Watch an aeroplane land at Mumbai on a gusty monsoon afternoon. It comes down the approach in a straight line towards the runway. Its nose does not point along that line. It points a little to one side, into the wind, and the aeroplane seems to slide sideways through the air. Just before the wheels touch, the pilot swings the nose round to line up with the runway.
The river of air
An aeroplane flies through the air the way a boat moves through water. Its engines and wings give it a velocity \(\vec v_{pa}\) relative to the air. The size of this velocity is the airspeed. The air itself moves over the ground with the wind velocity \(\vec v_a\). The velocity of the aeroplane relative to the ground is the sum.
\[ \vec v_p = \vec v_{pa} + \vec v_a \]
The size of \(\vec v_p\) is the ground speed. The nose points along \(\vec v_{pa}\), the heading. The aeroplane moves over the ground along \(\vec v_p\), the track. This is the river-boat rule with the air in place of the water.
A town B lies due north of an airfield A. A wind blows towards the east. The pilot keeps the nose pointing due north all the way. Where does the aeroplane pass when it reaches the latitude of B?
Play Fig 4.1 with the third chip. The nose points \(37°\) west of north, yet the aeroplane moves due north along the line AB. The aeroplane slides sideways through the air, exactly like the boatman's boat on the Ganga. The triangle on the right shows why. The westward part of \(\vec v_{pa}\) cancels the wind, and the northward part is left.
Heading into the wind
Let the course run due north and the wind blow towards the east at \(v_a\). The aeroplane must point west of north at an angle \(\theta\), so that the westward part of its air velocity cancels the wind. What is left of its air velocity carries it north.
\[ v_{pa}\sin\theta = v_a, \qquad v_p = v_{pa}\cos\theta = \sqrt{v_{pa}^2 - v_a^2} \]
This is the shortest-path crossing of the river, turned on its side. If the wind were faster than the airspeed, no heading could hold the course.
A town lies \(600\,\text{km}\) due north of an airfield. The aeroplane's airspeed is \(250\,\text{km/h}\), and a wind blows towards the east at \(150\,\text{km/h}\). Find the heading, the ground speed and the time of the flight.
Course: due north. \(v_{pa} = 250\,\text{km/h}\), \(v_a = 150\,\text{km/h}\) towards the east.
\(\sin\theta = v_a / v_{pa}\), \(v_p = \sqrt{v_{pa}^2 - v_a^2}\), \(t = 600 / v_p\).
\(\sin\theta = 150 / 250 = 0.6\), so \(\theta = 37°\) west of north. \(v_p = \sqrt{62500 - 22500} = 200\,\text{km/h}\). \(t = 600 / 200 = 3\,\text{h}\).
The pilot points \(37°\) west of north and reaches the town in \(3\,\text{h}\) at \(200\,\text{km/h}\) over the ground. With no wind the flight would take \(2.4\,\text{h}\).
An aeroplane with an airspeed of \(200\,\text{km/h}\) must fly due north. A wind blows towards the east at \(100\,\text{km/h}\). By how many degrees west of north must the pilot point the nose?
In a steady cross wind, a pilot holds a straight course over the ground. Which statement is correct?
A wind always costs time on a round trip
A wind helps an aeroplane one way and hinders it the other way. It is tempting to think that the two effects cancel on a return trip. They do not. The aeroplane spends longer on the slow leg than on the fast leg, so the slow leg counts for more.
\[ T = \dfrac{d}{v + w} + \dfrac{d}{v - w} \]
Why. the outward leg goes with the wind at \(v + w\), the return leg against it at \(v - w\).
\[ T = \dfrac{d(v - w) + d(v + w)}{(v + w)(v - w)} \]
Why. bring the two fractions to a common denominator.
\[ T = \dfrac{2dv}{v^2 - w^2} \]
Read it. the terms in \(w\) cancel on top, and \((v + w)(v - w) = v^2 - w^2\) below.
The still-air time is \(2d/v\), which is \(2dv/v^2\). The trip along the wind has the smaller denominator \(v^2 - w^2\), so it always takes longer. On the trip across the wind the pilot heads into the wind both ways at the ground speed \(\sqrt{v^2 - w^2}\). That trip takes \(2d/\sqrt{v^2 - w^2}\), also longer than in still air, but shorter than the trip along the wind.
An aeroplane with airspeed \(250\,\text{km/h}\) flies to a town \(600\,\text{km}\) away and back. Rank the round trips by time, longest first.
The flight from Delhi to Mumbai covers about \(1150\,\text{km}\) at an airspeed of about \(850\,\text{km/h}\). Suppose a wind of \(150\,\text{km/h}\) blows along the route. Roughly how many minutes longer is the flight against the wind than the flight with it?
In the nineteenth century, light was thought to travel through an invisible medium, the ether, and the Earth was thought to move through it. Michelson and Morley treated light like the aeroplane of this lesson. They split a beam of light into two halves. One half made a round trip along the supposed ether wind, the other a round trip across it, and the halves were brought back together. The formulas of this lesson predicted that the trip along the wind should take a little longer. Their instrument could measure a difference far smaller than the one expected. They found no difference at all. That famous null result became one of the foundations of Einstein's theory of relativity of 1905.
- Pointing the nose straight at the destination in a cross wind.The wind carries the whole aeroplane. Point into the wind at \(\sin\theta = v_a / v_{pa}\) to fly the straight course.
- Taking the ground speed across the wind to be the airspeed.Part of the air velocity is spent against the wind. The ground speed along the course is \(\sqrt{v_{pa}^2 - v_a^2}\).
- Believing that a wind helps as much one way as it hinders the other.The slow leg takes longer, so a round trip in any steady wind lasts longer than in still air.