Let’s hope it’s not, graduate of “Alan Jacobs Junior College.” Not a very good advertisement for the school! LOL
Simple to do with a tape measure…
- Measure the diameter of the tank.
- Slide the tape down until it reads 1/2 of the diameter and hold your finger on the center point (you could use chalk or some other marker, but it wouldn’t look as rediculous).
- Measure up from the bottom of the tank to the mark. Cut that length on the stick to indicate the 1/4 tank level.
Logic: Since the area of the circular tank is directly proportional to the diameter of the “circle”, then 1/2 of the diameter would be 1/2 of the area of the circle and, hence, the volume of the tank.
QED.
I know you said earlier that this is your story and you’re sticking to it, but it’s still wrong. There is no shortage of people on here showing ways to figure it out, and arriving at the correct answer. Or, you could try a tank simulation program, like http://www.arachnoid.com/TankCalc/. Input your answer in there and you’ll see it’s incorrect.
Here is an AutoCAD rendering of the circle with using the measurements previously calculated (the correct ones, not yours). The area given is what the program calculated the shaded area to be. As you can see, 78.5348 is 24.998% of 100*pi. I didn’t read through your sheet to see exactly where you’re wrong. These are the correct angles and measurements though.
The trucker question gives me a good excuse
to dust off my integral calculus. But here’s
a simpler empirical approach rather than
imposing on other truckers’ gas tanks.
Simply take a cylindrical can, metal or
plastic, from the kitchen. Slice it in half,
the long way. measure how much water is
needed to fill the half cylinder. Throw away
half the water and refill the half
cylinder.Then measure that height as a ratio
of the radius.
I used the equation for one coordinate of the centroid of a quarter circle = 4r/3pi (which is derived using a simple integral). If the tank has a radius of 10 inches, then you would mark the 3/4 line 4.24 inches above the center line, and the 1/4 mark 4.24 inches below the center line.
Kaliki
Given that the fuel tank on the rig is probably 100 gallons and you know that when filled to capacity it is 20" deep, then based on that we know that at half a tank it is 10"deep but we also know that half a tank is 50 gallons. So the simplest way to do this is when fueling the vehicle, stop at the half way mark then add 25 gallons, that is your 3/4 mark and the distance between the halfway mark and the 1/4 mark will be the same as from the halfway mark and the 3/4 mark. Its not rocket science, Rich just wants to know when hes down to his last 300 miles.
Cut/Paste from the link at mathforums.com…
( http://mathforum.org/library/drmath/view/61752.html )
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
Now, don’t get me wrong. I absolutely love the show as much as the next guy, but…
Notice the date at the top of the cut/paste?
The following shows how to compute how full the tank is based on the measurement. No calculus is needed, just some geometry and trig.
Given a cross-section of the gas tank, a circle of radius r (2*r=20?) and a depth of the liquid x where x <= r, label the following points:
a the point where the surface of the liquid touches one side of the circle.
b the point where the surface of the liquid touches the other side of the circle
c the center of the circle
f & g the points on the sides of the circle on a line parallel to the surface of the liquid.
We know the area below the f-g line (half the area of the circle). We will now break that area into four parts
-
The isoselese triangle a-b-c with two sides r in length and height r minus x. The triangle?s area (call it j) is:
1/2 the base heightj = sqrt( r**2 ? (r ? x)**2 ) * ( r ? x )
-
There are two sections of the circle f-c-a and g-c-b whose angle t we know because it is the same as c-f-g
cos( t ) = (r ? x)/r
so the area of the two sections together is (call it k):
% of circle two sections circle area
k = acos( 1 ? x / r )/360 * 2 * ? * r * * 2
-
So the area covered by the liquid is (call it z):
z = ? * r * * 2 / 2 ? j ? k
I have a pretty diagram, but this input control doesn’t let me paste it in.
OK. so I didn’t stick to it. It was fun, though.
The way your brother solved this problem was the way I remember doing similiar problems in high school. The only part I would still have trouble with is his solving for theta- trig transforms are easily forgotten.
Thanks for posting the white board picture, after about an hour of trying to figure this out myself using integrals it was nice to see it easily done with trig and algebra.
There are a lot of practical ways to solve this problem without math, (which Tom and Ray offered a few on the show) but I think the real challenge here was to find a mathematical solution. Really now that you have the ratio for 3/4’s full (~70% of the diameter) and 1/4 full (~30% of the diameter) you can use that same scale to any cylindrical gas tank regardless of the diameter (or length).
Thanks again for the picture- I would probably still be pulling my hair out writing equations all over my son’s coloring books if you hadn’t posted that.
Tom & Ray,
Great show ? love it!
FYI?the problem of accurately measuring a barrel-shaped object has been a puzzler for commerce since at least 1613, which is the year Johannes Kepler wrote the influential mathematical treatise Nova stereometria doliorum vinariorum, on measuring the volume of containers such as wine barrels. It?s the same Kepler who later developed the laws of planetary motion that rocket scientists still use today to define satellite orbits. Legend has it that he thought up the problem at his own wedding reception–not sure what that says about geeks and their patient wives (maybe he cast the die for all future mathematicians).
The math is complex; here are some good websites that give the tedious details:
http://www.karlscalculus.org/calc10_0.html
http://mathworld.wolfram.com/Barrel.html
I didn’t go through your whole write-up, but it was still open on my computer when I came back to it, so I glanced at it again. It looks like you might have gotten derailed early on when you started splitting up the area into right triangles. It looks like you’ve labeled the length of the base of each of those triangles “sin b” and the height as “sin theta.” The trig functions would tell us that the sin of an angle (in a right triangle) is equal to the length of the opposite side divided by the length of the hypotenuse. So you should have something like, sin b = x/10 (the hypotenuse is 10 - the radius). So the length of that side should be 10*(sin b), not sin b. Same thing with the height, it should be 10*(sin theta).
The 18 wheeler
What is the matter wich U’s guys? the poor 18
wheeler driver , i am surprised he did drive off
the road and hit the something solid on purpose . It is so simple. no need for math,
algebra calculus or differential equations… he knows the halfway point, right?
let us say it takes 400 gallons to fill the tank,
and he drives 286 mile to the half ways point, half way of the tank not delivering the turkey.
so a quarter of a tank will get him 143
miles. fill the tank, drive 143 miles, dip
the stick in, mark it. that would be 3
quarters full. drive 429 miles dip the stick
in, mark it that would be 1 quarter of tank
left.drive drive 71 miles you have used an
eight of a tank. and so on.
simple , hay
you guys went to M.I.T.?
Assuming he knows how much the capacity is, why not just siphon it to empty (or as close as he can get) then pour in 1/4 of the capacity then measure?
I have this problem completely solved, in that if he measures the depth, I can tell how many gallons are left. Now, I’m new here and not sure if the attachments will come through, so I’ve simply written the formula in this note. If interested in “pretty” copy, send e-mail to dgittinger@alamo.edu
I’ve attempted to attach two files, one with the math and the other with an Excel spread sheet. If you enter the values of h and r, it will tell you the # of gallons of fuel by finding the corresponding measured depth, d. Note, on the Excel spreadsheet, h is called “W.”
If the tank is shaped like a beer can on its side, (a right circular cylinder) and the horizontal length is h, the radius of the circle is r, the depth of the gas is measured to be d, then the volume in cubic inches is
V = h(r^2(pi/2 - arcsin((r-d)/r))-(r-d)sqrt(2rd-d^2))
Divide this result by 231 to convert to gallons.
Send me an e-mail and I’ll send you the cool math solution, along with an Excel spreadsheet that you can simply enter the width and radius and find the number of gallons remaining at various depths! dgittinger@alamo.edu
I can understand the dilemma of the Cat Talk Guys, as the determination of the volume of a cylinder is 8th grade math, not MIT-level math. The formula is:
Pi x r x r x h where pi = 3.1416, r = radius of the tank and h = height of the tank. That gives you the volume in cubic units (inches, feet, meters, etc.). The you must know the current level of the tank, which is difficult unless you have graduated markings on the inside of it or it is accessible and can somehow be measured. Either multiply that % times the cubic units or use the square units and multiple by that height to get cubic units. Finally, you must convert those cubic units into the appropriate weight using the density of the fluid. For gasoline, the weight-density of gasoline is about 740 kg/m? (6.175 lb/US gal; 7.416 lb/imp gal).
That took only 1 MIT grad to explain it. However, it is not something anyone could reel off on national radio on the spot. Give the Magliozzi brothers a break. Nobody would remember the weight-density of gasoline without looking it up!
Nice skimask… very nice. So now the question becomes, “Did Tom and Ray cut the audio from their previous show and paste it into their recent show or was it just a coincidence that another Ray who also just happened to be from Florida and who also just happened to be on the road in Missouri, and who also just happened to have a 20” tank, also just happened to call with the identical question?" I think Click and Clack have some explaining to do! Or maybe it was Rich who thought it would be funny to call again! The plot thickens!!!
Nice work Jon. I used the so-called Midpoint Rule to approximate the calculation and got 6". The midpoint rule basically divides the area to be measured into rectangles having one side of equal length (I divided it into 5 equal pieces). Taking the midpoint of the equals sides as the point to determine the height of the rectangle (where it intersects the edge of the circle), calculate the area of the rectangle.
By the way, Tom and Ray, you had this same question about 20 years ago. I seem to recall that you more quickly identifed that you needed calculus to solve the problem, but weren’t able to solve it. Seeing Jon’s formula, it isn’t a surprise why! Also, I don’t recall that at that time you had any practical solution like you did this time (i.e. find a trucker whose gauge says it is a quarter full and put the dipstick in that tank).