Hey everyone.

Just a quick question in relation to the practice question 22 on page 4.54:

"Heading and TAS have been kept constant from 2130 to 2140. If heading is changed by 9 degrees right at time 2140, at what time would you expect to regain the flight planned track?"

I understand that the closing angle must be 3 degrees, however all the questions make the closing angle either exactly half or exactly twice the track error. Will there ever be a situation where the closing angle isn't exactly half/double? And if so is there a way to work it out using the flight computer? I use the Jeppesen one if that's any help

Thanks in advanced

Hi Cessna,

Closing angles in the exam won't always be half or double the track error. In fact I'd be tempted to say, at least at CPL level, it is actually rather unlikely. For example, one kind of questions you can expect will be scenario style with you measuring off track distances on the WAC extracts in the CASA VFR exam workbook or, more often, using the 1-in-60 diagrams with values provided for the variables. These diagrams are also in the workbook.

Remember calculating the closing angle is no more complicated than working out the track error. The 1-in-60 principle still applies. The only difference is, the angle you calculate will be the angle with which you re-intercept FPT not the angle you were moving away from your FPT.

You want to use your flight computer? No worries. Let's look at an example:

At a fix, you realise you are 5 nm right of track after flying 40 nm and want to re-intercept FPT at a point 30 nm further along. What heading change is required to achieve this?

Find your TE using the flight computer: on the calculator side, align 5 on the outer scale with 40 on the inside scale. Now, on the inside scale look for the 60 marker. The computer is set up now and is saying "If you are 15 nm off after 40 nm, how far off will you be after 60 nm?" To answer that, read off the value opposite the 60 on the inside scale. There you find the value "75". Your TE has been 7.5 degrees.

To regain FPT in another 30 nm from this point, you will need to close the 5 nm off track distance in 30 nm. Again we can use the flight computer to work out the closing angle required. Align 5 on the outer scale against the 30 on the inside scale and look at the value opposite the 60 on the inside scale. The computer is now set up to answer the question "If I need to close 5 nm in 30 nm, how many would I close over 60 nm?". There you can read off the answer of "10".

So, if you would theoretically close 10 nm in 60 nm, according to the 1-in-60 rule, this must mean the closing angle is 10 degrees.

What change do you need in your heading? Well, it will be the 7.5 degrees (call it 8) to correct for the TE and then a further 10 degrees to get you heading back towards track.so you would need to turn left 18 degrees to get you back on track 30 nm from the point where you got your fix.

Once back on FTP you would turn back (i.e. right) by the CA i.e. 10 degrees to continue flying along FPT..

When working these problems it is always worth drawing a little sketch so you can get a feel for which way you need to turn and what is happening.

Cheers,

Rich

Thanks for that Rich,

Also another question I stumbled upon which I'm struggling to understand is on page 3.22 Q7. If the QNH is lower at the destination aerodrome than at the departure aerodrome, wouldnt the altimeter reading be lower since it's being read from a datum higher in the atmosphere, causing the aircraft to climb if it wants to maintain a constant altimeter reading? The answer is (a) but I think it's (b), am obviously misinterpreting something

Theres two other questions I have from another book however I'm not getting the answer exactly correct in comparison to the multiple choice answers:

Given the following conditions for planning a route segment:
Heading (M) 345 degrees
TR (M) 350 degrees
TAS 205kts
GS 190kts

The wind velocity being experienced is
(a) 299 degrees M/23kts
(b) 119 degrees M/23kts
(c) 310 degrees M/20kts
(d) 100 degrees M/25kts

I did it as accurate as I could and got approx 294 degrees at 23kts, the same happened with the examples in Bob Taits book. Do you think it's just due to the nav computer I'm using?

My final question is fairly long and requires a WAC so don't expect you to answer it however here it is if you have the time:

Refer (SYDNEY) WAC 3456
You are on a flight from COWRA (YCWR) (33degrees50S 148degrees38E) to BATHURST (YBTH) ( 33degrees24S 149degrees39E). you departed YCWR at 0645 EST. At 0710 EST you crossed the railway and road intersection at NEWBRIDGE (approximately 20nm southwest of BATHURST) and fixed your position at this time. What is the change of heading required to fly direct to YBTH from this position?

(a) 16 degrees left
(b) 16 degrees right
(c) 8 degrees left
(d) 8 degrees right

The track error I got was 4 degrees ( I measured a distance of 39nm and we were 2.5nm off track)
The closing angle I got was 8 degrees ( I measured a distance of 18nm and used the 2.5nm again)

So the total change required would be 12 degrees left from what I got, just wanted some reassurance to make sure I'm doing it correctly and the answers may be incorrect.


Thanks again for the help so far

Hi Cessna,

First Question: QNH
===============
A picture paints a 1000 words so let me save a few hundred.



Consider the situation at the sea level aerodrome, A, where the QNH is 1015. In other words the 1015 hPa isobar is passing through A (since it is at sea level). Now consider the destination, B, another sea level aerodrome but this time where the QNH is only 1005 hPa. That means the 1005 hPa isobar is passing through B. If that's the case, where is the 1015 hPa at B then? That's right, it must be lower, in fact, theoretically speaking (of course) it would be 300 ft below B.

That means the isobars must be sloping downwards towards the destination ( as you can see in the diagram). Now if you are flying steady at FL155, you are simply flying at a 15500 ft above the 1013 hPa isobar, We already saw all isobars are sloping down towards the destination so the 1013 hPa isobar must be sloping downwards as well.

Therefore, if you fly constant at FL155 towards B, your actual altitude above sea level will be decreasing.


Second Question: Winds in Flight
=========================
Make sure you set the ground vector up on the E6B when you do these problems. So, set Track of 350 against the index and line the grommet up with the ground speed of 190. Now, the heading, 345, is 5 degrees left of the track so find the drift line for 5 degrees to the left and mark the wind dot there, opposite to the ground speed of 190 kts on the vertical scale. Rotate the face to bring the wind dot onto the vertical scale and read off a wind speed of about 23 kt and a direction of about 299 degrees.

Here's some working:


Third Question: Cowra to Bathurst
=========================
I checked this one and I agree. You have 4 degrees of track error to Newbridge and a closing angle of 8 degrees to get to Bathurst. You will need to change your heading by 12 degrees Left to fly direct to Bathurst from the fix.

Cheers,

Rich

Thanks for clearing up the first and third question Rich.

With the second question these were the steps I took, if you could tell me what I was doing something wrong that would be great.
- Firstly I aligned the black arrow with the TAS (in this case 205kts)
- Secondly I aligned the heading on the top (in this case 345 degrees)
- I then determined the hw/tw componenet by comparing the TAS to the GS (in this case we have a 15kt hw) so I drew a horizontal line along the 15kts (if that makes sense)
- Lastly I compared the heading to the track, which gives us 5 degrees right drift meaning the wind is coming from the left (at a TAS of 205kts it's roughly 18 kts crosswind)
- I marked that point and turned the wheel until it was on the very top.

Apologies if that sounds confusing but thats exactly how I do it
You wouldn't happen to have an explanation with pictures for the Model CR-3 Computer? Am a bit lost with the diagram you linked

Hi Cessna,

Those diagrams are for the E6B type of flight computer and unfortunately I don't have my CR3 with me at the moment (it's in a container enroute to Australia at the moment in preparation for our move ) but here's a copy of the CR3 manual in case you don't have one:

www.mypilotstore.com/mypilotstore/Produc...s/crinstructions.pdf

Your steps in calculating the winds seem OK but double check you are using the effective TAS to calculate the headwind component (mind you with only 5 degrees of drift on it shouldn't be that critical). Check the working using the manual I linked to above and post again. Maybe include some shots of your set-up here. You should get the same answers with the CR3 as you would with the E6B so something's going wrong somewhere. Unless of course you left the CR3 on the dashboard of the car one Summer afternoon

Cheers,

Rich

Thanks for linking the manual it helped a lot. I sat my exam today and scored 85%, unfortunetely there were no questions on the transition layer or determining the wind (I spent alot of time on those type of questions as you could imagine with everything I've asked on this forum).

Thanks for the help Rich

You're welcome cessna and congratulations on your very respectable pass - well done!

Cheers,

Rich