AROTA
Meteorology
Weather related rule of thumbs
Estimating cloud base above ground level (ft)
(Surface air temperature - dew point temperature) x 400
If you are flying to an aerodrome without an automated weather information service you can estimate at what altitude above ground you would encounter clouds while climbing or break away from while descending. Example: Recorded surface temperature is 13° and dew point is 7°. Expected cloud base ≈ (13 - 7) x 400 ≈ 2400ft AGL *This rule is just a rough estimate, and it could also indicate the rough height of the cloud ceiling, depending on many factors, such as how many cloud layers there are or stability of the atmosphere, environmental factors, etc., so it may not be perfect. (Note: There is a difference between cloud base and ceiling)
Environmental lapse rate (ISA)
Temperature decreases 2° for every 1000ft
As defined by ICAO, the international standard atmosphere (ISA) temperature decreases 1.98° celcius every 1000ft up to 11km (Where the tropopause is located) According to ICAO, the standard temperature at mean sea level is approximately 15°C. This value is usually the datum to find the ISA temperature at a certain altitude. The actual atmosphere in the real world varies and is not always uniform, but this ROT is used to approximate ISA deviation for aircraft performance calculations such as density altitude, take-off roll, etc.
Barometric pressure (ISA)
Pressure changes 1hPa for every 30ft
Or
Pressure changes 1 inch for every 1000ft
In the international standard atmosphere, pressure changes with altitude/height, and this rule of thumb is usually the first step to deducing pressure altitude. According to ICAO, the standard pressure at sea level is approximately 29.92 inches or 1013 hPa. These two values are usually the datum to find the pressure at a certain altitude. Pressure altitude (PA) = Elevation +1000 (29.92-Altimeter setting(inches). Pressure altitude (PA) = Elevation +30 (1013-QNH(hPa).
Estimating crosswind
Method 1: Clock face method
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10° wind angle represents 10 minutes or 1/6 around the clock face.
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Crosswind = 1/6 x total wind speed
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15° wind angle represents 15 minutes or 1/4 around the clock face.
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Crosswind = 1/4 x total wind speed
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20° wind angle represents 20 minutes or 1/3 around the clock face.
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Crosswind = 1/3 x total wind speed
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30° wind angle represents 30 minutes or 1/2 around the clock face.
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Crosswind = 1/2 x total wind speed
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45° wind angle represents 45 minutes or 3/4 around the clock face.
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Crosswind = 3/4 x total wind speed
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60° wind angle represents 60 minutes or 100% around the clock face.
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Crosswind ≈ total wind speed
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You can visualize this rule of thumb and superimpose it on a clock. Example: You are approaching an airport and have a final approach course of 270°, and reported winds are 290° at 15 knots. The angle between your desired track and wind direction is 20°, which represents 20 mins and 1/3 of the clockface. Simply multiply 15 by 1/3 and you will find that you would encounter a 5 knot crosswind from the right. *If the angle between the wind and desired course is between 60° - 90°, assume full crosswind as the differences between them is negligible *Assuming headwind. If it is a tailwind, tailwind angle = 180 - wind angle. Then use the tailwind angle in the method above with the same results. Example: Approach course: 270° Wind from 110° at 15 kts Wind angle = 270 - 110 = 160° Tailwind angle = 180°-160° = 20° Clockface method: Crosswind = 1/3 x 15 = 5 kts.
Estimating crosswind
Method 2: Rule of Sixths
10° wind angle ≈ 1/6 of wind strength
20° wind angle ≈ 2/6 of wind strength
30° wind angle ≈ 3/6 of wind strength
40° wind angle ≈ 4/6 of wind strength
50° wind angle ≈ 5/6 of wind strength
>60° wind angle ≈ Full wind strength
This rule give a good approximation of the wind strength you would encounter during approach to land and is the easiest to remember. Example: You are approaching a runway with a final approach course of 180°, winds reported at 130° at 16 knots. Wind angle is 50°, and we can round up the wind strengths to the nearest sixth, ≈ 18 knots. So the crosswind wind is 18 x 5/6 ≈ 15 knots. Although the precise trigonomical solution is 12.256 knots, it is accurate enough for practical uses. *Assuming headwind. If it is a tailwind, tailwind angle = 180 - wind angle. Then use the tailwind angle in the method above with the same results. Example: Approach course: 270° Wind from 110° at 15 kts Wind angle = 270 - 110 = 160° Tailwind angle = 180°-160° = 20° Rule of sixths method: Crosswind = 2/6 x 15 = 5 kts.
Estimating crosswind
Method 3: Sine method
Estimating a Sine:
Drop the zero, add 2, move it to one decimal and then multiply wind strength
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As you would recall, the formula for crosswind is sin(wind angle°) x wind strength. This sine method is an arithmetic trick to quickly estimate the sine of any angle. Let's take a look at the angles and their sines: 10°: 0.17 20°: 0.34 30°: 0.50 40°: 0.64 50°: 0.75 60°: 0.86 70°: 0.94 80°: 0.98 As you can see, you can estimate the sine of any angle using this method. For example, let's say the wind angle is 30°, and the wind strength is 25 knots. to get the sine we remove the 0 and get the digit 3. Add 2 to 3 to get 5, and then move it to one decimal place to get 0.5. Therefore, the resulting crosswind component would be 0.5 x 25 ≈ 13 knots *This rule has a normalized accuracy towards the center, and is less accurate with angles less than 20° *Assuming headwind. If it is a tailwind, tailwind angle = 180 - wind angle. Then use the tailwind angle in the method above with the same results. Example: Approach course: 270° Wind from 110° at 15 kts Wind angle = 270 - 110 = 160° Tailwind angle = 180°-160° = 20° Sine method: Crosswind = 0.4 x 15 ≈ 6 kts.
Estimating headwind
Switching from Sine to Cosine
Estimating a cosine:
Reverse the sine, drop the zero, add 2, move it to one decimal and then multiply wind strength
As you would recall, the formula for crosswind is cos(wind angle°) x wind strength. In trigonometry, the angles of cosine and sine add up to 90°. For example, cos 40° = sin 50° (50 + 40 = 90) etc. Once we find the sine, we can use the same arithmetic method to find the headwind component Example: Reported winds are 150° at 20 knots. Runway in use is 180°. Headwind component = Cos 30° x 20 But we know that cos 30° is also = sin 60° (90 - 30). Headwind component is also therefore = 0.8 x 20 = 16 knots. (0.8 is derived by dropping the zero from 60 to get 6, + 2, and then moving it to one decimal place) *The exact calculation is a headwind of 17.3 knots *Assuming headwind
Estimating wind correction angle (WCA)
Step 1: Finding WCA max
WCA max(°) = Wind velocity(kts) ÷ speed number(NM/min)
Speed | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 |
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Speed number | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 | 5 |
WCA is the correction angle applied to the aircraft's course by pointing its nose towards the wind in such a way that it counteracts the effect of wind. WCA max is the maximum correction to apply, assuming the aircraft is experiencing direct 90° crosswind with no headwind or tailwind component. As mentioned in the 1 in 60 rule, memorizing the speed number is handy to deduce the WCA max. Recall 1 in 60 rule: Speed number = Airspeed(kts) ÷ 60 Example: An aircraft at flying at a true airspeed (TAS) of 180 knots is experiencing a 30 knot wind. Wind speed = 30kts Speed number = 3 Regardless of the angle the wind is coming from, we know that the maximum correction we need to apply for a TAS of 180 kts and wind strength of 30kts is = 30/3 = 10°.
Estimating wind correction angle (WCA)
Step 2: Finding WCA
WCA = (WCA max) x crosswind factor
Once we know the WCA max, we use also the crosswind factor we derived from either of the three rules of thumbs above (Clockface, rule of sixths, or sine method) Example: Aircraft is on a course of 350° with a TAS of 210 kts with winds at 330° 30 kts. WCA max = wind strength ÷ speed number = 30 ÷ (210/60) = 30 ÷ 3.5 ≈ 9° WCA for wind angle of 20° (Course - wind direction= 350 - 330): Using sine method: sine 20° ≈ 0.4 WCA = 9 x 0.4 ≈ 3° Since the wind is on the left, if the aircraft flies on a heading of 350° - 3°= 347°, it will be on course.