Have you ever seen a bicycle with square wheels? It is generally believed that only circular wheels can make our vehicles move forward smoothly, but this is only true for flat roads. Who says the road must be flat? As long as an appropriate road is paved, a bicycle with square wheels can still travel smoothly! In this article, let us pave a road for the square-wheeled bicycle.
In fact, square-wheeled bicycles are no longer a novelty; they have already appeared in many science and technology museums. As seen in the images, its special track is composed of many arc segments, where the length of each arc is equal to the side length of the square. When the wheel moves forward, the square remains tangent to the arc (ensuring no slipping). What kind of curve is the shape of such a road? Fortunately, it is not very complex, and surprisingly, it is the "catenary" we have studied before! It turns out that to design such a curved track, one does not need a sophisticated designer; one only needs to hold a chain and let it hang freely...
As shown in the figure, these are two special positions of the square wheel. Let the side length of the square be 2. From the figure, it is not difficult to see that if the x-axis is taken as the ground, the distance between the axle (R) and the ground is \sqrt{2}, and the distance between the middle vertex and the ground is \sqrt{2}-1.
Mark the points as shown below:
Let y=y(x) be the track equation, where I(x,y) is a point of tangency between the square wheel and the "road". According to the relationship between the road and the wheel, the length of segment AI must be equal to the length of arc KI. Denoting this length as s, it can be expressed using calculus notation as:
s=\int_{x_0}^x \sqrt{dx^2+dy^2}=\int_{x_0}^x \sqrt{1+\dot{y}^2}dx (x_0 is a zero point at the left end)
At the same time, we have the following relationships:
\begin{aligned}AI=s, \quad MI=1-s, \quad RM=1 \\ RM \cos\theta + MI \sin\theta + y = OP\end{aligned}
That is: \cos\theta + (1-s)\sin\theta + y = \sqrt{2}
Solving for s and substituting \tan\theta = \dot{y}, we get:
s = 1 + \dot{y}^{-1} + y\sqrt{1+\dot{y}^{-2}} - \sqrt{2} \cdot \sqrt{1+\dot{y}^{-2}}
Equating the two expressions for s, we obtain the integral equation:
\int_{x_0}^x \sqrt{1+\dot{y}^2}dx = 1 + \dot{y}^{-1} + y\sqrt{1+\dot{y}^{-2}} - \sqrt{2} \cdot \sqrt{1+\dot{y}^{-2}}
To transform this into a common differential equation, take the derivative of both sides.
The derivative of the left side becomes: \sqrt{1+\dot{y}^2}
The derivative of the right side becomes: -\dot{y}^{-2}\ddot{y} + (y-\sqrt{2}) \cdot \frac{-\dot{y}^{-2}\ddot{y}}{\sqrt{1+\dot{y}^2}} + \sqrt{1+\dot{y}^2}
After cancellation, it becomes: \sqrt{2}-y = \sqrt{1+\dot{y}^2}
Which can be simplified to: \frac{dx}{dy} = \frac{1}{\sqrt{(\sqrt{2}-y)^2-1}}
That is: \begin{aligned}-x &= \int \frac{1}{\sqrt{(\sqrt{2}-y)^2-1}} d(\sqrt{2}-y) \\ &= \operatorname{arccosh}(\sqrt{2}-y) + C\end{aligned}
Rewriting this as: y = -\cosh(-C-x) + \sqrt{2} = -\cosh(C+x) + \sqrt{2} (\cosh x is the hyperbolic cosine function, \cosh x = \frac{e^x+e^{-x}}{2})
According to the initial conditions, we can conclude that C=0. Thus, the final track equation is: y = -\cosh x + \sqrt{2} (Taking the part above the x-axis)
This is a catenary. The magical catenary!
After solving this problem, a natural extension arises: For any regular n-sided polygon wheel, what should the shape of the road be to ensure smooth travel?
The answer is surprising—it does not become much more complicated; it is still a catenary! As shown in the figure, let the side length of the regular n-gon be 2a and its height (apothem) be h. Then the final answer will be:
y = -h \cosh\left(\frac{x}{h}\right) + \sqrt{a^2+h^2}
If we take h=1 as the unit length, and given a = h \cdot \tan(\frac{\pi}{n}), we get:
y = -\cosh x + \sqrt{1 + \tan^2\left(\frac{\pi}{n}\right)}
This is simply the track of the square wheel shifted vertically!
When n approaches infinity, the regular n-gon becomes a circle. Common sense tells us that the road surface should be a straight line. However, at this point, the road equation becomes y = -\cosh x + 1. Readers might wonder why it is not a straight line. Those with such doubts should remember that we mentioned the track equation only takes the part above the x-axis (y \geq 0) and is then spliced horizontally. For this track equation, only the point (0,0) satisfies y \geq 0. Splicing such points infinitely results in a straight line!
Thus, our journey of paving the way for the square-wheeled bicycle is complete! This result once again makes me marvel at the mysteries of nature. The shape of a chain hanging naturally in a gravitational field actually paves the way for a square-wheeled bicycle. There seems to be a mysterious force behind the scenes, connecting things that appear completely unrelated. This is perhaps the most exciting part of mathematics.
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