Stuck on IMC VOR question
Pompey till I die
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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
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.
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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?
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?
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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?
Any help?
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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
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
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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.
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.
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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.
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.
Disclaimer: I would not have known how to work this out. Nobody flies with true bearings (in Europe)
And without true track, how do you work out your quadrantal or semicircular rule altitude?
G
Pompey till I die
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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
In the diagram the QDM is displaced 4 degrees from the required track
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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).
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Originally Posted by Genghis the Engineer
And without true track, how do you work out your quadrantal or semicircular rule altitude?
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.