Gen Nav Problem
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Gen Nav Problem
*An aircraft at FL350 is required to cross a VOR/DME facility at FL110 and to commence descent when 100 NM from the facility. If the mean GS for the descent is 335 kt, the minimum rate of descent required is:
Correct answer : 1340 FT/MIN
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I cannot seem to find the way to calculate this !! WHY ??
Can anyone point out the "how" in this example.
Thanks in advance.
Proxus
If you are interrested I have another one that I can't complete:
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An aircraft is flying TAS 180 knots and tracking 090°T. The W/V is 045/50. How far can the aircraft fly out from its base and return within 1 hour?
Correct answer : 85 nm
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>> here's yet another one I can't figure out:
*An aircraft at FL330 is rerquired to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required?
Correct answer : 1950 FT/MIN
Correct answer : 1340 FT/MIN
---------------------------------------------------
I cannot seem to find the way to calculate this !! WHY ??
Can anyone point out the "how" in this example.
Thanks in advance.
Proxus
If you are interrested I have another one that I can't complete:
------------------------------------------------------------------------------
An aircraft is flying TAS 180 knots and tracking 090°T. The W/V is 045/50. How far can the aircraft fly out from its base and return within 1 hour?
Correct answer : 85 nm
-------------------------------------------------------------------------------
>> here's yet another one I can't figure out:
*An aircraft at FL330 is rerquired to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required?
Correct answer : 1950 FT/MIN
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Problem 1 - First of all you need to find how many minutes it's going to take to get to your Vor.
100nm / 335kts * 60 = 17.9nm per minute
Then you need to calculate how much altitude you have to descend in that time.
35000' - 11000' = 24000'
Divide 24000' by 17.9nm per minute = 1340 Ft/min
Problem 2 is a psr question I think (can't work it out because I don't have my CRP with me).
Problem 3 is the same as the first problem
65nm / 330kts * 60 = 11.8nm per minute
33000' - 10000' = 23000'
23000' / 11.8nm = 1949 Ft/min
Hope this helps you out.
100nm / 335kts * 60 = 17.9nm per minute
Then you need to calculate how much altitude you have to descend in that time.
35000' - 11000' = 24000'
Divide 24000' by 17.9nm per minute = 1340 Ft/min
Problem 2 is a psr question I think (can't work it out because I don't have my CRP with me).
Problem 3 is the same as the first problem
65nm / 330kts * 60 = 11.8nm per minute
33000' - 10000' = 23000'
23000' / 11.8nm = 1949 Ft/min
Hope this helps you out.
Last edited by Dah Dah...Dit; 12th May 2004 at 13:53.
Northern Monkey
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ill give question 2 a shot...
An aircraft is flying TAS 180 knots and tracking 090°T. The W/V is 045/50. How far can the aircraft fly out from its base and return within 1 hour?
Using these figures in my crp, the aircraft will fly at a GS of 141kts outbound, and return with a ground speed of 211kts
If t is measured in hrs
141 x t = 211 x (1-t)
141t/211 = (1-t)
0.668 t = 1-t
1.668 t = 1
t = 0.599
so the ac flys at 141kts for 0.599 of an hr
so travels for 0.599 x 141 = 84.5 NM
we can check this as the aircraft will return for (1-0.599) hrs at 211kts
= 0.401 x 211
= 84.6NM (rounding errors as im only using 3 dp for calcs)
regards
NB
edited beacuse i cant read my CRP very well!!
An aircraft is flying TAS 180 knots and tracking 090°T. The W/V is 045/50. How far can the aircraft fly out from its base and return within 1 hour?
Using these figures in my crp, the aircraft will fly at a GS of 141kts outbound, and return with a ground speed of 211kts
If t is measured in hrs
141 x t = 211 x (1-t)
141t/211 = (1-t)
0.668 t = 1-t
1.668 t = 1
t = 0.599
so the ac flys at 141kts for 0.599 of an hr
so travels for 0.599 x 141 = 84.5 NM
we can check this as the aircraft will return for (1-0.599) hrs at 211kts
= 0.401 x 211
= 84.6NM (rounding errors as im only using 3 dp for calcs)
regards
NB
edited beacuse i cant read my CRP very well!!
Last edited by NinjaBill; 12th May 2004 at 10:56.
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unowho , I've now changed the typo error from 355kts to 335kts. I can see how this would have totally thrown you off track regarding the question Mr Simpleton. I hope this is satisfactory for you, if not, just let me know!
TightYorksherMan
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An aircraft is flying TAS 180 knots and tracking 090°T. The W/V is 045/50. How far can the aircraft fly out from its base and return within 1 hour?
Formula:
EOH / O+H
E = endurance which is 1hr
O = groundspeed out
H = groundspeed home
You need to work out the ground speed using your whiz wheel or CRP
Add it to the formula just to be on the safe side I would do E x O x H then press equals
Add O + H then divide first figure by second figure to avoid any calculator errors, good luck!!
An aircraft at FL330 is rerquired to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required?
Height to loose
____________
distance / speed x 60
height to loose = 330 - 100 = 23,000ft
distance is 65nm
Speed is 330kt
so,
23,000
--------
65 / 330 x 60
Answer is 1946ft/min
Formula:
EOH / O+H
E = endurance which is 1hr
O = groundspeed out
H = groundspeed home
You need to work out the ground speed using your whiz wheel or CRP
Add it to the formula just to be on the safe side I would do E x O x H then press equals
Add O + H then divide first figure by second figure to avoid any calculator errors, good luck!!
An aircraft at FL330 is rerquired to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required?
Height to loose
____________
distance / speed x 60
height to loose = 330 - 100 = 23,000ft
distance is 65nm
Speed is 330kt
so,
23,000
--------
65 / 330 x 60
Answer is 1946ft/min