Rigid Polygons from a Construction Set

What’s going on

A construction set has flat metal bars with a hole every unit. Two bars bolted through a hole can still turn, like a hinge. Which frames hold their shape?

A triangle is rigid. Once you know its three sides, its shape is fixed, so no angle can change. A square is not. Drag a top corner in the scene above: the square leans into a diamond, and every bar keeps its length.

Count the freedoms

There is a quick way to predict wobbling. Each joint can move left–right and up–down: two freedoms per joint. Three of these freedoms just slide or turn the whole frame without bending it. So a frame with j joints has 2 × j − 3 freedoms that could change its shape.

Each bar fixes one distance, so it can remove at most one freedom.

• Triangle: 2 × 3 − 3 = 3 freedoms, 3 bars. Nothing left: rigid.
• Square: 2 × 4 − 3 = 5 freedoms, 4 bars. One left: it can flex in one way.
• A polygon with n sides has 2n − 3 freedoms and n bars, so it bends in n − 3 independent ways. A hexagon bends in three.

The panel beside the frame does this count, and also checks that no bar merely repeats a condition that is already there.

Whole numbers only

The obvious fix for a square is a diagonal. Choose Square + diagonal: it is rigid. But look at the red label. A square with side 5 has a diagonal of 5 × √2 ≈ 7.07, and no construction-set bar is 7.07 holes long. In fact √2 is not a fraction, so no square with a whole-number side has a whole-number diagonal.

The way out: a bolt can go through any hole, not just the end of a bar. Choose Square + 3-4-5 brace. A bar of length 5 joins the hole 3 units along the bottom to the hole 4 units up the side. Since 3² + 4² = 5², that little triangle has a right angle, and it holds the corner at exactly 90°. The square is locked, using whole numbers only.

A bolt in a middle hole counts double: the joint must stay at one spot along the bar, which is two conditions.

Hexagons

A regular hexagon needs three freedoms removed. Two tricks:

• Hexagon + 2 long diagonals: two bars of length 6 (twice the side) bolted together at their middle holes. The bottom and top triangles, from the centre to two corners, are equilateral and rigid; the two remaining corners are then pinned in place. Two extra bars are enough, fewer than n − 3 = 3.
• Hexagon + three 3-5-7 braces: side 5, and a bar of 7 from the third hole of one side to the end of the next. By the cosine rule, 7² = 3² + 5² + 3 × 5 means the corner is exactly 120°, the angle of a regular hexagon. Three braces at every other corner lock it.

Use Add bar, Remove bar and Wiggle to invent and test your own frames.

Try it yourself

1. How many diagonals?

You may only add bars between corners, and you may use bars of any length. How many extra bars does a polygon with 10 sides need to become rigid?

Answer

Seven. It has 2 × 10 − 3 = 17 freedoms and 10 sides, so 7 freedoms are left, and each bar removes at most one. Seven diagonals from one corner cut it into eight triangles, which is rigid. In general a polygon with n sides needs n − 3 corner-to-corner bars.

2. A bigger square

Your square has sides of 10 holes. Find a single whole-number brace that makes it rigid.

Answer

Join the hole 6 units along one side to the hole 8 units along the other side, starting from the same corner, with a bar of 10. Since 6² + 8² = 10², the corner is a right angle. The 3-4-5 brace from the scene works too.

3. Why does the shared bolt matter?

Take the hexagon with two long diagonals, but bolt them only at their ends, not at the middle hole. Is it still rigid?

Answer

No: 6 joints give 2 × 6 − 3 = 9 freedoms, and 8 bars leave one, so it flexes. The middle bolt adds a joint (2 freedoms) but four conditions, since it sits at the middle of both bars. That net gain locks the frame. Check it: choose Hexagon, add the two diagonals with Add bar, press Wiggle.

Go deeper

• James Clerk Maxwell described this counting of freedoms and bars for frames in 1864. In 1970 Gerard Laman proved that for frames in general position the count tells the whole story: 2 × j − 3 well-spread bars make a flat frame rigid, as long as no part of it has more bars than that part needs.
• The physicist Gerard ’t Hooft, Nobel Prize 1999, wrote a series of puzzles called Meccano Math about building exact angles with construction-set bars.
• In 1876 Alfred Kempe showed that hinged bar linkages can draw a piece of any algebraic curve. His lecture How to Draw a Straight Line is still a delightful read.