# Today's puzzler

Why, when discussing the quarter-full mark in a cylindrical gas tank while sitting in a restaurant, not just fill/empty your cylindrical water glass to 1/4 full, put a plate over the top, tip it sideways, and see how far up the water level is by looking at the (bottom) end? water, fuel, diameter, length - no matter.

I heard today’s solution, and I don’t think it is correct. It seems to me that the center of mass does not mean that you have the same mass on both sides. Consider two people on a seesaw, one weighs 100 pounds and the other weighs 200. If the heavier person moves half-way to the center, the seesaw will balance – the center of the seesaw will also be the center of mass. Yet, one side weighs 100 pounds more than the other.

The solution assumes the density of the mass is equal throughout. In a gas tank, unless there’s something else in it besides gas, you can assume the density is uniform.

You’re introducing an additional entity into your thought experiment you don’t need: the Lever, one of the three Simple Machines of antiquity.

The center of mass means precisely “that you have the same mass on both sides.” It can be found through geometric means:

Step 1: An arbitrary 2D shape.
Step 2: Suspend the shape from a location near an edge. Drop a plumb line and mark on the object.
Step 3: Suspend the shape from another location not too close to the first. Drop a plumb line again and mark. The intersection of the two lines is the center of mass.

You are right – today’s solution is not correct. And the example with the seesaw exactly shows the problem.

In fact, center of mass does not mean “the same mass on both sides”, even if the density of the object is uniform. (The procedure described by tarcaulk will find the center of mass.)

Consider the following example which is a variation of the seesaw. If you have a 20 foot long board, then the center of mass is the same as the point where the mass is the same on both sides. It will be at the 10 foot mark.

But imagine that one end of the board has a very wide section adding extra weight to that end (at the 20 foot point) and there is another section that is very wide at the 5 foot point adding the same amount of extra weight. Then the center of balance will not be at the 10 foot mark (it will be a bit towards the 20 foot mark), but the “point where the mass is the same on both sides” will still be at the 10 foot mark. Note that in this example, the density of the shape is uniform.

The balance point/center-of-mass answer is definitely wrong today. Think about the center of the 1/4 line,and break up the masses into that above the line and that below the line. These two are, by definition, equal mass (both are 1/4 tank). But, just like two same size people on the teeter-totter, that doesn’t mean it balances, BECAUSE it also depends upon where each mass is located (how close to the fulcrum). And, the reason it won’t is because the mass above the line has an entirely different shape than the mass below the line. The center-of-mass of the upper portion is closer to the line than the center-of-mass of the lower. If you do the experiment with the pizza board, just find the balance point and cut it along that line. Then accurately weigh the two pieces. They won’t be the same. If you do the calculus, you will see that you are not solving similar problems. If you look at the posting I made last November (around #260), you will see the 4th grade solution I gave using a pizza board and a scales. I believe the center-of-mass and the 1/4 line were both calculated in those 300 submissions, but you have a lot to sort through to find the right answers.

brilliant!

that it for doitrrueit

You are all over-analyzing the problem. The solution posted on-air is correct, based upon assumptions. the density of the diesel fuel is uniform, therefore the uniform density of the cardboard “slice” is representative of the fuel mass. Teeter-totter examples are inappropriate because they don’t make the same assumptions. It’s not that hard to understand…just don’t pry too deeply into the problem…

use the stick as an anxiety meter. sure the ? tank is a bit above the ? stick mark. more interesting is to compare how much fuel on the stick is the minimum the driver will accept as a safe driving level. fuel down toward the bottom will be the ?devil may care? type. more toward the ? way mark: these drivers wear tightly whiteys and will have not a crumpled candy wrapper in the cab.

version 2: have different sticks for different regions: high marked stick for SW states. very low marked
stick for NE states where diesel pumps are every 1/8 mile.

The solution posted on air is not technically correct – the center of mass of an arbitrary shape of uniform density does not have equal mass on both sides (it has equal first moment, as others have stated).

For example, an equilateral triangle-shaped piece of paper has its center of mass such that 4/9 of the mass is on one side, and 5/9 is on the other side. More extreme examples can readily be concocted with very irregular shapes.

However, it makes little difference for the fuel tank problem. For a half-circle, the center of mass is found at 4/(3*pi) of a radius down from the center – so the (incorrect) “quarter-full” mark would be roughly 42% down the radius from the center line. The position of the correct “quarter-full” mark (with actually equal masses above and below) does not have an analytic solution, but it is found very close to 41% down the radius from the center line. Practically speaking, the difference in this case is too small to matter.

It's not that hard to understand...
Then why aren?t you understanding that the answer given is incorrect? I?ve attached a graphical representation. Using AutoCad, I created a semicircle as described in the problem, and used the program to find the centroid (center of mass), which can be found using the plumb-line method described earlier. Drawing a line through the centroid, and calculating the area of the regions created shows you that they are absolutely not equal. And, as suggested by another poster in the other thread on this matter, the difference becomes more apparent with a tank the shape of a diamond with corners 10 inches from the center.

The reason for not understanding the incorrectness is misunderstanding “center of mass”, which simply means that for force calculations all the mass may be considered to be at this point; it says nothing about area which is needed to calculate volume (multiplying by the length of the tank). Consider a gas tank with a cross-section formed with identical “T” shapes, and then just think about the bottom half. If the crossbar portion of the T has the same dimensions as the vertical, then this portion of the tank will be 1/2 full (complete tank 1/4 full) when the crossbar portion is empty, the vertical portion just full. Yet the center of mass of this portion is, with crossbar height=a, width=b, at (3b+a)/4 up from the bottom. This is less than b up from the bottom (unless a grows to equal b or more). The calculus is the tool to solve this problem specifically; what’s important is to hone intuitions so they work well enough to meet the given practicalities.

Consider a gas tank with a cross-section formed with identical "T" shapes...
It sounds like you're talking about the same thing we discussed in on of the other threads on this topic. This is the visual I attached to describe what I think you're talking about.

oops. who’d’a’thunk there’d be so many postings and threads. By the way the difficulty with integrating can be reduced by asking when is the portion of the unfilled quarter equal to half of the lower half. The empty space in the bottom portion, when half filled, is t/2 radians * r^2, plus a volume of r*sin(t)rcos(t)/2, figured for the triangular space below the volume of airspace swept by integral down from the horizontal. The identity sin(T)cos(T)= sin(2T)/2 simplifies expressing the equality where the empty space above the fuel in the right bottom quadrant is at the angle T. Half the quadrant = the swept area plus the triangular area: (pi/4)/2 =T/2 + sin(2T)(1/4). then play with tables, or give it to wolfram.

Yeah, though the center of mass/area is CLOSE in this case, it’s not quite right. The seesaw analogy given in the first reply is a good one for illustrating why center of mass does not necessarily divide the mass equally on either side.

Here is what I believe to be the correct answer, and a center of mass calculation in addition to show the difference (click on image for larger version):

http://img828.imageshack.us/img828/2978/cartalkpuzzlernotes2011.th.jpg

And here is a simpler form of the answer using just trigonometry:

http://img8.imageshack.us/img8/2978/cartalkpuzzlernotes2011.th.jpg

Great job! A picture (and a few equations) are worth a thousand words! I’m particularly impressed by the trig-only solution. I didn’t even imagine that one.

The calculus solutions are excellent for showing that you are starting with two different equations, so it is not surprising to get two different answers.

In this problem one needs to consider the linear density of the cardboard; i.e. how much mass is there at each coordinate along the radius. As the width of the cardboard is different at different distances from the balance point, the linear density is not the same. In fact, the linear density is much greater on the side with the straight cut along the diameter.

Center of mass means: mass times distance is the same on each side.

OK! There is a much simpler approximate solution - I’ll call it the jigsaw solution. So you have one half of the pizza platter as your half tank. Take the other half and cut it also in half to represent the quarter tank. Now place the radiused edges so that they coincide at the bottom. Then cut the quarter tank segment across the points of arcuate coincidence. Let’s call this the arcuate sector and keep in the same position. So you are left with a remnant which is a triangle. Now cut the triangle in several narrow strips/slats of equal width, and place the slanted edges together so that they form rectangular pieces. Then place the narrow strips/slats progressively above the arcuate sector on the half board. Progressively build up the level till all of the narrow slats are utilized. If there is overrun simply clip it and use the small segments in the final layer. Presto - you have an approximate height for the 25% tank level. PS I do have a PhD in Optical Physics (lasers). If I think of a laser solution I will write.