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Topic: Astrowrench (Read 4639 times) |
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william wu
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Astrowrench
« on: Jan 27th, 2003, 11:53am » |
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(physics knowledge helpful)
Two astronauts are standing on a spinning space station shaped like a disk. They are the same radial distance away from the disk's center, and standing opposite to each other across from the center (e.g., if you draw a line connecting the two astronauts, the line crosses the disk's center.) One astronaut wants to toss a wrench to the other. Among the infinitude of trajectories which will accomplish this goal, characterize one of the trajectories without writing a single equation.
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BNC
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Re: Astrowrench
« Reply #1 on: Jan 27th, 2003, 12:30pm » |
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I have a possible answer, but am not sure about it. Here goes
If the astronaut would that let go of the wrench, it will continue at a tangent to the circle the movement is “drawing”, at a speed determined by the rotation speed. So, if the astronaut would throw the wrench to the other direction, tangential from himself, at just the right force, the wrench would hover in place until the other astronaut gets around to pick it up.
// hidden by willywutang 8:43 PM 1/27/2003
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| « Last Edit: Jan 27th, 2003, 8:43pm by william wu » |
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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william wu
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Re: Astrowrench
« Reply #2 on: Jan 27th, 2003, 8:44pm » |
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Hmph. I was really hoping for a bunch of wrong answers first  In any case, I encourage people to post their answers even if they do not match BNC's, so their possible misconceptions about physics can be rectified. There are a couple common wrong answers to this problem.
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| « Last Edit: Jan 27th, 2003, 10:52pm by william wu » |
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BNC
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Re: Astrowrench
« Reply #3 on: Jan 28th, 2003, 4:39am » |
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OK. Lets initiate some discussion. Here’s a different solution (I won’t tell you if it’s correct):
Place the wrench on the disk, and kick it at a straight line to the other guy. Yes, angular momentum may have many affects, but due to symmetry, it will still end on the right place.
Right or wrong?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Icarus
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Re: Astrowrench
« Reply #4 on: Jan 28th, 2003, 3:31pm » |
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on Jan 28th, 2003, 4:39am, BNC wrote:
Place the wrench on the disk, and kick it at a straight line to the other guy. Yes, angular momentum may have many affects, but due to symmetry, it will still end on the right place.
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What symmetry?
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"Pi goes on and on and on ...
And e is just as cursed.
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When their digits are reversed? " - Anonymous
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SWF
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Re: Astrowrench
« Reply #5 on: Jan 28th, 2003, 7:08pm » |
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Yes, BNC your alternate suggestion of going straight over the center should be catchable by the other astronaut as long as it is thrown with with a velocity much greater than the tangential velocity. Although if you treat the astronauts as points, this velocity needs to be infinite. I think this problem hasn't generated much interest because considering only trajectories that require no equations makes the problem too easy. Here is a modification which will make the problem more challenging:
Suppose the astronaut is not strong enough to throw with infinite velocity or even the tangential velocity. What is the lowest velocity that the astronaut can throw the wrench so that it reaches the other astronaut? Use of equations is permitted.
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aero guy
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SWF, you weant from too easy to owe, that hurt. OK, I solved it. There is no way I am about to record all my work. The way I figured it, after you throw the ball it will have velocity Vnew. The ball is going to move in a straight line until it hits a side. Hopefully, the astronaut on the other side will have caught up to wherever the ball impacts through the rotation of the station by the time impact occurs.
So, if the angle through which the ball moves (with regard to an unmoving station-center-centered reference frame) is theta and the tangential velocity is V, then:
Vnew/V=2sin(theta/2)/(pi+theta)
What we are concerned with, though, is the minimum velocity you throw it at, Vt, not the minimum velocity it travels at, so:
Vt/V=sqrt(1+(Vnew/V)^2 -2(Vnew/V)cos(theta/2))
So now we combine these equations and minimize. After minimizing the equation was messy, assuming I did it right, so I numerically found the lightest you can throw it is about 99.2% of V in the backward/toward the center direction. This results in a catch 68.8 deg later, or when the station has rotated 248.8 deg.
There is likely some math error in the minimization, but this turned out to take much longer and be much less fun than I thought. e.g. using Vt and the angle you throw it at are bad choices for easy equations.
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SWF
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Re: Astrowrench
« Reply #7 on: Jan 29th, 2003, 7:40pm » |
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Sorry this problem has tormented you, aero guy, but thanks for trying. Your answer of 99.2% of tangential velocity is significantly higher than what I come up with.
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aero guy
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Yeah, SWF, I think it is a math error, are the equations what you got, is the approach the same? I also assumed that you would hit the other astronaut at his first pass around (rather than have the wrench floating there for several station rotations) but I think that was safe. If the concept was right I hope you don't mind my calling this one solved.
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SWF
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Re: Astrowrench
« Reply #9 on: Jan 30th, 2003, 8:40pm » |
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I wouldn't call this solved until somebody gives the right answer. Aero guy, your first equation is similar to one of mine, but I don't recognize the second one. I am sure there are many approaches to getting the right answer. Eventually, somebody will get it. This isn't nearly as difficult as many of the other problems in this riddle forum.
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aero_guy
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Re: Astrowrench
« Reply #10 on: Jan 31st, 2003, 7:34pm » |
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Hmm, think I found the error, using degree mode on the calculator when things were in radians. How about around 80%? I bashed away at it but was never able to get a closed form solution. Maybe my geometry is rusty. Were you able to? I still get that the station rotates through 240 or so deg between throw and catch.
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jabhiji
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Re: Astrowrench
« Reply #11 on: Feb 1st, 2003, 4:01pm » |
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I tried to do the minimization numerically as well. I end up with a somewhat different answer:
u = 66 % (magnitude)
thrown almost in the exactly opposite direction of the disc velocity. There was some funny limiting behaviour as the throwing direction approached the exact opposite direction.
amount of rotation = 320 deg.
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aero_guy
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Re: Astrowrench
« Reply #12 on: Feb 3rd, 2003, 3:36am » |
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Dang, if that is right than I need to rethink things. I tried out your numbers and they don't seem self-consistent, whatever angle you throw it at. I agree about the limiting behavior at the opposite of rotation direction. That was what was discussed earlier where by throwing it at the same speed in the opposite direction it will just hang there until the other astronaut rotates around to it. What was your general philosphy approaching the problem?
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jabhiji
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Re: Astrowrench
« Reply #13 on: Feb 3rd, 2003, 8:56am » |
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Take the coordinate system such that the disc of radius 1 is centered at (0,1). Assume anti-clockwise rotation. The astronaut A1 is initially at (0,0) and the other fellow A2 is at (0,2). The tangential velocity of A1 is 1 along +ve X.
Now A1 throws the wrench with speed u in a direction which makes an angle phi with the -ve X (This velocity points in the second quadrant, or NW). Then you compute the effective velocity and find the point where the wrench hits the circle. Let this point be H. After this, you equate the time taken by the wrench to reach H and the time taken by A2 to reach H.
I essentially get a quadratic equation for the speed u where the parameters are phi and theta (the angle subtended by H at (0,0)). I then minimize u w.r.t. phi and theta.
The equation for u I got looks something like this:
u = COS(phi) - SQR(term1 - term2)
term1 = 4 * (SIN(theta))^2 / (PI + 2 theta)^2
term2 = (SIN(phi))^2
Now that I think about it, there might be other solutions where u is even lesser than 66 % and disc rotations > 360 deg.
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SWF
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Re: Astrowrench
« Reply #14 on: Feb 3rd, 2003, 5:57pm » |
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Aero guy, that is the same result I got for minimum possible throw velocity. Now, do you take back what you said about this not being fun? I could not come up with an analytical solution, but phi, the angle of rotation of the astronaut who receives the wrench, is the solution of:
2=(phi2-2)*cos(phi)-2*phi*sin(phi)
That is easy to solve numerically: phi=4.3423012 radians or 248.8 degrees. The fraction of the tangential velocity that the astronaut throws is given by:
The value of this is 79.89% of the tangential velocity.
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jabhiji
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Re: Astrowrench
« Reply #15 on: Feb 5th, 2003, 7:22am » |
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I made a most embarassing mistake in my calculations. I now agree with the 80% solution. I completely neglected one of the two equations 
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aero_guy
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Re: Astrowrench
« Reply #16 on: Feb 5th, 2003, 9:46pm » |
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OK, it is fun. I just wasn't approaching it with the right attitude, it had been a long day. I will say, though, that though it was fun, it remains unsatisfying. I love closed form solutions. It is the same as getting an answer to a riddle that you say, "yeah I guess that would work," versus getting a "eureka".
Does anyone have access to a Mathcad type product where you might be able to find such an answer. I tried for quite a while by hand with no success. I highly doubt it would look like anything pretty, but you never know. I think the problem comes from oddly combining rotating and non-rotating reference frames. The math doesn't like it.
As an expansion to an existing riddle this was excellent. I knew that God-awful optimization class I took would come in handy some day... oh wait I didn't need it. Well at least I can pretend it came in handy.
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SWF
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Re: Astrowrench
« Reply #17 on: Feb 6th, 2003, 8:18pm » |
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Aero guy, I usually prefer closed form solutions, but I like it when a simple problem turns into something more complex than expected. I doubt if there is a closed form solution to this. After the first numerical approximations I was beginning to think there might be an analytical solution, because phi is within 0.06 percent of pi+6/5. Also, the normalized tangential component of velocity (relative to the thrower) is about 0.01 percent from -pi/4.
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