Untitled

1. Units of Measurement and Flight Safety
  • Significance of Unit Consistency:

    • In the 1983 Air Canada incident (the "Gimli Glider"), ground personnel computed fuel in pounds, while flight crew calculated requirements in kilograms, causing the aircraft to exhaust fuel mid-flight.

    • Unit consistency is a direct flight safety requirement, not an administrative detail.

  • Fundamental Quantities (SI Base Units):

    • Length: metre (m\text{m})</p></li><li><p>Mass:kilogram()</p></li><li><p>Mass: kilogram (\text{kg})

    • Time: second (s\text{s})</p></li><li><p>Temperature:kelvin()</p></li><li><p>Temperature: kelvin (\text{K})

  • Derived Quantities:

    • Velocity: metres per second (m/s\text{m/s})</p></li><li><p>Acceleration:metrespersecondsquared()</p></li><li><p>Acceleration: metres per second squared (\text{m/s}^2)

    • Force: newton (N\text{N})</p></li><li><p>Energy:joule()</p></li><li><p>Energy: joule (\text{J})

    • Power: watt (W\text{W})

  • Standard Aviation Units & Common Conversions:

    • Altitude: feet (\text{ft})

    • Speed: knots (kt\text{kt})</p></li><li><p>Distance:nauticalmiles()</p></li><li><p>Distance: nautical miles (\text{NM})

    • Mass: kilograms (kg\text{kg})orpounds() or pounds (\text{lb})

    • Key conversion factors:

    • 1 m=3.281 ft1\,\text{m} = 3.281\,\text{ft}

    • 1 kt=1.852 km/h1\,\text{kt} = 1.852\,\text{km/h}

    • 1 NM=1.852 km1\,\text{NM} = 1.852\,\text{km}

    • 1 kg=2.205 lb1\,\text{kg} = 2.205\,\text{lb}

    • Altitude calculation example: Bole elevation of 2334 m×3.281 ft/m≈7657 ft2334\,\text{m} \times 3.281\,\text{ft/m} \approx 7657\,\text{ft}.

  • Conversion Relay Exercises:

    1. Bole Elevation: Convert 7657 ft7657\,\text{ft} to metres: 76573.281≈2334 m\frac{7657}{3.281} \approx 2334\,\text{m}.

    2. Cruise Speed: Convert 850 km/h850\,\text{km/h} to knots: 8501.852≈459 kt\frac{850}{1.852} \approx 459\,\text{kt}.

    3. Route Distance: Convert 1360 NM1360\,\text{NM} to kilometres: 1360×1.852=2518.72 km1360 \times 1.852 = 2518.72\,\text{km}.

    4. Fuel Load: Convert 20000 lb20000\,\text{lb} to kilograms: 200002.205≈9070.3 kg\frac{20000}{2.205} \approx 9070.3\,\text{kg}.

2. Distance vs. Displacement
  • Distance (Scalar):

    • Total length of the path traveled without regard to direction.

    • Directly determines the amount of fuel required for an operation.

  • Displacement (Vector):

    • Net straight-line change in position from departure to arrival point, including directional orientation.

    • Directly influences directional tracking, navigation, and heading selection.

  • Aviation Case (Addis Ababa to Nairobi):

    • Path distance traveled: 1300 km1300\,\text{km}

    • Net displacement: 1160 km1160\,\text{km} South-Southwest (SSW)

3. Speed vs. Velocity
  • Speed (Scalar):

    • Defined as total distance traveled per unit time:
      Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}

  • Velocity (Vector):

    • Defined as total displacement per unit time, maintaining direction:
      vavg=ΔxΔt\mathbf{v}_{\text{avg}} = \frac{\Delta \mathbf{x}}{\Delta t}

  • Flight Application (Addis Ababa to Nairobi):

    • Given displacement Δx=1160 km\Delta x = 1160\,\text{km} and elapsed time t=2.5 ht = 2.5\,\text{h}:
      vavg=1160 km2.5 h=464 km/h\mathbf{v}_{\text{avg}} = \frac{1160\,\text{km}}{2.5\,\text{h}} = 464\,\text{km/h}

4. Acceleration and Deceleration
  • Definition & Formula:

    • The rate of change of velocity over time:
      a=v<em>f−v</em>ita = \frac{v<em>f - v</em>i}{t}

    • SI Unit: m/s2\text{m/s}^2

  • Takeoff Run:

    • Accelerating from rest (0 m/s0\,\text{m/s}) to rotation speed (80 m/s80\,\text{m/s}) in 35 s35\,\text{s}:
      a=80 m/s−0 m/s35 s≈2.29 m/s2a = \frac{80\,\text{m/s} - 0\,\text{m/s}}{35\,\text{s}} \approx 2.29\,\text{m/s}^2

    • Equivalent acceleration in terms of Earth's gravity (g≈9.8 m/s2g \approx 9.8\,\text{m/s}^2):
      Load Factor≈2.299.8≈0.23 g\text{Load Factor} \approx \frac{2.29}{9.8} \approx 0.23\,g

  • Landing Rollout:

    • Decelerating from touchdown speed (70 m/s70\,\text{m/s}) to a complete stop (0 m/s0\,\text{m/s}) in 25 s25\,\text{s}:
      a=0 m/s−70 m/s25 s=−2.8 m/s2a = \frac{0\,\text{m/s} - 70\,\text{m/s}}{25\,\text{s}} = -2.8\,\text{m/s}^2

    • The negative sign signifies deceleration (acceleration vector opposing the direction of forward flight).