NDB, ADF and RMI Question

[Yohan20]

Hello all, planning on writing my INRAT exam soon. Nee some help from all the smart people out there.

Question: Flying an aircraft with a Fixed Card ADF. To determine position, you decide to use 2 NDB stations.
NDB 1 - Select a RB = 120 Degrees
NDB 2 - Select a RB = 260 Degrees

Aircraft Heading = 220 Degrees
Local Variation = 23 Degrees West

In order to determine your position from the RBs, what would you plot on NDB 1 and NDB 2 on the chart?

Answers:
1. 160 and 300
2. 340 and 120
3. 137 and 277 (correct answer)
4. 83 and 323

I am unable to get to this answer. Will anyone now how this is being calculated?

Thank you very much in advance

[Mach Jump]

Question: Flying an aircraft with a Fixed Card ADF. To determine position, you decide to use 2 NDB stations.
NDB 1 - Select a RB = 120 Degrees
NDB 2 - Select a RB = 260 Degrees

Aircraft Heading = 220 Degrees
Local Variation = 23 Degrees West

In order to determine your position from the RBs, what would you plot on NDB 1 and NDB 2 on the chart?

Answers:
1. 160 and 300
2. 340 and 120
3. 137 and 277 (correct answer)
4. 83 and 323

Assuming that Variation is the same for all three locations, that the Aircraft heading is Magnetic (M), and that the answers are required in Degrees True (T):

Relative Bearings(RBs) are 'To' the beacons, so take the reciprocals to get bearings 'From' the beacons:

1 RB 120 = 300 'From'
2 RB 260 = 080 'From'

Bearing from Beacon 1 = 300 + Hdg 220 - 360 = 160M or - V 23 W= 137T
Bearing from Beacon 2 = 080 + Hdg 220 . . . . = 300M or - V 23W = 277T


Hope that helps.


MJ

[Yohan20]

Thanks Mach Jump. I should have read the question a bit better...totally missed the "from" part...

Do you have any recommendations regarding any books, CD's or reference material that will teach you how to tackle these questions. Most people at the flying school get stuck as well

[wanabee777]

I flunked...

[Mach Jump]

You don't really need any books or CDs to help you with this stuff, Johan..They say that a picture is worth a thousand words.

The problem with questions like this, is that it's difficult to visualise the situation from the text.

Draw the face of an RMI, on the heading, with the two needles showing the bearings given.

It's easy from that, to draw a diagram showing the position of the beacons, in relation to the aircraft.

Once you have that, the answers become self evident.

The worst part of this question, was that you had to guess the three assumptions I made at the beginning, There wasn't really enough information in the question to work it out.


MJ

[sdfarook4]

ADF numericals

Aircraft heading 130 (M), has an adf reading if 190 relative, to intercept the 170(M) track outbound from an ndb at 30. The relative bearing to the ndb that confirms track interception is?
A) 150 relative
B) 140 relative
C) 170 relative
D) 160 relative
Answer is B. Pls explain

[sdfarook4]

ADF numericals

Aircraft hdg 040 (M) has an ADF reading of 060 relative. The heading to steer to intercept the 120 track inbound to the ndb at 50 degrees is??
A) 080 M
B) 070 M
C) 060 M
D) 050 M
Answer is B, kindly explain

[Chris Scott]

Originally Posted by sdfarook4

Aircraft heading 130 (M), has an adf reading of 190 relative, to intercept the 170(M) track outbound from an ndb at 30. The relative bearing to the ndb that confirms track interception is?
A) 150 relative
B) 140 relative
C) 170 relative
D) 160 relative
Answer is B. Pls explain

They say there are many ways to skin a cat. (I'm a cat-lover, BTW.) Having started my IR training in the good-old Link Trainer, and completed it in a Cessna 310 - neither of which were blessed with RMIs - I propose to set about this in a slightly different fashion from the excellent one used by Mach Jump (above). Good thing is that we are sticking with magnetic bearings on this one, so the question can be done in your head (as you would normally do in the seat of an aeroplane).

Note that:
All tracks and headings in this exercise are magnetic (M);
QDR = magnetic bearing from the beacon;
QDM = magnetic bearing to the beacon;
the scenario seems to be in still air, so no drift - highly unusual.

Initial RB of 190 is equivalent to +10 as a bearing from the beacon, so the present QDR = 140.
We want to intercept a QDR of 170, so we must turn to starb'd (right). A 30-degree intercept will involve a track of 200. So, as there's no wind, apparently, HDG = 200.
On reaching the QDR of 170 on a HDG of 200, the RB will be -30 (from), which is an RB of 150.

So the correct answer, IMO, is (A), not (B). Is that why you asked?

PS
In case anyone is confused by the above method of interpreting the RB needle, I'll try to explain.
If the head of the needle is pointing forwards relative to the a/c:
readings left of the nose are minus; readings to the right are plus. When applied to the present HDG, you get the QDM (magnetic bearing to the NDB).
If the head of the needle is pointing aft:
readings left of the tail are plus; readings to the right of the tail are minus. When applied to the present HDG, you get the QDR (magnetic bearing from the NDB.)

[Chris Scott]

Originally Posted by sdfarook4

Aircraft hdg 040 (M) has an ADF reading of 060 relative. The heading to steer to intercept the 120 track inbound to the ndb at 50 degrees is??
A) 080 M
B) 070 M
C) 060 M
D) 050 M
Answer is B, kindly explain

Okay, I'll use the same method as in my previous post, again reliving my novice days in the Link trainer.
Once again, we are in the unusual situation of still-air.

Current QDM = 100 (QDR 280).
Desired QDM = 120 (QDR 300).
So we need to turn to port (left), and intercept the desired bearing from right to left.
50-degree intercept HDG = 120 - 50 = 070
So answer (B) is indeed correct.