E6b Flight Computer Exercises ◎ | WORKING |

The E6B flight computer is an analog tool used by pilots to perform aviation calculations. It consists of two sides: the Calculator Side (circular slide rule) and the Wind Side.

Here is a practical guide to the most common E6B exercises, broken down by the side of the computer used.

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Exercise 5: Descent Planning (Top of Descent)

The Scenario: You are at 9,000 feet. Pattern altitude is 1,000 feet (You need to lose 8,000 feet). You want a 500 ft/min descent at 140 knots groundspeed. How far out do you start down?

The Exercise:

1. Time to descend: 8,000 ft ÷ 500 ft/min = 16 minutes.
2. Distance needed: On the calculator side, set the Index (60) to your GS (140 kts).
3. Find 16 minutes on the inner scale. Read the distance on the outer scale. (Answer: ~37.3 NM).

Pro Tip: Add 5-10 NM to slow down before entering the pattern. e6b flight computer exercises

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Exercise 3: Total Fuel Required

Scenario: Your flight plan says 2 hours and 45 minutes en route. Your engine burns 9.5 gallons per hour (GPH).

• Task: Calculate the fuel required for the trip (not including reserves).
• Method: Set the Fuel Index (black triangle on the inner scale, often labeled “US GAL” or “LBS”) to 9.5 on the outer scale. Find 2:45 (which is 2.5 hours + 0.75 hours? No—careful: 2:45 is 2 + 45/60 = 2.75 hours) on the inner scale (time). Read the outer scale.
• Answer: 26.1 gallons.
• Exercise Variation: You have 18 gallons usable. At 8 GPH, how much flight time do you have? (Answer: 2 hours 15 minutes). Never fly to zero.

Part 5: Advanced / Cross-Country Planning

Exercise 14 – Compass Heading from True Heading
Given: True heading = 111°, Variation = 8° W, Deviation = +2°.
Find: Compass heading.

Answer: 121° (West variation is best, so add; then add deviation)

Exercise 15 – Off-Course Correction
Given: Planned to fly 150 NM. After 60 NM, you are 5 NM left of course.
Find: Correction angle to reach destination. The E6B flight computer is an analog tool

Method: Use 60-to-1 rule on E6B.
Answer: 5° right

Exercise 16 – Density Altitude
Given: Pressure altitude = 5,000 ft, OAT = +25°C.
Find: Density altitude.

Answer: ≈ 7,500 ft (use E6B’s density altitude window)

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1 — Groundspeed & Time En Route (circular side)

• Objective: Find groundspeed and time en route for a planned leg.
• Given: True airspeed (TAS) = 120 KT; distance = 150 NM; wind = calm.
• Steps:
  1. On the circular slide, align 120 (TAS) with the index.
  2. Read groundspeed = 120 KT (no wind).
  3. Time = distance / groundspeed = 150 / 120 = 1.25 h → 1 h 15 min.
• Answer: Groundspeed 120 KT; time en route 1 h 15 min.

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Exercise 5: The Decimal Problem

Scenario: You need to multiply $15 \times 3$. Exercise 5: Descent Planning (Top of Descent) The

1. Align 15 (Outer) with 10 (Inner) – which acts as the index for multiplication.
2. Find 30 (Inner) which represents 3.
3. Look at the Outer scale. The number is roughly 45.
4. Logic Check: The E6B gives you "45". It does not tell you if it is 4.5, 45, or 450. You must know that $15 \times 3$ cannot be 4.5 or 450. Therefore, the answer is 45.

Part 3: Fuel Consumption Problems

Exercise 7 – Fuel for a Leg
Given: Fuel burn = 8.5 GPH, Time = 2 hours 20 minutes.
Find: Fuel required.

Answer: 19.8 US gal

Exercise 8 – Endurance
Given: Usable fuel = 36 US gal, Burn rate = 9.2 GPH.
Find: Endurance (hours & minutes).

Answer: 3 hr 55 min

Exercise 9 – Gallons per Hour from Total Trip
Given: 240 NM trip, groundspeed 120 kt, total fuel used = 16 gal.
Find: Fuel burn rate (GPH).

Answer: 8.0 GPH

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7 — Drift & Ground Track (exercise for pronounced crosswind)

• Objective: Compute drift angle and ground track.
• Given: Course = 090°; TAS = 90 KT; wind = 350° at 35 KT.
• Steps:
  1. Wind relative angle = 350 − 90 = 260° → equivalent to −100° (cross from left).
  2. Crosswind component = 35 × sin100° ≈ 34.4 KT → WCA = arcsin(34.4 / 90) ≈ 22.6° (left).
  3. Heading = 090 + 22.6 = 112.6° (fly right to counter left drift).
  4. Headwind component = 35 × cos100° ≈ −6.09 KT → groundspeed ≈ 90 − 6.1 = 83.9 KT.
  5. Ground track will be 090° (desired) if heading corrected as above.
• Answer: Drift/WCA ≈ 23° left; heading ≈ 113°; groundspeed ≈ 84 KT.

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