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Topic: Glider contest (Read 1166 times) |
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BNC
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Glider contest
« on: Jan 18th, 2003, 3:28pm » |
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Jack and Jill went up the hill to participate at the national no-control glider contest, in which every contestant sends his / her glider down from a high place, to glide on its own.
Jill is a talented engineer, so she designed a brand-new type of glider. Jack lacks the design skills, so instead, he snuck to Jill’s room, and stole her design.
Jill had her glider manufactured to spec. Jack wanted his glider to be better, so he doubled all of the dimensions. So, for example, if Jill’s maximal measures were x,y,z, then Jack’s were 2x,2y,2z.
Who will have a better chance to win – Jack or Jill?
Note: I’m aerodynamics illiterate. This riddle needs only very basic physics knowledge – no more than high school level, maybe even middle school
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towr
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Re: Glider contest
« Reply #1 on: Jan 19th, 2003, 7:01am » |
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well, the surface area to mass ratio decreases as dimensions increase, so based on that Jack would probably loose..
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Speaker
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Re: Glider contest
« Reply #2 on: Jan 19th, 2003, 7:01pm » |
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If the talented engineer Jill designs a "Brick" then it is not likely to glide very far, and further exposing my outdated views on sexual equality, if Jack is a normal man with greater musculoskeletal development in his upper body, then he can throw the glider further, regardless of its doubled weight.
But, Jill may win, all things being equal.
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| « Last Edit: Jan 19th, 2003, 7:03pm by Speaker » |
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redPEPPER
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Re: Glider contest
« Reply #3 on: Jan 20th, 2003, 3:51am » |
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I don't think stopping at the surface to mass ratio is enough to be conclusive. Actually, I even think it leads to a wrong conclusion. The lift of a glider depends on the surface of its wings, but also on other factors, like its speed. Also, do you know that if you add weight to a glider without modifying its dimensions nor its center of gravity, the glider will have the same glide angle? It will follow the same trajectory, only faster.
So the surface to mass ratio is not really relevant.
I'm sorry but I'll have to use notions of aerodynamics to explain this, and I don't know how else you could come to a valid conclusion in this puzzle without them. I'll keep it simple anyway.
There are three forces that are applied on a glider in equilibrium: Lift, Drag and, opposing these two, Weight. The thing that matters for the glide angle of a glider is the Drag to Lift ratio, not the Weight. The Weight only influences the speed. And if you build a glider doubling all the measures, you will have the exact same Drag to Lift ratio. The bigger glider will follow the exact same trajectory.
When every dimension is doubled, Weight is 8 times as big while surface is only 4 times as big. I could use Lift and Drag formulas to show that the glider will have to go at sqrt(2) times the speed of the smaller glider for Drag and Lift to be 8 times bigger too. Short story is: Lift and Drag vary with Velocity2.
My conclusion is that the fly paths of both gliders will be the same, or at least parallel (if launched in the same direction, with no wind). Depending on the contest's specifics, Jake's glider might fly lower because it had to lose more altitude at "takeoff" to reach its higher cruise speed. On the other hand, if the gliders have a propelled launch and that Jack decides to double the launching speed too, then his glider will fly higher, as it only needs sqrt(2) times the speed, and will convert the excess speed in altitude.
But if we keep it simple and assume the gliders start at equilibrium, they'll both go as far, with Jake's glider going there faster.
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RexJacobus
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Re: Glider contest
« Reply #4 on: Jan 20th, 2003, 5:54am » |
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My answer is a combination of redPepper's and Speaker's.
Speaker thinks Jack, being a man, will be able to throw the glider farther even though its weight will be double. But I think Jack's glider will actually be 8x heavier than Jill's glider and even his manly physique won't be able to make up the difference.
For Jack to win he would have to launch his glider using more than 8x the force that Jill will use which I find improbable therefore I vote for Jill.
Rex
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towr
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Re: Glider contest
« Reply #5 on: Jan 20th, 2003, 7:25am » |
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on Jan 20th, 2003, 3:51am, redPEPPER wrote:Also, do you know that if you add weight to a glider without modifying its dimensions nor its center of gravity, the glider will have the same glide angle? It will follow the same trajectory, only faster.
So the surface to mass ratio is not really relevant. |
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I don't think that is true.. Suppose you add 1mm^3 piece of neutron star to the center of gravity.. I'm pretty sure any glider will go straight down, and not glide at any angle..
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James Fingas
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Re: Glider contest
« Reply #6 on: Jan 20th, 2003, 8:48am » |
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I think it depends on too many variables. If Jill makes her glider too small (picture throwing a very small, very light glider), the air resistance is too large for a controlled flight (like trying to throw a feather). Jack's design in that circumstance could be better from an engineering point of view.
Now if Jill's design is perfectly optimal, then obviously Jack's is not. But that doesn't say a whole lot...
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kenny
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Re: Glider contest
« Reply #7 on: Jan 20th, 2003, 12:13pm » |
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I have to go with Jill on this one.
Most of he previous answers ignore an important part of the puzzle: " Jill is a talented engineer".
Reason this way: if doubling the size makes for a better glider, then why wouldn't a talented engineer already have thought of that? The answer is that she would, and that she chose the optimal size. Thus, Jack's glider must be suboptimal.
-- Ken
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redPEPPER
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Re: Glider contest
« Reply #8 on: Jan 20th, 2003, 1:11pm » |
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on Jan 20th, 2003, 7:25am, towr wrote:| I don't think that is true.. Suppose you add 1mm^3 piece of neutron star to the center of gravity.. I'm pretty sure any glider will go straight down, and not glide at any angle.. |
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Well, I was thinking of less out-of-this-world weight modifications, of course. But just for fun...
Assuming the density of a neutron star is 1*1018 kg/m3, that's 1*109kg/mm3. The added piece weights a million tons.
Assuming the equilibrium speed of the initial glider is, say, 1m/s and that its weight is 1kg, the modified glider is 1*109 times as heavy. To make up for that increased weight, you need increased lift, by increasing the speed by a magnitude of sqrt(109) which is a little less than 31,623. 31,623m/s, that's about 70,740mph, or Mach 93.
In a perfect world, if you manage to get the modified glider at Mach 93, its wings will provide enough lift to carry the neutron star piece on the same trajectory than the original glider. This is of course ignoring problems with the sound barrier, material resistance, energy loss...
on Jan 20th, 2003, 8:48am, James Fingas wrote:| I think it depends on too many variables. If Jill makes her glider too small (picture throwing a very small, very light glider), the air resistance is too large for a controlled flight (like trying to throw a feather). Jack's design in that circumstance could be better from an engineering point of view. |
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If the air is perfectly still, the glider size doesn't really matter (okay, I don't mean atom-size). Let's say if it's bigger than 10cm or something.
If the air isn't perfectly still, that opens a whole new spectrum of possibilities!
Quote:| Now if Jill's design is perfectly optimal, then obviously Jack's is not. But that doesn't say a whole lot... |
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I've been thinking about that as a possible answer. But there's still the possibility of a tie.
Physics say the gliding angles at equilibrium are the same. So it all depends on what happens before equilibrium, i.e. how the gliders are launched. As RexJacobus pointed out, Jack's glider being 8 times as heavy, and requiring a speed sqrt(2) times as fast to reach equilibrium, it'll need more than 11 times the Force Jill's glider needs. the odds aren't in Jack's favor.
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James Fingas
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Re: Glider contest
« Reply #9 on: Jan 20th, 2003, 1:46pm » |
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redPepper,
I'm not sure I follow your reasoning perfectly, but it seems to me that your formulae don't take into account the Reynold's numbers of the two gliders. All other things being equal (I guess this means if Jack could actually throw 11 times as hard as Jill), then the glider with the higher Reynold's number would win every time (that would be Jack's glider). This could be important, especially if the gliders were released with zero velocity instead of being thrown (maybe they let them slide off the roof of a windmill  ).
Or do your drag considerations take this into account?
This is the same as a heavier and a lighter person racing bicycles down a hill--theoretically, they arrive at the same time, but the heavier person is more efficient in practise.
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| « Last Edit: Jan 20th, 2003, 1:58pm by James Fingas » |
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Chronos
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Re: Glider contest
« Reply #10 on: Jan 20th, 2003, 2:46pm » |
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Quote:| This is the same as a heavier and a lighter person racing bicycles down a hill--theoretically, they arrive at the same time, but the heavier person is more efficient in practise. |
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I'm not sure about that... All of the forces involved are approximately proportional to the mass, except for the air resistance, so air resistance should bethe deciding factor. Air resistance should be proportional to the cross-sectional area, which, assuming the people are about the same shape and density, should scale as the 2/3 power of the mass. So air resistance will be less significant for the heavier person, and so he should theoretically win the race. What do Reynold's numbers have to do with it?
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redPEPPER
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Re: Glider contest
« Reply #11 on: Jan 24th, 2003, 2:47am » |
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on Jan 20th, 2003, 1:46pm, James Fingas wrote:redPepper,
I'm not sure I follow your reasoning perfectly, but it seems to me that your formulae don't take into account the Reynold's numbers of the two gliders. All other things being equal (I guess this means if Jack could actually throw 11 times as hard as Jill), then the glider with the higher Reynold's number would win every time (that would be Jack's glider). This could be important, especially if the gliders were released with zero velocity instead of being thrown (maybe they let them slide off the roof of a windmill ).
Or do your drag considerations take this into account? |
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I was not aware of the Reynolds numbers before you mentioned them. I've been looking them up but I'm still confused, as the explanations I found are not very clear, and require some background in physics of aerodynamics.
In the Drag formula I used, there's a factor that takes that into account. It's called the Drag coefficient. From what I understood, this coefficient depends on the shape of the glider, but not on its scale (which is taken into account in the other terms of the equation, such as Area, Weight and Speed). I'm not so sure anymore that this coefficient is constant at all shapes but I'm not sure of the opposite either...
This riddle turned out to be much more of a physics problem than an actual riddle. I don't see how there could be a simple answer with no knowledge of aerodynamics. It comes down to finding if the Drag vs. Lift ratio is modified, and if it is, in which way.
Quote:| This is the same as a heavier and a lighter person racing bicycles down a hill--theoretically, they arrive at the same time, but the heavier person is more efficient in practise. |
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What you mean is that, without air friction (in a vacuum) they arrive at the same time, but the heavier person is more efficient with air friction, assuming that both people have the same shape and scale! If not, the lighter person could very well arrive first.
As chronos mentioned, this has nothing to do with the Reynolds numbers. It's simply about drag, which is taken into account in the gliders' problem (without drag, a glider would never come down).
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BNC
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Re: Glider contest
« Reply #12 on: Jan 24th, 2003, 9:42am » |
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“Riddle” origin:
Many years back, before my undergraduate studies, I tried entering an “excellence” program at the university I applied to. I was then called on an interview before a number of the local professors. The interview consisted with nothing but riddles. I must shamefully admit that I made very poor impression at the time – I was NOT into riddles back then.
Anyway, this was one of the riddles I was presented with – one I couldn’t answer. I asked the professor what the intended answer was, and the answer he gave me the original one by towr (surface area to mass ratio). He was a physics professor, but probably not aerodynamics specialist, so I don’t know if that’s the actual “official” answer.
I didn’t want to interrupt the discussion, as I was amazed at the various points of view. Thanks guys!
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James Fingas
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Re: Glider contest
« Reply #13 on: Jan 24th, 2003, 9:45am » |
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The Reynold's number is a single parameter that describes how a fluid flows. It measures the relative importances of viscosity and fluid inertia. For two bodies that have the same shape to have drag proportional to their frontal areas, they must also have equal Reynold's numbers, which means that the smaller object must travel faster than the larger object if they are in the same fluid (you can see this by examining the formula for the Reynold's number).
Therefore, if you make the smaller object travel slower, then the two no longer have the same Reynold's number, so your drag will not be exactly proportional to frontal area. I think the drag will be higher for the small one, but I'm not sure.
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