https://puzzling.stackexchange.com/questions/122232/is-this-duplo-train-track-under-too-much-tension Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange [ ] Loading... 1. + Tour Start here for a quick overview of the site + Help Center Detailed answers to any questions you might have + Meta Discuss the workings and policies of this site + About Us Learn more about Stack Overflow the company, and our products. 2. 3. current community + Puzzling help chat + Puzzling Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog 4. 5. Log in 6. Sign up Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. It only takes a minute to sign up. Sign up to join this community [ano] Anybody can ask a question [ano] Anybody can answer [an] The best answers are voted up and rise to the top Puzzling 1. Home 2. 1. Public 2. Questions 3. Tags 4. Users 5. Unanswered 3. Teams Stack Overflow for Teams - Start collaborating and sharing organizational knowledge. [teams-illo-free-si] Create a free Team Why Teams? 4. Teams 5. Create free Team Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Is this duplo train track under too much tension? Ask Question Asked 3 days ago Modified today Viewed 70k times 76 $\begingroup$ My kids were making this train track of duplo the other day, and this is what they have put together. They are still very young, and for them, this is something big: they were really proud that they have created this awesome track: duplo train track (click to enlarge) I liked it very much. It is build with 2 types of tracks, curved rails and straight rails (the road crossing is basically the same as 1 straight rail). However, as a father, I also don't want broken duplo pieces, so I wanted to make sure the track is not too much under tension. For example: If you want to make a closed loop, you could put 12 curved rails in a circle, and that's perfectly fine. However, if you are using brute force, you will be able to put 13 curved rails in a circle, but that way the track is too much under tension, and it's probably not good for the duplo pieces. I basically have 2 main questions: 1. Is there any way to quickly see if there is any tension, and why? (I know I could just take one piece out, and put it back in to feel it myself, but I am looking for a more logical way, so I am able to reason it.) 2. Suppose I want to update the track in the picture to have less tension. + If you have to take away exactly 1 rail piece (straight or curved), which one is the best, and why? + If you have to add exactly 1 rail piece (straight or curved), what is the optimal place to insert one? I am sure this could be calculated mathematically, but I prefer a more quick, practical way. I found this very interesting site about the duplo rails dimensions, which covers the basics of duplo rails. I think it is really useful for this question. --------------------------------------------------------------------- NOTE: This is the first time that I am asking a question, without knowing the answer myself. I am not entirely sure if this kind of question is on topic, feel free to give feedback. * mathematics * visual * geometry * optimization Share Improve this question Follow edited 2 days ago Kate Gregory's user avatar Kate Gregory 5,65311 gold badge2525 silver badges3434 bronze badges asked Sep 3 at 19:48 Lezzup's user avatar LezzupLezzup 4,36111 gold badge77 silver badges4343 bronze badges $\endgroup$ 9 * 1 $\begingroup$ I notice the Duplo link about curved rails says that Rot13(gjryir frpgvbaf qb abg znxr na rknpg pvepyr, gurer vf n fznyy bireync, naq fb znxr n pvepyr gur genpx zhfg or fgerffrq n yvggyr. Fb V nz hafher vs vg vf cbffvoyr gb unir n grafvba serr ynlbhg). $\endgroup$ - Weather Vane 2 days ago * 8 $\begingroup$ "I know I could just take one piece out, and put it back in to feel it myself, but I am looking for a more logical way, so I am able to reason it" and "I am sure this could be calculated mathematically, but I prefer a more quick, practical way" do not stand together. If you want the quick, practical way, you already stated it and have it. You're contradicting yourself. $\endgroup$ - Olorin 2 days ago * 7 $\begingroup$ You might consider asking at Bricks. $\endgroup$ - AakashM 2 days ago * 1 $\begingroup$ Please, for the love of whatever deity you believe in, don't let them become railway engineers without some more training. I can see a harmonic rocking derailment happening pretty quickly :-) <-- note the smiley, this is intended to be humour. $\endgroup$ - paxdiablo yesterday * 8 $\begingroup$ Not really an answer to the question as posted -- but I think the premise needs some good parenting advice: let your kids break bricks. They are pretty darn durable anyway and fabulously cheap to replace. So when they break one they will begin to learn about over-stressing materials through their own experiences. $\endgroup$ - Duthomhas yesterday | Show 4 more comments 7 Answers 7 Sorted by: Reset to default [Highest score (default) ] 106 $\begingroup$ First, we can check that there is no angular misalignment. Since 12 curved pieces are needed to make a full circle, the number of left pieces minus the number of right pieces must be a multiple of 12. Keeping track of the angle, starting at the crossing and going counterclockwise, I think that is satisfied in this setup: enter image description here Second, we look for positional misalignment. What we want to do is add up the vectors from the start to the end of each piece and check that the vector sum is zero. I haven't found a way to do this that ends up neater than just doing the trigonometry. However, things work out nicely in that all the coordinates end up being the sum of a rational number and a rational multiple of $\sqrt{3}$ (in math lingo, they are members of $\mathbb{Q}[\sqrt{3}]$). This means that the coordinates sum to zero only if their rational parts sum to zero and the coefficients of $\sqrt{3}$ sum to zero independently. This means that we can assign each piece type a list of 4 numbers (2 for the x-coordinate and 2 for the y-coordinate) and the pieces form a loop only if all 4 numbers sum to zero. Those numbers are: Curves: 0-1: 1 0 2 -1 1-2: -1 1 -1 1 2-3: 2 -1 1 0 3-4: -2 1 1 0 4-5: 1 -1 -1 1 5-6: -1 0 2 -1 Straights: 0-0: 1 0 0 0 1-1: 0 0.5 0.5 0 2-2: 0.5 0 0 0.5 3-3: 0 0 1 0 4-4: -0.5 0 0 0.5 5-5: 0 -0.5 0.5 0 Note that a 0-to-1 curve and a 1-to-0 curve have the same displacement so we only list the left curve. Also, a 180-degree rotation just negates the coordinates, so we can get away with only calculating the first half of the table. Adding up the pieces in the layout gives a result of 4.5 -3 -2 0.5, i.e. a displacement of $\left(4.5-3\sqrt{3},-2+0.5\sqrt{3}\right)$. This is not zero, so the layout is not free of stress. In fact, if we draw out the path that the shown pieces would ideally take, we get this shape: enter image description here No combination of fewer than 5 pieces will complete the loop, however the following combination of 5 will: 3-4 3-4 1-1 8-8 11-11. (There are 3 more combinations of 5 that give the correct displacement, but would add an odd number of curves and therefore cause angular misalignment.) Note that instead of adding a piece we can remove its 180-degree rotation. Therefore, I would suggest removing 10-9-10 and 5-5 and adding a 1-1 and 8-8: enter image description here That's only if you want an 'exact' solution, for only changing a single piece you'd want to add a 2-2, e.g. after the first gray curve, which would yield the following: enter image description here Share Improve this answer Follow answered 2 days ago 2012rcampion's user avatar 2012rcampion2012rcampion 18.1k33 gold badges6464 silver badges9696 bronze badges $\endgroup$ 3 * 9 $\begingroup$ Very thorough. Except I think that if you go around just the once, you'll need the sum to be exactly 12 (or exactly -12, if you go the "wrong way"). Other multiples would have to have some number of self-intersections, with 0, 24 and -24 all requiring at least 1, and the relative curvature of the two loops determining which one you get. Something something turning number, something something turning a sphere inside out $\ endgroup$ - No Name 2 days ago * 2 $\begingroup$ Duplo sells rail overpasses, so self-intersection is not necessarily problematic. $\endgroup$ - ioctl 2 hours ago * $\begingroup$ @2012rcampion Congrats with the golden batch! Didn't expect this question got this much attention :) $\ endgroup$ - Lezzup 57 mins ago Add a comment | 10 $\begingroup$ My answer: A basic strategy could be to count the curved sections: Suppose there are no straight pieces of track. The (number of inward curving) minus (number of outward curving) sections must be 12 to join up cleanly. That may not guarantee it to be free from stress, but without that equality it can't be. Now suppose there are straight sections. The sum of the direction vectors of the straight pieces should be 0. A simple way to achieve that if there are an even number, would be to arrange them in non-colinear parallel pairs. If there are 3 lengths, have them after 4 nett units of inward curving sections. In the picture, there are (20 - 8 = 12) nett inward curves, with 4 straights in random positions. An easy way to make it stress free, would be to * Remove 2 of the 4 straight sections, and reinsert them 6 nett units away (i.e. opposite) from another straight section, and rejoin the parts. One way to check the stress (if there are 12 curves), could be to * Measure the angle of each straight piece, and see if the sum of their sines is 0 and the sum of their cosines is 0. Share Improve this answer Follow edited 2 days ago answered 2 days ago Weather Vane's user avatar Weather VaneWeather Vane 12.4k11 gold badge1919 silver badges4646 bronze badges $\endgroup$ 0 Add a comment | 3 $\begingroup$ I would first check for track flatness. When locked in with extra effort, the loop will warp a little, basically going into 3d instead of flat 2d. As for "too much tension", obviously it's subjective. The links are made with certain tolerance for lateral misalignment and will distribute it over the loop. The "too much" will likely manifest itself at the joints, when some joint would appear somewhat bowed rather than straight. Also gaps at the joint rails may appear somewhat irregular - either wider, or stepped, or both. This can be felt by inspective touch. Yet the ultimate test is a train ride. If it's reasonably smooth and steady, then the tracks can handle the load... for now. Share Improve this answer Follow edited 4 hours ago answered 5 hours ago choochoodingding's user avatar choochoodingdingchoochoodingding 3122 bronze badges New contributor choochoodingding is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. $\endgroup$ 1 * $\begingroup$ A neat way to accentuate this - although obviously this could be considered "destructive" - would be to glue acetate sheets to the underside of each section of track. Any that are forced out of the plane will buckle the attached sheet. $\ endgroup$ - regularfry 3 hours ago Add a comment | -1 $\begingroup$ Just a theory: if straight sections are only ever installed as a pair of two exactly parallel sections, AND right and left curve pieces are installed in equal numbers, there should not be any tension in the track. Share Improve this answer Follow edited 2 hours ago answered 3 hours ago random guy's user avatar random guyrandom guy 111 bronze badge New contributor random guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. $\endgroup$ 2 * 1 $\begingroup$ Can you come up with a closed loop that satisfies your constraints? It seems to me that it is very difficult to make a closed loop that has the same number of left and right curves (unless you make an 8-shape and then it is not flat). You usually need 12 more of one type than the other... $\endgroup$ - wimi 2 hours ago * $\begingroup$ How does this add to my answer, which advised using straight sections in parallel pairs? Also the "right and left curve pieces are installed in equal numbers" is incorrect - they can't close a circuit. $\endgroup$ - Weather Vane 48 mins ago Add a comment | -3 $\begingroup$ Sorry, my brain was AFK. That only makes sense if you count right and left curved pieces by mentally driving along the track, and of course there is an additional number of curved pieces in one direction needed to make a full circle. Share Improve this answer Follow answered 2 hours ago random guy's user avatar random guyrandom guy 1 New contributor random guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. $\endgroup$ Add a comment | -3 $\begingroup$ I can help answer question one easily: If you take one of the /trains / away such that you have n - 1 trains for any n number of kids, you will quickly find out whether there is tension on the tracks. As for question two, it's hard to say without sight of the rest of the toy box but I'd probably give them their train back before finding out. Share Improve this answer Follow answered 1 hour ago Matthew's user avatar MatthewMatthew 1 New contributor Matthew is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. $\endgroup$ Add a comment | -3 $\begingroup$ Is not the correct answer to use clear tracks and then a polarized lens to see the stress directly? Share Improve this answer Follow answered 1 hour ago Dave Aitel's user avatar Dave AitelDave Aitel 1 New contributor Dave Aitel is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. $\endgroup$ 1 * $\begingroup$ This is not in the spirit of the question, which deals with the tracks as presented. $\endgroup$ - bobble 1 hour ago Add a comment | Your Answer [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] Thanks for contributing an answer to Puzzling Stack Exchange! * Please be sure to answer the question. 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