Fuel Gage Problem

[Jim Brier]

An 18-wheeler has a broken fuel gage. The driver wishes to measure the level of fuel in his cylindrical tank by using a stick. The tank diameter is 20 inches and the fill hole is directly on top. He knows that 10 inches is half a tank. He would like to know how many inches on his stick is one-quarter of a tank. Obviously, it is not 5 inches because the tank is cylindrical.

[Bozeman]

Are you just asking us to do trigonometry?

[Ed Miller]

I'm assuming you mean that the tank is "laying on its side" rather than standing on its base (or else 5 inches would be the answer).

If Theta is angle between the edges of the tank at gasoline level (at quarter tankness) and the center of the cylinder:

1/4 pi * R^2 = Theta/2 * R^2 - R^2 sin(Theta/2)cos(Theta/2)

Theta/2 - sin(Theta/2)cos(Theta/2) = 1/4 pi

The number of inches X for a quarter tank is then

X = R (1 - cos(Theta/2)) where R is 10 inches

Blah.. I have a meeting... /forums/images/icons/smile.gif

[JTG51]

Yes, I think he is.

Jim, it's too bad that most of your posts come in the form of math problems these days. Your contributions to the mid/high stakes forum are missed.

By the way, I really liked your most recent card player article (on playing high limits). I doubt I'll ever play that high, but I found it very interesting.

[Homer]

Does the trucker have a computer? I came up with two ways of doing the problem, but for both I ultimately needed a computer to reach the final solution.

Method 1:

I used some trig and came up with the following formula:

{5/9*pi*cos-1((10-h)/10)} - {2*(10-h)*sqrt(20*h+h^2)} = 25*pi

I used the Goal Seek tool in Excel (basically plug this formula into cell A2 and h into cell A1, then tell it to set A2 equal to 25*pi by manipulating the value of h in cell A1) to find a solution to this equation.

With this method I got an answer of 8.04.

Method 2:

A horizontal "slice" of a vertical (circular) cross-section of the cylinder has an length of sqrt(20*h + h^2). If these "slices" are integrated from zero to x (with x being the height filled with gas) and this is set equal to 25*pi, solving for x will provide the solution.

I plugged this function into "The Integrator" (http://integrals.wolfram.com/) and got some crazy ass function spit out at me. I guess I could plug this function into Excel and solve the same way as I did in my first method, but I don't feel like it.

...in any event, there is probably a ridiculously easy way to get the answer to this problem that I didn't think of. You probably didn't want us to use a computer anyway. /forums/images/icons/smirk.gif

-- Homer

[Mano]

Assuming the trucker knows how much his tank holds, fill the tank to halfway (which can be done using his stick), and then add another quarter tank, thus making the tank 3/4 full. Now measure the height on your stick, and the height for the 1/4 full tank will be 20 - the height for the 3/4 full tank.

[irchans]

I used majorkong's and simpson's methods to get an answer of 5.96 inches.

[Homer]

Your answer is correct. My formula had a slight error. It should have been:

{5/9*pi*cos-1((10-h)/10)} - {2*(10-h)*sqrt(20*h-h^2)} = 25*pi

instead of:

{5/9*pi*cos-1((10-h)/10)} - {2*(10-h)*sqrt(20*h+h^2)} = 25*pi

...oops.

-- Homer

[JTG51]

There's a problem with that solution.

After filling the tank half way, you want him to add another 1/4 of a tank. If he knew how to add 1/4 of a tank, he wouldn't need to go through the process you described. He could just add 1/4 to begin with and measure that.

[JTG51]

I imagine every trucker in the world knows how many gallons his tank holds. If not, it's easy enough to look up.

Divide the total capacity by 4. Add that much to the tank and measure it with the stick.

Nice and easy, only one simple division. There's always an easier way. /forums/images/icons/smile.gif I doubt this is the solution Jim was looking for, but I like it a lot more than the calc and trig.

[Mano]

The problem with that would be, either he would have to run the tank completely dry (which would be bad, and difficult to time), or there would be some undetermined amount of gas left in the tank.

[JTG51]

That is a good point. But, since we only seem to be interested in measuring quarters of a tank, it doesn't need to be extremely accurate.

Of course, I don't think any of this matters. I don't think Jim is looking for a common sense solution, he's looking for the math. At least I think so.

I still like our ideas better. /forums/images/icons/smile.gif

[irchans]

After reading JTG51's post, I wondered if there were solutions which did not require trig or calc. I thought of two. Maybe others can think of more.


Solution 1: Monte Carlo

a) Genereate 1 million points in a 20 by 20 area
b) Remove all the points that are more than 10 from the center to get a disk of points
c) Sort the points by height
d) Choose the point that is 1/4 of the way through the list

I did this with a simple program and got 5.95613 on the first run and 5.96997 on the second run.


Solution 2 (Integration with out calculus):

a) Split the 20 inch circle into horizontal strips of height 0.01 inches.
b) Approximate the area of each strip.
c) Add up areas starting from the bottom until you reach 78.54 square inches ( 1/4 of the area ).

Using excel, this method gave 5.97 inches.

This example shows that computer skills can sometimes serve as a subsitute for math skills.

[Homer]

Very nice irchans...

[JTG51]

Wow, very creative irchans. I like both ideas.

Although, I don't think anyone without strong math skills could have come up with either method, or solved them.

This is a great example of the fact that there's always multiple ways to solve any problem. A little thought beforehand can often times save you a lot of thought later.

[Mike Haven]

draw a circle of radius 10cm to represent the cross-section

draw a chord across below the centre to represent the liquid level

join the ends of the chord to the centre to form a triangle with angle A

the area of the liquid equals the area of the sector minus the area of the triangle

using radians as angle measure this is calculated as (A x radius squared / 2) minus (1/2 x radius squared x sinA)

this comes to 50A - 50sinA and is known to be a quarter of the tank cross-section or pi x radius squared / 4 or 25 x pi

this equation, because of the mixture of trig and non-trig can only be solved by calculator iteration (don't forget we are working in radians) and the answer is 2.31 radians

half this to get the half angle of the triangle and use 10 x cosine1.155 (calculator on radians) to find the vertical distance of the liquid level below the middle

subtracting this answer from 10 gives you the dipping measurement of 5.96

[Munga30]

You don't have to run it dry.

Fill the tank to 10 inches, which you know is half a tank.

THEN add the 1/4 tank, measured at the pump in gallons.

THEN measure with the stick and subtract 10 inches.

[Phat Mack]

</font><blockquote><font class="small">In reply to:</font><hr />
Solution 2 (Integration with out calculus):

a) Split the 20 inch circle into horizontal strips of height 0.01 inches.
b) Approximate the area of each strip.
c) Add up areas starting from the bottom until you reach 78.54 square inches ( 1/4 of the area ).


[/ QUOTE ]

How did you approximate the area of each strip?