You leave A to fly to B, 475 nm away, at 1000 hours. Your ETA at B is 1130. At 1040 you are 190 nm from A. What ground speed is required to arrive on time at B?

a) 317 knots

b) 330 knotsc) 342 knots <-- Correctd) 360 knots

Distance to go 475-190 = 285

Time available = 1130-1040 = 50 minutes

Speed required = 285/0.833 = 342 knots

You are flying at a True Mach No of 0.82 in a SAT of -45oC. At 1000 hours you are 100 nm from the POL DME and your ETA at POL is 1012. ATC ask you to slow down to be at POL at 1016. What should your new TMN be if you reduce speed at 100 nm distance to:

a) M .76

b) M .72

c) M .68d) M .61 <-- Correct

Required Speed = Distance / Time = 100 / 16min = 375 kts

Local Speed of Sound = 39 x Square root of (-45+273) = 589

Mach Number = 375/589 = 0.63 (0.61 is probably through CRP-5)

An aircraft at FL 310, M0.83, temperature -30oC, is required to reduce speed in order to cross a reporting point five minutes later than planned. Assuming that a zero wind component remains unchanged, when 360 NM from the reporting point Mach Number should be reduced to:

a) M 0.76b) M 0.74 <-- Correctc) M 0.78

d) M 0.80

Determine the TAS. Find the time to cover 360 nm at this TAS. Increase this time by 5 minutes. Find the TAS again for this increased time. Convert it back to Mach.

Mach = TAS/LSS. LSS = Local Speed of Sound which is 39 x square root of Absolute temperature (temp in centigrade + 273).

Given: Distance A to B is 475 NM, Planned GS 315 kt, ATD 1000 UTC, 1040 UTC - fix obtained 190 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at B?

a) 320 kt

b) 360 kt

c) 300 ktd) 340 kt <-- Correct

Planned Time to B = 475/315 = 1:30

Planned ETA = 1000 (ATD) + 1:30 = 1130

190 nm point reached in 40 min, therefore GS is 285

Time left to be on time (1130) at B = 0:50 mins (1130-1040)

GS required = Distance left / time = 285/.833 = 342 kts

An aircraft is planned to fly from position A to position B, distance 480 NM at an average GS of 240 kt. It departs A at 1000 UTC. After flying 150 NM along track from A, the aircraft is 2 min behind planned time. Using the actual GS experienced, what is the revised ETA at B?

a) 1203b) 1206 <-- Correctc) 1153

d) 1157

The point where it was observed that the aircraft is 2 min behind the planned time is 150 nm from A. Call it "X".

According to the planned GS of 240 ETA at X = 150/240 = 37.5 mins from A.

It was observed that at X the aircraft was 2 min behind planned ETA. That means that it took 37.5 + 2 = 39.5 mins from A.

Time at X = 1000 + 39.5 minutes = 1039.

Ground speed experienced = 150/0.658 = 228 kts.

Revised ETA at B will be:

GS = 228

Distance to go 480-150 = 330

330/228 = 1.447 or 1 hr and 27 mins from point X.

Time at X (1039) + Time to go to B (1:27) = ETA B (1206)

Half way between two reporting points the navigation log gives the following information:

TAS 360 kt

W/V 330o/80 kt

Compass heading 237 o

Deviation on this heading -5o

Variation 19o W

What is the average ground speed for this leg?

a) 360 kt

b) 354 kt

c) 373 ktd) 403 kt <-- Correct

Course is not given. On E6B, determine the tailwind component and apply to TAS.

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Good luck,

CaptainMFT.