Some question for you to have answers to

1) Imagine you have a car with a single penny in the cupholder or ashtray. How many miles do you have to drive before the coin's additional weight is equal to its value in extra gas consumption?

(edited for clarity)

2) If you could have one plastic surgery, done gratis by Sean McNamara and Christian Troy, what would you choose?

7 comments:

Sophist said...

As many miles as it takes to consume the weight of the entire car in gas?

J.Po said...

Most definitely, my hand skin. I would have them replace my hand skin (perhaps with skin off my supple ass).

Mister Vertigo said...

1) I have NO clue. One thing for sure though is that it would take me longer to figure out the answer to the question than it would be to actually drive that far.

2) That's a tough question. Its something I've never thought about before. I guess if I had to pick something I'd have some scars removed or something minor like that. I have no desire at all to change my physical appearance. I don't want to sound stuck up, but anything that would require surgery isn't something that I'd ever want to change.

Theo said...

So about (1), I might be wrong here, but I think it has a pretty simple solution provided one important assumption that the product of the car's weight and its efficiency is constant. That is, if W_c is the weight of the car, W_p the weight of the penny, M_p the efficiency (in miles per gallon) of the car with the penny on board, and M the efficiency of the car without the penny, then

(W_c + W_p) M_p = W_c M

Then we want to calculate the cost of an m-mile trip with a penny on board and without a penny on board and find the number of miles m such that the difference between them is 0.01. Is that correct?

The cost of the trip with the penny on board is:

( (W_c + W_p)/W_c ) * (1/M)* m * p

Where p is the price of gas in $/gallon. The cost of the trip without the penny is

(1/M)* m * p

Thus the difference in costs is

( (W_c + W_p)/W_c - 1 )*(1/M)*m*p

This can be simplified to be

(W_p/W_c)*(1/M)*m*p

If we set this equal to 0.01 and solve for m, we get

m = 0.01 (W_c/W_p)*(M/p)

And this makes some intuitive sense. Notice that as the weight of the penny becomes utterly insignificant with respect to the weight of the car, m approaches infinity. Thus, for weightless pennies, it costs nothing to carry them on board, so you can drive forever before you get to the point where you should have spent that penny putting fuel in your car. Additionally, as the penny becomes very heavy, m approaches 0. This also makes sense. If the penny is very heavy, it is so costly to carry it on board that you better trade it in for some fuel.

Note, however, that this analysis does not take into account the weight of the fuel. This is probably a significant effect. Fuel was left out for simplicity to demonstrate the method, but I think it shouldn't be too difficult to work through the problem including the extra fuel. At this point, without working out the problem, I'm just going to speculate that the answer will be something like:

m = 0.01 ((W_p - W_f)/(W_c + W_f))*(M/p)

Notice that there is a chance that the weight of the fuel purchased with a penny actually outweighs the weight of the penny. In this case, m is negative. That is, it is a cheaper trip with the penny than without it! We assume this is a pathological case.

Also notice that I am assuming that W_f of fuel is on-board through the whole time. In reality, as the fuel burns the car will get lighter. Thus, we really need to involve some simple calculus to get a better handle on the rates and consider the solutions above as sort of bounds on the real solution. I'm going to assume here that this amount of rigor is not necessary.

Theo said...

Note that this analysis might be made simpler with the use of credit cards. Thus the additional fuel would be the only difference in weight in the two cases. Of course, there would be no additional cost to keeping the penny as well, so I guess this would have a trivial solution.

Theo said...

Final note... If you examine the mathematical machinery that would actually be needed to solve the real problem here, a problem that involves the rate of fuel burn over time and distance, you can see that questions of fuel economy are complex ones. The speed of driving, the amount of weight in the car, even the dynamics of the fuel as it sloshes around due to the car's shakey movements are all factors that effect the car's performance.

Putting "spinnies" on the wheels of a pimped up Civic, for example, can greatly decrease the efficiency of the civic due to the extra energy sucked away due to storage aspects of the inertia in the spinners.

Traveling at faster speeds is good for the internal cumbustion engine, which cannot be optimized for slow speeds, but with the faster the speeds the greater the cost drag cost. Thus, there is an optimization problem there.

So after all of this... There are so many variables that a general solution is probably intractable. Thus the practice of efficient driving (and car weight management, etc.) is a bit of a black magic art -- it's a very inexact science.

towwas said...

1) I got into journalism so I wouldn't have to do math anymore. Ok, actually that's not true, and if that were my intention I would have failed, but I'm still going to ignore this question.

2. Dang...I actually can't think of anything either. I don't like surgery. And if I got, like, Botox, it only lasts a few months. And my wrinkles aren't there yet. Ask me again in 40 years, I'll probably want a face-lift.