You guys were close as you worked it in the radio; here’s a no-calculus solution that is 'close’
Yes, the area of the cylinder slice is 100pi, or 314. Yes, 1/4 of a tank will require an area of some shape and dimension that is 1/4 of 100pi in area, or 25pi, about 78 square inches, but call it 80 since we are approximating here.
Now, at the center of the tank, we need a shape of 80 sq inches were one side of the shape is the center line through the tank. This means this side is 20inches and straight. The easiest shape to assume is a rectangle, which because of the tank curvature near the center is just a little bit aggressive. Thus, we need a rectangle of 20 x 4 to represent the area consumed for the 1/4 of a tank from 1/2 to 1/4 full. That means that the remaining 1/4 tank will be the area from 4" below the center to the bottom, a distance of 6 inches. Therefore, to a reasonable approximation, the stick should be calibrated to read 1/4 tank when it shows 6 inches of fuel in the tank.
Of course the round off to 80 from 78, and ignoring the curvature of the tank near the center are offsetting errors, so the 6" remains a reasonable answer
I like the empty the tank solution . . . I can’t believe if some dumbo punctured one tank you couldn’t isolate the tank and use the other . . .If the tank holds 200 gallons, put in 50, measure the 1/4 level, put in another 50, measure (should equal 1/2 way up the stick)then put in another 50 and measure again for the 3/4 measurement.
Does the term ‘tongue in cheek’ sound familiar? It’s hard to transmit irony on a blog. Of course, using using all that cardboard and string and stuff is a pain, especially when various trig or geometric functions will yield a theoretically more accurate answer.
My point was to present an admittedly Rube Goldberg approach–one that I thought would appeal to the typical Car Talk enthusiast’s sense of humor–obviously I was mistaken.
Someone solved this before this way: http://mathforum.org/library/drmath/view/54105.html
Okay - here’s a no brainer… Rich knows what he gets in average miles per gallon - he also knows what his tank capacity is… SO… Stop the truck when he has gone through 75% of his total calculated milage and measure… Walla!
I have a very easy answer that doesn’t require a lot of complex math.
Get a cylindrical container (a round drinking glass) and fill it to 1/4. Tape the top so that turning it on its side will not make it spill, and then turn it on its side. Measure the diameter of the glass, and then how “deep” the water in the glass is. Divide that by the diameter, and you have the percentage of the diameter that constitues a 1/4 full cylinder.
For instance, if the glass was 5 inches in diameter and the water was 2 inches deep with the glass on the side, that would be 2/5 or 40%. Now since his tank is 20 inches in diameter, 40% of that is 8 inches.
I didn’t do the glass test, so someone would have to measure the depth of a glass on its side to get the actual percent.
No calculus, a little trigonometry.
Let?s assume that his gas tank is 100
gallons and that he gets 10 miles to the
gallon evenly throughout a tank full of
gas. (I know what happened when we assume,
but nonetheless…) That would be 1,000 miles
to a tank of gas. At 750, he could use the
dipstick, mark it, and that would be the
point where only a quarter tank remains.
Let me know if you think this will fly.
Actually I think some of this is pretty funny. All the math - geometry, trig, calculus - boil down to the same equation, something like theta + sin(theta) = something. There are in general no analytic solutions for this (in special cases there may be clever ways to do it, but if this is one of them I am not clever enough to find it).
Using computers or calculators to approximate the answer is an engineering solution, not a mathematical one, however dressed-up it might be in rigor. And, if you are going to use an engineering solution, there isn’t much purpose in playing with mathematical symbols first, at least after realizing that there is no analytic solution.
Here is a brief and inefficient program which figures it out from simple principles: breaking the unit square into N cells and seeing how many of them are inside the region. I’m sure a calculator could do this even easier, but I don’t have one.
[i]
#include <stdio.h>
#include <stdlib.h>
#define PI 3.14159265358979
float one_fourth(int radius) {
int x, y, xlimit;
int inside = 0;
int limit = PIradiusradius/4;
for (y = radius; y >= 0; y–) {
xlimit = radiusradius - yy;
for (x = -radius; x < radius; x++)
if ((x*x < xlimit) && (++inside >= limit))
return ((float) (radius - y))/radius;
}
return 0;
}
int main(int argc, char **argv) {
if (argc == 2) {
int radius = atoi(argv[1]);
if (radius > 0) {
printf("ratio of depth for 1/4 full is %f, estimating from circle with radius %d
", one_fourth(radius), radius);
exit(0);
}
}
printf("Invoke: ‘%s RADIUS’ : RADIUS: radius of circle to use for estimate
", argv[0]);
exit(1);
}
[/i]
Using a radius of 100 gets pretty close, using 1000 even close.
Is the math that hard? First, we only have to know the dimension of the end. It is a circle. The length of the tank does not matter. The tank is horizontal. We only want to know how far up from the bottom of the tank to make a mark on the stick.
For simplicity, assume the radius of the tank is 1 (units do not matter). The area of the top-half of the circle (allowing a function) is Pi times r squared. The value of the semicircle’s first derivative is pi, with radius = 1. The mark on the stick needs to be placed “pi/4 times the radius” of the way up the stick from the bottom of the tank to the centerline (top-half of the circle). I know, I skipped the “and then a miracle happens” part…, but -
If the tank is 20" in diameter, the radius is 10". Doing the above mentioned math, the 1/4-tank mark should be about 7-7/8 (10 times pi divided by 4) inches up from the bottom of the stick.
If the tank is not horizontal, the problem becomes harder.
If this is not the case, then I had more fun in college than I remember.
7-7/8 inches would be about 37% full.
Here is another solution done similarly, but different.
Tom & Ray came to the right conclusion, that the fuel dipstick problem could be solved with integral calculus. The function for the equation of a circle is f(x)=SQRT(r^2-x^2) where r is the radius of the circle (the cross section of cylindrical tank). The function can be integrated over the interval zero to any distance, a, from the center to yield an equation that relates the depth of the fuel (r-a) on the dipstick to the fraction of a full tank remaining. The resulting expression 0.5[aSQRT(r^2-x^2)+r^2arcsin(a/r)] can be evaluated for any value of radius r (10 inches in the example) and a (where the depth is r-a) to find the fraction of a full tank remaining. I have attached an Excel file for the r=10 inches case that contains a graph of fuel remaining versus depth of fuel. The graph shows that the quarter full tank would have a depth of about 6 inches, and one tenth of a tank would be about 3.2 inches, etc.
This was a fun puzzle!
No, integral calculus doesn’t solve it, all it does is provide an unsolvable equation, which then is approximated by numeric methods. Calculating with a computer is not really a mathematical solution, and as a technical one not as elegant as the Pringles can suggested earlier.
If resorting to numeric methods, they can be used directly without bothering with any calculus at all. The same unsolvable equation can be found using calculus, geometry, or trig, as has been done by many in this thread. but in any case needs to be approximated numerically. Integration is certainly the right first approach to take - it is in general the right tool for this type of job, but in this case does not work. If number crunching is to be used, there is not really a benefit of using calculus first - see my C program above.
If no exact solution is possible, I prefer approximations (my solution yesterday). Using something like Newton’s approximation to find a numeric value is OK. If I use a computer/calculator I don’t see the value-added of calculus.
Actually, the analogue solutions some have suggested are fine too. A Pringles can - could any truck driver resist?
When I e-mailed this suggestion last Saturday I didn’t know a blog forum was the standard place to talk about issues like this.
Here are 2 non-calc ways:
- Cut a circle out of some suitable material. Weigh it on an accurate scale. Cut slices off it perpendicular to its diameter until it has lost one forth its weight. The percentage of the circle’s diameter removed is the percent of 20 inches equaling 1/4 tank. (Weight is proportional to Area which is proportional to volume in a cylinder. I did this and got about 5.9 inches as a rough result.
- Draw a circle on graph paper. Measure the diameter and calculate the area in graph paper squares. Divide that by 4. Begin counting squares from the bottom until you arrive at that 1/4 circle area number and mark the point on the diameter. To be accurate you must estimate partial squares as you go. I did this and came out very close to 6 inches on his 20 inch deep tank.
It isn?t calculus, it?s trig and some algebra. High school stuff. You didn?t need 18.01 to solve this.
See attached doc for the diagram.
Bottom parallel line marks quarter of the tank. Figure out length of triangle height h. It?s distance below mid point marking a quarter of the tank. Add the radius and that?s how far down a quarter is.
a/360 is fraction of the arc below middle line. So
(a/360)?r2 is fraction of the area in that section of circle. Area of lower triangles
is ? ((cos a)®)((sin a)® )
So 2 of the sections of circle, two triangles equals quarter of the area of circle
The rest is left as an exercise for the reader.
hi,
this is kind of hard to read so I’ll attacha csv of the spreadsheet.
dry" wet" gals. percent
0 20 20 100
0.1 19.9 19.98 99.91
0.2 19.8 19.96 99.78
0.3 19.7 19.93 99.63
0.4 19.6 19.89 99.45
0.5 19.5 19.85 99.25
0.6 19.4 19.81 99.03
0.7 19.3 19.76 98.8
0.8 19.2 19.71 98.55
0.9 19.1 19.66 98.29
1 19 19.6 98.01
1.1 18.9 19.54 97.72
1.2 18.8 19.48 97.42
1.3 18.7 19.42 97.1
1.4 18.6 19.36 96.78
1.5 18.5 19.29 96.44
1.6 18.4 19.22 96.1
1.7 18.3 19.15 95.74
1.8 18.2 19.08 95.38
1.9 18.1 19 95
2 18 18.92 94.62
2.1 17.9 18.85 94.23
.
.
.
10 10 9.94 49.7
10.1 9.9 9.81 49.06
10.2 9.8 9.69 48.43
10.3 9.7 9.56 47.79
10.4 9.6 9.43 47.16
10.5 9.5 9.3 46.52
.
.
.
13.9 6.1 5.11 25.54
14 6 4.99 24.96
.
.
.
19.8 0.2 0.03 0.13
19.9 0.1 0.01 0.04
20 0 0 0
Try these number out.
Be safe!
For a 20 inch diameter tank, the 25% full mark is 5.96 inches high and the 75% mark is at 14.04 inches. Ref: http://mathforum.org/dr.math/faq/faq.circle.segment.html#12
Michael Simson MD & David McCarthy MD
contact: 215 662 2187 or 215 662 6802
Integral calc. is one fun way to solve it, but here’s what an hour and the ability to write computer programs gets you:
The question the trucker asked about his gas
tank can be simplified to problem of finding
the center of mass of a semicircle. Instead
of thinking about a cylindrical gas tank,
consider a semicircular object with radius r
that has a constant density (maybe it’s made
of wood). The center of gravity is the point
at which the semicircle balances. If we drew
the center of gravity on our wooden
semicircle, we could balance our semicircle
on a pencil at that point. This means that
the semicircle has the same area above and
below the point, which is what we want.
From first semester calculus, we know that
the formula for the y component of the center
of gravity is the moment about the x-axis
divided by the area of the semicircle. the
area is (1/2)pir^2. The moment is the
integral from -r to r of (1/2)f^2 where
f(x)=sqrt(r^2-x^2) (the formula of a
semicircle with radius r).
After some calculation Mx = (2/3)r^3
And the y-value of the center of gravity is
(4/3)(r/pi). So the trucker needs to draw
his 1/4 and 3/4 lines 80/(3pi) inches above
and below his center line. That’s about 8.5
inches. I can’t figure out how to send a
picture to you guys. Please let me know how
and I’ll draw a picture.