Time for some fun with Pythagoras!
Lego tends towards rectangular constructs, but sometimes we might want to make triangular connections. Technic beams especially lend themselves to this.
Rightangled pythagorean triangles are fairly easy to construct. Here’s a 3,4,5 triangle:
Notice that when you build it you build it as 4,5,6. That’s because it’s the distance between the pin centers that has to be in the 345 ratio. Two pins separated by 2 holes are a total distance (centertocenter) of three holes apart. Or said another way, two pins right next to each other are distance “one unit” apart, not two. Thus: A twohole beam would be a distance of “one unit” in this analysis.
There are many pythagorean integer triangles. Here’s the complete list up to a reasonable size:
 345 (also 6810, 91215, 121620 etc)
 51213
 72425
 81517
If you try to build these don’t forget the +1 factor for counting holes.
If we allow half unit offsets there are some more pythagorean triangles available. Obviously any of the above constructs can be divided in two; beyond those we also have:
 4.52020.5
 617.518.5
 7.51819.5
 1010.514.5
 1222.525.5
 13.51822.5
In the 345 family the angle opposite the “4” side is 53.13 degrees. The reason this is good to know is that the standard bent beams are at that same angle (53.130) and now you know why:
If you build this you will see that the angles match up perfectly.
If you combine a pythagorean triangle with its mirror you will get an isosceles triangle. Here, for example, is a 655 construction which is two mirrored 345 triangles
Notice, once again, you build 655 as 766 if you are counting the total number of holes in each side. This kind of construction will be very strong and also has the advantage of being able to attach things to the vertical riser in the spine of the triangle and have them be “on grid” whereas the holes in the angled beams are rarely ever “on grid” except at the vertices.
Trapezoid Extension – Pythagorean
Once you can make a triangle, you can make a corresponding trapezoid by extending it evenly in one dimension. This is easy to do for right (pythagorean) triangles:
Here I took a 345 triangle and extended out the base (by 4 in this case; “n” as shown in the diagram). I have to extend the connection point up top out the same amount. The geometry still works and everything lines up.
We’ll revisit this topic for nonpythagorean triangles in a moment; there are some triangles that you can extend into trapezoids and some that you can’t (easily).
System Brick Triangles
You can make triangles using system (“standard lego”) bricks too:
Note how I used two 1×1 bricks to create “pylons” for the angled brick. You have to do this, or something similar, because the neighboring studs will interfere with the angled piece.
General Triangles
Any three beams that satisfy the triangle inequality can be formed into a triangle. In general you will need to use four beams in the construction because you need to “jump” up/down one level to make the last connection work. Look at the construction of the 345 pythagorean triangle to see an example of why 4 beams are needed (and in that case since one of the joints is a 90 degree angle I used a T piece but in general you will want to use just straight beams when making angles that are not 90 degrees).
The lengths of the triangle sides will determine the angles according to standard triangle sidesideside (SSS) solutions. Just hook them together and let the angles fall where they may.
Of course, we might like to know what those angles actually are. I wrote a small python program that calculates Lego triangle angle information for a given configuration of sides. The source is available here: http://neilwebber.com/files/legotriangles
If you prefer, here is the table of output it generates for fullunit constructions and halfunits:
You can use the command line options (read the source) or just peruse the uploaded table and see that there are some other useful angles you can construct, though in some cases you will have to settle for close approximations. For example, there is no 45 degree construction but there are two that come close:
 51215 will generate an angle of 45.036 degrees.
 131717 generates 44.959
Any equilateral construction generates 60 degrees but so do these:
Here’s a 378 construction. The angle between the blue and red beams is 60 degrees.
This construction also more clearly illustrates the need to use four beams (not just three) to make a triangle because of how the “levels” at the vertices work out. One of the sides has to jump back down in order to join with the remaining side.
Because of that 60 degree angle, that implies you can build a 378 connection across an 888 equilateral triangle:
The angles opposite the blue beam (both above it and below it) are both 60 degree angles (by definition because the outer frame is an equilateral triangle). This construction is probably demonstrating some deep mathematical relationship between the 378 60 degree construction and the 578 construction; perhaps someone will point it out to me and I’ll add it back into this note.
If you have a larger equilateral triangle you can still make this connection at this point near the vertices or of course you could scale up the 378 or 578 as appropriate (possibly using half unit constructions too if necessary).
Don’t forget (in case you are confused again): building something with a 378 relationship takes hole counts of 489. You can count the holes in the above picture and verify this.
Nonpythagorean Trapezoids
Let’s look at a 4810 triangle (which would be equivalent to a 245 triangle but this is easier to construct):
As always: built as 5911.
The angles on this triangle are 22.332 (between blue and yellow), 49.458 (between blue and red), and 108.21 (between red and yellow).
Let’s say we want to make a trapezoid out of this. We can, for example, extend the blue beam to the right by making it longer, and finish out the rest of the trapezoid with right angles:
but, as you can see, we have a problem with the final joint at the red/yellow (and pink) vertex. That point isn’t on grid, whereas the pink beam, the black “riser” L piece, as well as the blue base are all on grid.
We can calculate the location of the red/yellow vertex using standard trigonmetry. Using the yellow/blue vertex as the origin, the X and Y coordinates (X along the blue beam) of the end of the yellow beam will be:
X = 8 * cos(22.332°) = 7.4
Y = 8 * sin(22.332°) = 3.04
NOTE: It’s “8” because that’s the length of the yellow side of the triangle.
The vertex doesn’t even line up vertically even though it looks like it might. It’s just close. It definitely doesn’t line up horizontally. We have no way to make that final connection.
Generally speaking you can’t easily extend nonpythagorean triangles into gridlegal trapezoid constructions.
A few constructions come very close; whether you want to use them in your models is up to you. They aren’t “on grid” but any given construction always has a certain amount of play in it and if they are close enough then things will appear to work. I believe the Lego corporation would refuse to accept these constructions as part of a design for a set, so in that sense they are “illegal”.
Here’s one that comes very close. Take the 378 triangle we saw before:
Taking the blue base as the X axis and the bluered vertex as the origin, we can calculate the location of the redyellow vertex. We already know that the angle between the red and blue is 60 degrees (you can reverify this using my tables or perform your own SSS triangle solution for 378; the angle opposite the 7 side is 60 degrees). Since the red leg is 8 units in length the location of the redyellow vertex is:
X = 8 * cos(60°) = 4.00
Y = 8 * sin(60°) = 6.93
The X coordinate works out evenly ongrid, thanks to the cosine of 60 degrees being exactly one half. The Y coordinate is off grid but close enough that the following construction will appear to work:
In reality the pink beam is at (relative) height 6.93 on its left end and 7.00 on its right, thus there is some stress/flexing in the construction to make it work. You might reduce that stress slightly by only pinning the pink beam with one pin at the right end (vs the two shown in this build).
This just happens to work out as a case where the offgrid amount is smaller than the inherent flex in the pieces. Your mileage may vary; personally I don’t use these nonaligned constructions in my models but I present this as being one that is very close to ok. You can find others generally by doing math or by just trying things to see what works within reason.
Going back to the 4810 (5911) construction, there is a way to make a trapezoid out of it, like this:
because what I did here is just extend the base, then insert a parallel corresponding extension (the black beam), and connect them with a beam (the second red beam) that is parallel to the original side of the triangle. This preserves all the geometry and gets everything lined up in the right spot; however, in this entire construction only the blue beam is “on grid”; everything else is at nonstandard locations.
This structure is quite stable and reasonably strong though I’d be cautious about putting too much weight or pressure on it from top (perhaps someone will work out the statics on this one; just going by feel it seems reasonably strong but I’m not sure exactly where the forces are going when it is put into compression that way).
Approximate “standard” angles
I already listed two constructions that are very close to a 45 degree angle; here is a more complete list of approximations to common/useful angles.
30 degrees (approximate)
a 
b 
c 
∠° 
Error 
Comments 
3 
5 
6 
∠A = 29.9264 
0.0736° 
should prefer 4 7 8 usually 
4 
7 
8 
∠A = 29.9947 
0.0053° 
closest / smallest 
7 
12 
14 
∠A = 29.9947 
0.0053° 
same angle as 478; larger 
Of course multiples of those triples works, so for example 81416 is another good 30 degree approximation (doubled from 478).
Halfunit constructions can get even closer:
a 
b 
c 
∠° 
Error 
Comments 
5.5 
9.5 
11 
∠A = 29.9996 
0.0004° 

5.5 
14.5 
19 
∠B = 29.9996 
0.0004° 
same angle as 5.59.511 which is easier to build. 
7.5 
13 
15 
∠A = 29.99997 
0.00003° 
nearly “exact” for all practical purposes. 
The 7.51315 triple can be looked at as half of a 151515 equilateral triangle. A point 7.5 along one leg is halfway; if we then went exactly 90 degrees up to the other vertex we should have a height of:
h = 15 * sin(pi/3) # (pi/3 = 60 degrees in radians)
which computes out to h = 12.9904. So our actual height of 13 won’t form a true half of a true equilateral triangle; it will resolve to angles that are slightly off in all spots, but based on how close 13 is to 12.99 you can see why building the triangle at 7.51315 gets so close to the 30 degrees.
If you really need an exact 30 degrees you can get it from the complementary angle on an equilateral construction or from any of the other exact 60 degree constructions (see table lower down).
45 degrees (approximate)
There aren’t any 45 degree approximations that are as close as the 30 degree ones unless we allow half unit constructions. Here is the table of some useful ones including halfunits:
a 
b 
c 
∠° 
Error 
Comments 
3 
19 
21 
∠B = 45.06135 
0.061° 

5 
12 
15 
∠B = 45.03565 
0.036° 
best/smallest integer construction. 
6 
12.5 
16 
∠B = 45.00612 
0.006° 
very good; see 88.512. 
8 
8.5 
12 
∠B = 45.00612 
0.006° 
Same ∠ as 612.516 but smaller. Second best known. 
9.5 
14 
19 
∠B = 45.00349 
0.0035° 
Best (but uses halfunits). 
11 
18 
24 
∠B = 45.05405 
0.054° 
51215 is a better integer construct. 
60 degrees
You can, of course, easily get 60 degrees from any equilateral triangle construction (all sides equal). You can also get it from the complementary angle on any of the above 30 degree constructions.
There are quite a few triples that lead to exact 60 degree angles. Here are a few of the simpler and more likely useful ones (including half units):
a 
b 
c 
Comments 
1.5 
3.5 
4 

2.5 
3.5 
4 

2.5 
9.5 
10.5 

3 
7 
8 
Just double 1.53.54 
4 
6.5 
7.5 

5 
7 
8 
Just double 2.53.54 
NOTE: In all of these, it is angle B that is the 60 degree (exact) angle.
41.4096 degrees
Say what? Well, the cosine of 41.4096 is 0.75, which means that a beam on this angle will generate a halfunit offset at two holes and every four holes thereafter.
Is this useful? I’m not sure. There is a problem that the sine (0.6614) is nothing useful, so this means that while your beam will have holes at halfunit offsets in one dimension, the other dimension will never be ongrid. Nevertheless, for your amusement, I present this table:
a 
b 
c 
cosine 
Comments 
2 
2 
3 
cos(∠A) = 0.75 exact 
Isoceles; angle B also works. 
3 
8 
10 
cos(∠B) = 0.75 exact 
Note this one is angle B 
4 
5 
6 
cos(∠A) = 0.75 exact 

Here’s an example 223 construction, with the orange lines showing which holes (in the blue beam) are on halfunit offsets from the adjacent beam. Any appearance of not being in alignment is an artifact of the 3D rendering; the math works. However, as already pointed out, the holes are definitely offgrid in the other dimension so I’m not sure how this sort of construction might be useful. I present it anyway just as food for thought.
If for some unusual reason you needed to get a very specific offset, you could use this cosine concept to figure out what angle you’d need to get you there. Let’s say you need an 0.6 offset and you’ve decided (usually for construction reasons) you want to go three units along the hypotenuse to get there. You need an angle that has a cosine of 0.2, which, after you’ve gone three units along the angled beam will have moved you 0.6 on the rectilinear grid. You find the angle using arccos; that angle would be 78.463 and you could build it using (per the table or the program) 567.
Admittedly, it’s going to be a very unusual set of circumstances that makes you want special offsets — probably involving integrating with other nonLego parts, or desparately trying to make a nonstandard trapezoidal construction work out. Regardless, you’ll still have the problem of the other dimension which just won’t be where you want it anyway. In reality you’ll probably just use an axle and one piece slid along the axle (thus taking on a nongrid position) to get there; thus all this math is probably a waste of space. But maybe it will inspire someone to do something particularly clever one day.
53.130 degrees
We already saw that this angle (more precisely it is 53.130102) appears in the 345 family of pythagorean triangles and is the angle of one of the standard lego “bent” beams.
Are there any other triples that generate this angle? Yes, there are. First of all, any isoceles triangle constructed using a mirrored pythagorean 345 (or multiples of this) will have this angle. We already saw that in the 655 example.
Other combinations (that are not 345 in disguise) that give a 53.1301 angle are:
a 
b 
c 
Comments 
2 
6.5 
7.5 

4 
13 
15 
Just double 26.57.5 
5 
8.5 
10.5 

6.5 
7 
7.5 
Angle A 
There are actually quite a few of these; I’ve only listed some above. All of those except the last form the 53.130 angle at B.
Half Unit Construction
There are many ways to generate halfunit offsets with technic; here’s one construction you may find handy when trying to make these triangles:
That’s probably the best one I’ve found so far. You can build this with the 1×2 bricks with pins, as shown, or the 1×2 technic bricks (with holes) and separate pins.
Here’s another technique that isn’t quite as elegant but might be handy at times:
The yellow beam and the holes the yellow connector (usually red in real life) form a beam of total length 7.5 (the two holes in the connector are at 6.5 and 7.5).
As shown above the half offset connector is held in place by an axle pin and isn’t necessarily completely stable. You can brace it in a variety of ways if necessary:
That’s it for today’s Lego rambling. Happy Building!
The drawings in posting were created with Bricksmith and rendered using LDView all made possible via LDraw.