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An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kt. The rate of descent of the aircraft is approximately:

a) 650 ft/min

b) 6500 ft/min

c) 4500 ft/min

d) 3900 ft/min

1) Speed x % Gradient = 540 x 12 = 6480.

2) Tan Angle = opp/adjacent

Angle = 6.84 deg (inv tan of 12/100)

Adjacent = 540nm/hr or 9nm/min

Rate of Decent (opp) = Tan 6.84 x 9 = 1.079 nm/min or 6556 ft/min

3) 1 in 60 rule: 6.84 x 9 / 60 = 1.026 nm/min or 6234 ft/min

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An aircraft at FL 350 is required to descend to cross a DME facility at FL80. Maximum rate of descent is 1800 ft/min and mean GS for descent is 276 kt. The minimum range from the DME at which descent should start is:

a) 79 NM

b) 69 NM

c) 49 Nm

d)59Nm

a) 79 NM

b) 69 NM <-- Correct

c) 49 NM

d) 59 NM

S = V x T

27000 to loose @ 1800 fpm = 15 minutes or 0.25 hrs

S = 276 x 0.25 = 69 nm

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An aircraft maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain; its height is approximately:

a) 680 ft

b) 2210 ft

c) 1890 ft

d) 3640 ft

a) 680 ft

b) 2210 ft <-- Correct

c) 1890 ft

d) 3640 ft

First convert gradient into angle. Tan=5.2/100 then do the rest (0.052x7x6076 = 2211). This time its flat terrain so there is no need to add the height above threshold.

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Given:

ILS GP angle = 3.5o

GS = 150 kt

What is the approximate rate of descent?

a) 1000 ft/min

b) 700 ft/min

c) 900 ft/min

d) 800 ft/min

a) 1000 ft/min

b) 700 ft/min

c) 900 ft/min <-- Correct

d) 800 ft/min

1) 1 in 60 rule:

3.5 = ROD/150 x 60

ROD = 3.5 x 150 / 60 = 8.75 nm/hr

or 8.75/60 = 0.145 nm/min or

0.145 x 6080 = 886 feet / min

2) By multiplying Speed with Gradient, its just that gradient needs to be determined first.

tan 3.5 = opp/adjacent

adjacent = 150nm/60 min or 2.5nm/min

opp = 2.5 x tan3.5 = 0.153

gradient = 0.153/2.5 x 100 = 6.1%

6.1 x 150 = 915 fpm

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You are on ILS 3o glideslope which passes over the runway threshold at 50 feet. Your DME range is 25 nm from the threshold. What is your height above the runway threshold elevation? (Use the 1 in 60 rule and 6000 feet = 1 nautical mile)

a) 8010 feet

b) 7450 feet

c) 6450 feet

d) 7550 feet

a) 8010 feet

b) 7450 feet

c) 6450 feet

d) 7550 feet <-- Corrrect

If the given distance (25nm) was from the point where the 3° glide started (beyond the threshold), then 7500 ft would have been the height from touch down. But since the distance is given from the threshold (i.e about a 1000 feet less), we have to add 50 feet (glide path height over the threshold) to get the right answer.

When it says by 1 in 60 then dont do trigonometry because through trigonometry, Height = tan 3 x 150,000 = 7861+50 = 7911 (close to a wrong choice). If its not specified then cross check for the best answer before selecting it. Many answers are observed to be accurate with 1 in 60 rule.

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Create quizAtpl general navigation calculation practical questions helpful to ATP students who are preparing for license exams

17 questions

English

12-09-2020

Hogeschool van Amsterdam / Aviation / General aviation

daniel kariuki nini

I joined Aviation industry in 2013 bringing with me a wealth of experience in flight dispatch and operations. I started my career as a junior licensed flight dispatcher and managed to step up and current i'm the head of flight dispatch and operations in one of the best airline in the world.

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