# Is this duplo train track under too much tension?

My kids were making this train track of duplo the other day, and this is what they put together. They are still very young, and for them, this is something big. They were really proud that they created this awesome track: (click to enlarge)

I liked it very much. It is built 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 under too much 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, with brute force you could put 13 curved rails in a circle, but the track would be under too much 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 duplo rail 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.

• "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. Sep 4 at 8:52
• You might consider asking at Bricks. Sep 4 at 9:25
• 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. Sep 4 at 23:24
• 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. Sep 5 at 0:53
• Could anyone please tell me why is this question so popular? Sep 7 at 12:04

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: 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 somewhat nicely.

For example, if we assume that a straight piece has length 1, then its vector at angle 0 (pointing to the right; I call this a 0-0 piece) is $$(1,0)$$. If we rotate this piece to angle 1 (30 degrees; I call this a 1-1 piece) then its vector is $$\left(\cos{{30}^\circ},\sin{{30}^\circ}\right)=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$$. A curve piece has a radius that is twice a straight piece so a curve that starts at angle 0 and ends at angle 1 (which I call a 0-1 piece) has a vector of $$\left(2\sin{{30}^\circ},2\left(1-\cos{{30}^\circ}\right)\right)=\left(1,2-\sqrt{3}\right)$$.

Note that the coordinates of these three vectors, and in fact the vectors for every piece orientation, are a fraction (rational number) plus $$\sqrt{3}$$ times another fraction (the fraction can be zero, e.g. $$\frac{\sqrt{3}}{2} = \frac{0}{1} + \frac{1}{2}\sqrt{3}$$. If we add two numbers of this form, the answer is just adding the pure fraction parts together and adding the $$\sqrt{3}$$ parts together; and because $$\sqrt{3}$$ is irrational, the only way the sum can be zero is for both parts to be zero. This is the same way that vector addition works: you add the coordinates to each other, and the result is the zero vector only if both coordinates are zero. Therefore we can treat our 2-d vectors of (possibly) irrational numbers as a 2 × 2 = 4-d vector of fractions.

To summarize, we can give each piece a list of 4 numbers, and the pieces form a loop only if summing the pieces' numbers with each other result in a list of four zeros. 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 0-1 curve and a 1-0 curve have the same displacement so we only list the left-hand 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: 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, so instead of adding 3-4 and 11-11 we can remove a 9-10 and 5-5. Finally, remember that swapping the beginning and ending of a curve has the same displacement, so we can remove one 9-10 and one 10-9 instead of two 9-10s. Putting this together, we can get s stress-free solution by removing a 10-9, 9-10 and 5-5 and adding a 1-1 and 8-8: 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: • 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 Sep 4 at 16:50
• Duplo sells rail overpasses, so self-intersection is not necessarily problematic. Sep 6 at 17:55
• @2012rcampion Congrats with the golden batch! Didn't expect this question got this much attention :) Sep 6 at 19:02
• So... If I really wanted to understand this answer (the second part, the 1st I get) where would I start? (past school level maths). Now I'll have to spend the weekend in Wikipedia & Khan Academy! :D Sep 7 at 5:22
• @marc.fargas If you can live with these 4-dimentional vectors as a shorthand for 2 dimentional vectors (i.e. , the vector (x=1+sqrt(3),y=-2*sqrtI(3)) is the vector (1,1,0,-2), 2021rcampion is just summing vectors. Notice that all pieces 2-3 start from the same angle and finish at the same angle, so they represent the same vector (also, notice that 2-3 is notation for had 2 curves, now have 3, this notation was probably invented by the answerer for this particular question) Sep 7 at 12:22

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.

• I like the simplicity of rearranging the straight sections to be in pairs 6 net curve sections apart. As long as each pair share an orientation, and there are net 12 inward curves, we're good. Sep 17 at 10:53

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.

• 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. Sep 6 at 16:02

My metric for this is fun. If you attach 2, 3 or 4 duplo wagons that can detach easily and make them go through the track, if they detach or derail, then it's too tense.

Just by looking at the track I can feel in my dad bones that the trains are going to derail or the wagons are going to detach.

My rule is, if a track goes at an angle, say left, it is forbidden for the next-up rail to go right, you have to add a straight line.

Here's a simple and interesting approach which may have some merit:

1. As discussed above the number of left pieces minus the number of right pieces must be a multiple of 12.
2. Each segment must have a non-consecutive (not joined to each other) mirror image. This mirror image does not have to be directly across from its mate. It just must have the same angle but facing in the opposite direction.
• That won't work. Try to build a track with 13 left and 1 right. Sep 12 at 19:37
• @FlorianF if you require that each piece has a mirror image then you need an even number of pieces of each type. In fact this condition is sufficient but not necessary (consider a triangle made of 3 straight pieces with 4 curves between each). Sep 14 at 12:13
• Oh, I see. I didn't get the "non-consecutive" condition. Sep 14 at 16:53