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Things below were written in 2025 and on 08/16/2026 I decided to move to a new page called Read A Proof, because I bought a really good book some days ago. (14/08/2026)
This page will be about Math. Maybe a little bit messy.
Consider a fountain that sprays water with a fixed initial speed \( v \), at an angle \( \theta \) that is uniformly distributed in the interval \( [0, \frac{\pi}{2}] \).
The horizontal landing distance \( R \) of a droplet, as a function of angle \( \theta \), is given by:
$$ R = \frac{v^2}{g} \sin(2\theta) $$
This function reaches its maximum when \( \theta = \frac{\pi}{4} \), which corresponds to the furthest possible landing distance:
$$ R_{\text{max}} = \frac{v^2}{g} $$
We solve for \( \theta \) in terms of \( R \) by the inverse function:
$$ \theta = \frac{1}{2} \arcsin\left( \frac{R}{R_{\text{max}}} \right) $$
Apply the change of variables in probability density:
$$ f_R(R) \cdot dR = f_\theta(\theta) \cdot d\theta $$ That is, $$ f_R(R) = f_\theta(\theta) \cdot \left| \frac{d\theta}{dR} \right| $$
We compute the derivative $$\frac{d\theta}{dR}$$ by the chain rule:
$$ \frac{d\theta}{dR} = \frac{1}{2 R_{\text{max}} \sqrt{1 - \left( \frac{R}{R_{\text{max}}} \right)^2}} $$
Substituting into the formula:
$$ f_R(R) = \frac{2}{\pi} \cdot \frac{1}{2 R_{\text{max}} \sqrt{1 - \left( \frac{R}{R_{\text{max}}} \right)^2}} = \frac{1}{\pi R_{\text{max}} \sqrt{1 - \left( \frac{R}{R_{\text{max}}} \right)^2}} $$
This is to say,
$$f_R(R) \propto \frac{1}{\sqrt{1 - \left( \frac{R}{R_{\max}} \right)^2}}$$
As \( R \to R_{\text{max}} \), the denominator tends to zero. This explains why droplets accumulate most densely at the farthest point.
This is beautiful.
No matter whether you go from the narrow top to the wide middle or from the wide middle to the narrow bottom of a sphere, forming a hemisphere in either case, the surface area remains the same:
$$ A = 2\pi r^2 $$
Seems a bit counterintuitive. It turns out the surface area of a spherical cap is:
$$ A(h) = 2\pi R h $$
This means that the spherical cap area increases linearly with height \( h \). If we divide a full sphere (height from 0 to \( 2R \)) evenly into \( n \) horizontal slices, each with height \( \Delta h \), then the increase in surface area from height \( h \) to \( h + \Delta h \) is the same for all layers.
That is, every spherical zone has the same surface area when the vertical height is divided equally.
It's counterintuitive! The top is steep and short, the middle part is flat and long, but they balance out perfectly. Equal areas, no matter how many slices.