Cannonball Competition

[Griffin]

Here's my guess at the order for Cannonball Competition. What do you people think?

Order of hitting the bottom (quickest to hit to slowest): 4, 1, 7, 6, 3, 5, 2.

It's based on higher temperture air creates less air resistence, the parachute will slow you down a lot, except on the moon, the bigger shpere will lose more speed on impact that the smaller ones, and moon gravity accelerates a lot slower than earths.

[This message has been edited by Griffin (edited 12-20-1999).]

[Ghost Post]

my solution:
3 5 6 7 1 4 2

reasoning:

no air resistance on moon, heaps faster than the rest... 3 beats 5 because of the parachute

the rest is all "terminal velocity" dependant... v^2 proportional to mass, inversly proportional to temp and Diameter squared (area)

using these proportionalities:
6 and 7 are both faster than 1, 6 is twice as fast, 7 is only 5/4 as fast (actually the sqare root of those numbers) so 6 beats 7.
then 4 is slower than 1.
2 has got no hope.

[araya]

hmm, boom, you're taking physics in university? Two things I notice immediately:

1. Parachutes function due to air resistance
2. Water is frozen at 25 F

Actually, my entire ordering is quite different (fastest to slowest):

4 1 7 5 3 2 6

4 - Earth, 75 F
1 - Earth, 50 F, slightly slower since water is slightly more dense. VERY slightly more dense - 1000 kg/m^3 vs 997.2 kg/m^3
7 - Earth, 500 kg, 1 m dia sphere into 50 F water. The sphere moves slowly once it hits the water (terminal velocity of 1.6 m/s vs 2.9 m/s for the smaller sphere), but it still hits the water faster than the moon dives (and sinks faster)
5 - Moon, 75 F, slower than earth since acceleration due to gravity is roughly 1/6 as large.. the parachute has no effect
3 - Moon, 50 F, slightly slower still since the water is denser
2 - Earth, 50 F, with parachute so the descent is very slow
6 - Earth, 25 F, that's gotta hurt

I'm not too sure about #7, I think it might be between 2,3 rather then 1,5. The drag force on an object is
Drag = 0.5*C*rho*A*V^2 where C is drag coefficient (1.2 for a sphere), rho is density, A is frontal area and V is velocity

so one could easily set up a differential equation for the velocity function (if one weren't as lazy as I am).

[This message has been edited by araya (edited 12-21-1999).]

[Ghost Post]

hello araya
will answer one by one:

yes i do do physics at university.

i thought the parachute might slow him down even a little in the water, depends on how long the chords are really, less than ten meters then the parachute will definately slow him in water

most importantly, i live in australia and have no idea of the farenheight system, so i didnt know water was frozen at 25F
so i have 6 wrong

you are still missing the point with moon/earth i think... in 1 kilometer you will quickly reach terminal velocity on earth, (those spheres have surface area) and on the moon they wont, so i stick by the moons winning out-right.

if parachutes are irrelevant in water, then it is 5 3 instead of 3 5

the rest of my logic looks sound to me, so six is all i change

high temp -> higher pressure -> higher air resistance -> slower terminal velocity

new solution:
5 3 7 1 4 2 (6) - (6 doesnt land)


[This message has been edited by boom (edited 12-21-1999).]

[skoobee47]

I'd have to agree with araya's choices. By the way boom, in the problem, it states that the parachute is gone once you hit the water so it can not be factored into that part of the equation. Also, no offense, but were you truly not aware that water is frozen at 25 F? With as much as we americans use the Celsius, I wonder how much of the world generally disregards the Fahrenheit system. Is there anyone one else out there who is generally unaware of even the basics of Fahrenheit? Once again, I do not mean to offend you boom, I am just wondering if that is fairly common throughout the world.

[Ghost Post]

sorry i thought the "detach" meant they were deployed... quite a stupid interpretation

from what i can see only 2's parachute comes into play and i had that because of the wording anyway.. so i stick by my answer.

in answer to your question: no i have never used/studied/done anything with farenheight.
seems like a watse of time to me, how long will it take the world to go metric and get with the program!!!!!!!!

[Griffin]

Yeah, that makes sense, Araya. I totally missed the fact that the water would be frozen for #6.

[kitakaze]

I did mine independently, and have exactly araya's answer. RATS, he beat me to it, but at least I did it.

By the way, I also studied physics, and there is an important point to remember:

A perfect sphere in a perfect fluid (gas or liquid) has no drag.

The moon's gravity will pull at 1/6th Earth's, and with d = 1/2 a*t^2, that is significant.

Mr. Blobbo at 500kg shouldn't move a lot slower until he hits the water, where his bouyancy will really slow him down (as well as the bigger splash), compared to #1.

[Ghost Post]

Hmmmm... I wonder how much $1,000,000 will be worth in 500+ years (in today's money)... the weekly shopping bill?

[kitakaze]

David T in the other thread raises an interesting point, but I think he gets it wrong.

The densities, as I work them out are 1527kg/m^3 for 1-6, and 954kg/m^3 for #7,

So # 7 may not reach the bottom either. I don't know how to work that out, though!

So, I'll revise my guess to 4, 1, 5, 3, 2, 7, 6

Where 6 never makes it, and 7 may or may not.

[Ghost Post]

kitakaze: perfect fluid?
when did we start discussing perfect fluids?????
this is earth and the moon, i honestly dont understand why ppl think that the earth ppl will beat the moon ppl... the moon ppl NEVER reach terminal velocity.. they just keep accelerating

some of that water stuff is a bit dodgy... but at those speeds for only 10 m... i'm not sure its gonna make much difference

mabye my answer will revise to
5 3 1 4 7 2 (6) or 5 3 1 4 2 (7,6)

[Ghost Post]

I got 4 1 7 3 2 5 6, as well. I was thinking as did the person in the other thread, that the difference in accelleration would more than make up for the changed terminal velocity. But I was just too slow to beat all of you folks!

Unlike some, however, I noticed that number 7 has the least density of the bunch, and figured that he would be slowed the most by his impact with the water. Also, even though they all have a density lower than water, I assumed they'd all hit with enough velocity to be driven to the bottom of the pool. The impact, however was fast enough that I didn't think standard fluid resistance equations would apply. You're still going to splash just about all of the water out of the pool with that impact!

[araya]

hi boom.. I have noticed that the rest of the world tends to ignore the fahrenheit scale, which is definitely a good thing, hopefully it will be entirely phased out in the near future. I'm sure you're aware of the general conversion formula F = 9C/5 + 32

On the moon you never reach terminal velocity, that's true, but look at the acceleration.. In 1 km, accelerating at roughly 1.6 m/s^2, it takes 35 seconds to hit the water at a max velocity of 56 m/s. On the earth, the terminal velocity for the 0.5 m spheres is roughly 80 m/s. If there were no air resistance on earth, they would hit the water in 14.3 seconds. Air resistance doesn't make THAT much difference on a streamlined object. The terminal velocity for the 1m sphere is 49 m/s.

kaze: a sphere still has drag.. drag simply refers to the amount of work it takes to displace a fluid as you move through it. In any case, water and air are not perfect fluids. Your point about #7 is well taken, though, I'm still not sure about what happens when they hit the water surface at that speed. I think they should all have enough speed to get to the bottom of the pool.. the 0.5 m spheres sink anyway, and the 1m sphere is only slightly buoyant, but then it would sink so slowly that it may not make it. I remember that a sphere breaking the surface of a fluid does so much easier when the fluid is turbulent, but I don't think that is the case for the pool. How much speed do they lose on impact?

[Ghost Post]

ok, i dont lkike calculations and actual numbers much, i would of though 1 kilometer would be enough to let the moon ball catch up, obviously not.

no i didnt know the conversion formula, and i hope it does get phased out.
it also gives me the shits when puzzles have inches and feet and stuff... i dont know any of that stuff.. 12 inches to a foot? thats a guess
it annoys me that america is too stuck up to give up their crappy measurements systems

[kitakaze]

I guess my point about the perfect fluid was made because I don't like dealing with drag and terminal velocity (I don't have the equations or tables). But air is close to a perfect fluid for this problem, and araya showed that with his calculation: The .5 m spheres don't reach terminal velocity. And the drag force is negligible.

Drag is not caused by displacing the fluid, but by the turbulence behind the object. A perfect sphere and a perfect fluid would generate no turbulence, so the displaced fluid would actually push the object from behind as it fills back in. We did these rough experiments at UVic, where you put a cylinder in a very slow moving stream of fluid. The cylinder doesn't get pushed.

That's an aside.

So, do you think that a 500kg, 1 meter sphere traveling at 49 m/s has enough energy (1/2 mv^2 or 600250 J) to displace a cylindrical column of water 10m long (mass = 8000kg?) Simplistically, you could lift the whole column 10m, which would take 80,000J. So it seems it could hit the bottom, but I don't know how much water gets sucked into the air from surface tension... But there is a lot of energy in that falling sphere to allow for 7 times the water to get splashed out of the pool.

[Ghost Post]

Oops, I think that in my first post I reversed the numerators and denominators in my density equations. The density of #1-6 is 1.527g/cm^3 and the density of #7 I get .91g/cm^3.
I also converted from Kg/m^3 to g/cm^3 so I could work with numbers greater than 1. I'm still sticking with my choices though.
4 7 1 5 3 2 6

[Ghost Post]

to find the bouyancy force on the spheres when they're in the water, you just find the mass of the water displaced by the sphere and find mg... (multiply that mass by 9.8)

that does your "perfect fluid" trick and ignores the splashing etc

would be a constant force all the way down and make simple calculations.

has anyone double checked the air resistance calculations done by araya?

[araya]

boom - I don't live in the US either, but you have to be aware of different systems of units, if only because a lot of the time you're dealing with apparatus or problems which have originated in the US, or Britain, or wherever. inch, btu, oz, stone..

kaze - UVic? I didn't know you lived around here too. Can you elaborate on that experiment a bit? It sounds interesting.

DavidT - I noticed that you calculated specific volumes instead of densities, but overall I like your answer, because it's almost the same as mine I don't understand why you have #7 in between 4 and 1 though. The only difference between 4 and 1 is a slight temperature difference in the water, which would make an absolutely tiny difference in the falling time.

The pool is 10m deep, there's no way all that water is getting displaced. I'm just wondering how much speed you lose when you break the surface of the water. Here are some quick calculations for the terminal velocities of the balls in the water:
100kg, 0.5m in water on earth: 1.7 m/s
500kg, 1.0m in water on earth: -1.4 m/s (upwards)
100kg, 0.5m in water on moon: 0.68 m/s
So the smaller balls will sink anyways, but I wonder if the large ball will make it to the bottom of the pool before it stops and begins to return to the surface. The drag force on the ball is proportional to V^2, so it's a pain because of solving the differential equation, which I haven't bothered to try yet. Maybe I'll save that super-fun activity for christmas morning.

[kitakaze]

araya,

I studied at UVic, though I moved to the Yukon since then. I thought you might pick up on that... I think you said you're from Vancouver.

(So, Boom, there are at least 2 Canadians on this forum, but we have to be bilingual in units, with all those Americans so close!!) <gr>

To elaborate a bit on the experiment - you can simulate a perfect fluid with very slow velocities, and very thin (low viscousity) fluids. Drag is determined by how much turbulence is produced behind the object (energy lost through heat production by 'mixing' the fluid.)

The experiment we did was with a cylinder, so I may be wrong about the sphere. Have a slightly inclined plane with a thin fluid running down (I think water works). You can place a cylinder in the flow (with a force measuring device on it) and measure the push.

In theory, with an infinitly long cylinder (avoid edge effects) there is no net force on the cylinder. In practice, there is a small force due to viscosity, and turbulence around the sharp edges.

That's all I remember. It was over 10 years ago.

Kaze

[The Kernal]

I think the answer is 4 1 7 2 5 3 6

I think that 2 will beat 5 and 3 just for the fact that the gravity is so small at that distance from the moon. 6 will land on ice. That leaves 4 1 7 and 2 to battle it out. 4 is fastest because of the decreased resistance in the higher temp. 1 is slightly slower but still not slower than 7 whose increased drag in the water would slow it down even more. 2 is obviously slower.

I'm quite an amateur with the physics but that seems pretty believable.

[araya]

oh yeah kaze, I remember that you live in the Yukon now. So many people to keep track of!

I've studied a few viscosity/fluid flow experiments but I've never tried that one. The water is flowing past the cylinder, which is floating in the water, but the cylinder is not moving. Or at least it shouldn't, ideally.. I guess that sounds plausible. The fluid must be moving very slowly indeed.

Kernal: Good ordering, but I guarantee the moon drops are faster than the earth parachute drop. btw, do you mean "kernel"?

[hank]

Hey all you smart ###es(my wife installed net nanny), READ THE QUESTION! The prize goes to the one who predicts the order in which the participants reach bottom. No where is it stipulated they all jump at the same time. Obviously, then, the order must follow the sequence provided to identify each of the participants. Number 5 wouldn't jump before Number 3. So the answer is as follows:

1,2,3,4,5,6,7

[Andy]

Hank - you make a very interesting point. It's also not stipulated that the participants jump in the order in which they are numbered, nor whether each participant waits for the previous participant to reach the bottom. In fact, the problem says nothing about when each participant will jump, except "as part of the end of year celebrations." I don't recall reading anyone else's statement assuming any particular timing. You've at least identified a significant issue, and your reasoning appears to me to be as sound as anyone's.

[araya]

Someone always has to point out some inadequacy in the wording of the question. Yes, often questions are not set up perfectly, mentioning every detail. We can't expect the Minotaur to sit there for hours proofreading the puzzles to make sure he's covered every possible interpretation. Sometimes we have to use common sense to interpret what the question is asking. It has nothing to do with being a smart ####er - if you want to solve the puzzle, you don't settle for the "shortcut" solution, because you KNOW it's not what was intended.

[Andy]

Araya - the nit-picking isn't intended to belittle the intended puzzle (at least mine isn't). It's actually playing on another dimension. As you point out, the intent of this puzzle seems clear. That's not always the case, however; some puzzles really are 'trick questions.' It's not always obvious what the puzzler intends. I don't expect the Minotaur to spend hours searching for unintended but possible interpretations. I do expect that ambiguous wording can be pointed out and the resulting possible solutions explored. In this particular puzzle, several people have posted calculations and reasoning based on either simultaneous starting times, or measuring elapsed time for each jump rather than order of finishing. Griffin (in his first post) specifies his interpretation as "Order of hitting the bottom (quickest to hit to slowest)" and everyone else follows implicitly. Since this interpretation has been covered thoroughly, and it's not the only possible one, it seems to me (and apparently to Hank) to be a good time to explore the alternatives.

[This message has been edited by Andy (edited 12-29-1999).]

[araya]

It seems to me that the question is perfectly clear, but I guess other interpretations aren't so much wrong as they are inane.

[hank]

Thankyou Andy, for your moral support. I always enjoy coming up with alternative solutions, and in good faith,needle my fellow labyrinthers. I certainly wasn't nitpicking, as I found the omission with regard to simultaneous jumpstarts startling. I actually believe the omission was deliberate on the part of the minotaur. He could be hi-lighting a common trait amongst us, that we love to use our math and logic skills, diving hellbent for solutions and willing to ignore an anomaly in the question and make assumptions to justify our method. I've been caught by this many times in the past and refer to a statement made by one of my teachers from the distant past. "READ THE QUESTION"

Inane, Araya? Doesn't this whole problem appear inane(silly) to you? I would suspect that is intended by the minotaur to support his joke on us.

By the way , Araya, I noticed you are the person responsible for the notion that water freezes at 25 degrees fahrenheit. As far as I know, 99% or thereabouts of water on earth freezes at 0 degrees Fahrenheit or lower. So I wouldn't be so confident to place number 6 last. After all, would he be stupid enough to attempt the jump if the pool was frozen?
[This message has been edited by hank (edited 12-31-1999).]

[This message has been edited by hank (edited 12-31-1999).]

[This message has been edited by hank (edited 12-31-1999).]

[mithrandir]

Water freezes at 0 degrees CELCIUS or lower, which is equivalent to 32 degrees Farhenheit. 25 degree water is ice.

[Griffin]

I think Hank was refering to ocean water, which would freeze at a lower temperature due to all the salt dissolved in it. But I would assume the pool is filled with fresh-water.

[Andy]

Fresh water is a reasonable assumption. The puzzle does refer to "water temperature" and "water pressure" so we are assured that the pools are indeed filled with water in some form, and there's no mention of any additives except the divers.

[hank]

Okay then, assume fresh water. But has anyone concerned themselves with the problem facing the moon jumpers! Their pools are boiling and rapidly depleting. The saturation vapour pressure of water at 50 degrees fahrenheit is 1230 newtons /meter squared!( Like everyone else of course, I am assuming 0 atmospheric pressure.)

[hank]

Wait a minute! A thin force field keeping moon water at the surface at the same pressure as at sea level on earth! Forget about vapour pressure, but I can't think of anywhere where fresh water exists at sea level! You guys cannot assume fresh water. It must be sea water.

[mithrandir]

From Here:

"The freezing point for 35o/oo average seawater is -1.9oC."

-1.9oC is around 29oF, so seawater is still frozen at 25oF.

[This message has been edited by mithrandir (edited 12-31-1999).]

[mithrandir]

Besides, hank, it says a pool, probably refering to a man-made pool with fresh water pumped in, it is 2525 afterall.

[hank]

Thanks for that info mithrandir. You are correct and I am partially wrong. If you check out www.unidata.ucar.edu/staff/blynds/tmp.html you can see an example of where I got the idea that sea water freezes at 0 degrees fahrenheit. Obviously, the degree of salinity is very significant at sea level, and would have been specified along with temperature and force fields If the intended solution was meant to be credible, in a mathematical sense.

[This message has been edited by hank (edited 01-01-2000).]

[araya]

yeah, all the problems we waste our time on are pretty silly.

hank, if I understand, your solution is that the contestants jump not at the same time but in sequence, so of course they finish in that sequence as well. And we can assume that the sequence is 1,2,3,4,5,6,7. It seems highly doubtful that a cash prize would be offered for predicting this sequence, although in 2525 a million dollars will probably be worth about a nickel.

"Can you correctly guess the order in which divers will arrive at the bottom of their respective pools? If you can, that $1,000,000 is yours, because the rest of humanity is out of shape and bad at science."

Everything in the question points to the usage of logic and physical principles to solve the question. I suppose it could be a trick by the evil minotaur, but I'm confident that when the answer goes up, it will not be 1,2,3,4,5,6,7.

I also don't see why you think it should be sea water. Why in the world would they use sea water?

[hank]

There's more Araya, but first lets justify the sea water assumption. If we expect that the solution requires pure physics then we do have to be careful of using correct information. In order to make conditions on the moon at the surface of the water up there the same as earth, we have been offered specific information. To quote, " moon pools have a thin force-field over them that doesn't affect the cannonballer, keeping water pressure equal to earth sea level." It follows then that all pools used in this competition on earth have surfaces that occur at sea level. It just occurs to me that fresh water is very rare at sea level. Some people may have a hard time visualizing a sea pool, but they are quite common. They are usually pens just off the shore, containing farm fish, or aquatic mammals, or enclosures for the protection of people.
Now I will ask you this question, Araya. I can visualize the platforms on earth for a one km jump as planes, and of course , the cannonballers on earth have some opportunity to direct themselves toward the pools using atmospheric resistance, but what about on the moon. Planes don't fly up there, and there is no use jumping off a moon satellite. It would take some highly sophisticated space ship that could position itself at the perfect spot for the cannonballer to jump, because afterward he has no opportunity for corrections.

Now, I know very little about what takes place during the impact of a metal sphere of half a meter diameter and 100 kgs, on the surface of the water but I'm quite sure you guys haven't addressed it. I would imagine a significant trauma could occur resulting in several pieces of metal being ejected from the pool while some settled to the bottom.It is at this point that I think a clean, clear logical solution is beyond our grasp.

Now I take you back to the quotation you provided in your last posting. Notice that we are invited to guess, not to solve, calculate or determine.I don't think the minotaur is evil at all.If I turn out to be right,(1,2,3,4,5,6,7),would you think the minotaur is evil?

[araya]

Extremely evil. I'll tell you what though, hank, I'm tired of talking about this problem, because I know you're wrong, but I'm never going to convince you of that. I'm glad that you found a solution that pleases you. We'll see what answer was intended soon enough.

[hank]

Actually Araya, I prefer to call it a guestimate. If I am wrong, I'm not going to lose any sleep over it. I admit that I have been having a little fun with it. Good work on the pendulum topic.

[Wonko the Sane]

Okay, I tried this, got the same answers at Araya, and read the posts. There are a few things everyone is fighting about, so I did some calculations to find approximate velocities and times to test this terminal velocity idea.

x = .5(vf + vi)*t
vf = sqrt(vi^2 + 2*a*x)

I used 1000 (1 km) for x and 9.8 for a on Earth, 1.633 for a on the Moon and got...
Earth: 14.3 seconds with a final velocity of 70 m/s
Moon: 34 seconds with a final velocity of 35 m/s

I don't have any idea what terminal velocity for the spheres is, but it doesn't matter. The person on Earth has 20 seconds on the guy from the moon anyway. Even if they guy on Earth hit terminal velocity in just a few seconds, he'd still win. There's no contest, terminal velocity can be disregarded.

Second, doesn't matter what kind of water it is. Since salt water is frozen at 25º, 6 will still lose. Also, since the fall threshholds will be relative and everyone is using the same kind of water (we can at least ASSUME that since it isn't stated, and Minotaur ALWAYS states every bit of required information). I agree with Araya...it's 4,1,7,5,3,2,6.

[This message has been edited by Wonko the Sane (edited 01-06-2000).]