Go Back  PPRuNe Forums > Non-Airline Forums > Private Flying
Reload this Page >

Stuck on IMC VOR question

Wikiposts
Search
Private Flying LAA/BMAA/BGA/BPA The sheer pleasure of flight.

Stuck on IMC VOR question

Thread Tools
 
Search this Thread
 
Old 18th Apr 2012, 12:49
  #1 (permalink)  
Pompey till I die
Thread Starter
 
Join Date: Nov 2006
Location: Guildford
Age: 51
Posts: 779
Likes: 0
Received 0 Likes on 0 Posts
Stuck on IMC VOR question

21) Your VOR (shown below) is correctly tuned with the OBS set at 150, your desired track. If the variation is 4 degrees west, what is the true bearing from the VOR to your current position?



A 152
B 326
C 146
D 322
You chose B, but the correct answer was D.

I kept ending up with 324 degrees and worked through it several times but it still came back wrong. Where is my mistake?

1. I am wanting to go 150 to the VOR but I am to the right (needle is left so telling me to go left) which means I am 148
2. Because we want to be FROM the vor I add 180 = 328
3. Variation east, compass least. Variation west, compass best = 328M is 324T which is not correct

Last edited by PompeyPaul; 18th Apr 2012 at 14:47.
PompeyPaul is offline  
Old 18th Apr 2012, 13:15
  #2 (permalink)  
 
Join Date: Dec 2011
Posts: 2,460
Likes: 0
Received 0 Likes on 0 Posts
Nobody flies true bearings

(sorry I can't help).
peterh337 is offline  
Old 18th Apr 2012, 13:29
  #3 (permalink)  
 
Join Date: Dec 2006
Location: .
Age: 37
Posts: 649
Likes: 0
Received 0 Likes on 0 Posts
I also got the same as you did Paul.

150 TO = 330 FROM. 1 dot deflection left = 2 degrees left, so 148 TO (328 FROM). Subtract Mag Variation to convert to True = 324 Deg T. Am I doing something daft?

Perhaps an instructor or Instrument Ace can enlighten thicko Smithy?
Captain Smithy is offline  
Old 18th Apr 2012, 13:39
  #4 (permalink)  
Pompey till I die
Thread Starter
 
Join Date: Nov 2006
Location: Guildford
Age: 51
Posts: 779
Likes: 0
Received 0 Likes on 0 Posts
Deflections

1 dot = 2Deg on VOR & 0.5Deg on ILS
PompeyPaul is offline  
Old 18th Apr 2012, 13:41
  #5 (permalink)  
enq
 
Join Date: Jan 2002
Location: Essex, Innit
Age: 55
Posts: 90
Likes: 0
Received 0 Likes on 0 Posts
Does the answer lie in the image? - I recall seeing a VOR OBS question where the first "dot" was actually the outer radius of the centre circle so that the first actual dot represented 4 Deg off - if this is the case with your Q then your reasoning as applied becomes 150 - <4> +180 = 326M less 4 = 322T

Any help?
enq is offline  
Old 18th Apr 2012, 13:49
  #6 (permalink)  
 
Join Date: Aug 2007
Location: Back in the real world
Posts: 196
Likes: 0
Received 0 Likes on 0 Posts
I think I had exactly the same question in my exam recently and got it wrong as well.

Could the answer be that the picture is missing the first (inner) dot

I might add the same paper which asked a question about RIS and RAS
Nibbler is offline  
Old 18th Apr 2012, 14:47
  #7 (permalink)  
Pompey till I die
Thread Starter
 
Join Date: Nov 2006
Location: Guildford
Age: 51
Posts: 779
Likes: 0
Received 0 Likes on 0 Posts
Update

I've posted the picture now in the original question
PompeyPaul is offline  
Old 18th Apr 2012, 14:55
  #8 (permalink)  
 
Join Date: Dec 2011
Posts: 2,460
Likes: 0
Received 0 Likes on 0 Posts
That looks like a 4 degree deviation to me.

The little circle is 2 degrees.

Very confusing though.
peterh337 is offline  
Old 18th Apr 2012, 15:09
  #9 (permalink)  
enq
 
Join Date: Jan 2002
Location: Essex, Innit
Age: 55
Posts: 90
Likes: 0
Received 0 Likes on 0 Posts
Yep, that's the one wot I said.

Never accuse the CAA of failure to obfuscate.

Regards, enq.
enq is offline  
Old 18th Apr 2012, 15:22
  #10 (permalink)  
Pompey till I die
Thread Starter
 
Join Date: Nov 2006
Location: Guildford
Age: 51
Posts: 779
Likes: 0
Received 0 Likes on 0 Posts
Thanks!!

I still don't understand why it's 4 degree deviation though. Although that DOES make the answer work.
PompeyPaul is offline  
Old 18th Apr 2012, 15:27
  #11 (permalink)  
 
Join Date: Dec 2011
Posts: 2,460
Likes: 0
Received 0 Likes on 0 Posts
As Mad Jock would say this is what separates the real pilots from



Disclaimer: I would not have known how to work this out. Nobody flies with true bearings (in Europe).
peterh337 is offline  
Old 18th Apr 2012, 15:29
  #12 (permalink)  
 
Join Date: Feb 2006
Location: Ireland
Posts: 53
Likes: 0
Received 0 Likes on 0 Posts
full deviation is 10 degrees - so in this case inner circle must represent 2 degs
neilr is offline  
Old 18th Apr 2012, 15:45
  #13 (permalink)  
 
Join Date: Dec 2011
Posts: 2,460
Likes: 0
Received 0 Likes on 0 Posts
This stuff should not be examined like that. It is appalling.

This



is fairly obvious. Obvious 2 degree divisions 0 2 4 6 8 10 degs.

But what about this



The first dot is "half scale" and if you reach that on an ILS you go around.
peterh337 is offline  
Old 18th Apr 2012, 15:51
  #14 (permalink)  
 
Join Date: Jul 2004
Location: On the wireless...
Posts: 1,901
Likes: 0
Received 0 Likes on 0 Posts
The circle represents +/- 2 degrees (total 4 degrees) course deviation. The left or right circumference of the circle represents 1 dot = 2 degrees.
In the diagram the QDM is displaced 4 degrees from the required track therefore the QDM is 146M.
The QDR is therefore 326M
Applying the Magnetic Variation of 4W the QTE is therefore 322T = Answer D.

Modern CDIs display simply 5 dots either side of centre.
At 2 degrees per dot the full scale deflection is at the 5th dot = 10 degrees.
Beyond that fifth dot Course Deviation is not measurable.
Talkdownman is offline  
Old 18th Apr 2012, 16:08
  #15 (permalink)  
Moderator
 
Join Date: Feb 2000
Location: UK
Posts: 14,222
Received 48 Likes on 24 Posts
Disclaimer: I would not have known how to work this out. Nobody flies with true bearings (in Europe)
So did you not finally pass your JAA IR?

And without true track, how do you work out your quadrantal or semicircular rule altitude?

G
Genghis the Engineer is offline  
Old 18th Apr 2012, 16:32
  #16 (permalink)  
Pompey till I die
Thread Starter
 
Join Date: Nov 2006
Location: Guildford
Age: 51
Posts: 779
Likes: 0
Received 0 Likes on 0 Posts
Strange

The left or right circumference of the circle represents 1 dot = 2 degrees.
In the diagram the QDM is displaced 4 degrees from the required track
Thanks, I'm obviously thick though. It looks like the needle is displaced by one dot to me, therefore it's 2 degrees off? I don't understand why it's 4 degrees off of track? Surely that should be 2 dots off?
PompeyPaul is offline  
Old 18th Apr 2012, 16:38
  #17 (permalink)  
 
Join Date: Nov 2004
Location: UK, mainly
Age: 39
Posts: 423
Likes: 0
Received 0 Likes on 0 Posts
The circle encompasses 4 degrees of deviation - 2 left, 2 right. So the needle has deviated by the circle (2 degrees off-centre) and it's also hovering over a single dot (a further 2 degrees).
madlandrover is offline  
Old 18th Apr 2012, 16:40
  #18 (permalink)  
Pompey till I die
Thread Starter
 
Join Date: Nov 2006
Location: Guildford
Age: 51
Posts: 779
Likes: 0
Received 0 Likes on 0 Posts
AHHhhhh!!

Thanks, I get it! The inner circle is a dot itself, plus another dot as well. Damn, crafty bar stewards!
PompeyPaul is offline  
Old 18th Apr 2012, 16:51
  #19 (permalink)  
 
Join Date: Feb 2006
Location: London
Posts: 307
Likes: 0
Received 0 Likes on 0 Posts
Originally Posted by Genghis the Engineer
And without true track, how do you work out your quadrantal or semicircular rule altitude?
1) On magnetic track, as same as everybody else, since the quadrantal / semicircular rules are based on magnetic tracks

2) For en-route airways, you plan the directions the airways chart or Eurocontol allow. You fly what ATC clears you to. Only outside controlled airspace quadrantals are realy relevant.
Cobalt is offline  
Old 18th Apr 2012, 16:59
  #20 (permalink)  
Moderator
 
Join Date: Feb 2000
Location: UK
Posts: 14,222
Received 48 Likes on 24 Posts
Ooops, so we do.

G
Genghis the Engineer is offline  


Contact Us - Archive - Advertising - Cookie Policy - Privacy Statement - Terms of Service

Copyright © 2024 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.