In part since I’m living with a seven-year-old, I’ve been thinking about laser-cutting a construction set based on a design Matt Heck showed me at TechShop something like a decade ago. If I recall, he had some things laser-cut from Masonite. I can’t find anything he’s put online about it.
Although the original was laser-cut from I think high-density fiberboard, you could realize this design with any sheet-cutting process, including scrollsaw cutting, plasma cutting, oxy-fuel cutting, laser-cutting of plastic or metal, waterjet cutting, hot-wire cutting, wire EDM, or sheet-metal shearing.
The basic Heckballs unit is a square with rectangular slots cut halfway from its edge to its center, with the width of the slot being the same as the thickness of the square. By slotting squares into each other’s slots (edge-lap joints), so that their edges match up with each other’s centers, you can easily assemble a somewhat free-form 3-D lattice. 80 mm is a reasonable size for the squares; 1.5 mm is a reasonable thickness.
To be somewhat concrete about costs, in the most basic Heckballs form, you can cut an 810 mm × 450 mm sheet of 1.5 mm MDF into 10×5 = 50 squares using 10×5 cuts totaling 10·400 mm + 5·800 mm = 8 m of cutting; each slot is 20 mm + 1.5 mm + 20 mm = 41.5 mm, and there are 200 of them, for another 8.3 m of slot cutting, a total of 16.3 m. At 30 mm/s, this is 543 s ≈ 1 hour ÷ 6.6, which is probably about US$6 of laser cutting time, or about 12¢ per Heckballs square.
MDF costs AR$179 ≈ US$12 for 1830 mm × 2600 mm of 3 mm-thick MDF, which works out to about US$2.50/m², or AR$40 for 300 mm × 600 mm of 1.5 mm thick MDF, which works out to about US$14/m². So the cutting is probably the bulk of the expense, which suggests it might be worth trying to cut stacked MDF.
(Steel is supposedly around US$300 per tonne right now, which would be US$7.20/m² at 3 mm, which rather surprisingly suggests that ASTM A36 steel only costs about twice what MDF does per unit volume, despite having 210 GPa of Young’s modulus, which is 50× MDF’s stiffness, and 290 MPa yield stress, which is 16× MDF’s strength.)
In this form, this has a few drawbacks:
The lattice is entirely made of perpendicular and parallel planes, which is the worst possible configuration for rigidity. More liberty of form would allow much more efficient material use.
There’s no way to make a flat surface without stuff sticking out of it, like for a shelf, chair seat, table top, or foot.
The slot width is unforgiving of imprecision; if the material is slightly thicker or thinner than spec, or if the slot is cut slightly narrower or wider, it’s easy to get parts that won’t fit together or don’t stay together reliably.
If you fill all the slots of a square, the other squares collide in the middle.
The tensile strength of the joint is low because it is entirely dependent on friction.
The square corners of the slot create stress risers in the material, resulting in much lower resistance to cracking than is necessary.
Covering a long distance demands a lot of material, because you can’t make a “beam” of joined squares that’s less than 80 mm wide in both dimensions.
Covering a long distance creates a lot of angular slop, because there are a lot of serial degrees of freedom in all the slot joints.
There’s no way to make joints that are free to turn or slide, only rigid connections.
To provide tensile strength, the joints can be provided with tapered clips (i.e. snap joints). To eliminate stress risers, the internal corners of the slots can be replaced with rounded divots, or better for laser cutting, chamfered divots. The tapered clips should also allow making the slots a bit wider, eliminating the sensitivity to precision, but curving the slots slightly might also solve the problem without introducing more play. Curving the tapered clips could make it easier to fit a fingernail or whatever between them and what they’re clipping onto so that you can unclip them.
To prevent the collisions at the centers, it probably makes sense to chop off the edges of the squares, so that the distance between the centers of two edge-lapped squares is still 40 mm, but the squares are only 77 mm wide.
To permit covering a longer distance, I think probably the right solution is to have more than one kind of unit. For example, a long beam with 20 mm slots only at its ends, could hold the centers of the two squares at its ends 320 mm apart instead of 80 mm apart. This gives it a total length of 320 mm (or 317 mm to avoid collisions).
If the beams are 8 mm wide, we can cut an 80 mm × 320 mm rectangle into either 4 squares (roughly) or 10 beams. We probably want to have more beams than squares, because even in a 2D square grid, you have twice as many beams as vertices. The 4 squares require 320 mm + 4·80 mm + 4·4·21.5 mm = 984 mm of cut (although adding clips might add another 300 mm or more to that). The 10 beams require 10·320 mm + 80 mm + 2·4·21.5 mm = 3452 mm of cutting, which suggests a certain cost disequilibrium, pressuring the scale to increase to avoid wasting expensive laser to conserve cheap MDF.
Solving the rigidity problem involves using a different lattice. The incremental change is to add diagonals: convert the squares into octagons with eight slots and add beams of 320√2 mm ≈ 452.5 mm. Alternatively, you could try to switch to a sphere-close-packing lattice.
Once you have 8-mm-wide beams, you could enable rotation by cutting circular holes through the centers of some octagons or squares. Curves are much slower to cut on typical laser cutters, so you probably don’t want to do this for all the octagons, just some of them. The diameter of the hole should probably be about 8.2 mm.
Making flat panels that can fit snugly onto the coplanar edges of some octagons shouldn’t be very difficult, although they won’t be watertight. You just cut slots for the edges of the octagons to peek through.
These beams will be somewhat prone to buckling, since their slenderness ratio 320:1.5 is over 200; although the column effective length factor K is probably 0.5, this is still plenty long enough to enable buckling. If you could strengthen the beam into a T-beam with another 8 mm beam, the problem would probably be solved.
The aforementioned pressure to increase the scale adds more pressure to do the T-beam thing, since at a larger linear scale, buckling becomes a bigger problem.
What would the face-centered-cubic close packing solution look like? The unit cell is a cuboctahedron; in different planes, the number of lines coming out of a vertex are 0, 2, 4, or 6, all evenly spaced. If you think of the face-centered cubic arrangement as being a stacking of square layers, the connections within a given layer are at 0°, 90°, 180°, and 270°, while the connections to the layer above or below are at 45°, 135°, 225°, and 315°. Unfortunately, both the layer above and the layer below are at exactly the same angle, so they can’t both connect. Still, you could enable a double-layer close-packed truss with this arrangement by simply having octagon joints and extra unit-length beams whose end slots are angled at 45°. For the moment I think I’ll give up on fcc and hcp.
So, what does the most basic Heckballs kit look like? Let’s say I want to be able to make a cube with diagonals on most faces. Four vertices, four regular beams, and one long beam make a square (320mm side plus an extra 40mm stickout). But then there’s no immediate way to extend the lattice in a perpendicular direction. By adding more joints onto the edge, we could get a perpendicular direction and thus a sort of cube, but now some of the diagonals are the wrong length.
So here’s the part where Heckballs actually become balls. We eliminate rotational symmetry from the octagon; one of its slots now extends to its center, and the opposite one is eliminated. Two such octagons can be edge-lapped together to form something approximating a sphere with 12 available valences, 8 of which are among the 12 directions needed for the fcc crystalline structure. To make the cube with some diagonals, we can use the three valences along one edge of one octagon to make the two sides and diagonal of one square, and the valence in the center of the side of the other octagon to make the side of a perpendicular square. Two corners of the top square must be made in this way, with their axes coplanar with and at 45° to the square’s sides. The other two corners could be made with their axes coplanar to the perpendicular square and perpendicular to its diagonal, thus enabling four of the six cube squares to have diagonals. The other two faces are not thus diagonalizable.
If it is our priority to have three adjacent square faces diagonalized, neither this structure nor any similar structure can solve that problem, because connecting to the ends of the axis is impossible. A possible solution is to make some beams integral with joints, enabling them to connect axially.
Another possibility is to avoid connecting at vertices and instead connect around vertices, forming a polygon or polyhedron enclosing the vertex. For example, beams could penetrate into one another at some angle near their ends. It’s easy for me to imagine how this would work for 90° in a single plane around a vertex, but harder for me to imagine other angles.
Oh, wait, it’s not that hard. You can get an arbitrary angle with the plane of each beam rotated by some amount with respect to the plane of the angle; as long as the beam’s plane is neither parallel nor perpendicular to the angle plane, it’s workable (though more so if it’s not nearly so). Hmm, I have to think about that some more.
Anyway, back to the cube. Here we have eight vertices, each of which needs two half-slitted octagonal joints, sixteen octagons in all, plus 12 short beams and 4 long beams. Let’s suppose we scale up a little bit to respond to the cost-optimization thing to ⅛m (125 mm) for the octagon nominal width, and add another 4 long beams for extra flexibility, and cut down the beam length to half the octagon width. Now we have:
That all adds up to about 0.42 m², slightly more than a single 810 mm × 450 mm cut at Max58, even before worrying about packing. Maybe to get quicker feedback I should scale down by a factor of √2 to 88.4 mm as the basic unit? This is getting fiddly, I should probably just use 100 mm, and go back to 4× for the beams. Also I don’t need to leave a whole 3 mm off at the end; I only need ¾ mm off each end, or 1.5 mm in total:
That’s 0.39 m², still slightly too big, argh.
All right, let’s bite the bullet and go down to a basic unit of 62.5 mm:
That’s 0.147 square meters, so I should be able to nearly double it.
It’s about 41% octagons, 31% short beams, and 28% long beams. We could probably do with a higher proportion of beams: maybe 24 octagons, 28 short beams, and 20 long beams.
(Actually the octagons could still be squares maybe, although the slot ends should still be radially symmetric.)
A thing to somewhat worry about with these small octagons is that the slot ends will be awfully close together, and especially close to the deepened slot. The distance from the center of the joint to the end of the slot is 31.25 mm, with another 29.75 mm to the edge of the octagon. The clip hole needs to be at least 1.5 mm further in, probably better 3 mm further in, and the clip hole and slot probably need to be 1.7 mm across, and the clip hole probably needs to be 3 mm long, so the innermost part of the clip hole will be (31.25 - 3 - 3) mm = 25.25 mm from the center. But the slots at 45° from the deepened slot will have the center of the innermost end of their clip hole only √½ 25.25 mm = 17.9 mm from the deepened slot, and the 0.85 mm diagonal reduction from the width of the clip hole reduces that by 0.6 mm further down to 17.3 mm. Actually that’s the distance the clip hole corners will be from one another, but the deepened slot will also have its own clip on one side.
What should the clip dimensions look like?
The maximum tensile strength we can expect is the compressive strength of the 1.5 mm × 1.5 mm maximum possible contact area, 10 MPa · (1.5 mm)² = 22.5 N, or 45 N when we take into account the clips on both sides, which is pretty wimpy. Ideally this thing would withstand my harshest pulls without breaking, and with my legs I can pull close to twice my body weight, so about 2000 N. So this is about 44 times weaker than it needs to be.
It’s possible to get some improvement there with multiple clips in multiple holes, but there isn’t room for 44 sequential holes. At 6 mm per hole, there’s really only room for 3 at most. So this is probably unattainable. It does rather strongly suggest that 1.5 mm MDF is suboptimally thin and will result in assemblies that are far too easily broken. Indeed, probably even 3 mm is too thin. The next size up stocked by Max58 is 5.5 mm, giving us about 13× as much contact area.
So what if it’s 5.5 mm thick? 10 MPa · (5.5 mm)² · 2 = 600 N, about a third of what we’d like it to withstand. So maybe a double hole in 5.5 mm and call it a day at 1200 N of compressive strength.
But for that double hole, we probably need 22 mm from the center to the inside of the slot, even if the hook holes are absolutely square. So we probably need to make our octagons bigger, more like 150 mm. 15 such octagons would occupy our entire 810 mm × 450 mm sporting area, so we don’t have room for anything else on the first pass. Which is fine, because this is already taking me forever.
What about tensile strength? MDF’s UTS is only 18 MPa, so to withstand the 600 N of tension on either of the double hooks, we need 6.1 mm of width on the 5.5-mm-thick clip at the first hook to keep it from simply breaking from tension. That’s going to make it kind of tough to unclip by hand, I think.
(I have a new design to keep the clip head from tearing off due to a tension concentration on the hook side of the clip; it involves a bar on the other side sliding into a slot, preventing the clip from rotating. But the clip still needs to withstand the tensile stress.)
What about the force and energy needed to unclip the clip? Patrick Fenner tells us that if we don’t taper,
the force to deflect the tip by a set distance is F = dEta³/(4l³), where d is deflection, E is the Young’s modulus of the material, t is beam thickness and a is the depth of the beam.
In this case, d is at least 5.5 mm, E is 4 GPa, t is 5.5 mm, and a is 6.1 mm. So if l = 75 mm, which is kind of an absolute limit if the diameter of the thing is 150 mm, then we need 16 N, which is doable without tools. For a more reasonable value of 40 mm, we get 107 N, which is going to require a screwdriver or something.
We may be able to improve the situation by tapering the clip down to 3.1 mm at the second hook, which should still be enough to resist the tensile force there, starting from a thicker root.
The clips and clip holes shouldn’t extend all the way to the center, to avoid colliding with the deepened slot or excessively weakening the overall joint, nor should they extend all the way to the edge, in order to keep the angle properly controlled. The clip beam can extend deeper than the slot, but not by too much.
I don’t know, maybe I’m being too demanding of the tensile strength of a clip joint. Suppose that instead of designing for 1200 N we settle for 360 N, 90 N per hook, which we can get with 3 mm MDF instead of 5.5 mm, so we don’t need 11 mm of depth per hook hole, only 6 mm; we can now get 180 N tensile strength out of <3.4 mm of beam. And maybe we go back down to truncated 100 mm squares for our octagons. If the hook holes end 12 mm after the end of the slot, and we extend the clip beam by those same 12 mm, then its total length is 24 mm. So dEta³/(4l³) = 3 mm · 4 GPa · 3 mm · (3.4 mm)³ / (4 (24 mm)³) = 25 N, which you can operate with your fingers. It should get even better with tapering.
(ETA: actually it's totally reasonable to have arbitrarily low unclipping force with arbitrarily high tensile strength. Among the things you can do are to fold up the spring serpentine-style or to cut lengthwise slits in it to make it less stiff, in the limit becoming like the pages of a book.)
Plasma-cutting clips like these in steel might be feasible, but they’d have to be much thinner, since steel’s stiffness is 50× higher; about 3.7 times thinner to retain the same operating force they had in MDF. Plasma has tolerances in the 300-micron range, so if your steel clip's nominal thickness was 1.5 mm, it might come out as 1.2 mm (and have 51% of the desired operating force) or 1.8 mm (and have 173%). So the 920-micron clip width indicated ought to be doable, and its 290 MPa yield stress would give it a tensile strength of 800 N (or as low as 540 N if it came out thin). The steel’s 0.13% elongation at yield is only a fourth the MDF’s 0.45% or so elongation at break, so steel flexures have to be quite a bit thinner in order to work at all.
Anyway, back in MDF-land, the whole clip/slot perforation extends to 13 mm from the center, which is only 8 mm in the case of the corners of the 45° and 315° diagonal slots next to the deepened slot. I’m getting less worried about that, though.
Bending strength of the joint should be good. If you have octagon A and octagon B edge-lapped and B tries to rotate around its axis, it’s pressing against A at almost A’s center (like uh 2 1.5 mm √2 = 4.2 mm from the center I think?) and has one of A’s hooks in it at 25 mm + 12 mm = 37 mm out. It’s not totally easy to predict how that’s going to end up, but let’s say we have a 15 mm lever arm from where the fulcrum ends up being and something like 30 mm² effective surface area; then the surface will start to be damaged at about four or five newton-meters. Hmm, that suggests that maybe the double clips are overkill?
Trying to squish the two octagons against each other may break them more easily. The surface only has a 1.5 mm lever arm at most to resist the 50 mm lever arm where you can apply torque.
The 400 mm × 400 mm area devoted to the 16 100 mm octagons in the starter kit leaves a spare 400 mm × 10 mm area along one side and another 450 mm × 410 mm area for beams. No long beams will fit that way! We could fit long beams if we have a 200 × 400 area with 8 octagons, and then 8 more octagons in a 400 × 200 area to the right, leaving a 610 × 250 area above it into which we pack 10 564.2 × 25 long beams, leaving a 45 × 250 area to its right, and a 200 × 200 area below, or something. So then one short beam can fit vertically on the right. Blegh.
Alternatively the second group of octagons could be 600 × 100, cutting the total octagons to 14, and leaving space for plenty of short beams above it. Actually no none of this makes any sense. I should drag and drop stuff or some shit.