/2002 at 13:30:36
From: HydroJ
Subject: Dipstick problem...

This past week on the National Public Ratio radio program Car Talk, 
Tom and Ray Magliozzi got a call from Rich from Florida, who was 
actually on the road in Missouri. 

Here's how they state his intriguing mathematical conundrum: 

The gauge on his 18-wheeler is kaput. So he uses a dowel to measure 
the diesel in his tank - which just so happens to be cylinder-shaped 
and 20 inches in diameter, and which sits on its side. 

Now, any moron (that would be us) could tell Rich that his tank was 
half empty when the stick measured 10 inches of fuel. 

But when is the tank exactly three-quarters empty? Or a quarter empty? 

We tried, but got lost in the calculus. 
Thanks...

- Hydro

Date: 12/10/2002 at 23:18:49
From: Doctor Ian
Subject: Re: Dipstick problem...

Hi Hydro,



Looking at the tank from the side, we see a circle:


                *   *          |
            *           *      |  h
                               |
         A.................B   |
                                  
        *         C         *  

         *                 *

            *           *
                *   *
                
                
What we'd like to do is find h such that the area above the chord AB 
is some fraction of the total area, so that we can make an appropriate 
mark on the dipstick. 

Our FAQ on segments of circles,

   http://mathforum.org/dr.math/faq/faq.circle.segment.html  
    
explains how to take any two of the standard attributes of a segment 
and use them to find all the others.  

Case 13 tells how to proceed when we know the radius (r) of the circle 
and the central angle (theta) the subtends the segment. Of course we 
don't know theta yet, but it will be useful to pretend for a moment 
that we do. 

(In the illustration above, theta is the angle ACB.) 

The area of the segment is given by
  
        theta - sin(theta)
    r^2 ------------------
              2

In the general case, we would like for the area of the segment to be 
some fraction a/b of the area of the entire circle. For illustrative 
purposes, let's set a/b equal to 1/4. 

Now we can set the area of the segment equal to 1/4 the area of the 
entire circle:

      theta - sin(theta)   pi r^2
  r^2 ------------------ = ------
             2               4  

We can cancel r^2 from both sides to get 

      theta - sin(theta)    pi 
      ------------------ = ----
             2              4

and multiply both sides by 2 to get 

                            pi
      theta - sin(theta) = ----  (approximately 1.57)
                            2
                             
Note that pi/2 radians is 90 degrees, so theta is going to be larger 
than 90 degrees. How much larger? Could it be 120 degrees? That gives 
us

  2 pi/3 - sin(2 pi/3) = 2.09 - 0.87

                       = 1.22
                       
                       
which is a little low. How about 150 degrees? That gives us

  5 pi/6 - sin(5 pi/6) = 2.62 - 0.5

                       = 2.12
                       
which is a little high. So it's somewhere between 120 and 150 degrees, 
and you can narrow this down as precisely as you want by guessing. (If 
you're in a hurry, you can use the numerical method described at the 
bottom of the FAQ on segments of circles to make more efficient 
guesses.)

A little fooling around with a spreadsheet says that 132.3 degrees is 
pretty close to the actual value. 

In any case, suppose we find the value of theta that makes the area of 
the segment equal to 1/4 (or any other fraction) of the area of the 
circle. Then what?  

Back to Case 13:

   h = r - r cos(theta/2)
   
     = r(1 - cos(theta/2))
     
     = r(1 - cos(132.3/2))
     
     = r(1 - 0.404)

or about 6/10 the radius of the tank.  

By changing the fraction a/b, you can mark the dipstick any way you 
want, although you'll have to go through a fresh round of guessing for 
each new fraction. But note that you only have to consider fractions 
up to 1/2, because you can use symmetry to generate the others.  

Anyway, for a tank with a 10-inch radius (20-inch diameter), 3/4 full
is about 6 inches from the top of the tank, and 1/4 full is about 6
inches from the bottom. 

For variations on this problem, see:

   Trisecting a Circle with Parallel Cuts - Dr. Math archives
   http://mathforum.org/library/drmath/view/60807.html 

   Sphere Slices - posted April 29, 2002
   Trigonometry and Calculus Problem of the Week - The Math Forum
   http://mathforum.org/calcpow/solutions/solution.ehtml?puzzle=141

   The Rock Garden - posted November 26, 2001
   Lucent-Rutgers Problem of the Week - The Math Forum
   http://mathforum.org/lucentpow/solutions/solution.ehtml?puzzle=57

For solutions sent to Tom and Ray, posted on their site, see:

   The Great Gas Tank Math Solved
   http://cartalk.cars.com/Radio/call/dipstick/proof.html

   The Non-calculus Approach (PDF file)
   http://cartalk.cars.com/Radio/call/images/click_and_clack.pdf

and see

   Slightly Skewed Solutions
   http://cartalk.cars.com/Radio/call/dipstick/mail.html

- Doctors Ian and Sarah, The Math Forum
  http://mathforum.org/dr.math/