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The effect of wind
Rev. 23 — page content was last changed October 12, 2009
consequent to editing by RA-Aus member Dave Gardiner
|Flight Planning and Navigation|
An aircraft in flight is airborne and subject to the movement of the air mass in relation to the surface; i.e. the wind. The relatively low cruising speed of light aircraft makes them particularly affected by wind velocity. Consequently the calculation of the wind effect on aircraft movement relative to the ground is a major part of light aircraft flight planning and navigation.basic forces module of the flight theory section we said it is common practice to estimate resultant forces non-mathematically by drawing scaled, arrowed lines to represent each vector quantity; this produces the resultant of two vector quantities in a vector triangle or parallelogram. The lengths of the lines represent the magnitude of each force, and the placements indicate the application points and directions. We also know that an aircraft in flight is airborne, and consequently both the path it projects over the ground and its speed relative to the ground are the resultant of the aircraft velocity and the wind velocity.
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 cruising at 75 knots TAS. At the time of departure, the wind velocity at the cruise altitude is 135°/20 knots; i.e. the 20-knot 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 miles 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.
The wind triangleSo, if we want to track over the direct route from Alpha to Beta we will have to ascertain both the wind velocity at the time of flight and a heading to fly that will provide the necessary crosswind correction angle. Remember that velocity vectors have both speed and direction. In the wind triangle we have only one completely known vector — the forecast wind velocity. We know part of the heading vector — the true airspeed — but not the direction. We also know part of the resultant vector — the direction (required track) from Alpha to Beta — but not the ground speed.
We can determine the two unknowns — the heading and the ground speed — by plotting scaled vectors on paper. You will need some drawing instruments, a protractor and ruler, but a pair of compasses or 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 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 knots (Figure 2).
• Using the scale, open up the dividers or compasses 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/compass 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. Its orientation with true north is the heading (060°T) and its length is the TAS. 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 track (045°T) and the heading (060°T) — is 15°, and the drift will be to the left — also known as port drift.
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 were a full crosswind, 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.
Direct headwind/tailwindIf the wind is aligned directly with the required track then of course it is not possible to construct the triangle, as there is no wind correction angle and the ground speed is the TAS ± wind speed. However, just as an illustration that the wind triangle still provides the correct answers, I have repeated the previous Alpha to Beta plot with winds that are only 10° off the required track; i.e. nearly full headwind and tailwind components.
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 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.
Plotting the wind vector triangle is the most accurate method for ascertaining heading and ground speed, but there are two other methods that are quite accurate enough for light aircraft cross-country navigation.
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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.
The trigonometrical relationships of the two wind components — crosswind and headwind/tailwind — is shown in a wind triangle (Figure 5). In this example the wind angle is 30° relative to the required track and the wind speed is 20 knots. The sine of an angle = opposite side/hypotenuse, while the cosine of an angle = adjacent side/hypotenuse. In this wind triangle the hypotenuse represents the wind velocity vector, the side opposite to the angle represents the crosswind component of the wind velocity vector and the adjacent side represents the headwind component of the wind velocity vector. An abridged trigonometric table is contained in the Flight Theory manoeuvring forces module. Reading from that table, sine 30° is 0.5 and cosine 30° is 0.866 — near enough to 0.9.
Using 1-in-60 to estimate WCAThe two/three-step technique described below approximates the sine/cosine relationships and produces results near enough to the trig calculations.
• 1. First find the crosswind component of the forecast wind velocity by estimating the (acute) angle at which the wind meets the required track, divide that by 60 and multiply the result by the wind speed. However, if the 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 angle = 30 = 30/60 × 20 = 10 kn crosswind.
• 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.
• 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.
You should also read the meteorology module dealing with southern hemisphere winds and particularly section 6.3.
Using tables for ground speed and WCAThe third and simplest method for estimating WCA, heading and ground speed is to use tables such as those following. Table 1 is for wind speeds up to 30 knots in 5-knot intervals, and for wind angles relative to either side of the required track between 0° and 180°. In the table you will see that headwinds have a negative adjustment and tailwinds a positive adjustment for ground speed. However if the calculated WCA exceeds about 10° the inbuilt crab problem becomes apparent and a small additional calculation to derive a more accurate ground speed has to be made (Table 2).
*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.
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The Jeppesen CR2, available from the Airservices Australia online store navigation and planning accessories for about A$50 is okay and will fit into your pocket — together with a small folding rule — and can be operated with one hand for time and distance calculations.
There are hand-held E-6B calculators or 'computers', costing around A$150, which do much the same job as the whiz wheels. I think that all the in-flight variables, to which light aircraft flying at comparatively low levels are subject, negate any potential cost/benefit advantage of such expensive single purpose devices. However, E-6B software utilities for PALM OS and POCKET PC handheld computers are readily available for about US$20 — or possibly as freeware. To find sources, google 'E6b software'.
It is my opinion that the whiz wheel gives a navigator a better grasp of the essentials of the wind triangle and thus makes it easier to mentally envisage in-flight corrections/adjustments without the need to fiddle with the whiz wheel or an electronic E-6B device. Money available for flight planning and navigation aids is well spent if you install a top quality magnetic compass and spend some time measuring, adjusting and recording the compass deviation in situ — and recheck deviation once or twice a year.
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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. Safety audit | 7. Airmanship & flight discipline |
| 8. En route adjustment | 9. Supplementary navigation technique | 10. Global Positioning System |
| 11. Using the ADF | 12. Electronic planning & navigation | 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|
Copyright © 2001–2009 John Brandon [contact information]