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Discussion in 'Pilot Training' started by Drecula, Nov 27, 2018.
I couldn’t recall how to calculate this. Can someone jog my memory.
If I were attempting to solve it, I’d first (roughly) convert km to nautical miles to determine separation at “maximum visual range,” since the closure rate is given in knots. 4 km is roughly 2 nautical miles. Then solve for closure rate. 280 knots is roughly 4.7 nautical miles per minute. What’s the closest correct answer?
BTW, where did you find this question?
I am going to guess Canada because of the use of metric system and worded in English..
Two other clues that this is not a USA FAA exam question is there are four possible answers instead of three, and there are small case answer choices (a., b., c., d.) instead of upper case (A., B., C.)
Do the Canuks allow 4 km VFR? Lol
1 nautical mile is about 1.85 km. So, 4 (km)/1.85 (km/NM)= 2.16 NM. They are about 2.16 NM apart at the initial visual contact.
Knots is NM/h. Thus, they will collide in 2.16 (NM) / 280 (NM/h) hours, which is 0.0077 hours. Therefore, 0.0077*3600 = 27.8 seconds.
Thats correct. The answer is Alpha.
It’s a question in the Oxford CBT for PPL.
Not Canada. UK
TROLL...….. just getting that out of the way, lol.
That is a PIA question, you have two different unit of measures used, so you have to convert one to the other. I'll convert km to nm, using the googles, 4km = 2.16nm, then I would divide 2.16 nm by 280 nm/hour which would equal, 0.008 which is actually 0.008 hours. Now you need to convert that to seconds, so knowing there are 3,600 seconds/1 hour I would multiply 0.008 hours times 3600 sec/hour which equals 29 seconds. So A is correct.... (I rounded some numbers)
Or I'm sure someone sells a whiz wheel that would figure it out for you.
I think I need an E6C.
By the time I figure that one out, traffic will be "no factor".
So, is 4 km flight visibility legal VFR in the UK?
I’m not from UK. But as far as I know it’s 5km over there.
This CBT is a pretty old one.