How to Calculate an Aircraft’s Required Heading from Wind Data

An aircraft needs to fly due north towards a checkpoint, but the wind is carrying it east. If the pilot points the nose north, the aircraft will miss the checkpoint.

What heading should the pilot choose instead?

The answer is to point slightly west of north, into the crosswind. The aircraft needs enough westward velocity relative to the air to cancel the wind’s eastward velocity. The exact heading depends on the aircraft’s airspeed and the sideways component of the wind, not simply the total wind speed.

There’s a second question worth asking: once the aircraft is heading in the correct direction, how fast will it actually approach the checkpoint? A wind blowing partly along the route can change the ground speed even when the sideways drift has been completely cancelled.

01Which velocity determines the aircraft’s heading?

Imagine you’re watching an aircraft from the ground. You see it travelling north-east, even though its nose points north.

Now imagine sitting inside the aircraft. Relative to the surrounding air, the aircraft is moving straight in the direction its nose points.

Neither observation is wrong. The observers are measuring motion relative to different things.

There are three velocities to distinguish.

The aircraft is our moving object. The ground is the reference frame of the observer tracking its journey. The air provides another moving reference frame.

For a simple HSC Physics model, assume the aircraft flies steadily without sideslip. Its velocity relative to the air then points along its heading.

The direction of its ground velocity is its ground track, which may be different.

We connect the three velocities using vector addition:

\[ \vec v_{PG}=\vec v_{PA}+\vec v_{AG} \]

Here:

• \(\vec v_{PG}\) is the aircraft’s velocity relative to the ground.
• \(\vec v_{PA}\) is the aircraft’s velocity relative to the air.
• \(\vec v_{AG}\) is the air’s velocity relative to the ground, or wind velocity.

All velocities are measured in metres per second (\(\mathrm{m\,s^{-1}}\)) unless stated otherwise.

A useful way to picture this is a swimmer crossing a river.

The swimmer moves relative to the water, while the water moves relative to the riverbank. To swim directly across, the swimmer may need to aim upstream.

The aircraft’s air velocity plays the role of the swimmer’s velocity through water. The wind plays the role of the river current. The ground observer sees their vector sum.

The analogy describes how velocities combine. It doesn’t describe the forces keeping an aircraft flying or the forces acting on a swimmer.

There’s also an important mathematical detail: airspeed and ground speed are scalars, but the corresponding velocities are vectors.

A wind of \(30\ \mathrm{m\,s^{-1}}\) doesn’t automatically add or subtract \(30\ \mathrm{m\,s^{-1}}\) from the aircraft’s ground speed. Its direction determines which components change.

For a broader explanation of these reference frames, see Aircraft in a Crosswind: Heading, Ground Track and Wind. Our task here is more specific: work backwards from the desired ground track to calculate the heading the pilot must use.

02The method: cancel the sideways wind first

Suppose an aircraft must travel due north.

Choose:

• East as the positive \(x\)-direction.
• North as the positive \(y\)-direction.
• Angles measured clockwise from true north when giving compass bearings.

A true bearing of \(000^\circ\) means north, \(090^\circ\) means east, \(180^\circ\) means south, and \(270^\circ\) means west.

For our calculations, let \(V\) be the aircraft’s true airspeed, \(w_E\) the wind’s eastward velocity component, and \(w_N\) its northward velocity component. Components can be negative.

Let \(\theta\) be the aircraft’s heading angle measured from north, positive towards the east and negative towards the west.

Step 1: Identify the wind’s sideways component

If the wind carries the air east at \(30\ \mathrm{m\,s^{-1}}\), the aircraft must provide a westward air-velocity component of \(30\ \mathrm{m\,s^{-1}}\).

The target is zero east-west ground velocity:

\[ v_{PG,x}=0 \]

Using vector addition:

\[ v_{PA,x}+w_E=0 \]

Therefore:

\[ v_{PA,x}=-w_E \]

This is the condition that keeps the aircraft on the north-south route.

Step 2: Calculate the required heading

The aircraft’s eastward air-velocity component is \(V\sin\theta\).

For zero sideways ground velocity:

\[ V\sin\theta=-w_E \]

So the required correction angle is:

\[ \boxed{\theta=\sin^{-1}\left(-\frac{w_E}{V}\right)} \]

For a normal forward-flying solution, choose the angle between \(-90^\circ\) and \(90^\circ\).

A negative angle means west of north. A positive angle means east of north.

For example, if \(\theta=-15^\circ\), the aircraft must head \(15^\circ\) west of north, or on a true bearing of \(345^\circ\).

Step 3: Find the resulting ground speed

Once the sideways components cancel, only the north-south components remain.

The aircraft’s northward air-velocity component is:

\[ v_{PA,y}=V\cos\theta \]

Therefore:

\[ \boxed{v_{PG,y}=V\cos\theta+w_N} \]

Equivalently, for the forward-pointing heading:

\[ v_{PG,y}=\sqrt{V^2-w_E^2}+w_N \]

This result gives the signed northward ground velocity. It is a northward ground speed only if the result is positive.

A sufficiently strong headwind could make it zero or negative, even though the sideways wind has been cancelled.

Step 4: Check whether the heading is physically possible

The aircraft cannot produce a sideways air-velocity component greater than its total airspeed.

Therefore, exact sideways cancellation requires:

\[ |w_E|\leq V \]

If the crosswind exceeds the airspeed, no heading can cancel it.

At equality, the aircraft must point completely sideways. It has no air-velocity component towards the checkpoint, although a separate tailwind component could still carry it north.

Even when cancellation is possible, you must check that the resulting ground velocity actually points towards the destination.

A quick decision table

For a desired track other than north, the same method works. Resolve the wind parallel and perpendicular to the required track. Cancel the perpendicular component first, then calculate the ground speed along the route.

04How could you check the heading using measurements?

Suppose the wind forecast was wrong. The pilot calculates the expected heading, flies it, and the GPS still shows the aircraft moving east of its intended route.

What has the calculation actually established?

Only the heading required for the assumed wind.

We can test the model with an illustrative flight experiment.

Imagine an aircraft travelling at constant altitude with three measurements available:

1. A heading readout showing the direction the aircraft points.
2. A true airspeed measurement showing its speed relative to the surrounding air.
3. GPS position measurements showing its displacement over the ground.

The heading and airspeed determine the aircraft’s air-velocity vector. Successive GPS positions determine its ground-velocity vector.

Assume the instrument values are accurate and the wind is approximately constant during each short test.

The first measurement has a weakness

Suppose the aircraft holds a steady heading, but the pilot records only its GPS ground speed.

That scalar measurement is not enough to identify the wind vector.

A tailwind could increase the GPS speed. A crosswind could also increase the magnitude of the ground velocity by adding a perpendicular component.

To distinguish them, the revised test records the direction of the GPS displacement as well as its magnitude.

Then we can rearrange the velocity equation:

\[ \vec v_{AG}=\vec v_{PG}-\vec v_{PA} \]

In words:

Wind velocity equals ground velocity minus air velocity.

This is vector subtraction, not subtraction of two speeds.

Why the fastest GPS reading isn’t necessarily the best heading

Keep the wind from that second example: \(20\ \mathrm{m\,s^{-1}}\) east and \(30\ \mathrm{m\,s^{-1}}\) north.

Compare pointing the aircraft due north with applying the correct wind correction.

Pointing due north gives a slightly greater GPS ground speed. But the aircraft drifts east, away from the required route.

Pointing west of north produces a lower GPS ground speed but keeps the aircraft on the correct line.

The task is not to maximise the magnitude of the ground velocity. It is to make that velocity point towards the checkpoint.

05Assumptions and common errors

These calculations describe horizontal motion using a simplified, constant-wind model. They are useful for HSC Physics vector problems, but they are not complete operational flight-planning procedures.

In particular, we assume:

• The supplied airspeed is true airspeed, not an uncorrected indicated-airspeed reading.
• Wind velocity is uniform and steady over the relevant part of the flight.
• The aircraft maintains the calculated air velocity without sideslip.
• All bearings use the same true-north reference.
• The aircraft can establish the required heading, and changes of heading take negligible time in the calculation.
• Vertical motion, magnetic variation, instrument errors, gusts, and changes in wind with altitude are neglected.

A real aircraft may encounter changing wind, so a calculated heading can be a prediction that needs updating rather than a permanently correct value.

Three mistakes are particularly worth avoiding.

Mistake 1: Treating a wind reported as “from” a bearing as though it moves towards that bearing.

A wind from \(270^\circ\) moves east, not west. Always establish the direction the air actually travels.

Mistake 2: Adding wind speed directly to airspeed.

Scalar addition only works for velocities along the same line with the appropriate signs. For an oblique wind, resolve the velocity into components first.

Mistake 3: Finding a mathematical heading without checking whether it produces progress.

Cancelling the crosswind only makes the ground velocity parallel to the desired route. It does not guarantee that the aircraft is moving towards its destination.

The final two questions below explore why that last check can be surprisingly difficult.

06Questions and solutions

All wind data in these questions are illustrative. Assume a flat horizontal plane, true bearings, constant wind within each specified region, and accurate true airspeeds. Ignore the time taken to turn unless stated otherwise.

Question 1

An aircraft has a true airspeed of \(100\ \mathrm{m\,s^{-1}}\).

The wind is reported as from \(090^\circ\) true at \(20\ \mathrm{m\,s^{-1}}\).

The aircraft must follow a ground track due north.

Calculate:

a. The required true heading.

b. The resulting ground speed.

Solution 1

The aircraft must head \(011.5^\circ\) true, and its ground speed will be approximately \(98.0\ \mathrm{m\,s^{-1}}\) due north.

The wind is from the east, so the air is moving west at \(20\ \mathrm{m\,s^{-1}}\).

To cancel this, the aircraft needs an eastward air-velocity component of \(20\ \mathrm{m\,s^{-1}}\).

Let \(\theta\) be the correction angle east of north.

\[ \begin{aligned} 100\sin\theta&=20\\ \sin\theta&=0.20\\ \theta&=11.5^\circ \end{aligned} \]

The required true bearing is \(011.5^\circ\).

The ground speed is:

\[ \begin{aligned} v_{PG} &=\sqrt{100^2-20^2}\\ &=98.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

The aircraft travels due north at \(98.0\ \mathrm{m\,s^{-1}}\).

A tempting error is to subtract the \(20\ \mathrm{m\,s^{-1}}\) wind speed from the airspeed and obtain \(80\ \mathrm{m\,s^{-1}}\). That would treat the wind as a headwind, even though it is perpendicular to the required track.

Question 2

An aircraft maintains a true airspeed of \(100\ \mathrm{m\,s^{-1}}\) while conducting two short test flights in what is believed to be a steady wind.

On the first flight, it points due north. GPS gives a ground velocity of:

• \(30\ \mathrm{m\,s^{-1}}\) east.
• \(110\ \mathrm{m\,s^{-1}}\) north.

On the second flight, it points due east. GPS gives a ground velocity of:

• \(130\ \mathrm{m\,s^{-1}}\) east.
• \(10\ \mathrm{m\,s^{-1}}\) north.

a. Determine whether these observations are consistent with a single steady wind. Calculate its components.

b. Find the required heading and ground speed for a subsequent due-north ground track.

c. Explain why knowing only the GPS ground speed from the first flight would not uniquely determine the wind.

Solution 2

The two flights are consistent with a steady wind of \(30\ \mathrm{m\,s^{-1}}\) east and \(10\ \mathrm{m\,s^{-1}}\) north. The required heading is \(342.5^\circ\) true, giving a northward ground speed of approximately \(105.4\ \mathrm{m\,s^{-1}}\). The first GPS speed alone would not identify the wind direction.

For part a, use:

\[ \vec v_{AG}=\vec v_{PG}-\vec v_{PA} \]

On the first flight, the aircraft’s air velocity is \(100\ \mathrm{m\,s^{-1}}\) north.

The inferred wind components are:

\[ \begin{aligned} w_E&=30-0=30\\ w_N&=110-100=10 \end{aligned} \]

Both are in \(\mathrm{m\,s^{-1}}\).

On the second flight, the aircraft’s air velocity is \(100\ \mathrm{m\,s^{-1}}\) east.

Therefore:

\[ \begin{aligned} w_E&=130-100=30\\ w_N&=10-0=10 \end{aligned} \]

The independent estimates agree. That supports the steady-wind model, although two matching observations do not prove the wind will remain constant indefinitely.

For part b, the aircraft needs a westward air-velocity component of \(30\ \mathrm{m\,s^{-1}}\).

\[ \begin{aligned} 100\sin\theta&=-30\\ \theta&=-17.5^\circ \end{aligned} \]

So the true heading is:

\[ 360^\circ-17.5^\circ=342.5^\circ \]

The resulting northward ground speed is:

\[ \begin{aligned} v_{PG,y} &=\sqrt{100^2-30^2}+10\\ &=95.4+10\\ &=105.4\ \mathrm{m\,s^{-1}} \end{aligned} \]

For part c, the first GPS ground speed has magnitude:

\[ \begin{aligned} v_{PG} &=\sqrt{30^2+110^2}\\ &=114.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

But a purely northward tailwind of approximately \(14.0\ \mathrm{m\,s^{-1}}\) could also produce almost the same GPS speed for an aircraft heading north at \(100\ \mathrm{m\,s^{-1}}\).

The difference would be its ground-track direction.

This is the experimental trap: agreement in a scalar measurement does not establish agreement in a vector model. Recording displacement direction makes it possible to distinguish the two explanations.

Question 3

An aircraft travels from airport A to airport B, which lies \(60.0\ \mathrm{km}\) due north of A.

The aircraft maintains a true airspeed of \(100\ \mathrm{m\,s^{-1}}\).

Throughout the journey, the wind velocity has components:

• \(30\ \mathrm{m\,s^{-1}}\) east.
• \(40\ \mathrm{m\,s^{-1}}\) north.

The aircraft must follow the north-south line on both the outward and return journeys.

a. Calculate its required true heading from A to B.

b. Calculate its required true heading from B back to A.

c. Calculate the total flying time for the round trip.

d. Explain why simply reversing the outward heading by \(180^\circ\) would not work.

Solution 3

The required headings are \(342.5^\circ\) true outbound and \(197.5^\circ\) true on return. The round trip takes approximately 25.4 minutes. The return heading is not simply the reciprocal of the outward heading because the wind continues to blow east on both legs.

For part a, the eastward crosswind requires a westward air-velocity component of \(30\ \mathrm{m\,s^{-1}}\).

\[ \begin{aligned} 100\sin\theta&=30\\ \theta&=\sin^{-1}(0.30)\\ &=17.5^\circ \end{aligned} \]

The outward heading is \(17.5^\circ\) west of north:

\[ H_{\mathrm{out}}=342.5^\circ \]

For part b, the aircraft must now travel south, but the wind still moves east.

Its air velocity must still contain a westward component of \(30\ \mathrm{m\,s^{-1}}\).

So the aircraft points \(17.5^\circ\) west of south:

\[ H_{\mathrm{return}}=197.5^\circ \]

This is the important distinction. The crosswind doesn’t reverse merely because the aircraft reverses its journey.

For part c, first calculate the air-velocity component along either route:

\[ \begin{aligned} v_{\parallel} &=\sqrt{100^2-30^2}\\ &=95.4\ \mathrm{m\,s^{-1}} \end{aligned} \]

On the outward journey, the wind contributes \(40\ \mathrm{m\,s^{-1}}\) north.

\[ v_{\mathrm{out}}=95.4+40 \]

\[ v_{\mathrm{out}}=135.4\ \mathrm{m\,s^{-1}} \]

On the return journey, the same wind opposes the aircraft’s southward motion.

\[ v_{\mathrm{return}}=95.4-40 \]

\[ v_{\mathrm{return}}=55.4\ \mathrm{m\,s^{-1}} \]

Convert \(60.0\ \mathrm{km}\) to \(60\,000\ \mathrm m\).

Using unrounded ground speeds for the division:

\[ \begin{aligned} t_{\mathrm{out}} &=\frac{60\,000}{135.394}\\ &\approx443\ \mathrm s\\[4pt] t_{\mathrm{return}} &=\frac{60\,000}{55.394}\\ &\approx1083\ \mathrm s \end{aligned} \]

Total time:

\[ \begin{aligned} t_{\mathrm{total}}&=443+1083\\ &=1526\ \mathrm s\\ &\approx25.4\ \mathrm{min} \end{aligned} \]

For part d, the reciprocal of the outward heading would be:

\[ 342.5^\circ-180^\circ=162.5^\circ \]

That heading points east of south. It would give the aircraft an eastward air-velocity component of \(30\ \mathrm{m\,s^{-1}}\).

Combined with the wind, its eastward ground-velocity component would be:

\[ v_{PG,x}=30+30 \]

\[ v_{PG,x}=60\ \mathrm{m\,s^{-1}} \]

The aircraft would drift east rather than follow the southward route.

The return heading must be determined from the wind again, not obtained by reversing the first heading.

Question 4

An aircraft must travel from A to B, \(120\ \mathrm{km}\) due north.

Its true airspeed is fixed at \(100\ \mathrm{m\,s^{-1}}\).

The route passes through two wind regions. The boundary between them is an east-west line \(60\ \mathrm{km}\) north of A.

The winds are uniform within each region and change abruptly at the boundary.

Unlike the previous questions, the pilot must maintain one fixed heading throughout the flight. The aircraft is allowed to drift away from the direct route, provided it reaches checkpoint B.

a. If the aircraft heads due north, will the opposite crosswinds cancel over the full journey? Calculate its final east-west displacement.

b. Determine a single fixed true heading that allows the aircraft to reach B.

c. Explain whether the aircraft remains on the original north-south route and why another mathematical heading solution must be rejected.

Solution 4

Heading due north would miss B by \(7.50\ \mathrm{km}\) west. The only physically valid fixed heading that reaches B is approximately \(003.45^\circ\) true, slightly east of north. The aircraft will not remain on the original line throughout the journey.

For part a, a due-north heading gives \(100\ \mathrm{m\,s^{-1}}\) north relative to the air.

In the first region, ground velocity is \(120\ \mathrm{m\,s^{-1}}\) north and \(30\ \mathrm{m\,s^{-1}}\) east.

Time in the first region:

\[ t_1=\frac{60\,000}{120}=500\ \mathrm s \]

Eastward displacement:

\[ x_1=30(500)=15\,000\ \mathrm m \]

In the second region, ground velocity is \(80\ \mathrm{m\,s^{-1}}\) north and \(30\ \mathrm{m\,s^{-1}}\) west.

Time in the second region:

\[ t_2=\frac{60\,000}{80}=750\ \mathrm s \]

Taking east as positive, its east-west displacement is:

\[ x_2=-30(750)=-22\,500\ \mathrm m \]

The final east-west displacement is:

\[ \begin{aligned} x_{\mathrm{total}} &=15\,000-22\,500\\ &=-7500\ \mathrm m \end{aligned} \]

So the aircraft finishes \(7.50\ \mathrm{km}\) west of B.

The tempting prediction is that equal and opposite crosswinds must cancel.

But the aircraft spends longer in the second region because the wind opposes its northward motion. Equal sideways wind speeds acting for unequal times produce unequal sideways displacements.

For part b, let \(\delta\) be a fixed heading angle east of north.

Define the aircraft’s eastward and northward air-velocity components as:

\[ s=100\sin\delta \]

\[ c=100\cos\delta \]

Both are measured in \(\mathrm{m\,s^{-1}}\).

In the first region, the ground-velocity components are:

\[ v_{1,x}=s+30,\qquad v_{1,y}=c+20 \]

In the second region:

\[ v_{2,x}=s-30,\qquad v_{2,y}=c-20 \]

Each region requires \(60\,000\ \mathrm m\) of northward displacement.

For the aircraft to reach B, the total east-west displacement must equal zero:

\[ 0=\frac{s+30}{c+20} +\frac{s-30}{c-20} \]

The common \(60\,000\ \mathrm m\) distance has cancelled from this equation.

Multiplying through by the denominators gives:

\[ \begin{aligned} 0&=(s+30)(c-20)\\ &\quad +(s-30)(c+20)\\ 0&=2sc-1200\\ sc&=600 \end{aligned} \]

Substitute the definitions of \(s\) and \(c\):

\[ 100^2\sin\delta\cos\delta=600 \]

Using \(2\sin\delta\cos\delta=\sin2\delta\):

\[ \sin2\delta=0.120 \]

This gives two mathematical candidates between \(0^\circ\) and \(90^\circ\):

\[ \delta=3.45^\circ \]

or

\[ \delta=86.55^\circ \]

For part c, the second angle must be rejected.

At \(\delta=86.55^\circ\), the aircraft’s northward air-velocity component would be only about \(6.01\ \mathrm{m\,s^{-1}}\).

But the second region has a southward wind component of \(20\ \mathrm{m\,s^{-1}}\).

The aircraft’s ground velocity there would therefore point south. It couldn’t continue north across the second region towards B.

The valid heading is:

\[ \boxed{003.45^\circ\text{ true}} \]

At this heading, the eastward air-velocity component is approximately \(6.01\ \mathrm{m\,s^{-1}}\), and the northward component is \(99.82\ \mathrm{m\,s^{-1}}\).

The time in the first region is:

\[ t_1=\frac{60\,000}{99.82+20} \]

\[ t_1\approx501\ \mathrm s \]

The aircraft’s first-region eastward displacement is approximately:

\[ x_1=(6.01+30)(501) \]

\[ x_1\approx18.0\ \mathrm{km} \]

In the second region, the westward ground-velocity component brings the aircraft back towards the north-south line.

Its second-region east-west displacement is approximately \(-18.0\ \mathrm{km}\), cancelling the first displacement.

So it reaches B, but only after drifting about \(18.0\ \mathrm{km}\) east of the intended route at the boundary.

This question separates two requirements that are easy to confuse:

• Reaching the correct endpoint.
• Remaining on the correct ground track throughout the journey.

The single fixed heading satisfies the first requirement, not the second. Remaining on the original route would require a heading change when the wind changes.

Question 5

An aircraft is at airport A. An emergency beacon B is exactly \(12.0\ \mathrm{km}\) due south of A.

The aircraft’s maximum true airspeed is \(100\ \mathrm{m\,s^{-1}}\).

The wind is uniform everywhere, with velocity components:

• \(30\ \mathrm{m\,s^{-1}}\) east.
• \(96\ \mathrm{m\,s^{-1}}\) north.

The aircraft may change its heading as often as it likes, fly in any horizontal direction, and take as long as necessary. It cannot exceed its maximum true airspeed.

A pilot notices that pointing the aircraft due south produces a small southward component of ground velocity.

The pilot argues:

“If we can make progress south, we should eventually be able to reach the beacon. We might just need to fly a dogleg to cancel the eastward drift.”

a. Calculate the ground-velocity components when the aircraft points due south at maximum airspeed. Find its displacement after \(100\ \mathrm s\).

b. Determine whether any fixed heading can produce a ground velocity directed due south.

c. Decide whether a journey involving multiple headings could reach B. Give a proof that does not assume the aircraft flies in a straight line.

d. Calculate the threshold true airspeed for southward progress on a due-south ground track. Explain whether reaching exactly this threshold is sufficient. Then find the southward ground speed if the aircraft’s available true airspeed increases to \(101\ \mathrm{m\,s^{-1}}\).

Solution 5

The pilot’s argument is incorrect. Pointing due south produces a ground velocity of \(30\ \mathrm{m\,s^{-1}}\) east and \(4\ \mathrm{m\,s^{-1}}\) south, but neither a fixed heading nor any sequence of doglegs can reach B at the available airspeed. Southward progress along the required line needs a true airspeed strictly greater than \(100.58\ \mathrm{m\,s^{-1}}\). At \(101\ \mathrm{m\,s^{-1}}\), the correctly aimed aircraft can travel south at approximately \(0.442\ \mathrm{m\,s^{-1}}\).

For part a, if the aircraft points due south at \(100\ \mathrm{m\,s^{-1}}\), its air-velocity components are:

\[ v_{PA,x}=0 \]

\[ v_{PA,y}=-100\ \mathrm{m\,s^{-1}} \]

Add the wind:

\[ \begin{aligned} v_{PG,x}&=0+30=30\\ v_{PG,y}&=-100+96=-4 \end{aligned} \]

The aircraft really does move south relative to the ground.

After \(100\ \mathrm s\):

\[ \begin{aligned} \Delta x&=30(100)\\ &=3000\ \mathrm m \end{aligned} \]

\[ \begin{aligned} \Delta y&=-4(100)\\ &=-400\ \mathrm m \end{aligned} \]

It has travelled \(3.00\ \mathrm{km}\) east and \(0.400\ \mathrm{km}\) south.

This confirms the pilot’s observation. The aircraft is getting further south, but it is also drifting rapidly east.

For part b, a due-south ground track requires zero eastward ground velocity.

The aircraft must produce \(30\ \mathrm{m\,s^{-1}}\) west relative to the air.

Its maximum possible southward air-velocity component is then:

\[ \begin{aligned} v_{PA,S} &=\sqrt{100^2-30^2}\\ &=95.394\ \mathrm{m\,s^{-1}} \end{aligned} \]

But the wind moves north at \(96\ \mathrm{m\,s^{-1}}\).

Therefore the net northward ground-velocity component would be:

\[ \begin{aligned} v_{PG,y} &=96-95.394\\ &=0.606\ \mathrm{m\,s^{-1}} \end{aligned} \]

The aircraft would move north, not south.

No fixed heading can produce a due-south ground velocity at \(100\ \mathrm{m\,s^{-1}}\).

The crucial distinction is that the aircraft can move south while drifting sideways, but cannot both cancel the eastward wind and overcome the northward wind.

For part c, could a dogleg solve the problem?

Suppose the aircraft flies for \(T\) seconds and eventually arrives at B, exactly \(12\,000\ \mathrm m\) south of A.

During those \(T\) seconds, the wind carries the surrounding air:

• \(30T\) metres east.
• \(96T\) metres north.

To finish exactly at B, the aircraft’s total displacement relative to the air would therefore need to be:

• \(30T\) metres west.
• \(96T+12\,000\) metres south.

Here \(T\) is measured in seconds, so the displacement values are in metres.

The magnitude of that required air-relative displacement is:

\[ \begin{aligned} d_{\mathrm{required}} &=\sqrt{(30T)^2}\\ &\quad\overline{\phantom{=}\!} \end{aligned} \]

Combining the perpendicular displacement components gives:

\[ d_{\mathrm{required}} =\sqrt{(30T)^2+(96T+12\,000)^2} \]

Even if the beacon were at the starting point rather than south of it, merely cancelling the total wind displacement would require:

\[ \begin{aligned} d_{\mathrm{cancel}} &=T\sqrt{30^2+96^2}\\ &=100.578T\ \mathrm m \end{aligned} \]

Because B is further south, the actual required displacement relative to the air is even greater:

\[ d_{\mathrm{required}}>100.578T\ \mathrm m \]

Yet the aircraft’s maximum airspeed is \(100\ \mathrm{m\,s^{-1}}\).

In \(T\) seconds, it can travel at most:

\[ d_{\mathrm{flight}}=100T\ \mathrm m \]

through the surrounding air.

Its net displacement relative to the air cannot exceed the total distance it travels through that air. Turning only makes the net displacement smaller than or equal to that total distance.

But reaching B would require:

\[ d_{\mathrm{required}}>100.578T \]

while the aircraft is limited to:

\[ d_{\mathrm{flight}}\leq100T \]

These conditions cannot both be satisfied.

Therefore no sequence of headings can bring the aircraft to B, even with unlimited flight time.

The pilot’s local observation was correct: the aircraft can temporarily make southward ground progress. The failed inference was assuming that this progress could be combined with enough westward motion to return to the required line.

The aircraft does not have enough total airspeed to achieve both.

For part d, let \(V\) be the new maximum true airspeed.

A due-south ground track requires a westward air-velocity component of \(30\ \mathrm{m\,s^{-1}}\).

For positive southward ground speed, the aircraft’s remaining southward air-velocity component must exceed \(96\ \mathrm{m\,s^{-1}}\).

Therefore:

\[ \sqrt{V^2-30^2}>96 \]

Squaring both positive sides:

\[ \begin{aligned} V^2-30^2&>96^2\\ V^2&>30^2+96^2\\ V&>\sqrt{10\,116}\\ V&>100.578\ \mathrm{m\,s^{-1}} \end{aligned} \]

So the threshold is approximately \(100.58\ \mathrm{m\,s^{-1}}\).

Equality is not sufficient. At exactly the threshold, the aircraft could cancel the entire wind velocity but would have zero ground speed. Positive progress south requires an airspeed strictly above it.

If the aircraft’s true airspeed becomes \(101\ \mathrm{m\,s^{-1}}\), the greatest available southward air-velocity component while cancelling the eastward wind is:

\[ \begin{aligned} v_{PA,S} &=\sqrt{101^2-30^2}\\ &=96.442\ \mathrm{m\,s^{-1}} \end{aligned} \]

The resulting southward ground speed is:

\[ \begin{aligned} v_{PG,S} &=96.442-96\\ &=0.442\ \mathrm{m\,s^{-1}} \end{aligned} \]

The aircraft can now make progress directly towards B, but extremely slowly.

The deeper result is that producing a component of velocity towards a target does not guarantee the target is reachable.

Every movement towards the destination must also satisfy the sideways-displacement constraint. Over an entire journey, those requirements must hold together.

06What to do after calculating the heading

For an HSC question, the strongest final check is to substitute your calculated heading back into the velocity equation.

Calculate the predicted east-west ground-velocity component. For a due-north route, it should be zero.

Then calculate the northward component. It must be positive if the aircraft is actually to approach a checkpoint to the north.

This checks both parts of the problem: direction and progress.

If the wind changes, the required heading may also change. If GPS measurements disagree with the predicted ground track, use the measured ground velocity and known air velocity to estimate what wind would explain the discrepancy.

That is the useful progression from this method: first predict the heading from the wind, then use the aircraft’s observed motion to test the prediction.

The same reference-frame reasoning also works when the observer is moving. Instead of comparing an aircraft with the stationary ground, you can compare two moving objects using vector subtraction, as in calculating one boat’s velocity relative to another.

The principle remains the same: establish who is observing what, keep directions attached to velocities, and use components before interpreting the result.