Help needed with some Gen Nav questions!!
Thread Starter
Joined: Jul 2002
Posts: 36
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From: UK
Help needed with some Gen Nav questions!!
I’ve got my Gen Nav exam next week. I’m rather stuck with some feedback questions. They are probably dead obvious and im dead thick but I would appreciate any help from the experts out there.
1. An aircraft is fling TAS 180kts and tracking 180 T. the w/v is 045/50. how far can the aeroplane flyout of its base and return within 1 hour?
2. By what amount must you change your rate of descent given a 10knot increase in headwind on a 3 glideslope?
3. an aircraft starts at position 0410S 17822W and heads true north for 2950nm, then turns 90 degrees left and maintains a rhumb line track for 314 km. What is its finial position?
4. 5hours 20 minutes and 20 seconds hour time difference is equivalent to which change of longitude?
5. on a 12% glideslope your groundspeed is 540knots what is your rate of descent
6. An aircraft departs a point 0400N 17000W and flies 600nm South, followed by 600nm east then 600nm north then 600nm west. what is its finial position?
7. At what lat does maximum difference between geodetic and geocentic lat occur??
8. TAS 240knots. The relative bearing from an NDB is 270 R at 1410. At 1420 the bearing has changed to 315R. What was your distance from the NDB at 1410?
Many thanks!!!!
1. An aircraft is fling TAS 180kts and tracking 180 T. the w/v is 045/50. how far can the aeroplane flyout of its base and return within 1 hour?
2. By what amount must you change your rate of descent given a 10knot increase in headwind on a 3 glideslope?
3. an aircraft starts at position 0410S 17822W and heads true north for 2950nm, then turns 90 degrees left and maintains a rhumb line track for 314 km. What is its finial position?
4. 5hours 20 minutes and 20 seconds hour time difference is equivalent to which change of longitude?
5. on a 12% glideslope your groundspeed is 540knots what is your rate of descent
6. An aircraft departs a point 0400N 17000W and flies 600nm South, followed by 600nm east then 600nm north then 600nm west. what is its finial position?
7. At what lat does maximum difference between geodetic and geocentic lat occur??
8. TAS 240knots. The relative bearing from an NDB is 270 R at 1410. At 1420 the bearing has changed to 315R. What was your distance from the NDB at 1410?
Many thanks!!!!
Joined: Sep 2001
Posts: 33
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From: I don't know!
Kapooley,
Here are the answers to your questions:
1. This is a PSR calculation. First thing to do is to work out your Speed out and your speed home by using the windside of your CRP-5 using the given track for the speed out and the reciprocal for the speed home.
Speed out on 180 track is 212kts GS
Speed home on 360 track is 141kts GS
Then simply the equation is:
time x speed home
------------------------
speed out + speed home
= 1x141
-------- = 0.399 (time) x speed out to give distance = 85nms
212+141
2. This question is quite easy to answer as you may know that on a 3 degree glideslope the ROD = Ground Speed x 5, so imagine you're doing 100kts, which equals 500ft/min ROD. Then do the same taking into account your 10kt headwind (90x5), which is 450ft/min ROD so the rate of descent must decrease by 50ft/min to maintain the required 3 degree glide slope.
3.
4. So that you understand this one better: Imagine the sun moves once round the earth every 24 hours. As there are 360 degrees around the earth, if you divide 360 by 24 this gives you the apparent movement of the sun per hour, which is 15 degrees/hr. Therefore, 5 hours 20 mins and 20 secs x 15 (5.338 x 15) equals 80 degrees and 5 minutes.
5. Careful on this one. It's saying 12% and not 12 degrees.
On the CRP-5 align the inner blue 10 with 12 on the outer. Then against the black 60 triange read 7.2, which is your degrees. Now align the black 60 triangle with your speed (540) and read off the 7.2 degrees glide slope on the inner and the outer is your required ROD = approx 6500ft/min. Rough way to make sure that you've got the right amount of zero's. 540x5=2700 (on a 3 degree glide slope) times by 2 and a bit to get it close to the 7.2 degrees, which is roughly 6210ft/min. This will save you ticking the box that says 650ft/min!
6.
7. 45 degrees N/S. Don't try to understand it, just remember it!
8.
I'll try and answer the other ones a bit later. It's just that I'm a bit short on time and they may take a bit of typing to explain. Just wanted to give you some help.
Good luck on Wednesday mate!!
Barnaby
Here are the answers to your questions:
1. This is a PSR calculation. First thing to do is to work out your Speed out and your speed home by using the windside of your CRP-5 using the given track for the speed out and the reciprocal for the speed home.
Speed out on 180 track is 212kts GS
Speed home on 360 track is 141kts GS
Then simply the equation is:
time x speed home
------------------------
speed out + speed home
= 1x141
-------- = 0.399 (time) x speed out to give distance = 85nms
212+141
2. This question is quite easy to answer as you may know that on a 3 degree glideslope the ROD = Ground Speed x 5, so imagine you're doing 100kts, which equals 500ft/min ROD. Then do the same taking into account your 10kt headwind (90x5), which is 450ft/min ROD so the rate of descent must decrease by 50ft/min to maintain the required 3 degree glide slope.
3.
4. So that you understand this one better: Imagine the sun moves once round the earth every 24 hours. As there are 360 degrees around the earth, if you divide 360 by 24 this gives you the apparent movement of the sun per hour, which is 15 degrees/hr. Therefore, 5 hours 20 mins and 20 secs x 15 (5.338 x 15) equals 80 degrees and 5 minutes.
5. Careful on this one. It's saying 12% and not 12 degrees.
On the CRP-5 align the inner blue 10 with 12 on the outer. Then against the black 60 triange read 7.2, which is your degrees. Now align the black 60 triangle with your speed (540) and read off the 7.2 degrees glide slope on the inner and the outer is your required ROD = approx 6500ft/min. Rough way to make sure that you've got the right amount of zero's. 540x5=2700 (on a 3 degree glide slope) times by 2 and a bit to get it close to the 7.2 degrees, which is roughly 6210ft/min. This will save you ticking the box that says 650ft/min!
6.
7. 45 degrees N/S. Don't try to understand it, just remember it!
8.
I'll try and answer the other ones a bit later. It's just that I'm a bit short on time and they may take a bit of typing to explain. Just wanted to give you some help.
Good luck on Wednesday mate!!
Barnaby
SpaceRanger
Joined: May 1999
Posts: 213
Likes: 0
From: Samsonite
Re #5 you can use the departure MDR that says v/s = GS * climb gradient.
12*540 = 6480 ft/min.
This rule works when using the units of climb gradients (as well as descent gradients), knots and feet/min ONLY.
This MDR rule is normally used for departure (because SIDs have 'climb gradients' wheras glide slopes are in degrees), but since the angle is the same whether you climb or descent, it can be used here, because it is a gradient.
12*540 = 6480 ft/min.
This rule works when using the units of climb gradients (as well as descent gradients), knots and feet/min ONLY.
This MDR rule is normally used for departure (because SIDs have 'climb gradients' wheras glide slopes are in degrees), but since the angle is the same whether you climb or descent, it can be used here, because it is a gradient.
Joined: Jul 2002
Posts: 128
Likes: 0
From: Behind You
Just to add to barnabys solutions. For question 2 (ROD) you can halve the ground speed and add a zero for the answer. (3 deg glide slope only)
ie required rod on a 3 deg glide, gs 100 kts. 100/2=50, add a zero...500 fpm
or for a change in head/tail wind ie 10kts. 10/2=5, add a zero...50 fpm
May save you a few seconds in the exam
ie required rod on a 3 deg glide, gs 100 kts. 100/2=50, add a zero...500 fpm
or for a change in head/tail wind ie 10kts. 10/2=5, add a zero...50 fpm
May save you a few seconds in the exam
