# Interesting Questions

Actually, many of the questions Marilyn vos Savant got asked in her column are positively ingenious. They also provide a lot of food for thought. Here are a few that caught my eye. As always, feel free to hash them out in the comments:

• I need glasses to see things at a distance. When I look in the mirror, I can see my face clearly. When I look at the things reflected in the room behind me, they appear fuzzy, yet the distance is an illusion, and all objects reflected are an equal distance from my eyes. Can you tell me why this is so?
• If you had a completely enclosed truck with birds inside, would the truck weigh the same if the birds were in flight?
• Joe plays a lottery on a regular basis. The odds are 1000 to 1. He has played 999 times and never won. Moe plays for the first time on Joe’s 1000th try. Are Joe’s chances of winning better than Moe’s?
• I’ve heard that if you’re in a falling elevator, you should jump up and down repeatedly so that when the elevator touches the ground, there’s a good chance you’ll only be a few inches in the air and will land harmlessly. Is that true?

Discuss.

1. #1 Michael Ralston
December 26, 2006

Glasses: Because the objects really are different distances. The light by which you see the object bounces off of the object, then off of the mirror, and then hits your eyes. As the blurriness is determined entirely by the distance the relevant light rays are travelling, the objects behind you will be blurry.

Truck: Yes, as a bird stays in flight by pushing air down, which would then push against the truck. However, it might weigh less as a bird descends and more as one ascends. Assuming the flock is in continual flight, though, it seems safe to assume the net effect of altitude changes will be negligable.

Lottery: No. The chances of winning the lottery are independant of how many times one has played them.

Elevator: No, that’s not true. Jumping right before the elevator hits will reduce the velocity at which you hit slightly, but the effective fall distance will only be reduced by the height which you can jump by – which is unlikely to be meaningful.

2. #2 Chris
December 26, 2006

The elevator jumping was tested by Mythbusters a while back, and their answer was a conclusive “No”. The velocity at which you push yourself upwards is tiny compared to the downward velocity of the fall, so you’ll still hit the bottom at a high speed and likely die.

3. #3 steve s
December 26, 2006

1 it’s not possible to give a satisfactory answer to this question, to someone who hasn’t learned some optics. Bottom line is, the light reflected from a mirror does not have the same geometrical properties as light from the object would have at that spot. It has the properties as if the object were on the other side of the mirror, equally far away (in the case of a flat mirror). I know this won’t be satisfactory, but it’s not possible to be satisfying here without a chapter or two of Optics 101.

2 yes, at equilibrium.

3 No, the lottery has no memory.

4 A quick internet search doesn’t give me any figures on ‘human jump velocity’, but assuming the timing was perfect, your velocity upward would be subtracted from the elevator’s downward velocity. I suspect a human in good shape would have a t=0 velocity of (order of magnitude) about 5 meters per second upward, and the elevator’s probably falling at something like 30-50 meters per second downward, so it would make very little difference. BTW, somebody told me elevators only fall like this in the movies, real ones have safety systems nowadays.

4. #4 Tom
December 26, 2006

I’m no engineer, let alone a physicist, but I disagree about the birds in the truck. True, the birds in flight will create downward “wind pressure” on the floor of the truck, but even with a single bird what you will really have is wind turbulence, not a unform downward pressure. There will even be some pressure on the roof of the truck. I would expect the net vertical pressue to be downward, but less than the pressure of the bird(s)resting on the floor.

5. #5 Jonathan Vos Post
December 26, 2006

I think the column of hers that got the most response, including wrong ones from mathematicians who should have known better, was on the Monte Hall problem.

en.wikipedia.org/wiki/Monty_Hall_problem

The Let’s Make a Deal Applet
Despite a very clear explanation of this paradox, most students have a difficulty understanding the problem.

“Smartest woman in the world?” No way! My wife is MUCH smarter. And Marilyn Vos Savant looked like a fool when she hit on David Letterman, on his show, maybe 5 yerars ago.

See the 7 pages citing Marilyn Vos Savant at the Online Encyclopedia of Integer Sequences:

http://www.research.att.com/~njas/sequences/?q=vos+savant&language=english&go=Search

Also, I am NOT related to her, despite the one of those:

A124146 U.S.A. currency denominations in dollars.

1, 2, 5, 10, 20, 50, 100, 500, 1000, 5000, 10000, 100000

COMMENT
The only currency denominations in dollars ever produced by the U.S. Treasury Department. Some are historical, no longer printed or circulated. The largest denomination was only used between banks, not individuals. The sum of this set is \$116,688, as pointed out as the solution to a Math puzzle in the Gorbiski reference.

REFERENCES
Jeff and Lynn Gorbiski, “Ask Marilyn”, Marilyn Vos Savant, syndicated column, 10 Dec 2006.

AUTHOR
Jonathan Vos Post (jvospost2(AT)yahoo.com), Dec 13 2006

6. #6 steve s
December 26, 2006

The downward force created by the birds has to equal m*g for them to be stationary. Of course there’ll be fluctuations, but we’re talking on average here.

7. #7 Michael Ralston
December 26, 2006

Tom: The net pressure must be vertical downwards equal to the weight of the birds, or the birds will not remain in flight.

If it is more (and thus the truck weighs more) then the birds must (overall) go up.

If it is less (and thus the truck weighs less), then the birds must (again, overall) go down.

Yes, there will be a bunch of forces due to the turbulent nature of air, and some of them will even be up – but the net force exerted on the truck as a whole has to be equal to the net force exerted on the birds. And that force must, if they are staying at the same altitude, be equal to their weight.

8. #8 Mr. Econotarian
December 26, 2006

If the birds are flying at a reasonable height above the truck floor, I think much of the energy applied to move the air will be dissipated as heat due to turbulent friction in the air.

9. #9 Mark C
December 26, 2006

A real physicist can call me names at any time, but if you leapt in an elevator in free-fall, the downward force of your jump would cause an increase in acceleration (a=F[gravity]+F[jump]/m). You might float above the elevator floor for a while, but go splat just the same.

10. #10 joec
December 26, 2006

Mirror: The mirror reflects things perfectly, therefore things at a distance from the mirror appear to be the same distance “behind” the mirror.

Truck: Well, the simplest explanation is that the truck is a closed system, therefore its mass and therefore its weight cannot change. Remember, mass and weight are not the same thing. The more complicated explanation would involve the fact that a bird flies because the motion of air over its wings created a net upward pressure at the surface of its wings. However, that net reduction in pressure also acts on the roof of the truck containing the birds causing the truck to be pushed downward by the same amount that the bird is lifed. Ergo, no change in truck weight.

Lottery: The odds on the lottery are the same no matter how many times you play it. Ditto for flipping a coin. The odds are 50/50 each time you flip.

Elevator: Someone else correctly answered this. The force you exert on the elevator when you jump exactly cancels out the amount you rise off the floor. Ergo, you’re screwed no matter whether you’re Michael Jordan or a fifty-something white guy like me who can barely get his feet off the ground.

11. #11 Michael Ralston
December 26, 2006

Econotarian: It likely will. But all that means is that the bird is wasting energy.

For a bird to fly level, the net upwards force exerted on it by the air around it MUST be precisely equal to the force gravity exerts upon it. And for that to be the case and for the air as a whole not to move (which it cannot – closed system), the air must exert a downward force on the truck precisely equal to the upward force it exerts on the bird.

Now, turbulence means that at any given instant, the air might have a small net acceleration, but this is negligable and effectively random. (indeed, it must average out to be zero in the long run.)

12. #12 Pseudonym
December 27, 2006

The truck with the flying birds will weigh less, but not for the reason that most people think.

If you put the truck on a set of scales, the force that the truck exerts on the scales (which is what scales measure) would not be constant over time. The “turbulence” that some have mentioned is no different than if you had a toddler in the truck that wouldn’t keep still, or if you jiggle around on your bathroom scales. The truck would bounce around and exert more or less force on the scales over time.

What the question is really asking is what would the scales read, when averaged over time? If it were only the fact that the birds are airborne that was important, then the mass of stuff in the truck is the same regardless of whether or not the birds are flying. So the average measured weight is the same as the weight of the truck when the birds aren’t flying.

However, the fact that you’re computing a time average is important when you consider energy loss. The flying birds are expending more energy than the still birds. Even in a perfectly contained truck which doesn’t let heat escape, the fact that the flying birds are moving the truck itself means that some energy will get loss in friction and heat loss in the truck’s (and the scale’s) suspension system. So the truck with the flying birds will lose actual mass more quickly than than the truck with the still birds.

In summary, the time average of the measured weight will be slightly less for the flying birds than for the stationary birds. And if you persuade the birds to stop flying, you’ll find that the mass of the truck + birds has slightly decreased.

13. #13 Mustafa Mond, FCD
December 27, 2006

I need glasses to see things at a distance. When I look in the mirror, I can see my face clearly. When I look at the things reflected in the room behind me, they appear fuzzy, yet the distance is an illusion, and all objects reflected are an equal distance from my eyes. Can you tell me why this is so?

14. #14 Michael Ralston
December 27, 2006

Pseudonym: True, but that’s also a very small effect.

15. #15 Kevin
December 27, 2006

I’m not entirely convinced on the birds one….what if they aren’t flapping? What if they are just gliding? the force comes from air moving over their wings…not from downward exertion…

16. #16 Kevin
December 27, 2006

If we consider air as a low density fluid, that means it should weigh the same. For example, if you fill a truck with water and penguins, it wont weigh any different if the penguins are swimming around or if they’re laying on the truck bed. the same should be true with a truck full of air – since it’s an entirely closed system, you’re just rearranging the contents when the birds are flying or at rest, but it should always weigh the same.

yes?

17. #17 MikeM
December 27, 2006

But since the truck is not airtight (a reasonable expectation), the penguin example is not quite apt. Pressure changes due to flight are negated by equalized pressure from outside. A flock of birds flying over an open-air truck bed would not increase that truck’s weight, would it?

18. #18 Kevin
December 27, 2006

MikeM-

No, a real truck wouldn’t be airtight, but I assumed “If you had a completely enclosed truck with birds inside,” where “completely enclosed” meant “closed system” for this thought experiment…

19. #19 MikeM
December 27, 2006

My oversight. In that case, the penguin example is terrific.

20. #20 Michael Ralston
December 27, 2006

Kevin: It won’t matter if they’re flapping their wings or gliding, as long as they’re going level at the moment.

Either way, the air is exerting a net upwards force on the birds. The reason why this is the case may vary, but the fact that it is the case will not.

Further, for the air to not be, overall, moving downwards itself, it must be exerting a net force downwards on the truck precisely equal to the force upwards it exerts on the birds.

And, as anyone who’s ever leaned on a scale knows, that’ll increase the weight. ðŸ˜‰

21. #21 Jon H
December 28, 2006

Any downward force applied by the birds’ wings to the air would not be restricted to a downward direction. There’s no guarantee that the force expended by the wing to the air will be transferred to any particular surface – a lot could happen to divert it before it hits the floor.

This would reduce how much force, if any, is actually applied to the bed of the truck. A fair portion of the energy would be transferred to the walls, instead. Or even to the ceiling.

Consider an alternate scenario with a truck full of flying gnats. In that case, the wing motion would merely generate heat, and there probably wouldn’t be a discernable flow of air. I suspect the same would be true of hummingbirds.

The amount of pressure applied to the floor of the truck by the wing beats would probably depend on the amount of air moved and the size of the birds. With tiny critters like gnats or hummingbirds, not much air needs to be moved to stay airborne.

22. #22 Pseudonym
December 28, 2006

Michael Ralston: Yes, a very small effect. But in the world of perfectly sealed trucks and perfect scales, such a thing should obviously be measurable!

23. #23 Michael Ralston
December 28, 2006

Jon: The amount of air that needs to be moved goes up as the number of critters goes up, though ðŸ˜‰

The point is that it doesn’t matter how much energy is wasted, if there is not a net downward force on the truck precisely equal to the weight of the birds (when averaged over time), they will eventually either crash into the floor or ceiling.

Pseudonym: True enough! ðŸ˜‰

24. #24 Jon H
December 28, 2006

“The amount of air that needs to be moved goes up as the number of critters goes up, though ;)”

True, but the air doesn’t need to be moved very far.

I honestly don’t think 100lbs of gnats in a sealed box would weigh 100 lbs if they were all airborne, especially if the volume of the box were significantly larger than the volume taken up by the gnats themselves.

In a box the size of a WalMart, any force exerted by the gnats’ wings would certainly be dissipated as heat long before reaching any surfaces.

I suppose the answer to the question depends on the ratio between the size of the container and the size of the birds. Eagles in a sealed U-Haul would move enough air to definitely push against the floor. Eagles flying in a sealed domed stadium, probably not.

25. #25 AJS
December 29, 2006

Glasses
The light reflected in the mirror has travelled further, but that’s not really what’s important. What matters is that it has changed direction, so that it looks as though it’s coming from somewhere else — in the case of a flat mirror, from as far behind the mirror as the original object was in front of it. That is to say, if you stand 0.5m. away from a mirror, your reflection appears to be 0.5m. behind the mirror; something 3.5m. behind you would appear to be 4m. behind the mirror. The rays of light which make up the reflection are entering your eye from exactly the same direction as though they had come from 4m. behind the mirror. So things far from the mirror will look fuzzy if you are short-sighted.

Birds
My first guess (and here we’re getting onto the same dangerous ground where Marilyn fell) is that each bird is experiencing an upward force equal to its own weight (otherwise it would fall) and the equal and opposite reaction to that force must manifest itself downwards onto the truck. So the truck should weigh the same even though the birds are airborne.

Lottery
Joe and Moe have identical chances of winning. Just because you have played the lottery 999 times and lost, doesn’t affect your chances of winning the next time. Remember, probability is defined as Lim (attempts -> ∞) (successes / attempts). A mere thousand attempts is nothing, unfortunately.

This reminds me of my personal Bingo strategy: only ever play one card at a time. If you’re playing 6 cards, you cover all 90 numbers; but even if one of your six cards wins, you’ve also paid for five losing cards. As a strategy, mine doesn’t improve your probability of winning, but it certainly makes your money last longer.

Elevator
If you jump, by the time you get back down to where you started you are going to be travelling downwards just as fast as you set off travelling upwards. But by this time, the lift floor will have moved down; so you will actually be going faster than you took off when you eventually hit it.

Not that this really matters much, though, because the speed at which a lift (assuming we’re talking real-world ones and not TV/film lifts, which have their own version of the laws of physics) can fall under fault conditions is determined by the rate at which fluid is being lost from the hydraulic cylinder. The least unlikely failure mode is for a hose to rupture; even so, the aperture in the union where it joins the cylinder is probably no bigger than 30mm. across. Friction between the escaping fluid and the walls of the aperture will limit the rate of descent.

Note that if the cylinder wall fractures, allowing fluid to leak freely, all bets are off. Assuming the lift fell from rest, its speed after falling h metres is v = sqrt(2 * 9.8 * h) ms-1. Assuming the building has six floors each 3.0m. high and you fall from the top floor, you will be doing no more than sqrt(2 * 9.8 * 15) = 17.15 ms-1 by the time you reach the ground floor. That is 61.7 kmh-1, and so should be survivable (TV presenter Richard Hammond survived a rapid deceleration from over 400 kmh-1). The spring-loaded buffers at the bottom of the lift shaft would absorb much of the impact.

26. #26 Michael Ralston
December 29, 2006

Jon: No, the ratio doesn’t matter at all. The only way the ratio matters is the mesurability of the effects, and we’re assuming perfect measures (and perfectly airtight containers), so … no relevance.

You can resolve the problem with nothing more than Newton’s laws – the forces /must/ add up to zero for the fliers to not crash into the top or botton of their container eventually. The only way this can happen is for there to be a normal force of some sort; if they were resting on the floor it would be molecular repulsion. As they’re flying, it must be the force of their wings. Either way, that force must be applied downwards to the container or the forces can’t add up.

Draw a force diagram if you don’t believe me.

27. #27 j a higginbotham
December 31, 2006

A124146 U.S.A. currency denominations in dollars.
1, 2, 5, 10, 20, 50, 100, 500, 1000, 5000, 10000, 100000
COMMENT
The only currency denominations in dollars ever produced by the U.S. Treasury Department.
———————
When considering whole dollar denominations only. http://home.earthlink.net/~icepick119/page8.html
Not to mention interest bearing and compound-interest notes.