Weighing a 747. In an Interview.
In a programming interview, you might be asked how you’d weigh a 747 aircraft without using a scale of any sort. What follows is the correct answer:
Empty the plane’s gas tank. Fill up the plane’s gas tank. Note how many gallons of gas you used (calculate the mass of of the gas, as well, as you might not have time later). Get in the plane, take off, climb to altitude. Put the plane into a nosedive (turning off the engines will make the estimate more accurate). Measure the terminal velocity of the plane. Now dump all of the gas. Measure the new terminal velocity.
The ratio of the square of the two velocities is equal to the ratio of the two masses (plane plus gas versus plane). From there it is simple algebra. Depending on your algebra skills (and the initial height of the aircraft) you should be able to complete the calculation before hitting the ground.
October 28th, 2008 at 3:56 pm
I am confused. Please clarify. Terminal velocity should not be proportional to mass. Galileo showed this with dramatic experiments of dropping balls off the Leaning Tower of Pisa. Moreover, (v_f)^2 = (v_0)^2 + 2ad. There is no m in the equality. Great blog, btw.
October 28th, 2008 at 4:21 pm
Galileo’s experiments and the kinematic equation you’ve given assume no air resistance. Without air resistance, there is no terminal velocity.
Terminal velocity is proportional to mass because terminal velocity is the velocity at which your weight equals your air resistance. If you put this second force into your equation, you’ll find mass starts to show up.
October 29th, 2008 at 5:37 pm
“calculate the mass of of the gas, as well[...]”
Isn’t this cheating if you can’t use any sort of scale?
In any case, if the answer can be fanciful enough to involve destroying the plane, why not envision a solution in which you put the plane on a series of lifters, each of which can lift a given weight before they fail? The weight of the plane will be between the max weight supported by the best failed lifter, and the first lifter that didn’t fail. Sort of a variant on the snarky way to determine how much weight a bridge can take…
October 29th, 2008 at 5:40 pm
Nah, Bob! The fuel pump will tell you how many gallons you pulled. You do need to know or look-up the density of the fuel, though, to figure out the mass. Or, you could just go with an estimate as if it was water, and say 1 m^3 is 1000 kg.
October 29th, 2008 at 5:57 pm
Hmm, perhaps I see constraints that aren’t there. To me, a scale means anything you can use to determine mass. The fact that your scale is in the form of a fuel pump, measuring flow over time, makes it still a scale, since it’s a mass counter. I think it’s the same reason some people do poorly on true/false tests; they can see unintended angles implied or allowed by wording.
October 29th, 2008 at 6:00 pm
Well, it’s also that my answer borders on the absurd. I think if an interviewer asked this question he wouldn’t mind it if you embraced the absurdity and attacked the problem in a novel way. Presumably, that’s what he really cares about.
October 29th, 2008 at 6:26 pm
I see what you’re saying, but to get novel solutions from some candidates, you may have to pose a similar problem without constraints, such as “Tell me all the ways you might determine the weight of a 747.” One of those would be “scale” and the rest would be the novel solutions you seek. I think it depends on how people process constraints and whether they accept them at the surface or look for implied meaning.
I used to ask people “Tell me why chocolate fire hydrants are a bad idea,” so that I could see how they thought about engineering, manufacturing, material science, robustness, etc.
April 15th, 2009 at 9:38 pm
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March 26th, 2010 at 8:14 am
If knowing the mass of the fuel is an acceptable assumption, then why not just assume that you know the mass of the 747 to start with?