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 Post subject: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 2:53 am 
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Hi Twisty Puzzles fans,

Bram's Fortress is an idea by Bram Cohen (from 2009), inspired by several gear designs by Oskar. This twisty puzzle has eight Rubik's Cubes that are mutually connected by gears. If you turn one face, then four other faces turn with it. The video shows three failed prototypes. There is too much force and manual alignment needed to make the puzzle turn. I am sorry ...

Watch the YouTube video.
Watch the puzzle at my Shapeways Shop.
Check out the photos below.

Enjoy!

Oskar
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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 4:00 am 
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Thanks for posting this. There is absolutely nothing wrong with failure: your many, many great successes prove that it is all just part of the process (the path to success), and worth it...

It seems the issue is the gearing ratio not being able to overcome friction due to insufficient tolerance and/or excessive contact between sliding surfaces, however I understand that the gears have to transmit the force smoothly between the cubes in all directions, so perhaps you could make each gear as a double gear, with both high and low diameters, or increase the tolerances, or place metallic or PTFE washers/sheets between the sliding surfaces? However, I also understand if you feel that you've already invested too much into risky prototypes for this particular puzzle.

BTW, did you ever give Spherigears another try with stiffer and longer hinges all on the same side?

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 4:30 am 
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I'm just trying to make constructive criticism here: I think you should work out your tolerances, because you master gears. The gears can't work with the tolerances you use because they are too low. I think you should try giving 0.5 on the vmech to make sure it's smooth, and maybe remove springs if there are any, and just adjust the screw tension right as to make it smooth without springyness (to avoid gear skips).

On another note, I can't wait to see a V2, because i've always waited to see the forteress cube and now it's finally out :D

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 5:23 am 
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RubixFreakGreg wrote:
I think you should work out your tolerances...
Increasing the tolerances might help, but part of the problem is that any turning force applied to one cube ultimately has to be split between four cubes, so I think a combination of solutions will be required to be on the safe side. Just my opinion. :wink:

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 5:59 am 
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KelvinS wrote:
RubixFreakGreg wrote:
I think you should work out your tolerances...
Increasing the tolerances might help, but part of the problem is that any turning force applied to one cube ultimately has to be split between four cubes, so I think a combination of solutions will be required to be on the safe side. Just my opinion. :wink:
Thank you for the opinions. In this case, the cubes turn very smoothly without the six big gears. The problem is that they skew/jam a bit when driven by the big gear. Increasing the gap would only aggravate that problem. For now, I see no other option than making the puzzle a bit (e.g. 2x) larger. However this also makes it a bit^3 more expensive to print. Other suggestions are welcome.

Oskar

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 8:11 am 
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Stupid suggestion maybe, but how about dividing the force that is all on the one gear onto .. say.. three?

I have no engineering insight or anything, but I do believe in "vele handen maken licht werk", lots of hands make light work.

Not a clue if that can be or even is applicable here.


Mind I LOVE your builds!

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 8:24 am 
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Kattenvriendin wrote:
Stupid suggestion maybe, but how about dividing the force that is all on the one gear onto .. say.. three? ... lots of hands make light work.
You are right, when using four hands, this puzzle turns very smoothly. :D

Oskar

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 9:10 am 
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ROFL!!

I meant on the gears :lol:

Instead of one big gear per face, maybe more smaller gears per to spread the force that is needed over them?

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 1:15 pm 
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I saw a comment on the video that might help. What if you added some kind of crank or a way to turn the big gears directly?

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 2:27 pm 
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What if you made the gears thicker so you could use a herringbone design? That should help keep everything flat in the same plane.

It may be better to make a "key" tool to drive the central gear directly rather than turn an outer gear and expect the tension to transfer to all of the others perfectly without binding.

I assume there is some propensity for a gear to want to push away from the gear driving it and the only thing stopping that from happening in the screw at the axis. What if you have a large cylindrical groove on the gear and the platform it is resting on so the gear moves in that channel and doesn't put any binding force on the axis. This is also done in some dihedral puzzles like Luke's Hexoid and Eric's "Bram's Brick" to keep the two hemispheres of the core nicely aligned through turns.

Also, I think you could try adding a "click" mechanism to the gears with a piece that bumps out and falls into a dimple at appropriate intervals to keep the gear turning modulo whatever it needs to.

Finally, you might be able to add a click mech to the 3x3x3 edges into the 3x3x3 centers which might make each face of each puzzle snap at perfect intervals to keep one sides aligned after turning another side.

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 2:31 pm 
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For the theorists out there:

Here we have eight times the FusedCube interconnected.
A single FusedCube has 170659735142400 permutations.
BramsFortress has 44970586918084054475602454040030532394593344045352013301519502539635304443033153306864467165839360000000000000000 permutations.

More interesting is the quotient:
Size(Fortress)/Size(FusedCube)^8
which is 1/16

Any comments?


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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sat Aug 25, 2012 3:03 pm 
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Andreas Nortmann wrote:
For the theorists out there:

Here we have eight times the FusedCube interconnected.
A single FusedCube has 170659735142400 permutations.
BramsFortress has 44970586918084054475602454040030532394593344045352013301519502539635304443033153306864467165839360000000000000000 permutations.

More interesting is the quotient:
Size(Fortress)/Size(FusedCube)^8
which is 1/16

Any comments?

Well I'm not using Gap but here goes.

A single cube should have:
? (7! * 9!) / 2 * (3^6 * 2^8)
% = 170659735142400


I've created a 3-cycle for the edges of just one cube and a pure 3-cycle for the corners of just one cube. In this way, you can perform any even permutation on a cube without affecting any other cube.

So the question really is, how are the parities of the fused cubes tied to each other? The parity of 4 cubes is connected. Make a turn and you change the permutation parity of 4. By symmetry you will always have either 0, 4, or 8 cubes with in an even permutation (and the rest odd). (EDIT: this is faulty logic, see my post below)

So the total total number of positions is:
? (((7! * 9!) / 2 * (3^6 * 2^8)) ^ 8) / (2^4)
% = 44970586918084054475602454040030532394593344045352013301519502539635304443033153306864467165839360000000000000000


So we get the same number :D

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Last edited by Brandon Enright on Mon Aug 27, 2012 2:58 pm, edited 1 time in total.

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Sun Aug 26, 2012 4:41 pm 
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I say put 4 gears in a line(if you can) so it looks like one big elongated gear in the Mechanism(so the gears don't get so stress out) or/and maybe put a handle some how over the screw so that all four of them would turn when you turn the handle.

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Mon Aug 27, 2012 2:44 pm 
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bmenrigh wrote:
So the question really is, how are the parities of the fused cubes tied to each other? The parity of 4 cubes is connected. Make a turn and you change the permutation parity of 4. By symmetry you will always have either 0, 4, or 8 cubes with in an even permutation (and the rest odd).
[...] (2^4)

I've realized this logic is faulty. It produces the right answer for the wrong reason.

The real question is not how many "parities are connected" but rather how many independent parity patterns are reachable (versus how many we assumed there would be).

So we have 8 cubes, each could be even or odd so without any restrictions there would be 2^8 == 256 possible cases. When you make one turn the parity of the cubes on the face you turn toggles.

So the question is, what fraction of the 2^256 states are actually reachable?

This is pretty easy to answer: each turn toggles the parity of 4 cubes. This means that there will either be 0, 4, or 8 in an even permutation and 8, 4, 0 in an odd, respectively. Also, for a given face, it can either have 0, 2, or 4 even and 4, 2, or 0 odd.

Now it's easy to enumerate all the states:

2: The all even and all odd states
2: The two checkerboard states
6: One face all even, the other all odd (six of these, one for each face)
6: 2 columns of two even cubes, 2 of two odd (two of these per axis, 3 axes, six total)

2 + 2 + 6 + 6 == 16.

So only 16 of the 256 possible parity configurations exist. 16 / 256 == 1/16.

So the correct calculation is:

? (((7! * 9!) / 2 * (3^6 * 2^8)) ^ 8) * ((2 + 2 + 6 + 6) / 2^8)
% = 44970586918084054475602454040030532394593344045352013301519502539635304443033153306864467165839360000000000000000


All that matters is that our calculation assumed all parity configurations were reachable so we need to multiply by the number that actually are reachable over the assumed total.

It was easy to enumerate all 16 parity patterns manually for this puzzle but if this a more complicated solid with stranger restrictions I don't know how I'd enumerate them / count them. There must be a way to enumerate these using something like Burnside's Lemma but this is an area of counting I need more experience with.

EDIT: Andreas I'd love to hear your thoughts on how to count how many parity configurations are possible (in a more generic way).

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 Post subject: Re: Bram's Fortress by BRAM and OSKAR
PostPosted: Tue Aug 28, 2012 3:13 am 
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bmenrigh wrote:
Andreas I'd love to hear your thoughts on how to count how many parity configurations are possible (in a more generic way).
Burnside's Lemma wouldn't have been of great help in this case. Burnside's Lemma would have told you that there are 6 configurations (without symmetric duplicates) to distribute 4 odd parities on 8 positions.
In this case I would have used the samy way as you did because of the additional restrictions. To find a more generic way a not so simple example would be helpful.


Andreas


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