Car Talk and the gas tank again, but wrong

Here is the original puzzler about how to measure 1/4 th full mark in a sideways cylindrical tank.

Sometime ago you entertained a caller who wanted to know how to measure the fuel level in the cylindrical tank of his diesel truck. These tanks are cylinders that lie on their side and the filler is on the top. Specifically, he wanted to know if he inserted a broomstick through the tank’s filler opening, where on the stick should he put the ¼-full mark?

The ‘answer’ (although wrong) to this puzzler is essentially to take a circular piece of cardboard, cut it in half. Then use a pencil to find where this cardboard balances. This (they claim) will be the 1/4 th mark. They even have a video of this technique.

So, that is their answer. It is wrong. Wait. Let me remind you how much I love Car Talk. Actually, I suggested the names “Car” and “Talk” for two of our children. These names were rejected from the Allain-family naming committee.

Ok, let me get on with this. Why is this wrong. First, let me skip to the problem of finding the point on a flat circle such that one fourth of the area below that point is one fourth of the area. Can we all agree that this is the real problem and that it is equivalent to finding the height where the volume of the cylindrical tank is one fourth full? Great.

Here is my main problem Ray (from Car Talk) found the center of mass (center of area) of a half circle. I suspect his reasoning was something like this:

“Ok, so the half circle is balanced on this pencil. This means that half of the cardboard (and thus half the area) are on each side of that point. Expanding this to a full circle would mean that the location is the fourth-full mark.”

The mistake is thinking that the center of mass means that equal masses (or areas) are on each side of this point. BOOOGUS. (Ray likes to say that). Ray is confusing torque and weight. Let me give an example of where Ray’s method works.

Here a line going through the center of mass would also be a line splitting the object into two equal areas. Suppose the above shape is cardboard. Suppose also I have an extra piece of cardboard that I attach to each side with a coat hanger wire like this:

For this case, the dotted line still splits the object into two equal areas. However, it would not balance here. If something balances, what does that mean? That means that the net torque on the object about that balance point is zero (technically a vector). You could say that the torque from the stuff on the left of the balance point is equal and opposite to the torque on the right side. Here is the key: torque depends on the weight AND its distance from the balance point.

Let me write torque like this. The torque about some point is:

The vector r is from the balance point to the mass (the center of the mass) and F is obviously the force. θ is the angle between these two, for simple cases (like here) θ is π/2. But how does this relate to the half-circle cardboard thing. Suppose I find the balance point and then fold it in half along the radius. This would be a side view.

I drew those rectangles so that you could imagine them as individual masses. On the left, you need more of these rectangles because they get shorter (however, they are also farther away). The point is that just because it is balanced does not also mean equal areas.

One more point. This is probably close to the correct answer. However, taking 1/4 th the diameter is pretty close to the correct answer also.

For completeness, let me calculate the center of mass (even though this is in just about every single calculus textbook) and compare that to point to indicate a fourth of the tank.

Center of mass (area) for a half circle

Here is my object and my coordinate system:

Clearly, I just need to look at the x-direction for the center of mass (the y-coordinate of the center of mass would be zero). The x-coordinate of the center of mass is:

This just says that the center of mass is the weighted average of the masses of these rectangles that I drew. They are weighted by the distance from the origin. The dm i is the mass of these rectangles and x > is the location the x-coordinate of the center of these rectangles. Since it is an average, I have to divide by the total mass (M). In the limit the the width of the rectangle goes to zero, this becomes the following integral (or you could leave it like that and do a numerical integration with python).

Here I have the variable x, but an integration variable dm. That needs to be fixed. So, what is the mass of the little tall rectangle in terms of x? Well, suppose the surface area density is:

This means the area and mass of the rectangle is:

(The 2 comes from the height of the rectangle) Great, I removed the dm but now I have a y. Well, there is a relationship between x and y since it is the equation of a circle. I can write:

Putting this together, I get the following integral:

This isn’t too difficult of an integral. It can be evaluated by doing a substitution. Anyway, if you do that, you get (or you can try this on Wolfram Alpha). Actually, Wolfram Alpha will even show the steps in this integration and even let you save it as an image. Good job. Here is that image. (but don’t cheat and use this for your homework)

Now, I just need to evaluate the limits of integration. I get:

Check in your calc book or google it. This is the same answer. Also, it has the right units (distance) and it is negative (for this case).

Comparing values

There are three answers for this problem. First, the real answer (determined using calculus). This gives the area as a function of distance from the bottom as:

Note, this is the area for a half-circle partially filled. Put in h = R and you get the area of half of a circle. But, what I want is the h that gives half of a half of a circle. This means I need to solve for h in the following:

Solving that for h does not look like fun. Good thing I already did this (see previous post). For 1/4 th full mark, it is 0.298 times the diameter from the bottom. Let me call this 0.596R

The next method is the Car Talk Balance Method. From above this gives a distance from the bottom of the tank for 1/4 th as: (remember the x-center of mass from above was from the center of the circle)

Putting in values for π this gives a height of 0.5756 R.

There is a third method. What if I just measure 1/4 th of the height of the tank? This would give a height of 0.5R.

To summarize: here are the percent differences from the actual answer

• Correct method = 0.596R. This is a 0% difference from the correct answer.

• Balancing pencil method = 0.5756R. This is a 3.4% difference from the correct answer.

• The fourth is a fourth method = 0.5R. This is a 16.1% difference from the correct answer.

I still love Car Talk and it is still a very clever method that gives a fairly close approximation for a fourth full tank. This doesn’t work for any other measurements though (well, I guess you would have to think of some other clever method).