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Few explanations and useful calculus
by Koala
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Definitions :

• QDR : angle between the magnetic north and the axis station to plane.
Also called radial, the QDR enables to know your position to the tuned station.

• QDM : angle between the magnetic north and the axis plane to station.
The QDM is the heading to fly to turn towards the station.

It is important to remember that : QDR +/- 180° = QDM

In navigation we use foot (ft) for vertical informations, nautical mile (NM ) for distance and knots (kts) for speed.

Useful calculus formulas :

In the vertical axis :

Rate (ft / NM) = slope (degrees) x 100 = slope (%) x 60

D (NM) = Flight level difference (FL) / rate (ft / NM)

Vertical speed (ft / mn) = rate (degrees) x 0,6 x ground speed (kts) = rate (%) x ground speed (kts)

Other formulas :

Base factor FB = 60 / Air speed (AS) (kts) or K (NM/mn) = 1/ FB = AS / 60

T (mn) = D (NM) x FB = D / K

AS= Indicated air speed ( IAS) + FL /3 + temperature variations (°C) (for IAS < 200 kts)

Drift d = lateral wind speed (LWs) x FB = Vt / K

Vt = Wind speed ( Ws) x sin ß (ß is the angle between the plane axis and the wind axis).
We (effective wind) = Ws x cos ß

T (sec/mn ou mn/ H) = We x FB = We / K this is the flying time correction depending on the wind.

Ground speed Gs = As + We

Standard temperature = 15°C at mean sea level and -2°C change per 1000 ft or -6,5°C per 1000 m

To have a better understanding of these formulas, here are some exercices :

  1. The normal slope of descent on final is 5,2% or 3°. Calculate the rate of descent.
  2. At IAS= 180 kts, FL120 and T°=-3°C, calculate the AS and the time to fly 15 NM.
  3. At IAS=150 kts, FL150, wind 280° 12kts, I'm flying on heading 235. Calculate the drift, the heading correction to apply, and the time to fly 15 NM.
  4. At IAS=180kts, FL150, I must fly at 5000 ft within 35 NM. How far from my arrival point should I begin descent with a rate of descent at 1000 ft/NM ? What will my vertical speed be without wind ?
  5. When taking off from Chateaudun (altitude 500 ft moreless), T=16°C. If I climb i the vicinity of the airport, at what altitude, may I ice ?

Answers :

  1. 5,2 x 60 = 312 ft / NM or 3 x 100 = 300 ft/NM.
  2. As= 180+ ( 150 / 3) - 3 = 180 + 50 - 3 = 223 kts.
  3. FB = 60 / 150 = 0,4 or K = 150 / 60 = 2,5. 280-235 = 45°
    LWs = 12 x sin 45 = 8,5 kts and We = 12 x cos45 = 8,5 kts d = 8,5 x 0,4 = 8,5 / 2,5 = 3,4 i.e. 3°.
    The wind comes from the right : 235 + 3 = hdg 238. T = 15 / 0,4 = 15 x 2,5 = 6 min. t = 8,5 x 0,4 = 8,5 / 2,5 = 3 sec / min
    Conclusion T final = 6 min 3 sec.
  4. 15000 - 5000 = 10000 ft of difference. So D = 10000 / 1000 = 10 NM I will have to decent 25 NM from my position.
  5. 16°C at 500 ft therefore at meansea level, We will have 17°C. And so, the 0°C level will be at (17 / 2 ) x 1000 = 8500 ft theorically !

Other exercices are, of courses, easy to find depending on your use or needs.

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