Hexagonal Honeycombs

When bees build honeycombs, they do so in a pattern of tiled hexagons.  They create the borders of the hexagon with wax, leaving hexagonal holes that they can use to store honey, pollen, or eggs.

But have you ever wondered why bees use hexagons?  Why not use a simpler shape, like squares, triangles, or circles?

The quick answer is that the hexagon is the single tileable polygon that has the least perimeter for a given area.  This is important because bees have to spend a lot of energy to make the wax for the borders of each honeycomb cell.  Less perimeter means less wax, and less wax means less spent energy.

Let’s examine the math behind the claim that the hexagon is the single tileable polygon that has the least perimeter for a given area.  First of all, we must show that there are only three regular polygons that can be tiled: the triangle, the square, and the hexagon.  Secondly, we must show that out of those three polygons, the hexagon is the one with the least perimeter for a fixed area.

Tileable Regular Polygons

First of all, we need to show that there are only three regular polygons that can be tiled.  In all regular polygons with n sides, the interior angle θ is θ = ^{(n – 2)180°}/_n.  So for a triangle (the polygon with the least amount of sides possible), n = 3 and θ = ^{(3 – 2)180°}/₃ = 60°; for a square, n = 4 and θ = ^{(4 – 2)180°}/₄ = 90°;  for a pentagon, n = 5 and θ = ^{(5 – 2)180°}/₅ = 108°; for a hexagon, n = 6 and θ = ^{(6 – 2)180°}/₆ = 120°; for a heptagon, n = 7 and θ = ^{(7 – 2)180°}/₇ = 128⁴/₇°; for a octagon, n = 8 and θ = ^{(8 – 2)180°}/₈ = 135°; and so on.  We can see that the values of the interior angles of a regular polygon are greater than or equal to 60° (for the triangle) but less than 180° (which represents a straight line, because otherwise it won’t close to make a polygon) and so 60° ≤ θ < 180°.

However, in order for a regular polygon to be tileable, its interior angle must also divide evenly into 360°.  The largest factor of 360° is 360° ÷ 1 = 360°, which is too big to be an interior angle of a regular polygon.  The second largest factor of 360° is 360° ÷ 2 = 180°, which is also too big.  However, the third largest factor of 360° is 360° ÷ 3 = 120°, which is the interior angle of a regular hexagon; the fourth largest factor is 360° is 360° ÷ 4 = 90°, which is the interior angle of a square; the fifth largest factor of 360° is 360° ÷ 5 = 72°, which is not the interior angle of any regular polygon; and the sixth largest factor of 360° is 360° ÷ 6 = 60°, which is the interior angle of an equilateral triangle.  After this, the factors of 360° are less than 60°, which are too small to be an interior angle of any regular polygon.  So the triangle, square, and hexagon are the only regular polygons that can be tiled.

Least Perimeter

Secondly, we must show that out of the three tileable regular polygons of the triangle, square, and hexagon, the hexagon is the one with the least perimeter for a fixed area.  Consider a regular polygon of n sides inscribed in a circle with a radius of r.   The polygon can be divided into n triangular slices, and each of those slices can be bisected at each central angle, making 2n symmetrical right triangles with an angle of ^π/_n and a hypotenuse of r.  The two legs of each right triangle are then r sin (^π/_n) and r cos (^π/_n), which makes the area of one of those right triangles ½r²sin(^π/_n)cos(^π/_n), and making the area of the whole regular polygon 2n times this or A = nr²sin(^π/_n)cos (^π/_n).  Also, since the opposite leg is r sin (^π/_n), 2 times this would give the length of one side of the whole regular polygon, and n times that would give the perimeter of the whole regular polygon or P = 2nr sin (^π/_n).

Solving the area equation for r gives us

and substituting this into the perimeter equation gives us

which simplifies to P = 2√A·√n·√tan(^π/_n).

So for a triangle, n = 3 and P = 2√A·√3·√tan(^π/₃) = 2√(3√3)√A ≈ 4.559√A.  For a square, n = 4 and P = 2√A·√4·√tan(^π/₄) = 4√A.  And for a hexagon, n = 6 and P = 2√A·√6·√tan(^π/₆) = 2√(2√3)√A ≈ 3.722√A.

Shape	Perimeter
triangle	4.559·√A
square	4.000·√A
hexagon	3.722·√A

Therefore, the hexagon has the least perimeter for a fixed area.

Wax Width

We have shown that the hexagon is the single tileable polygon that has the least perimeter for any given area.  But in a honeycomb grid, the perimeter of each cell is actually a wax border with its own width that is shared by adjacent cells, and this may affect our assertion that the hexagon is the most efficient shape.  So to be thorough, we need to consider the wax to honey ratio for each tiled shape.  We will consider the potential candidates of the triangle, square, hexagon, and circle.

Triangular Honeycombs

Let’s say there is a type of bee called the Triangle Bee that builds its honeycombs with equilateral triangles.  The Triangle Bees make wax walls that have a width of w that will separate each honey cell with an area of A and sides of s, and by the nature of the triangular pattern, the wax walls will meet in hexagon shapes.  The wax to honey ratio for the whole honeycomb would then be the same as one individual triangular tile, as colored below.

The wax walls for one tile consist of 3 sixths of a hexagon and 3 rectangles, and so its area W would be W = 3·¹/₆·^3√3/₂w² + 3·½sw or W = ^3√3/₄w² + ³/₂sw.  Since the area of an equilateral triangle is A = ^√3/₄s², solving for s would give us s = ^{2√(3√3)√A}/₃, and substituting back into W would give us W = ^3√3/₄w² + ³/₂(^{2√(3√3)√A}/₃)w or W = ^3√3/₄w² + √(3√3)√A·w.  The wax to honey ratio for the Triangular Bee is then (^3√3/₄w² + √(3√3)√A·w) / A ≈ (1.299w² + 2.280√A·w) / A.

Square Honeycombs

Now let’s say there is a type of bee called the Square Bee that builds its honeycombs with squares.  The Square Bees make wax walls that have a width of w that will separate each honey cell with an area of A and sides of s, and by the nature of the square pattern, the wax walls will meet in square shapes.  The wax to honey ratio for the whole honeycomb would then be the same as one individual square tile, as colored below.

The wax walls for one tile consist of 4 quarters of a square and 4 rectangles, and so its area W would be W = 4·¼·w² + 4·½sw or W = w² + 2sw.  Since the area of a square is A = s², solving for s would give us s = √A, and substituting back into W would give us W = w² + 2√A·w.  The wax to honey ratio for the Square Bee is then (w² + 2√A·w) / A.

Hexagonal Honeycombs

Now let’s say there is a type of bee called the Hexagonal Bee that builds its honeycombs with hexagons (like regular bees).  The Hexagonal Bees make wax walls that have a width of w that will separate each honey cell with an area of A and sides of s, and by the nature of the hexagonal pattern, the wax walls will meet in triangular shapes.  The wax to honey ratio for the whole honeycomb would then be the same as one individual hexagonal tile, as colored below.

The wax walls for one tile consist of 6 thirds of an equilateral triangle and 6 rectangles, and so its area W would be W = 6·¹/₃·^√3/₄w² + 6·½sw or W = ^√3/₂w² + 3sw.  Since the area of a regular hexagon is A = ^3√3/₂s², solving for s would give us s = ^{√(2√3)√A}/₃, and substituting back into W would give us W = ^√3/₂w² + 3(^{√(2√3)√A}/₃)w or W = ^√3/₂w² + √(2√3)√A·w.  The wax to honey ratio for the Triangular Bee is then (^√3/₂w² + √(2√3)√A·w) / A ≈ (0.866w² + 1.861√A·w) / A.

Circular Honeycombs

Finally, let’s say there is a type of bee called the Circular Bee that builds its honeycombs with circles.  The Circular Bees make wax walls that have at least a width of w that will separate each circular honey cell with an area of A and a radius of r.  The wax to honey ratio for the whole honeycomb would then be the same as one individual hexagonal tile, as colored below.

The area of the wax walls for one tile would be equivalent to the whole hexagon minus the circle, and so its area would be W = 2√3(½w + r)² – A.  Since the area of a circle is A = πr², solving for r would give us r = ^√π√A/_π, and substituting back into W would give us W = 2√3(½w + ^√π√A/_π)² – A = 2√3(¼w² + ^√π√A/_πw + ^A/_π) – A = ^√3/₂w² + ^{2√3√π√A}/_πw + (^2√3/_π – 1)A.  The wax to honey ratio for the Circular Bee is then (^√3/₂w² + ^{2√3√π√A}/_πw + (^2√3/_π – 1)A) / A ≈ (0.866w² + 1.954√A·w + 0.103A) / A.

Summary

The wax to honey ratio with respect to wax width and area are summarized for each shape below:

Shape	Wax to Honey Ratio
triangle	(1.299w2 + 2.280√A·w + 0.000A) / A
square	(1.000w2 + 2.000√A·w + 0.000A) / A
circle	(0.866w2 + 1.954√A·w + 0.103A) / A
hexagon	(0.866w2 + 1.861√A·w + 0.000A) / A

Since both w and A must be positive, the shape with the smallest wax to honey ratio coefficients would have to have the least wax to honey ratio, which is the hexagon.  Therefore, the hexagon is the most efficient shape to use for building a honeycomb.

Real honeybees build honeycomb cells that are an average of 4.85 mm wide (http://www.bushfarms.com/beesnaturalcell.htm), which gives a honey area of 81.484 mm², and with wax walls that have an average thickness of 0.5 mm (http://keepingbee.org/bee-honeycomb/).  Using these values, we can use the formulas above to calculate the numerical wax to honey ratio for each shape:

Shape	Ratio
circle	21.357%
triangle	13.025%
square	11.385%
hexagon	10.575%

Which means that the average honeycomb (made up of hexagons) has a 10.575% wax to honey ratio.

Conclusion

There are generally two opposing trains of thought concerning the fact that all bees happen to use the most efficient shape to build their honeycombs.  Evolutionists attribute the bees’ efficiency to natural selection.  They would say that at some point there may have been bees that tried to make square or triangular honeycombs, but they were not as efficient as the bees that made hexagonal honeycombs and were eliminated by the rules of the survival of the fittest.  On the other hand, creationists attribute the bees’ efficiency to Intelligent Design.  They would say that the instinct to build hexagonal honeycombs was put there by a God who not only created but also upholds all the geometrical laws of the universe.

Unfortunately, there are a few problems with using the theory of natural selection to explain why bees make their honeycombs in a hexagonal pattern rather than some other pattern.  First of all, although we have mathematically proved that the hexagon is the most efficient shape for a honeycomb of a fixed area, the square is not that far behind, and the difference of efficiency would be too small for natural selection to take place.  In fact, given that a real honeycomb has hexagon widths that vary from 4.6 mm to 5.1 mm (http://www.bushfarms.com/beesnaturalcell.htm), which would make honey cell areas vary from 73.300 mm² and 90.101 mm², a high end honey area made up of squares is actually more efficient than a low end honey area made up of hexagons! (A square honeycomb with w = 0.5 mm and A = 90.101 mm² would have a wax to honey ratio of 10.812%, whereas a hexagonal honeycomb with w = 0.5 mm and A = 73.300 mm² would have a wax to honey ratio of 11.165%.  Note that this does not contradict our above assertion because the areas are not the same.)  According to evolutionary natural selection, a mutation of bees that make large square honey cells should eliminate bees that make small hexagonal honey cells, but this of course has not happened.  Which brings us to the second problem of using the theory of natural selection to explain why bees make hexagonal honeycombs: out of the millions of beekeepers worldwide, and over three thousand years of beekeeping, there is not one recorded instance of a mutated colony of bees trying to make a honeycomb with something other than a hexagonal pattern.  The theory of evolution relies on these mutations to take place, both now and in the past, but there is just no evidence for this in bees.

If the reason bees make their honeycombs in a hexagonal pattern rather than some other pattern is not because of natural selection, then it must be because of some innate instinct.  But where did that instinct come from?  The only plausible explanation is that it came from an Intelligent Designer.  Bees build their honeycombs in the efficient hexagonal pattern because of an instinct that was put there by the same God who created the geometrical laws of the universe.

Astroids

Imagine a ladder falls down so that the top of the ladder slides down the wall and the base of the ladder slides along the ground away from the wall.  If you were to record the falling ladder from the side and overlay each frame onto a single picture, the ladders from the different frames would form a curve.

Falling Ladder

At first glance it may appear that this curve is circular.  After all, the top of the curve is measured when the ladder is leaning straight up against the wall, and the bottom of the curve is measured when the ladder is lying flat on the ground, which would appear to show a radius that is the same length as the ladder.  However, if we compare our falling ladder curve with a curve from a circle, we can see that the circular curve is slightly steeper.

Falling Ladder Curve vs. Circular Curve

Obviously, some more calculations will be needed to find an equation for this curve.  We can start by calling the wall the y-axis and the ground the x-axis.  Then if we call the length of the ladder w, and the angle between the ladder and the ground θ, then the distance from the top of the ladder to the origin (along the wall) is w sin θ and the distance from the bottom of the ladder to the origin (along the ground) is w cos θ.  We can also label a point (x, y) somewhere on the ladder.

There are then two similar triangles that we can set in proportion to each other: the large triangle with sides w sin θ and w cos θ, the smaller triangle with sides y and w cos θ – x.  Therefore:

y / w cos θ – x = w sin θ / w cos θ	(similar triangles)
yw cos θ = w sin θ (w cos θ – x)	(cross multiply)
yw cos θ = w2 sin θ cos θ – wx sin θ	(distribute)
y = w sin θ – x tan θ	(divide by w cos θ)

This means that at any vertical line at x, the ladder intersects it at a height of y = w sin θ – x tan θ.  The maximum height of this intersection would be when the derivative ^dy/_dθ = w cos θ – x sec²θ equals zero.  Therefore:

dy/dθ = w cos θ – x sec2θ = 0	(derivative equal to zero)
w cos θ – x/cos2θ = 0	(sec θ = 1/cos θ)
w cos3θ – x = 0	(multiply by cos2θ)
w cos3θ = x	(add x)
cos3θ = x/w	(divide by w)
cos θ = 3√x/w	(cube root)

If cos θ = ³√^x/_w, we can use trigonometric identities to show that sin θ = √(1 – cos²θ) = √(1 – (³√^x/_w)²) and tan θ = sin θ / cos θ = √(1 – (³√^x/_w)²) / ³√^x/_w = ³√^w/_x√(1 – (³√^x/_w)²).  Combining this with our previous equation y = w sin θ – x tan θ, we get:

y = w sin θ – x tan θ	(previous equation)
y = w√(1 – (3√x/w)2) – x 3√w/x√(1 – (3√x/w)2)	(substitute sin θ and tan θ)
y = w√(1 – x2/3w-2/3) – xw1/3x-1/3√(1 – x2/3w-2/3)	(rewrite as exponents)
y = w√(1 – x2/3w-2/3) – x2/3w1/3√(1 – x2/3w-2/3)	(x1x-1/3 = x2/3)
y = w2/3w1/3√(1 – x2/3w-2/3) – x2/3w1/3√(1 – x2/3w-2/3)	(w = w2/3w1/3)
y = w2/3√(w2/3(1 – x2/3w-2/3)) – x2/3√(w2/3(1 – x2/3w-2/3))	(w1/3 = √(w2/3))
y = w2/3√(w2/3 – x2/3) – x2/3√(w2/3 – x2/3)	(distribute, and w2/3w-2/3 = 1)
y = (w2/3 – x2/3)√(w2/3 – x2/3)	(combine like terms)
y = (w2/3 – x2/3)(w2/3 – x2/3)1/2	(rewrite as exponents)
y = (w2/3 – x2/3)3/2	(add exponents)
y2/3 = w2/3 – x2/3	(law of exponents)
x2/3 + y2/3 = w2/3	(add x2/3)

Therefore, the path of the curve that a falling ladder makes is x^2/3 + y^2/3 = w^2/3, where w is the length of the ladder.  In mathematics, this curve is called an “astroid”, which is derived from the Greek word for “star”, because when you graph it in all four quadrants you get a star shape.

Astroid

When you were kid, you may have played with a Spirograph kit, where you can create different designs by rotating geared circles inside another geared circle.

Spirograph Kit

In the same way, an astroid can also be created by following the path of a point on the edge of a circle that is rotated inside another circle that is four times its side.

Creating an Astroid with Circles

To prove this, we can call the radius of the big circle w, the radius of the little circle r, the angle between the x-axis and the segment that contains the two centers of the circles θ, and the angle between segment that contains the two centers of the circles and the segment that contains the center of the little circle to the moving point α. 

Since the big circle is four times the size of the little circle, the center of the little circle is always w – r = 4r – r = 3r away from the origin, so the coordinates of the center of the little circle can be expressed as (3r cos θ, 3r sin θ).  Also since the big circle is four times the size of the little circle, α = 4θ, so the standard angle for (x, y) with respect to the little circle would be 2π – (4θ – θ) = 2π – 3θ.  Putting these two details together, x = 3r cos θ + r cos (2π – 3θ) and y = 3r sin θ + r sin (2π – 3θ).  Simplifying:

x = 3r cos θ + r cos (2π – 3θ)	(x component)
x = 3r cos θ + r cos 3θ	(cos (2π – x) = cos x)
x = 3r cos θ + r (4 cos3θ – 3 cos θ)	(cos 3x = 4 cos3x – 3 cos x)
x = 3r cos θ + 4r cos3θ – 3r cos θ	(distribute)
x = 4r cos3θ	(simplify)
x = w cos3θ	(w = 4r)
x/w = cos3θ	(divide by w)
cos θ = 3√x/w	(cube root both sides)

and:

y = 3r sin θ + r sin (2π – 3θ)	(y component)
y = 3r sin θ – r sin 3θ	(sin (2π – x) = -sin x)
y = 3r sin θ – r (3 sin θ – 4 sin3θ)	(sin 3x = 3 sin x – 4 sin3x)
y = 3r sin θ – 3r sin θ + 4r sin3θ	(distribute)
y = 4r sin3θ	(simplify)
y = w sin3θ	(w = 4r)
y/w = sin3θ	(divide by w)
sin θ = 3√y/w	(cube root both sides)

Finally, substituting cos θ = ³√^x/_w and sin θ = ³√^y/_w into the trigonometric identity sin²θ + cos²θ = 1:

sin2θ + cos2θ = 1	(trigonometric identity)
(3√x/w)2 + (3√y/w)2 = 1	(substitute cos θ and sin θ)
(x/w)2/3 + (y/w)2/3 = 1	(rewrite as exponents)
x2/3 + y2/3 = w2/3	(multiply by w2/3)

which is the equation of an astroid.

There are two remarkable points that should be made about astroids.  First of all, it is amazing that the same curve can be used to describe two completely unrelated situations: a curve made by a falling ladder and a Spirograph design made between two circles.  Secondly, this is yet another geometric formula in the form of xⁿ + yⁿ = zⁿ, where n = ²/₃ for the astroid, n = 2 for Pythagorean’s Theorem, and n = -½ for the radii of three kissing circles and a line (see here).  As mentioned in a previous article, we must admit that so many different formulas of the same form does not describe a chaotic universe of random chance, but rather an orderly universe of intelligent design.

Similar Formulas and Intelligent Design

When you study science or geometry, you come across several formulas in the form of a = kx and b = ½kx².  For example, the velocity of an object starting at rest with a constant velocity can be determined by multiplying acceleration and time, or v = at, and the distance of the same object can be found by taking half of its acceleration multiplied by the time squared, or d = ½at².  For another example, the momentum of an object can be determined by multiplying its mass times its velocity, or p = mv, and the kinetic energy of an object can be found by taking half of its mass times its velocity squared, or E = ½mv².

Sometimes the equations need to be manipulated to be in the form of a = kx and b = ½kx².  For example, the circumference of a circle is C = 2πr and the area of a circle is A = πr², but if we substitute τ = 2π, then C = τr and A = ½τr².  For another example, the area of a triangle is A = ½bh, but if height is defined as a ratio of the base, then h = kb and A = ½kb².

Other equations in the form of a = kx and b = ½kx² include angular velocity and displacement, angular momentum and rotational energy, force and potential energy of a spring, electric flux and energy density, and electric charge and electric energy:

a = kx	b = ½kx2
Velocity v = at	Distance d = ½at2
Angular Velocity ω = αt	Angular displacement θ = ½αt2
Momentum p = mv	Kinetic Energy E = ½mv2
Angular Momentum L = Iω	Rotational Energy K = ½Iω2
Force of a Spring F = kx	Potential Energy of a Spring E = ½kx2
Electric Flux Density D = εE	Electric Energy Density Q = ½εE2
Electric Charge Q = CV	Electric Charge Energy E = ½CV2
Circumference of  a Circle C = 2πr = τr	Area of a Circle A = πr2 = ½τr2
Height of a Triangle h = kb	Area of a Triangle A = ½bh = ½kb2

In fact, if we lift the requirement that k must be a constant, we can include even more formulas to the form of b = ½kx².  Another pair of similar yet unrelated formulas is gravitational force (F = ^Gm1m2/_r^2) and electric force (F = ^kq1q2/_d^2).  Gravitational force can be rearranged to ½Gm₁m₂ = b = ½Fr² and electric force can be rearranged to ½kq₁q₂ = b = ½Fd².  Even Einstein’s famous theory of relativity E = mc² can be manipulated to ½E = ½mc².

Those familiar with calculus may recognize that the derivative of b = ½kx² is a = kx, which means a is a rate of b with respect to x.  For example, since d = ½at² and v = at, velocity is a rate of distance with respect to time.  A good way to visualize this is to use the two geometry formulas of the circle and triangle.  Since A = ½τr² and C = τr for a circle, the circumference is a rate of its area with respect to its radius, which means that the area of a circle is the sum of all the circumferences it contains. 

Area of a Circle = Sum of the Circumferences

Since A = ½kb² and h = kb for a triangle, the height is a rate of its area with respect to its base, which means that the area of a triangle is the sum of all the heights it contains.

Area of a Triangle = Sum of the Heights

Although calculus can be used to explain the relationship between b = ½kx² and a = kx, it cannot explain the high frequency of which these formulas appear in nature.  If there were some relationship between all these formulas we might be tempted to attribute the high frequency to a coincidence, but these equations appear in completely unrelated and different fields of math and science, from velocity to circles, from springs to electric fields, and from momentum to triangles.  So many different formulas in the form of b = ½kx² and a = kx cannot be a result of random chance, but rather must be the result of an Intelligent Designer of an orderly universe.