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The effect of wind
Rev. 25a — page content was last reviewed 22 April 2015
|Flight Planning and Navigation|
|An aircraft in flight is airborne and thus subject to the movement of the air mass in relation to the surface i.e. the wind velocity. Consequently, the calculation of the wind effect on aircraft movement relative to the ground is a major part of very light aircraft flight planning and navigation. In addition the relatively low cruising speed of very light aircraft makes them particularly affected by the significant variations to the forecast wind velocity that are normally encountered during flight.|
For example, waypoint Beta is 150 nautical miles north-east (045° true) of waypoint Alpha and an aircraft departs overhead Alpha for Beta, maintaining a heading of 045° true while flying at the aircraft's normal cruising speed of 75 knots TAS. At the time of departure, the wind velocity at the cruise altitude is 135° true at 20 knots; i.e. the 20 kn wind is coming from the south-east. Where will the aircraft be after two hours flight? Certainly not over Beta, as it will have moved 150 nm north-east within the air mass while the air mass has moved 40 nm north-west. So we might surmise that after two hours flight its position will be about 40 nm north-west of Beta, and this is shown in Figure 1. The aircraft has drifted from its intended path or track over the ground and the 'track made good' is about 15° to the left of the 'required track'.
We should note that, relative to the aircraft's course, the wind velocity normally has both a crosswind component and a headwind or tailwind component, and that headwind or tailwind component will also affect the aircraft's speed relative to the surface — the ground speed.
(I have used the ISO standard symbol 'kn' for knot in the diagrams; the symbol 'kt' sometimes seen is the standard symbol for kilotonne.)
Note: in the USA the term 'course' is synonymous with 'track' for air navigation, but the International Civil Aviation Organisation (ICAO) preferred usage is 'track'. The heading is the direction with which the longitudinal axis of the aircraft is aligned at any given time. This heading may be expressed as relative to true north — the true heading; or if adjusted for magnetic variation — the magnetic heading; and if adjusted for variation plus compass deviation — the compass heading.
We can determine the two unknowns — the heading and the ground speed — by plotting scaled vectors on paper (figure 2). You will need some drawing instruments, a protractor and ruler, but a pair of dividers can be useful.
• First draw a vertical line labelled 'true north' and mark a position on the line as waypoint Alpha.
• Using a protractor centred on Alpha and aligned with true north, mark the bearing to waypoint Beta; e.g. from the above, 045° true. Rule a line of appropriate length from Alpha through the bearing, marking it with two arrows to indicate it as the track direction and annotate that bearing.
• Wind velocity is given as the direction the wind is coming from and we need to plot the direction it is moving to — the reciprocal bearing. The reciprocal is the stated direction ±180°. Using a protractor centred on Alpha and aligned with true north, mark the reciprocal wind bearing: 135° ±180° = 315° true.
• Rule a line of appropriate length from Alpha through the wind bearing mark. Decide the scale to be used and mark off a distance along that line that equals the air mass movement during one hour; i.e. 20 nm (20 knots wind speed). Label that distance mark as the wind vector — v1. The convention is to add three arrows to the vector indicating direction, and annotate the wind velocity — 135/20 kn.
• Using the scale, open the dividers to the distance equalling the air distance the aircraft travels in one hour; i.e. 75 nm at the cruise true airspeed of 75 knots. With one divider point on v1 mark the track line with the other divider/compass point and label that v2 (Figure 3). Or just use the ruler to accomplish the same task.
• Draw a line connecting v1 and v2, marking it with one arrow to represent the heading vector and measure the line's orientation with true north with the protractor to determine the heading (060°T). Thus we have the first unknown — the direction in which to point the aircraft. Annotate the heading (060°T) and TAS (75 kn). Also note the wind correction angle [WCA] — the difference between the required track (045°T) and the heading (060°T) — is 15°, and the drift will be to the left — also known as port drift.
Note: the wind correction angle is the angular difference between the required track and the heading, intended to ensure that the track made good will equate with the required track. Note that the terms 'crab angle' and 'drift angle' are very often used instead of 'wind correction angle'. But the latter term is more precise; crab angle and drift angle do have slightly different meanings or associations. Drift angle is measured in flight, and is the angle between the heading and the track made good. Crab angle is the preferred term when associated with crosswind landing.
• Now measure the distance between Alpha and v2, which is the distance (72 nm) moved over the ground during one hour. This is the second unknown — the ground speed. Annotate the ground speed (72 kn) adjacent to the bearing (Figure 3).
• We can now calculate the sector flight time from overhead Alpha to overhead Beta; this time is called the estimated time interval [ETI].
ETI (minutes) = Distance (nm) / ground speed (kn) × 60 = 150/72 × 60 = 125 minutes.
It is interesting to note that even though the wind is a full crosswind to the track required, the ground speed is less than TAS and thus the ETI is a bit greater than you may have expected. This is because the heading of 060° would now include a small headwind component.
It may be thought that if an out-and-return trip is flown where the wind is directly aligned with the required track, the headwind encountered in one direction will be offset by the tailwind in the reverse direction; thus the total flight time will be equivalent to that in nil wind conditions. Not so — the greater the wind speed the greater the flight time on an out-and-return flight, no matter what the wind direction. Imagine a flight Alpha–Beta–Alpha in nil wind conditions. The ground speed on both the 150 nm outward and return legs would equal the TAS (75 kn) and each leg would take 120 minutes for a total flight time of 240 minutes. Now let's factor in a 25-knot north-east wind. The ground speed on the outward leg would be 50 kn and the ETI would be 180 minutes, whereas the ground speed on the return leg would be 100 kn and the ETI 90 minutes for a total flight time of 270 minutes.
The trigonometrical relationships of the two wind components — crosswind (that component of the wind velocity that acts at right angles to the track) and headwind/tailwind (that component of the wind velocity that acts inline with the track) — is shown in a modified wind triangle (Figure 5). The sine of an angle = opposite side/hypotenuse, while the cosine of an angle = adjacent side/hypotenuse.
In this example the wind angle is 30° relative to the required track of 045° true and the wind speed is 20 knots; the hypotenuse represents the wind velocity vector, the side opposite to the wind angle is drawn from the start of the wind vector so that it forms a right angle with the track so representing the crosswind component of the wind velocity while the side adjacent to the angle represents the headwind component. Reading from the abridged trigonometric table below, sine 30° is 0.5 and cosine 30° is 0.866 — near enough to 0.9, thus the crosswind is 0.5×20=10 kn and the headwind is 0.9×20=18 kn.
In wind triangle plots we assume that the forecast wind velocity is accurate and constant, the aircraft's magnetic compass is accurate, and the pilot will maintain a constant heading in flight. However, there will be considerable variability in each (for example read the boundary layer turbulence paragraphs in the microscale meteorology module), so there is no reason to try for absolute accuracy in the initial calculation of heading, ground speed and ETI.
So, rather than plotting the wind triangle we can introduce a few shortcuts to the process by using some simple mental arithmetic to estimate the crosswind and headwind/tailwind components of the wind velocity relative to the required track. Even so, it is wise to become familiar with plotting the wind triangle; the experience makes it much easier to mentally envisage the relationship between the vectors thus avoiding flying entirely in the wrong direction — which is remarkably easy to do.
• Step 1. First find the crosswind component of the forecast wind velocity by estimating the acute angle (i.e. less than 90°) at which the wind meets the required track, divide that by 60 and multiply the result by the wind speed. However, if that relative angle exceeds 60° just use 60.
(a) track = 045° w/v = 075/20 kn: relative angle = 30 = 30/60 × 20 = 10 kn crosswind.
(b) track = 045° w/v = 135/20 kn: relative angle = 90 [use 60] = 60/60 × 20 = 20 kn crosswind.
(c) track = 045° w/v = 195/20 kn: relative acute angle = 30 = 30/60 × 20 = 10 kn crosswind.
• Step 2. Then use the 1-in-60 rule to estimate the wind correction angle by dividing the crosswind component by the TAS and multiplying the result by 60.
(a) and (c) crosswind = 10 kn; TAS = 75 kn: 10/75 × 60 = 8° WCA.
or (b) crosswind = 20 kn; TAS = 75 kn: 20/75 × 60 = 16° WCA.
But combining steps 1 and 2 simplifies the calculation:
WCA = relative angle [60 max] x wind speed / TAS
Example (a) track = 045° TAS = 75 kn; w/v = 075/20 kn: relative angle = 30
WCA = 30 × 20/75 = 8°
And remember that the wind correction is applied in the direction the wind is coming from so that the aircraft crabs along the required track.
• Step 3. Then to estimate the ground speed, deduct the (acute) angle at which the wind meets the track from 115 (for angles up to 60°, use 105 for greater angles) and apply that as a percentage of the wind speed.
(a) track = 045° w/v = 075/20 kn: angle = 30; 115 – 30 = 85% of 20 = 17 knots headwind.
or (b) track = 045° w/v = 135/20 kn: angle = 90; 105 – 90 = 15% of 20 = 3 knots headwind.
or (c) track = 045° w/v = 195/20 kn: angle = 30; 115 – 30 = 85% of 20 = 17 knots tailwind.
Subtract the result from TAS if wind is coming from ahead to abeam, otherwise add. If you like to try a quick mental calculation with the two plots in Figure 4, you will find the arithmetic will produce much the same results as the plots.
You may think it wrong that if the wind is at 90° to the track the ground speed calculation will still come up with a headwind component. This is because the track and the wind velocity are relative to the ground, not to the aircraft's heading. With a wind at 90° to the required track the aircraft must take up a heading having some into-wind component, so that it crabs along the required track; try it by plotting a full wind vector triangle incorporating a wind at 90° to the required track. All the short-cut techniques described are not ultra-precise but they are quite okay for most cross-country navigation in visual meteorological conditions.
You should also read the meteorology module dealing with southern hemisphere winds and particularly section 6.3.
*If the WCA exceeds 10° then reduce the ground speed by an additional value that is a percentage of the TAS, as shown in Table 2. You will note that the adjustment to ground speed really only becomes particularly significant at WCAs above 20° and then, in such conditions, it is probably unwise for light aircraft to be engaged in cross-country flight.
Example 1. The track required is 090°, the wind velocity is 060°/15 knots and TAS is 70 knots. Then the wind angle relative to track is 30° left and, reading from Table 1, the headwind component is –13 and the crosswind component is 7. Thus the ground speed will be 70 –13 = 57 knots, the wind correction angle will be 7/70 × 60 = 6° (to the left) and the heading = 084°.
Example 2. The track required is 300°, the wind velocity is 075°/15 knots and TAS is 70 knots. Then the wind angle relative to track is 135° right and, reading from Table 1, the headwind component is +10 and the crosswind component is 10. Thus the ground speed will be 70 + 10 = 80 knots, the wind correction angle will be 10/70 × 60 = 8° (to the right) and the heading = 308°
Example 3. The track required is 360°, the wind velocity is 075°/20 knots and TAS is 70 knots. Then the wind angle relative to track is 75° right and, reading from Table 1, the headwind component is –5 and the crosswind component is 20. Thus the ground speed will be 70 – 5 = 65 knots, the wind correction angle will be 20/70 × 60 = 16° (to the right) and the heading = 016°. However, because the WCA exceeds 10°, Table 2 is consulted. This shows for a WCA of 16° the ground speed should be further reduced by 3% of the TAS — about 2 knots, so the adjusted ground speed is 63 knots.
The Jeppesen CR2 — available from the Airservices Australia online store navigation and planning accessories — for about $60 is okay and will fit into your pocket — together with a small folding rule. It can be operated with one hand for time and distance calculations.
There are E-6B software apps for smartphones, tablet computers and other personal electronic devices, readily available for a few dollars or possibly as freeware. To find sources, google 'E6b software'. However it is my opinion that the whiz wheel does provide a better grasp of the essentials of the wind triangle.
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Groundschool — Flight Planning & Navigation Guide
| Guide content | 1. Australian airspace regulations | 2. Charts & compass | 3. Route planning |
| [4. Effect of wind] | 5. Flight plan completion | 6.Pre-flight safety and legality check | 7. Airmanship & flight discipline |
| 8. En route adjustment | 9. Supplementary navigation techniques | 10. En route navigation using the GNSS |
| 11. Using the ADF | 12. Electronic flight planning & the EFB | 13. ADS-B surveillance technology |
| Operations at non-controlled airfields | Safety during take-off & landing |
|Section 5 of the Flight Planning & Navigation Guide discusses flight plan completion|
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