9780991975808

I would just visualize (on a map) where I am and where I want to be in relation to the VOR, and the approximate course to get there. I'd then look at my heading and turn the shortest way to get on that course.

It helps to understand the difference between accurate and precise and to always seek accuracy. Accurate (here) means correct, not erroneous: precise means exact, not approximate. Fixation on precision is often the cause of gross errors. My $0.02.

MikeF16

Except that it's not a simple trig question. At least not without a calculator. Unless of course you're Rainman and have 100% of the sine/cosine/tangent tables memorized.

This would work real world, however in this test question the available answers were within 5 degrees of each other which would require real math.

I'm hoping this is just bad poop and there is no such question on the test. If the question is identical, then I'm also good.

Here's another from the Sheppard Air prep: You lose your right engine after V1 while taking off from Runway 30. Which of the following wind conditions would make a continued takeoff most challenging?

2 answers are throw aways, the other two are:
180/30
360/30

So what is worse, air over the dead engine or a quartering tailwind over the good engine? My engine out takeoff scenarios have been pretty binary for most of my career. And who the hell would takeoff on RWY 30 with 180/30 winds anyway? Yet there's the question...

Flamer

Wind from failed engine side is harder. More rudder required. In this case 360/30.

MikeF16

TYVM. My noncenterline thrust knowledge is definitely on the low side.

9780991975808

Sometimes you don't have a choice. Lots of airfields are restricted to one way in, one way out operations in mountainous areas.

galaxy flyer

You always have a choice and that TW would exceed any reasonableness.

9780991975808

Explain that to the families of the pax and crew whose names are on the crosses at the north end of CWOB runway - for example. A pilot will take-off into wind towards the lee side of a towering mountain at the end of a runway only once.

KLM pilot

I fully agree with adlerdriver, this is how it's taught in JAA country(at least in mine).

To refine it further, if you have the winds aloft, you can calculate the drift angle very easily.

For a crosswind, you can calculate the crosswind component as follows:

For a wind from straight ahead, there is no crosswind component , crosswind from 30 degrees, multiply by .5, for a crosswind from 45 degrees multiply by .7, for a crosswind from 60 degrees multiply by .9 and for a crosswind from 90 multiply by 1.0.

To translate the crosswind component to a drift angle divide the TAS by 60! You then divide the crosswind component by this factor to give you a drift angle. For simplification purposes just divide the crosswind component by to for approaches(done around 120), by three for holdings( for speeds around 180) or by 4 for higher holdings, or enroute below 10.000.

For example, you have figured out that the track you have to fly is 180. The wind is from 210/30 and your TAS is 180. The crosswind component is .5 times 30=15, 180(TAS)/60=3. 15 divided by 3 is 5 so you have to steer 185.

Method is great to make sure you don't exceed crosswind limitations on landing for example too.

Just remember 0, .5, .7, .9 and 1.0 for crosswinds from 0, 30, 45, 60 and 90 degrees!

The same works for headwinds, just turn it around 1.0, .9, .7, .5 and 0! In the above mentioned example multiply 30 by .9 and you get a headwind component of 27. Try it on an e6B!!

The former is simple cosines calculus, the later sinus calculus.

If done correctly, you can nail a holding pattern on your first lap!!

galaxy flyer

One of the choices always available is......not going.

GF

Cubdriver

Thanks and I heard this one a few years ago from an airline guy and forgot about it. Finding the 30 deg. x-wind is easy because it's half. The 0, 60, and 90 ones are obvious so don't waste time calculating those. That leaves you needing the 45 deg. item which is a little tricky, how do you find 0.7 of something. What you can do with that one is think in units of ten and add 7 knots for each 10 knots of wind. Ex. 35 knots at 45 degrees, that's 3 wind units x 7 knots = 21 plus some change, say half the original value, so 24.5 which is very close to the correct answer (24.7). I would be happy if I quit thinking about it when I got the 21 personally.