About Ray's Recent Booogus Puzzler


#1


Just how bogus was Ray?s recent puzzler? Well, pretty darned bogus, it turns out.

Ray, in case you?re wondering, claims he was busy having an out-of-body experience. He doesn?t recall spewing such bogosity. (You can hear his explanation right here.)

Fortunately, we have a few thousand alert mathematicians in our audience. Two of our favorites, our old pal Wolfgang from the World?s Greatest University, and David Vogan, from our recently humbled alma mater down the street, e-mailed us with great explanations of the non-bogus solution. We?ll share them right here, below.

Both David and Wolfgang will be joining the discussion, when they?re not laboring to repair the damage we?ve done to their students? minds, that is.

Yours in future non-bogus puzzlers. We hope.

Tom and Ray Magliozzi

Click and Clack, the Tappet Brothers


#2

Hi Tom and Ray –

A dope slap is in order. Ray said there are equal masses on opposite sides of the center-of-mass, and that’s usually not true.

The COM (center-of-mass) is where we can pretend all the mass of the object is located. The COM doesn’t even have to be located on the object itself! If you suspend an object by its COM, it will hang balanced, not because there’s the same amount of mass on each side of the COM but rather, the torque due to the mass on one side is equal to the torque due to the mass on the other side. (The integral of mass times distance is equal on both sides of the COM.) To convince yourself, try this: Take one of your hammers and balance it to find its COM. Cut the hammer apart at the COM and note that the two pieces have different masses. I don’t know what you’re going to do with the two pieces except save them to remind you of the error of your ways. Thus, finding the COM of the semicircle and saying that’s the depth which defines the 1/4 level of the tank because the same amount of gas is above that point as below is technically wrong.

Having said that, I’m throwing you a lifeline. I decided to do the experiment you described and found that the COM of a semicircle is located (0.590)R from the circle’s edge. The numerical answer to the 1/4 Tank Problem is (0.596027)R, pretty darn close to the measurement! The fortuitous close agreement is because the distribution of mass in the semicircle is pretty uniform and the semicircle is not some “weird” shaped object (like a hammer). All logical thinking aside, your method works!

Your physicist friend,

Wolfgang


#3

Dear Tom and Ray,

I don’t really listen to your show. Our cats do many things to annoy us, and one of them is walking on the “on” switch on the clock radio while your show is on.

So I heard the “solution” to the cylindrical gas tank puzzle. I’m afraid I have to differ with you. The center of mass need NOT have half the mass on one side and half on the other: it also matters how far from the center the mass is. A 200 pound man (no offense) sitting three feet from the middle of a see-saw will balance his 120 pound significant other sitting five feet from the middle; but 5/8 of the mass is on one side, and 3/8 on the other.

The center of mass of the 20" diameter half disk is at a distance of 40/(3 pi) from the center, or about 4.244 inches.

To see how much area is below this point is a matter of 11th grade trigonometry. On your pizza half disc, draw two radii from the center to the chord through the center of mass (the gas level). Those radii make an angle theta with the top of the disc; you know that sin(theta)=4/(3 pi), so you can figure out that theta (in radians) must be arcsin(4/(3 pi)) = .4383137. So the chords make a big slice of pie, whose area is

100((pi/2) - theta) = 113.248.

That’s the piece below the center of mass, EXCEPT that it has on top of it an extra triangle; so you have to subtract the area of the triangle. Its height is 10sin(theta) = 4.383137, and its base is 20cos(theta)=18.10937. So the area of the triangle is half the base times the height, or 39.687. So the area below your gas line is 113.248 - 39.687 = 73.561.

The total area of the full pizza disk is 100pi = 314.159, so your method marks the point where the tank is at

73.561/314.159 = .23415:

barely 23.4% full!

You may say that this is, as the saying goes, “good enough for government work.” vehicle which is supposed to arrive with the 25% fuel load needed to return, when in fact there is only 23.4%. If somebody has to get out at the moon and walk, it won’t be me.

Faithfully,
David Vogan
Member, Department of Mathematics
Your Alma Mater

P.S. Actually finding the right place for the quarter-tank mark involves solving some transcendental equation that I don’t know how to type. Doing this numerically is a fine exercise for Newton’s method, which I assume that you enjoyed in MIT course 18.01, long, long ago in a galaxy far, far away.


#4

You don’t need a physicist, or any other professor to solve this puzzler.

Starting with a full tank of gas, (20 inch on stick),record odometer reading, drive until fuel measures 10 inch on stick. Record odometer and subtract from first reading. Drive half the difference and mark the stick at your current fuel level. This is the 1/4 level.

Don


#5

With all respect to Prof. Vogan and Wolfgang, their points about the center of mass, while correct, simply don’t matter in solving the problem. The hammer illustration is important when the balanced object is not uniform in its weight distribution, but the pizza cardboard IS uniform — enough so to be useful for dealing with the practical challenge, unless one side is saturated with pizza grease. And it’s continuous, so that the see-saw example, while correct as well, doesn’t bear on the problem.

But a quicker solution, right at the pump — fill the tank to the halfway mark on the stick. The owner knows the volume of the WHOLE tank, and thus can now proceed to pump exactly one-fourth of that capacity — into the half-full tank. Sounding the tank with the stick now locates the 3/4 full mark, and thus determines the number of inches between the halfway point and BOTH the 3/4 full mark and the 1/4 full mark, so the stick can be marked accordingly, and the problem is solved.

Regards,
Capt. Mike from Nahant


#6

I think the professors are trying to say is that because the pizza box is curved, some of the cardboard is going to be farther from the balancing point than other cardboard, so the weight of the cardboard from one side of the balancing point to the other is slightly different. But your “at the pump” solution sounds great.
Ron


#7

The pizza cardboard method really doesn’t work, even though the cardboard is uniform and continuous. It’s a little easier to see with an equilateral triangle of cardboard, sitting on its base and H inches high. (So now the gas tank is shaped like a prism, and the question is where half full is.) The center of mass is at height H/3 = .333333 H. The height with half the mass above and half below is H(1 - root(2)/2) = .292893 H. This time the error is about 12%, compared to about 5% in the puzzler answer.

Capt. Mike is absolutely right about the “at the pump” solution, which sounds like it’s as precise as the pump, and way easier than anything else that’s been proposed.
David

P.S. My letter, quoted by Tom and Ray, lost a little bit in translation. The last paragraph originally read

You may say that this is, as the saying goes, “good enough for government work.” In that case I hope not to ride to Mars with you in a vehicle which is supposed to arrive with the 25% fuel load needed to return, when in fact there is only 23.4%. If somebody has to get out at the moon and walk, it won’t be me.


#8

Apologies and a tip o’ the hat to Prof Vogan, and his illuminating equilateral illustration, which has caused light to dawn on Marblehead.
The cardboard method would work IF one were to slice the cardboard along a line meant to replicate the level in the tank (i.e. perpendicular to the radius described in the puzzler) and then place the two sections in a balance (as used by the the statute of Justice). But if the two sides didn’t balance at first (likely) one would have to take small slices off the heavier side, and shift them to the lighter side until balance was attaned, which is pretty clumsy. But it does remove the distorting effect of the location of the components of the weight.

Capt. Mike from Nahant


#9

Folks,

The Math Forum has posted your solution to this Puzzler at

http://mathforum.org/library/drmath/view/61752.html.

I’ve read it. As far as I’m concerned, “Forum” is “Twoum” plus “Twoum.”

Until a more artful solution surfaces, may I suggest:

  1. Begin with an empty (or nearly-empty) tank.
  2. Turn the truck up on its end - vertical orientation
  3. Fill tank to 1/4 full (i.e., 1/4 way up the height of the tank).
  4. Return truck to normal horizontal (level) position.
  5. Strike a line at the fluid level. That’s your 1/4 level mark! Make a reciprocal line for the 3/4 capacity (no extra charge)

Quid Est Demonstratum

No matter the size of the truck, I maintain this method is easier to accomplish than the Math Forum’s geometric gymnastics. And it’s better for you than eating pizza, just to get the box.

Rick


#10

My problem with the answer came way before the bit about the center of mass. It was with Step 1: Using 10th grade geometry, draw two chords to find the center of the circle. Now I?ll admit that it?s been a long time since 10th grade, but the only way I know that two chords determine the center of the circle is if they are intersecting diameters. But how do you find a diameter? Well, you could draw a chord through the center of the circle, but that reasoning seems a bit, uh, ?circular.? I know another method for finding a diameter (and its midpoint, the center of the circle), but it involves the use of a compass, and there was no mention of a compass in the puzzler. Maybe the pencil that was borrowed from the waitress was actually mounted in a compass.

If there?s a 10th grader out there, please tell me how to find the center of a circle using two chords.


#11

David – my measurement of the COM being 0.590R from the circle’s edge is not in disagreement with your calculation (0.576R). A good experimentalist always accompanies a measurement with uncertainties, and I neglected to do that. My quick and dirty measurement was good to within an 1/8", or 4%. It agrees within the error. I repeated the measurement more carefully and got 0.572R with 1% uncertainty, in closer agreement with your calculation. – Wolfgang


#12

You’re not looking at the tools at hand. You’re in a restaurant, so there is a pen or pencil available. You have a circular piece of cardboard.

Fold the cardboard in half. Unfold, then fold again at 90 degrees (ideally, but it doesn’t matter) to the first fold. Use the pen or pencil to draw the chords formed by the folds on the cardboard, marking the center of the circle.

The only thing you need from 10th grade geometry is the knowledge that any two diameters will intersect at the center.

Alternatively, you could use the pizza box as a straightedge. It’s a little more work to mark the diameter on the pizza box, but still trivial.


#13

This is why all fuel tanks should be rectangular, easy to calulate. I am now hungry for Pizza!


#14

Here is a very precise answer:

  1. Measure the diameter of the tank. Divide by 2 to get the radius.
  2. Mark the 1/2-full line horizontally on the stick, 1 radius up from the bottom.
  3. Multiply the radius by .4067. (.4067 is the sine of 24 degrees. I will explain soon.)
  4. Mark the 1/4-full line .4067 of a radius down from the 1/2-full line.
  5. Mark the 3/4-full line .4067 of a radius up from the 1/2 full line. Finished!

Here’s how I came up with the crucial .4067:
Get two pizzas with the same cross-section diameter as the fuel tank. Cut one pizza into 15 equal slices. Each slice will be 24 degrees. (15 slices time 24 degrees equals 360 degrees of pizza.) Eat 7 slices, and keep 8 slices. That equals 8 fifteenths of a whole pizza left to play around with.

Arrange four of the pieces along the diameter of the cardboard circle that came with the pizza, points in, two pieces on top of the diameter, and two on the bottom of the diameter line, like a bow-tie in the middle of the pizza circle. (2B Continued - I think I am running out of posting space.)


#15

Part II
Now you have four pizza pieces arranged in a bow-tie along the diameter of the pizza circle. Take your other four pieces, and place them backwards (points out at the circle edge, crusts toward the middle), above and below the pieces that form the bow-tie. Two pieces above, and two pieces below.

Note that the crusts will overlap a bit in the middle. Cut off the overlapping parts, so that you have one layer of pizza, with boundary lines (chords) parallel to the diameter, one line above and one line below. These are your 1/4-full and 3/4-full lines. They are .4067 of a radius below and above the diameter (1/2-full) line.


#16

Part III
Now, take a look at the overlapping pieces you cut off. (I hope you did not eat them.) See that each piece is about one quarter the size of a single 1/15th pizza slice, total one quarter plus one quarter equals one half of 1/15 of a pizza.

You started with 8 slices (= 8/15 of a whole pizza). You have cut off one half of 1/15, so you now have seven and a half fifteenths of a pizza. That is one half of a whole pizza! And, that half of the area or volume, is arranged right in the middle.

So, the rest of the pizza circle (above and below) also must total seven and one half fifteenths. The top quarter is three and three quarters fifteenths, and so is the bottom quarter!

You now may eat the cut-off overlaps. And, if you want to complete this creation, get a second pizza, and cut the top and bottom straight across, to fill in the top and bottom quarters of the pizza circle. Those pieces will be .5933 of a radius from top and bottom to middle. (1r minus .4067r equals .5933r.)

There is another way to do this using 60ths of the circumference, but not now. This thing has grown a beard. Cheers!


#17

Why stand the truck on end. Put 25 gal. in the 100 gal. tank put stick in and mark it.


#18

That’s a great solution except for the fact that many semis do not allow you to run on only one tank.

IOW you can’t run one of the tanks dry, ‘disconnect’ that tank from the fuel pump and then run on the other tank.

That means that you would have to arrange to run both of your tanks dry at the momemnt that you got to the pumps.

Yeah, it’s possible but not many drivers are going to want to do that.


#19

“Quid Est Demonstratum” means “What has been shown?” (IF you got the participle right, but it is clearly a question.)

The phrase you’re looking for is “Quod erat demonstrandum” which means “That which was to be shown.”

This is perhaps why most people stick to the much safer “Q.E.D.”


#20

The puzzle is really about volume, and equivalently area, not mass. So the solution should be only trigonometric, not trigonometric plus center of mass.

Find g, the height of the gasoline level when tank is ? full. Diameter of tank is 20".

Let W = area of circular end of tank
Let D = the left end point on the circle of a horizontal chord representing the gasoline level
Let B = the right endpoint of above chord
Let C= center of circle
Let A = the bottom point on the circle
Let G = the point where the chord DB intersects the radius AC
Let S = area of sector ABC
Let T = area of right triangle BCG
Let K = area of half-segment ABG
Let ? = the angle subtended by the arc AB
Designate the line CG as ?a?
Designate the line AG as ?g? (the above height of gasoline level)
Designate the line GB as ?b?

W = ? r^2 = ? 10^2 = ? 100 = 314.16

S = ? /360 ? 314.16
cos ? = a/10 (10 = the radius, and is the hypotenuse of AGB)
arccos (a/10) = ?
Substituting the value of ?,

T = ? height ? base = ? ab
a^2 + b^2 = 10^2 (pythagorean theorem)
b^2 = 10^2 ? a^2
b = (100 ? a^2)^?
Substituting the value of b,
T = ? a (100 ? a^2)^?

K = S ? T
2K = ? W (as specified by puzzle)
K = ? W = ? ? 314.16 = 39.27
S ? T = 39.27
Substituting the values of S and T,

    314.16/360 ? arcsos (a/10 - ? a(100 - a^2)^? = 39.27

Rather than try to simplify that equation to solve for ‘a’, I plugged the equation into a spreadsheet and varied ‘a’ until left side of the equation equaled 39.27. A value for ‘a’ of 4.04 approximates this quite closely, a value of 4.0396 even more closely

g = 10 ? a = 5.96 or more accurately, 5.9604