with about six feet left in a tank, that gives approximately a quarter tank...6 feet (72 inches) is a quarter tank that's only 20 inches in diameter... I can't wait to hear the explanation.
Hi Burke, I like your thinking. Much cleaner than calculus approach.
I used exactly the same procedure as you did and also did a spreadsheet to check my formula. I got 5.92 in stead of 5.96, but close enough. However, I think that is the distance from the center, so the 1/4 mark will be 10 - 5.92 = 4.08.
By the way, the end formula I got with theta in radians was:
Depth = r ( 1+sin{theta) )
Fraction of tank = (pi + 2theta + sin( 2theta) ) / 2pi
This is with theta = pi/2 at full, 0 at half, -pi/2 at empty
yes the tanks are joined by a fuel line but he gauge is reads one tank not both it is assumed the other is at about the same amount, which is not always the case. as for taking it to the shop for an expensive repair or using a dipstick i’d go with the dipstick
The correct answer was posted in the third top-level comment. See http://mathworld.wolfram.com/Chord.html
Interesting alternative approach, though.
Dear Messrs. Magliozzi,
I was delighted to hear "The Great Fuel Tank Problem"
of determining when a cylindrical gas tank is one
quarter full. But Rich Randall should learn a little
more mathematics before rescinding people’s slide rules.
The centroid of a region is not the point on each side
of which there is equal area. It is the point on each
side of which there is equal torque. Thus, if a small
child and a fat man are balanced on a seesaw, the fulcrum
is positioned at the center of mass (the centroid in the
constant density case), but there is certainly not equal
mass on each side.
Sincerely,
Greg Marks
------------------------------------------------
Greg Marks
Associate Professor
Department of Mathematics and Computer Science
St. Louis University
St. Louis, MO 63103-2007
U.S.A.
Phone: (314)977-7206
Fax: (314)977-1452
Web: http://math.slu.edu/~marks
------------------------------------------------I’m afraid you didn’t understand the problem. The dipstick is not inserted along the height of the cylinder. The cylinder is lying on its side, and the dipstick is inserted, if you will, from one side to the other of the cylinder. So the problem boils down to finding the area of a “circular segment”. See http://mathworld.wolfram.com/Chord.html
I was excited when I first saw your approach, because it offers a closed-form solution. However, your answer is wrong! Numerous solutions posted in this forum show the correct answer of 5.96" fuel depth when 1/4 of the fuel remains in the tank.
The reason your approach yields an incorrect answer is because the centroid depends on shape (both the area and the distance of an incremental area). In this problem the only relevant parameter is the cross-sectional area below the fuel surface, and the shape does not matter.
Hey, Chet.
Please tell your friend to carefully redo the calculation.
He gave you a number (7.008) to the thousandth of an inch,
but is off by a full inch. Lots of people here got the correct answer = 6.0 inches.
In another post (one day before the two above were posted)
you said you (or your friend?) got the answer from a book.
I don’t have that book handy, so please tell us: what do the numbers mean?
I mean 35.04% which is 7.008 inches for a 10 inch radius;
it must be the answer to a different question, and I am curious to know.
Hey, Bill.
I am delighted to see what Peter did, although he didn’t quite say
how he found the numerical value of approx 4.04inches; just curious
to know if he did it by hand with a basic calculator and trial-and-error,
or if he plugged it into some kind of smart solver, etc.
Either way is fine; whatever floats your boat, as we say.
Before I give Peter suggestions for other similar problems,
I want to say that you can tell him the calculus approach is
no better than his geometrical one. I actually used calculus first,
because I was too lazy to go with the geometrical way (which is the
more “clever” way, while calculus is a bit more machine-like).
Both approaches give the same “ugly” equation that can only be solved
numerically or graphically.
Also, I like that Peter has the common sense to do something
about the accuracy (number of digits). In fact, he could’ve rounded
his answer to 4.0 inches, but I’m glad to see how comfortable he is
with fractions. Many people posted here answers with up to nine or ten digits,
and didn’t bother to round them off, which is ridiculous.
Now, finally, to new problems for Peter (and his friends).
But I’ll post them separately.
Cheers,
–Ion
This solution does not require integral calculus. I used it to calculate the amount of gas in my airplane gas tanks. Materials.
- A sheet of graph paper and a pencil set up a graph of say 5 squares equals 5 gallons. on the horizontal axis and one square equals 1 inch on the vertical axis of the graph.
2, A yardstick and a black sharpie. - A 5 gallon fuel can and a means to pour fuel into the fuel tank.
Procedure:
Run the tank until nearly empty.
- dip the tank and mark that point on the graph paper.
- add 5 gallons of fuel to the tank dip the tank. Plot that point on the graph.
- Add another 5 gallons to the tank, dip the tank, and plot that point on the graph, this time at the 10 gallon mark.
- repeat step 3 until the tank is nearly full.
At this point in time you should have a graph that forms a curve. Extend the curve at each end until one end of the line intersects the zero gallon line and the other end of the line intersects a point where the tank would be full, i.e. 100 gallons.
You can now interpolate to find the point on the graph where there are 25 gallons in the tank. Mark that spot on your yardstick or transfer it to a dipstick and you are finished.
Let’s change the shape of the tank from cylindrical to one that
has a diamond-shape cross section instead of circular. The length
of the tank remains irrelevant, so it’s still a two-dimensional problem.
This problem is actually much easier, and a truly exact answer
can be found quickly. I hope Peter doesn’t feel insulted;
this is just to keep him warmed up for what’s coming next.
To make it fully clear, imagine that the original cylindrical tank
was made by welding together two semicircular troughs. Now, we use
two V-shaped troughs (one pointing down and one pointing up, facing
each other).
The question is the same:
What is the fraction of the height of the V-shaped trough
for which we have a quarter of the volume left in the tank?
The height of the “V” is similar to the radius before, and we
could still call it R. Actually, Peter called it "h"
as if anticipating my new problem. Anyway, it doesn’t matter which
letters we use; we shouldn’t fall prisoners of any particular notations.
Now, to make sure Peter doesn’t lose interest, lets step from
two to three dimensions. We a have a couple of choices;
first we pick “pointy” shapes, and then we go back to rounded shapes.
For pointed shapes we can either pick cones or pyramids.
So let’s take two identical cones (or pyramids) of height h,
and we’ll weld their bases together; this is our tank now.
We position the tank such that one apex is pointing down and the other up.
As before, we ask the same question:
What is the fraction of the height h which marks
a quarter of total volume of the tank.
This problem may well be an incentive for Peter to start learning calculus;
indeed, the integral needed here is actually easier than the one we did
for the original problem. Sounds a bit paradoxical, but it’s just one
of the quirks of nature.
Anyway, let’s take a (perfectly) spherical tank,
and let’s ask the same question:
What is the fraction of the radius (height) which marks
a quarter of the volume.
The equation that will have to be solved this time is much simpler,
but still a bit unpleasant because it’s cubical. So, we again do it
numerically. I never use those obscure formulas for the roots of cubical
equations.
I hope this last one will not make Peter turn away in disgust.
To bribe Peter, I pay him another compliment.
I actually meant to say this in my first reply:
Peter will surely make an excellent scientist, or engineer, or mathematician.
So, back to the future.
This is for Peter, a few years from now.
Or, maybe, only a few days from now; as I said about calculus,
Peter doesn’t have to wait until college. He can just grab a book now,
and he can learn all kinds of “advanced” things.
Now, to the tank problem in N-dimensions:
First, let’s do the pointy case: (hyper)cone, or pyramid.
Find the fraction of the height which marks a quarter of the (hyper)volume.
Then, the hyper-sphere.
Hint:
Mind you, Peter, after doing cones and spheres,
you’re already half-way to N dimension. What I mean
is that going from circle to sphere is how you will
go from sphere to hyper-sphere. There really isn’t anything new.
After all, for the original cylindrical tank, its length was irrelevant.
So, we weren’t really talking about volume; we were talking about areas.
And we simply think of “area” as “volume” in two dimensions.
Just don’t let this jargon (fancy words) get in your way.
Tom & Ray,
The attached is from “Pocket Ref” by Thomas J. Glover.
Great show. I subscribe to the weekly podcast.
Tom Oakey
My solution is in the attached file. Enjoy!
fix the gas gauge
It doesn’t have a gas gauge because this machine doesn’t use gas. It has a fuel gauge.
Your answer is incorrect; I believe you when you say you’re not a rocket scientist. Your answer would be at approximately 28% full, not 25%.
Get a 1 quart mason jar , pour 1/2 cup water into jar put top on , lay the jar on the side , check and mark bottom for distance , adding 1/2 cup again adds 1/8 increments of the tank’s volumn , convert to truck tank by this caculation (20in (tank heighth ) /6 in(Jar Heighth ) *( distance to each mark from the Bottom .