ATPL MET and RVR
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ATPL MET and RVR
Question in a sample paper:
The visibility at an aerodrome is reported as 900 RVR. As you pass vertically above this airfield at 3000 ft you can clearly see the runway below. You make an ILS approach using standard three degree glide slope.
The distance from the threshold should you expect to first sight the runway is:
a) 1 nm (300 ft on final)
b) 3/4 nm (250 ft on final)
c) 1/2 (180 ft on final)
d) 1/4 (120 ft on final)
Just wondering how to work this out.
Cheers
The visibility at an aerodrome is reported as 900 RVR. As you pass vertically above this airfield at 3000 ft you can clearly see the runway below. You make an ILS approach using standard three degree glide slope.
The distance from the threshold should you expect to first sight the runway is:
a) 1 nm (300 ft on final)
b) 3/4 nm (250 ft on final)
c) 1/2 (180 ft on final)
d) 1/4 (120 ft on final)
Just wondering how to work this out.
Cheers
Thread Starter
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Hey thanks bud.
I'm just worried about not knowing how to solve this question.
I've googled, looked in books and the AIP but of course nothing...
I'm sure one of two things will happen
I'm just worried about not knowing how to solve this question.
I've googled, looked in books and the AIP but of course nothing...
I'm sure one of two things will happen
- Some one is going to show how easy it is to solve it
- Someone is going to give a complex answer
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Cloudhigh I'd say figure out what 900mtrs is in NM and choose that answer. It sounds like they're trying to throw the "standard 3deg slope into confuse you".
I think it's the answer is c)1/2mile and 180ft on final.
Hope this helps
Rocket
I think it's the answer is c)1/2mile and 180ft on final.
Hope this helps
Rocket
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I think what they are asking for is a little pythagoras...
This is my take on it -
1/2nm is approximately 925meters
180ft is approximately 55meters
Plug that into - a^2+b^2=c^2
855,625+3,025=858,650
Sqrt 858,650 = 926 or approx 900 meters
Answer (c)
So by taking the distance we are away from the runway and how high we are the hypotenuse should equal the RVR.... I think, I am completely guessing so don't blame me when you're wrong...
This is my take on it -
1/2nm is approximately 925meters
180ft is approximately 55meters
Plug that into - a^2+b^2=c^2
855,625+3,025=858,650
Sqrt 858,650 = 926 or approx 900 meters
Answer (c)
So by taking the distance we are away from the runway and how high we are the hypotenuse should equal the RVR.... I think, I am completely guessing so don't blame me when you're wrong...
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The bloke at the end of the runway can see 900M down the strip which is roughly 1/2 Nm. With 1/2 nm vis on the ground, you would hope that you have the same going back the other way. So they are trying to get to Answer C.
However its a typical question where there are many factors left out and so you just have to pick the best answer to a poor question
However its a typical question where there are many factors left out and so you just have to pick the best answer to a poor question
