On a still, rainy day the rain at a bus stop falls straight down. Board the bus, and the streaks on the side window slant. The faster the bus goes, the more they slant, and they always slope back from the front of the bus. Nothing has happened to the rain. Only the observer has started to move.
The rain as a moving person sees it
Let the rain fall with velocity \(\vec v_r\) relative to the ground, and let a person move with velocity \(\vec v_m\). The velocity of the rain relative to the person is the difference of the two, by the rule of the first lesson. This is the apparent velocity of the rain, the velocity the person sees and feels.
\[ \vec v_{rm} = \vec v_r - \vec v_m = \vec v_r + (-\vec v_m) \]
Take rain that falls vertically at speed \(v_r\), and a person who walks horizontally at \(v_m\). Reversing the person's velocity adds a horizontal part \(v_m\) that points backward, towards the person. The apparent velocity is the hypotenuse of a right triangle. So the rain seems to come from the front, slanting down at an angle \(\theta\) to the vertical.
Where to point the umbrella
An umbrella keeps a person dry when its shaft points along the rain that reaches the person. That rain is the apparent rain, not the rain seen from the ground. So the umbrella must tilt forward, in the direction of walking, by the angle \(\theta\) of the apparent rain.
Rain falls vertically. A person starts to run with an open umbrella. How should the umbrella be held to keep the person dry?
Fig 3.1 moves with the person, so the person stays in the middle and the rain streams past. The dry region behind the umbrella always runs along the streaks. With the umbrella upright during a run, that dry region slants behind the runner, and the legs are wet. Tilted forward by \(37°\), the dry region covers the whole runner. The triangle on the right is \(\vec v_r + (-\vec v_m) = \vec v_{rm}\), drawn to scale.
Rain falls vertically at \(8\,\text{m/s}\). A student runs to the hostel at \(6\,\text{m/s}\). At what angle to the vertical should the umbrella be held, and at what speed does the rain meet the student?
\(\vec v_r = -8\,\hat j\), \(\vec v_m = 6\,\hat i\), with \(\hat i\) the direction of running and \(\hat j\) upward.
\(\vec v_{rm} = \vec v_r - \vec v_m = -6\,\hat i - 8\,\hat j\).
\(\tan\theta = 6/8 = 0.75\), so \(\theta = 37°\). \(v_{rm} = \sqrt{36 + 64} = 10\,\text{m/s}\).
The umbrella should lean \(37°\) forward from the vertical. The rain meets the student at \(10\,\text{m/s}\), faster than it falls, which is why rain stings the face of a runner.
Rain falls vertically at \(12\,\text{m/s}\). A cyclist rides at \(5\,\text{m/s}\). At what speed does the rain meet the cyclist, in m/s?
Rain falls vertically. From a train running at \(20\,\text{m/s}\), the streaks on the window make \(60°\) with the vertical. Find the speed of the rain relative to the ground, in m/s.
Rain driven by wind
On a windy day the rain itself slants. Its velocity \(\vec v_r\) has a vertical part, the fall, and a horizontal part, carried by the wind. Nothing changes in the method. The apparent velocity is still \(\vec v_r - \vec v_m\), but now the horizontal parts may add or cancel.
If a person moves in the direction of the wind at exactly the rain's horizontal speed, the two horizontal parts cancel. The person then sees the rain fall vertically, and can hold the umbrella upright. Moving faster than this, the person sees the rain come from the front. Moving slower in the same direction, the person sees it come from behind. Moving against the wind, the person sees it come from the front with an even larger slant.
A man walking east at \(3\,\text{km/h}\) finds that the rain falls vertically on him. What can you say about the rain?
Two observations find the true rain
The rain's true velocity has two unknown parts, horizontal and vertical. One observation by a moving person gives one equation. A second observation at another speed gives a second equation. Two equations fix the two parts, and so the true speed and direction of the rain.
A boy cycling at \(2\,\text{m/s}\) finds the rain falling vertically. When he cycles at \(8\,\text{m/s}\) in the same direction, the rain meets him at \(45°\) to the vertical. Find the speed of the rain and its direction.
Let the rain be \(\vec v_r = a\,\hat i - b\,\hat j\), with \(\hat i\) the direction of cycling. The unknowns are \(a\) and \(b\).
At \(2\,\text{m/s}\): \(\vec v_{rc} = (a - 2)\,\hat i - b\,\hat j\) is vertical, so \(a - 2 = 0\). At \(8\,\text{m/s}\): \(\vec v_{rc} = (a - 8)\,\hat i - b\,\hat j\) is at \(45°\), so \(|a - 8| = b\).
\(a = 2\,\text{m/s}\), then \(b = |2 - 8| = 6\,\text{m/s}\). The speed is \(\sqrt{4 + 36} = \sqrt{40} = 6.3\,\text{m/s}\). The angle with the vertical is \(\tan^{-1}(2/6) = 18°\), leaning forward.
The rain falls at \(6.3\,\text{m/s}\), slanting \(18°\) from the vertical in the direction the boy cycles. At \(8\,\text{m/s}\) he meets it at \(\sqrt{36 + 36} = 8.5\,\text{m/s}\), the second chip of Fig 3.2.
Read the angle carefully. A question may give the angle of the rain with the vertical or with the horizontal. \(\tan\theta\) is \(v_m / v_r\) for one and \(v_r / v_m\) for the other. Draw the triangle, mark the angle on it, and only then write the tangent.
Rain falls vertically at \(6\,\text{m/s}\). Rank these people by the angle of the apparent rain with the vertical, largest first.
Rain reaches a person in two ways. Drops land on the head and shoulders at a rate set by how hard it rains. So this part grows with the time spent in the rain. Drops are also swept up by the front of the body, and this part depends on the distance covered, not on the speed. Running cuts the time and leaves the swept-up part the same. So on the whole, running from the bus stop to the college gate does keep you drier, and a forward-tilted umbrella helps with the front.
- Holding the umbrella along the rain's ground velocity.The rain that reaches you is the apparent rain, \(\vec v_r - \vec v_m\). Tilt the umbrella along it, forward when you move.
- Adding the walker's velocity to the rain's velocity.Reverse the walker's velocity first. The apparent rain is \(\vec v_r + (-\vec v_m)\).
- Taking \(\tan\theta = v_r / v_m\) for the angle with the vertical.The side opposite the angle with the vertical is the horizontal part, \(v_m\). So \(\tan\theta = v_m / v_r\).