To prove the mathematical equation, and to rule out any unknown variables inside the fuel tank the equation needs to be backed up by a scientific experiment required for verification.
What a great question! I knew the boys were in trouble as soon as I heard it. I was a joy to hear their confidence disintegrate second by second with the caller on the line. I will use it next time I teach calculus. Their solution of measuring the depth of the gas in truck at truck stop with a gas gauge reading 1/4 full was much more “street smart” than a mathematician would ever come up with, but here’s the mathematicians answer.
The depth of the gas in a 1/4 full cylindrical gas tank 18 inches in diameter is 5.36424522… inches which equals 5 and 3/8 inches to the nearest sixteenth of an inch.
Geometry is more help than integral calculus here. The depth of the gas equals 9(1 - sin x) inches where x is the solution of the equation x + (sin x)(cos x) = ?/4. Using an Excel spread sheet to approximate the value of x, I got x = .41585597 radians. Plugging this value of x into 9(1 - sin x) gives 5.36424522… inches.
UPDATE: I mistakenly remembered the diameter of the gas tank being 18 inches. (I must have confused it with “18” wheeler.) For a 20 inch diameter tank, the formula for the 1/4 level becomes 10(1 - sin x). The value of x is the same as above. Plugging x into 10(1 - sin x) gives 5.96027247 inches which equals 5 and 15/16 inches to the nearest sixteenth of an inch.
Here’s the calculus for a 53 GMC short school bus with an OVAL tank. 1. Drive into a level gas station on fumes as usual. 2. Fill tank with exactly five gallons. 3. Mark stick. 4. Repeat steps 2 & 3 until tank is full. You are now the proud owner of a fuel gauge more accurate than the original. Super persnickety people could even mark their stick at each gallon if they had all the time in the world or were waiting for their wife to get out of the restroom.
Of course once you have all those marks on a stick, you have to be able to count them and divide by four to get your quarter divisions.
The integral of an equation for a smooth, continuous curve yeilds the area under that curve (See http://www.mathwords.com/a/area_under_a_curve.htm). Math is a very pure science and a very powerful tool. My solution was efficient in relation to both time and money.
My results are in line with the stated assumption that the fuel tanks are horizontal cylinders. Therefore, all we need to concern ourselves with is the cross-sectional area of the tank which is a circle with a 20" diameter.
Which solution do you really think Rich would be willing to employ?
I hope you have all kept in mind that:
A) The truck will stop running before the tank is truly empty. The fuel line doesn’t go all the way to the bottom of the tank. This allows sludge and dirt to collect at the bottom of the tanks instead of going through the engine.
B) The ends of the cylindrical tanks are usually rounded, not flat.
With that said, I think Tom and Ray already found the best solution. Find another truck that is the same model and measure the fuel when it is at 1/4 full according to its fuel gauge.
Definitely the best answer. What flavor Pringles though?
Strangely this is exactly the same problem you get when you try to find the position of the earth in it’s orbit on a random day. The rule “sweeping out equal area in equal times” sounds simple but you end up with an insoluble equation.
The truck will stop running before the tank is truly empty. The fuel line doesn't go all the way to the bottom of the tank. This allows sludge and dirt to collect at the bottom of the tanks instead of going through the engine.But we?re trying to calculate fuel left in the tank, not usable fuel. How much unusable fuel could there be at the bottom anyway? An inch? Two? The tank probably isn?t more than 100 gallons (at 20 inch diameter, it would be more than 6 feet long). So it?s actually probably closer to 60 or 70 gallons. 2 inches of space in a 70 gallon tank 20 inches in diameter is a little more than 3.5 gallons. That would get him 30 miles down the road; hopefully he?s not cutting things that close (although if he?s got dual tanks that would double that). An inch would be about a gallon and a third.
The ends of the cylindrical tanks are usually rounded, not flat.I don?t know if they ?usually? are, but they sometimes are. I had to refer to a picture of my dad?s KW T800?s that he uses to haul grain ? they appear to have flat ends. However, I don?t think it makes a huge difference because the ends are not rounded out very much. Out of curiosity, I simulated a tank on this program: http://www.arachnoid.com/TankCalc/, that was 20 inches in diameter, 51.5 inches long, with elliptical rounded end caps with radius 3 inches. It looks very much like a tank that you might find on a semi, although the 3 inch radius is maybe a bit much, but it being too large only strengthens my point. At 20 inches, the tank holds 75.48 gallons. At 5.96 inches (what I previously calculated the 1/4 mark to be), it holds 18.67 gallons ? 24.74%. That?s pretty darn close to 1/4, close enough for the purpose of estimating remaining fuel, wouldn?t you say?
I was able to solve the problem without calculus. You just have to break the problem into two simpler problems (finding the area of a circular sector, and the area of a triangle).
The area of the sector (red) is easy = (PIr^2)(Theta/360)
The area of the triangle (blue) requires some trig:
-The height = Sin(Theta)*r
-The base = Cos(Theta)*r
-The area of the triangle = (base * height)/2
So, the area of the fuel = (half the area of the whole tank) - (2 sectors) - (2 triangles). You could solve for Theta, but I just put the formula in a spreadsheet to find the value. I came up with Theta = 23.83 and a fuel depth of 5.96"
Come on, I studied math in school, but you need to be practical. What accuracy are you going to get off a dipstick? Simplify the picture to get a pretty good estimate.
Call the distance from the center to the surface of the gas x. You can get a reasonable approximation of the area of the region bounded by the horizontal diameter and the surface of the gas by taking the rectangle with height x and width 20. This should be 1/4 the area of the circle:
b/4 (is about equal to) 20*x[/b]
Use 22/7 for pi (we are estimating, after all)
2522/7 (is about equal to) 20x
5/4*22/7 (is about equal to) x
110/28 (is about equal to) x
Call this 4, since it is pretty close to that. The height of the dipstick should then be about 6.
Looking at some other answers where people figured it out, the exact answer is nearer 5.96. I’d call 6 a pretty good ballpark figure.
Feynmann’s physics lecture notes emphasize understanding the problem and choosing the most appropriate method of solution, whether it be mathematical or numeric. For this I’d vote for a reasonable approximation.
Would it be easier to go bacm and listen to the 2002 CAR TALK show where I think this first showed up?
PS" (i really like the Pringle-can method!)
OK, I think I’ve figured it out. If the Diameter (D) is 20", then the Radius ® is 10". The volume of this circle is calculated as V=(pi)®^2, or V=3.14X(10X10), or V=3.14X100, or V=314 sq. in. Now, 1/2 the volume of the circle- or the volume of 1/2 the circle - is 157 sq. in.
Here is the tricky bit - I figured the volume of a rectangle of 20" (the diameter) by 10" (the radius). The volume of the this rectangle is 200", while the volume of the 1/2 circle with a diameter of 20" and a radius of 10" is 157, or 78% of the rectangle.
Take the 10" radius and multiply by .78, and you get 7.8". This should be the distance from the bottom of the tank to equal 1/4 of the total area. Or do I have it completely wrong?
But we?re trying to calculate fuel left in the tank, not usable fuel.
Um, I think if you ask the truck driver who called in to the show, he will disagree. I think he whould be quite concerned about running out of USABLE fuel. If you disregard the unusable fuel, this whole exercise is academic. The driver called because he has a real practical problem, not an academic one.
How much unusable fuel could there be at the bottom anyway? An inch? Two?
It’s more than you think, and it varies by manufacturer, model, and year. I am betting this guy’s truck is old enough that it was manufactured before low sulfur diesel was in use, which would mean it is an appreciable amount.
If you choose to ignore what I have said, you might leave this driver, who came to Tom and Ray for help, stranded on the side of the road. Taking a chance of doing so out of intellectual laziness wouldn’t be very nice.
[i]Like so many of the questions in this forum, this one can’t be answered without knowing the year, make, and model,[/i] but even if we did, I wouldn’t know the answer. Someone would have to look up the relevant specs.
I think you made a mistake when you took the antiderivative of (1-x^2)^(1/2): if you differentiate your answer to check, because of the chain rule you get an extra 2x multiplier that is not in the original integrand.
I did the integral by making the substitution x=sin(u). You end up taking the integral of (cos u)^2 du = (1+cos(2u))/2 du (by a trig identity). Switching back to x, the antiderivative is:
(1/2)*arcsin(x) + (1/2)xsqrt(1-x^2) + C
Evaluating, setting the two integrals equal, I end up trying to solve this equation:
arcsin(a) + a*sqrt(1-a^2) = pi/4, or
pi/4 - theta + cos(theta)*sin(theta) = 0, where theta is the angle such that a = cos(theta), or equivalently:
theta - (1/2)sin(2theta) - pi/4 = 0.
I can’t think of a way to solve this exactly; I’d be interested to see if anyone else can. You could get an approximate answer by iterated educated guessing, or Newton’s method, or other numerical method (I haven’t done this yet).
Very nice: I had the feeling there was a geometric way to do this without calculus, but was too dense to see it. I still can’t use this method to get an exact answer, however. I end up trying to solve the same equation I got using calculus:
theta - cos(theta)*sin(theta) = pi/4
You can get an approximate answer, but does anyone see a way to solve this exactly?
I think this (stated equivalently by “so far so good” below) is the best method.
Nice. Certainly the most creative (and hunger-satisfying) method (although scaling up from the scale of a Pringles can to the scale of a big gas tank will probably introduce some error).
LarryCoregon described this same method above, and I think you’ve both got the best answer.
There are numerous web sites just for calculating TANK VOLUME. They are used for various liquids. Rich can input the width & length of his horizontal tank & print out the results & carry it with him. Some websites will even break it down to .125inches. All he would need is a yardstick that he can use for just this purpose. We use them frequently for DIESEL FUEL tanks. Here’s an excellent site: http://www.odayequipment.com/Support/TankChart/tankchartcalculator.shtml
My brother whiteboarded it for me.
Never mind. I screwed it up. Damn that liberal arts degree!