Rain Accumulation Simulator

The Problem

Rain falls vertically at terminal velocity $v_y$, with spatial density $\rho$ drops per m³. A person — modeled as a rectangular prism with human proportions — walks forward at $\dot{x}$. Does speed matter? Which surfaces capture the rain?

Two surfaces face the rain: the TOP $A_t$ and the FRONT $A_f$. The total accumulation is:

$$N(t) = \rho\,t\,\bigl(A_t v_y + A_f \dot{x}\bigr)$$

Body Dimensions (m)

Simulation Controls

Walking speed $\dot{x}$1.5 m/s

Rain terminal velocity $v_y$9.0 m/s

Rain density $\rho$200 drops/m³

TOP Horizontal Surface

In the ground frame, rain passes through the horizontal area $A_t$ at a flux $\Phi = \rho v_y$ drops per m² per second. Accumulation over time $t$:

$$N_t(t) = \Phi\, A_t\, t = \rho\, v_y\, A_t\, t$$

Counter-intuition: horizontal velocity $\dot{x}$ does not appear. The top sweeps into new rain at exactly the rate it leaves old rain.

Vector Derivation

Relative rain velocity $\vec{v}_{rel} = -\dot{x}\hat{i} - v_y\hat{k}$. Top normal $\vec{A} = A_t\hat{k}$.

$$\text{Rate} = \rho(\dot{x}\hat{i}+v_y\hat{k})\cdot A_t\hat{k} = \rho v_y A_t$$

The $\hat{i}$ term vanishes because the top's normal has zero horizontal component.

FRONT Vertical Surface

The frontal silhouette $A_f$ sweeps horizontally at speed $\dot{x}$, capturing every drop in its wake:

$$V_{swept} = A_f\,\dot{x}\,t \;\Rightarrow\; N_f(t) = \rho\, A_f\, \dot{x}\, t$$

Since $d = \dot{x}t$, this simplifies to $N_f = \rho A_f d$ — depending only on distance traveled, not on speed. Running does not reduce frontal accumulation.

Vector Derivation

Front normal $\vec{A}_f = A_f\hat{i}$.

$$\text{Rate} = \rho(\dot{x}\hat{i}+v_y\hat{k})\cdot A_f\hat{i} = \rho\,\dot{x}\,A_f$$

The $\hat{k}$ (vertical) term vanishes — vertical rain speed is irrelevant to the front.

Combined Accumulation

Summing both contributions gives the full expression for a moving rectangular prism:

$$N(t) = \rho\, t\,\bigl(A_t v_y + A_f \dot{x}\bigr)$$

Using flux $\Phi = \rho v_y$:

$$N(t) = \Phi\, t\left( A_t + A_f\,\frac{\dot{x}}{v_y}\right)$$

Unified Vector Form

Total outward area $\vec{A}_{tot} = A_f\hat{i} + A_t\hat{k}$. The single dot product:

$$N(t) = \rho\, t\,(\dot{x}\hat{i}+v_y\hat{k})\cdot(A_f\hat{i}+A_t\hat{k})$$

Each surface pairs naturally with the velocity component aligned to its normal.

Fixed-Distance Analysis

For a fixed distance $d$, exposure time is $t = d/\dot{x}$. Substituting:

$$N(d) = \rho\, d\left(\frac{A_t v_y}{\dot{x}} + A_f\right)$$

The front term is independent of speed. Only the top term decreases as $\dot{x}$ increases. Conclusion: faster is drier — but you asymptotically approach $\rho A_f d$, never zero.

Total Drops vs Walking Speed

Top

Front

Total

$\dot{x}^{*}$

Current $\dot{x}$

Distance traveled $d$50 m

The Role of $v_y$ in Optimal Speed

Strictly speaking, in no-wind rain, there is no true optimum — $N(\dot{x})$ over fixed distance decreases monotonically. But a characteristic "diminishing-returns" speed emerges where top and front accumulation rates are equal:

$$\dot{x}^{*} \;=\; v_y \cdot \frac{A_t}{A_f}$$

At $\dot{x} = \dot{x}^{*}$ the total is exactly twice the infinite-speed asymptote $N_\infty = \rho A_f d$ — half top, half front. Below $\dot{x}^{*}$ the top dominates. Above, you approach $N_\infty$ asymptotically.

Linear Proportionality to $v_y$

$\dot{x}^{*}$ scales linearly with rain's terminal velocity. Faster rain ⇒ higher critical speed.

$$\dot{x}^{*} = \ldots$$

Curves for Varying $v_y$

$v_y$=4 (light)

$v_y$=7 (medium)

$v_y$=10 (heavy)

$v_y$=14 (downpour)

Critical Speed by Rain Intensity

Rain

$v_y$ (m/s)

$\dot{x}^{*}$ (m/s)

Drizzle

—

Medium

—

Heavy

—

Downpour

—

Beautifully, for human proportions $(A_t/A_f \approx 0.14)$, $\dot{x}^{*}$ falls in the 0.5 – 2 m/s range — essentially walking speed. That's why running helps far less than intuition suggests: a brisk walk already pushes you past the knee of the curve.