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# How would you cross puzzle land? Part V

## Question

You come upon the grounds where a traveling circus is setting up camp, and notice what appears to be a fight about to break out. The ringleader is facing the wrath of one of the strongmen, who has just lifted the proprietor of the circus overhead in an apparent prelude to a smashing. Amidst the ringleader's begs for mercy, you ask what the trouble is. The strongman explains that he's sure the ringleader is paying some of the performers with counterfeit coins. He points toward several sacks of coins in a corner of a small tent and explains that one of them is bound to be full of counterfeits that are either slightly heavier or slightly lighter than genuine coins. The circus of course has a balance with which you can weigh the coins, but naturally the balance is on the brink of collapse and will only be able to provide a few more weighings to infinite precision before its warranty expires and it disintegrates. They ask if you might help them figure out if any of the bags are full of counterfeit coins, and if so, which.

Using your problem solving skills from the first few "puzzle land"s, since you can't find any noticeable difference between the coins on close inspection without using the balance, you haul all the cash to the local bank and deposit the coins in exchange for bills. The exchange proceeds without a complaint from the teller, and you win the circus crew's gratitude for giving them pay that's slightly less likely to be bogus, as well as a ticket to see their show.

But now they have another question for you. They're setting up the trapeze act and are wondering how far apart the platforms can be placed.

Those who have seen a circus are probably familiar with a typical trapeze setup. There's a high platform at one end of the circus, a trapeze with a bar that the acrobat can grasp from the platform, a second trapeze (often carrying someone who catches the first acrobat), and another platform after the second trapeze. You would naturally want the platforms to be as far apart as possible to put on a good show, but also don't want the acrobats to be doomed to fail.

How far apart can the two platforms be? Assume that the platforms are at the same height as the axis of the trapezes, and the ropes on both trapezes are equal in length. And as usual with "How would you cross puzzle land" riddles, either solving the riddle legitimately or cheating is acceptable, but cheating strategies must be creative and funny.

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As the trapeze platform is the same height is the the same as the axis of the rope, for a situation where the floor is level, the length of the ropes needs to be less than the height of the platform. This is to avoid the acrobat face-planting at the lowest point of their swing. This means that the maximum distance between platforms, to avoid falling would need to be less than the height * 4.  However if the floor is indeed level you can have any distance you like between platforms and any size you like of ropes and cross puzzle land on foot from the base one platform to another. That is assuming that the acrobat can walk that distance on level ground without falling over.

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I came up with a solution where the distance between the platforms could be greater than height * 4, even with a level floor.

Having the platforms as far apart as you want and making the acrobats hoof it across the circus grounds counts as a legitimate cheating (and thus paradoxical) answer, and is so far the best for others to try to top.

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I can too, but it involves tall acrobats (stilts maybe) who can varying their height (or somersault to above the bar) at the bottom of their swing. Extra meters could also be obtained if they can build up enough angular momentum (by swinging around and around the bar) during the swing to somersault when it reaches the apex. In other words it would be quite a show.

Edit That or exchange the trapeze for a flying fox

Edited by phaze
Other idea
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The non-cheating solution I have doesn't involve the sorts of things you speak of. The acrobats can be considered as points that can attach to and release from a trapeze at a distance rope length from the trapeze's axis, and that just follow Newtonian mechanics aside from being able to grab onto the trapezi (or whatever the plural of trapeze is).

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Short of adding additional trapezi for an increase in 2 times the platform height with each one added I am at a loss for ideas.

Edited by phaze
spelling
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An acrobat can release the trapeze at any point and follow the usual laws of physics (of a point mass without wings or rocket engines or such) until coming into contact with the other trapeze bar. Will add a drawing next if that would help.

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initial velocity 0? Gravity 9.8 ms2?

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Correct, initial velocity is zero. The answer is independent of whatever value you choose for gravitational force, but you could use 9.8 m/s2 if you like.

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Hmm, let's put the axes of the trapezae on each of two motorized pinions which would travel the full length of each of two racks in the time the swinger travelled from platform one to trapeze two or from trapeze two to platform two.  Assuming the length of each rack is infinity/2, the platforms could be infinity apart.

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There may be a way if during the upswing on the trapeze the acrobat lets go and circumscribes a perfect parabolic curve (much like performed by ballistics) and catches the next trapeze on the down swing.  Couldn't be bothered doing the math (probably trig) to figure if it would actually gain a little more  distance though.

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assuming zero initial velocity, If the platforms are the same height, then the acrobat leaving platform 1 will not be able to rise above the height of platform 2, whether still holding on to the trapeze or not, without additional input of energy (as plainglazed's motorized pinions could provide).

However, as we all know from childhood, additional input CAN be added by "pumping"--standing up when near the platform height, and lowering the center of gravity when near the ground. I have no idea how to calculate how much additional energy (hence height, distance, etc.) can be added that way. For one thing, if you can get the trapeze to go higher than the platform, it will tend to fall closer to the platform...

Edited by CaptainEd
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oops, pumping involves raising the CG near the ground, and lowering it near the platform.

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On 6/3/2016 at 5:40 PM, phaze said:

There may be a way if during the upswing on the trapeze the acrobat lets go and circumscribes a perfect parabolic curve (much like performed by ballistics) and catches the next trapeze on the down swing.  Couldn't be bothered doing the math (probably trig) to figure if it would actually gain a little more  distance though.

That phrase was posted on Brainden?!?

Regarding pumping, I'm not sure how that would work out in practice. If the starting situation has the acrobat on a platform with the trapeze rope essentially going horizontally, then if you add more force to the swing you'll end up going above horizontal on the other side before gravity pulls you back down, and you wouldn't be swinging like a pendulum but more like a weight on a rope that was just held out somewhere in space and dropped. So you could pump exactly once before letting go.

And motorizing the pivot points of the trapezi is certainly enough of a cheat to qualify as a Puzzle Land answer.

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15 hours ago, plasmid said:

That phrase was posted on Brainden?!?

I'm here all day folks...

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After the movie comes the stage show and after the stage show "Trapeze on Ice"

If friction is counted as minimal

Spoiler

at the bottom of a trapeze swing most of the vertical velocity is converted to horizontal velocity,  It the length of the trapeze is nearly the same as the height of the tower and friction can be negated (i.e on something infinitely viscous)  they could slide to the next trapeze.  in theory with no friction the acrobat could pick up the second trapeze at the lowest point even if it had no current inertia and reach the platform.

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On 6/15/2016 at 1:10 PM, phaze said:

After the movie comes the stage show and after the stage show "Trapeze on Ice"

If friction is counted as minimal

Reveal hidden contents

at the bottom of a trapeze swing most of the vertical velocity is converted to horizontal velocity,  It the length of the trapeze is nearly the same as the height of the tower and friction can be negated (i.e on something infinitely viscous)  they could slide to the next trapeze.  in theory with no friction the acrobat could pick up the second trapeze at the lowest point even if it had no current inertia and reach the platform.

I believe this gets honors as the best "cheating" answer thus far.

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