FlightCalculators

Worked Examples

Real flight planning rarely uses one calculator in isolation. These six scenarios show how several tools work together — from a mountain departure to an IFR fuel plan — with every step shown so you (or your students) can follow the reasoning and reproduce it. They make good CFI ground-lesson material.

The numbers below are illustrative and rounded to show the method. Aircraft-specific figures (weights, arms, performance distances, demonstrated crosswind) always come from your aircraft’s POH/AFM, and weather values from an official briefing. Use these as teaching examples, not as flight data.

1. High-Density-Altitude Mountain Departure

Density AltitudePressure Altitude

You are departing a mountain airport at 9,900 ft field elevation. The altimeter reads 30.10 inHg and the OAT is 20 °C — a warm day. How thin is the air really, and what does that mean for takeoff?

  1. 1Pressure Altitude
    Given
    Field elevation 9,900 ft, altimeter 30.10 inHg
    Method
    PA = elevation + (29.92 − altimeter) × 1,000 = 9,900 + (29.92 − 30.10) × 1,000
    Result
    Pressure altitude ≈ 9,720 ft
  2. 2Density Altitude
    Given
    PA 9,720 ft, OAT 20 °C (ISA at this level ≈ −4 °C)
    Method
    DA ≈ PA + 120 × (OAT − ISA temp) = 9,720 + 120 × (20 − (−4))
    Result
    Density altitude ≈ 12,600 ft
Takeaway: The airplane will perform as if it were at roughly 12,600 ft. Expect a much longer ground roll and a weak climb rate. Consult your POH high-density-altitude performance charts, lean for best power, and consider a cooler departure time.

2. Complete Cross-Country Planning

Weight & BalanceWind CorrectionTime-Speed-DistanceFuel Burn

A 130 nm cross-country in a four-seat trainer. You need to confirm you are loaded within limits, work out the heading and groundspeed in wind, then confirm you have the fuel with a legal VFR reserve. (Weight & balance figures below are illustrative — always use your own aircraft’s POH.)

  1. 1Weight & Balance
    Given
    Empty 1,600 lb @ 39.0 in; front seats 340 lb @ 37.0; fuel 40 gal (240 lb) @ 48.0; baggage 30 lb @ 95.0
    Method
    Total weight = 2,210 lb; total moment = 89,350 lb-in; CG = 89,350 ÷ 2,210
    Result
    Gross weight 2,210 lb and CG ≈ 40.4 in — within a typical envelope
  2. 2Wind Correction Angle
    Given
    True course 090°, TAS 110 kt, wind from 040° at 18 kt
    Method
    WCA = arcsin(18 × sin 50° ÷ 110); GS = 110 × cos(WCA) − 18 × cos 50°
    Result
    WCA ≈ 7° right → heading ≈ 097°; groundspeed ≈ 98 kt
  3. 3Time-Speed-Distance
    Given
    Distance 130 nm, groundspeed 98 kt
    Method
    ETE = distance ÷ groundspeed = 130 ÷ 98
    Result
    En-route time ≈ 1 h 20 min
  4. 4Fuel Burn
    Given
    Burn 8.5 gph, ETE 1.33 h, plus 30 min VFR day reserve (91.151)
    Method
    Trip fuel = 1.33 × 8.5 = 11.3 gal; reserve = 0.5 × 8.5 = 4.3 gal
    Result
    Fuel required ≈ 15.6 gal — well under the 40 gal on board
Takeaway: Loaded within limits, a 097° heading holds your course, and 40 gal gives a large margin over the 15.6 gal required. Chaining the calculators turns four separate numbers into a single go decision.

3. IFR Fuel Planning with Alternate

Fuel BurnReserve FuelEndurance

An IFR flight that requires an alternate. Under 14 CFR 91.167 you must carry fuel to the destination, then to the alternate, then 45 minutes at normal cruise. Do you have it?

  1. 1Fuel Burn
    Given
    Cruise burn 11 gph; destination ETE 2 h 00 min; alternate ETE 0 h 40 min
    Method
    Destination = 2.0 × 11 = 22 gal; alternate = 0.667 × 11 = 7.3 gal
    Result
    Destination + alternate ≈ 29.3 gal
  2. 2Reserve Fuel
    Given
    91.167 IFR reserve = 45 min at normal cruise, burn 11 gph
    Method
    Reserve = 0.75 × 11
    Result
    Reserve ≈ 8.25 gal → total required ≈ 37.6 gal
  3. 3Endurance
    Given
    Usable fuel 50 gal, burn 11 gph
    Method
    Endurance = 50 ÷ 11
    Result
    Endurance ≈ 4 h 33 min — comfortably above the required 37.6 gal
Takeaway: With 50 gal usable you satisfy the 37.6 gal IFR requirement and still hold roughly an hour beyond legal reserves. If the alternate were farther, re-run the reserve and endurance steps before committing.

4. Gusty Crosswind Landing Assessment

Crosswind Component

Landing on Runway 27. The wind is 310° at 22 kt gusting 28 kt, and your aircraft lists a 15 kt maximum demonstrated crosswind. Is this within your and the airplane’s limits?

  1. 1Crosswind Component
    Given
    Runway heading 270°, wind 310° (40° off), steady 22 kt
    Method
    Crosswind = 22 × sin 40°; headwind = 22 × cos 40°
    Result
    Steady crosswind ≈ 14 kt; headwind ≈ 17 kt
  2. 2Crosswind Component
    Given
    Same 40° angle applied to the 28 kt gust
    Method
    Gust crosswind = 28 × sin 40°
    Result
    Gust crosswind ≈ 18 kt — above the 15 kt demonstrated value
Takeaway: The steady-state crosswind (14 kt) is within limits, but gusts drive it to ~18 kt, beyond the demonstrated 15 kt. Consider a different runway, waiting for the gusts to ease, or diverting — the demonstrated value is not a limitation, but exceeding it demands proven skill and honest self-assessment.

5. Top-of-Descent and Rate Planning

Top of DescentRequired Rate of Climb/Descent

Cruising at 9,500 ft, you want to arrive at a 1,000 ft MSL pattern altitude. Where do you start down, and what descent rate keeps it comfortable at 120 kt groundspeed?

  1. 1Top of Descent
    Given
    Altitude to lose = 9,500 − 1,000 = 8,500 ft; descent rate 500 fpm; GS 120 kt (2 nm/min)
    Method
    Descent time = 8,500 ÷ 500 = 17 min; distance = 17 × 2
    Result
    Begin descent ≈ 34 nm from the airport
  2. 2Required Rate of Climb/Descent
    Given
    Prefer a 3° path at 120 kt groundspeed
    Method
    Rate ≈ 5 × GS = 5 × 120; 3° distance ≈ altitude ÷ 300 = 8,500 ÷ 300
    Result
    About 600 fpm, starting ≈ 28 nm out for a 3° profile
Takeaway: A 500 fpm descent means starting down ~34 nm out; a steeper 3° profile needs ~600 fpm from ~28 nm. Pick the profile that fits terrain, airspace, and passenger comfort, and set your target before you get busy in the pattern.

6. VFR Weather Go / No-Go Check

Cloud BaseFreezing LevelFlight Category

Surface temperature is 18 °C with a 10 °C dewpoint at a 1,000 ft field, and reported visibility is 8 sm. Will clouds and freezing levels cooperate for a VFR flight?

  1. 1Cloud Base
    Given
    Temperature 18 °C, dewpoint 10 °C (spread 8 °C)
    Method
    Cloud base AGL ≈ (T − Td) ÷ 2.5 × 1,000 = 8 ÷ 2.5 × 1,000
    Result
    Estimated cumulus base ≈ 3,200 ft AGL
  2. 2Freezing Level
    Given
    Surface 18 °C at 1,000 ft, standard lapse ≈ 2 °C per 1,000 ft
    Method
    Height to 0 °C ≈ 18 ÷ 2 × 1,000 = 9,000 ft above the surface
    Result
    Freezing level ≈ 10,000 ft MSL
  3. 3Flight Category
    Given
    Ceiling ≈ 3,200 ft AGL, visibility 8 sm
    Method
    Compare against VFR/MVFR/IFR thresholds
    Result
    Ceiling > 3,000 ft and visibility > 5 sm → VFR
Takeaway: Conditions are VFR with a comfortable cloud base and a freezing level well above your planned altitude — icing is not a concern for a low-level VFR flight. Always confirm with a full preflight weather briefing; these estimates supplement, not replace, official products.

Want the underlying formulas and sources? See our methodology page, and the blank printable worksheets to run your own numbers.