Small Nav Q
Thread Starter
Joined: Apr 2000
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From: West Midlands, UK.
Small Nav Q
I would be most grateful for an explanation as to the solution of the below. Thank you in advance for your help.
Given ETA to cross a meridian is 2100 UTC, GS is 441 kts, and TAS is 491 kts. At 2010 UTC ATC requests a speed reduction to cross the meridian at 2105 UTC. The reduction to TAS will be approximately:
a) 90kt
b) 40kt
c) 60kt
d) 75kt
Given ETA to cross a meridian is 2100 UTC, GS is 441 kts, and TAS is 491 kts. At 2010 UTC ATC requests a speed reduction to cross the meridian at 2105 UTC. The reduction to TAS will be approximately:
a) 90kt
b) 40kt
c) 60kt
d) 75kt
Joined: Feb 2004
Posts: 113
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From: SE UK
You were expecting to do 441 kts GS, so at 2010 you had 50/60 x 441 = 367.5 nm to go.
You now have to do this in 55 mins, so required GS is 367.5 / (55/60) = 401 kts
GS must reduce by 40 kts. Wind velocity hasn't changed so the reduction in TAS is also 40kts. Answer b.
You now have to do this in 55 mins, so required GS is 367.5 / (55/60) = 401 kts
GS must reduce by 40 kts. Wind velocity hasn't changed so the reduction in TAS is also 40kts. Answer b.
Joined: Mar 2002
Posts: 87
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From: UK
Not the easiest to explain, but here goes!
You basically need to start by working out the differences in distance that would have been travelled had the aircraft not slowed down:
ATC ask you to reduce speed at 2010 to cross the meridian at 2105. However, if you don't adjust your speed you will cross it at 2100. Work out how far past the meridian you would fly. In 5 minutes at 441kts you would travel (d=s x t) d=441 x 5/60= 36.75 nm
Therefore the reduction in speed you need to make is calculated by re-arranging the formula, (s=d / t).
Reduction in speed = reduction in distance / time to reduce by distance
= 36.75 / (55/60) [2010 to 2105]
= 40kts
Although you have been using the ground speed to make the calculations, you can assume that the TAS remains constant in relation to the GS, so the speed reduction, (40kts) would be true for both the GS and TAS. If the question asked for the actual TAS required, then you would simply deduct the answer of 40kts from the original TAS.
Hope this makes sense to you and is indeed the correct answer! Instructors please correct me if it's wrong!
You basically need to start by working out the differences in distance that would have been travelled had the aircraft not slowed down:
ATC ask you to reduce speed at 2010 to cross the meridian at 2105. However, if you don't adjust your speed you will cross it at 2100. Work out how far past the meridian you would fly. In 5 minutes at 441kts you would travel (d=s x t) d=441 x 5/60= 36.75 nm
Therefore the reduction in speed you need to make is calculated by re-arranging the formula, (s=d / t).
Reduction in speed = reduction in distance / time to reduce by distance
= 36.75 / (55/60) [2010 to 2105]
= 40kts
Although you have been using the ground speed to make the calculations, you can assume that the TAS remains constant in relation to the GS, so the speed reduction, (40kts) would be true for both the GS and TAS. If the question asked for the actual TAS required, then you would simply deduct the answer of 40kts from the original TAS.
Hope this makes sense to you and is indeed the correct answer! Instructors please correct me if it's wrong!
Jet Blast Rat
Joined: Jan 2001
Posts: 2,081
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From: Sarfend-on-Sea
Pushapproved's answer might be a little more complicated, but it has the advantage that the technique still works when they reverse the question, by telling you that you are to reduce speed by 40 knots and asking you when that should take place.
! You getting on OK?
