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.