Rain and Wind Vector Simulation
Determining the optimal umbrella angle when wind blows
Theory
When rain is falling vertically and wind starts blowing horizontally, the apparent direction of rain changes due to the vector addition of the rain's velocity and the wind's velocity.
Given in the problem:
Vertical rain velocity (\( v_r \)) = 35 m/s downward
Horizontal wind velocity (\( v_w \)) = 12 m/s (east to west)
Resultant velocity (R):
Magnitude: \( R = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2} = 37 \, \text{m/s} \)
Direction: \( \theta = \tan^{-1}\left(\frac{v_w}{v_r}\right) = \tan^{-1}\left(\frac{12}{35}\right) \approx 19^\circ \) west of vertical
The boy should hold his umbrella at an angle of about 19° with the vertical towards the east (opposite the wind direction) to protect himself from the rain.
Interactive Simulation
Example
Question:
Rain is falling vertically with a speed of \(35\,\mathrm{m\,s^{-1}}\). Winds start blowing after some time with a speed of \(12\,\mathrm{m\,s^{-1}}\) in the east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella?
Solution:
Let the velocity of rain be \(\vec{v}_r\) and the velocity of wind be \(\vec{v}_w\). By vector addition, \[ R = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2}\,\mathrm{m\,s^{-1}} = 37\,\mathrm{m\,s^{-1}} \] The direction \(\theta\) that \(R\) makes with the vertical is \[ \tan\theta = \frac{v_w}{v_r} = \frac{12}{35} = 0.343 \] \[ \theta = \tan^{-1}(0.343) = 19^\circ \] Thus, the boy should hold his umbrella in the vertical plane at an angle of about \(19^\circ\) with the vertical towards the east.



