Gen Mutley's Gen Nav Questions

[General Mutley]

I have a couple of questions but will start with this one -

What is the final position after the following rhumb line tracks and distances have been followed from position 60degrees 00minutes N 030degrees 00minutes W
South for 3600nm
East for 3600nm
North for 3600nm
West for 3600nm

Answer - 60degrees 00minutes N 090degrees 99minutes W

Could someone explain simply how this answer is reached - thanks.


Question 2!!

The rhumb line track between 45degrees 00 minutes N, 010 degrees 00 minutes W and 48degrees 30 minutes N, 015degrees 00minutes W is approximately -

Answer - 315degrees.

A simple explanation would be much appreciated.

Thankyou all :ugh:

[downwind24]

The first answer is the departure formula , a quick look shows that it puts you down on the equator where as you travel east and then back up the meridian you wouldnt use departure (ch long x cos lat) BUT as you come back west you would then use the departure formula and so you will not end up where you started due to the earths spherical shape (hence using departure)

The second question i remember ( and had on my exam in March) but cant recall the answer , Bristol do have the full explaination on the databank tho , it just involves you drawing it out , the other 3 answers are obviously wrong when you draw it.

Hope this isnt too garbled im having 2 mins from Met and Instruments but both explanations are on the DB :bored:

DW

[General Mutley]

Thanks for taking time to reply. Im trying to decipher the depature formula etc but it really isnt too clear to me!!!!
thanks again

[General Mutley]

Sorry to ask you all again but....

Given -
A is 55degrees N 000 degrees
B is 54degrees N 010 degrees E
The average true course of the great circle is 100 degrees. The true course of the rhumbline is....?

Answer - 104 degrees.

Can anyone explain what I am supposed to do please?

Thankyou :{

[paco]

Yes, the second question just needs a simple diagram. Don't think too deeply!

The first:

S for 3600 nm takes you to the Equator
E For 3600 nm takes u to 30E
N for 3600 nm takes u back to 60N
W for 3600 nm at .5 (cosine 60 degs) is twice the number of degs, or 90W

Phil

BTW - that was all in the notes ;)

[General Mutley]

question 2 is sorted, question 1 has me stumped!!!

[paco]

Again, draw a diagram. You can eliminate 2 answers from the choices see below) without working anything out due to the relative positions of great circles and rhumb lines. For the audience, the choices are:

107
100
096
104

phil

[downwind24]

Working back dont forget to rearrange the formula..........

3600 divide by 60 = 60 divide by cos 60 = 120

30 degrees east travelling west for 120 degrees = 90 west

When you have the departure distance you have to divide , i really dont know how i passed with 86% but i did , make sure you nail the crp-5 tho that will get you over 50% of the marks !

hope this helps

[wheelbrace]

You go south to the equator from 60 degrees north (3600/60 = 60 degrees; on a meridian, so no complications).
you go east by another 60 degrees (3600/60 = 60 degrees; on a horizontal meridian, the equator, so no complications).
you go north by 60 degrees (3600/60 = 60 degrees; on a meridian, so no complications).

If you draw this (just rough, nothing exact) you will see that you are on the same latitude as the start point, but 60 degrees to the east of your original longitude, ie 30 east.

Head west 3600 nm at 60 degrees north? You are looking for the change of longitude, so apply departure: (3600/60)/(cos60) = 60/0.5 = 120 degrees west from 30 degrees east = 90 degrees west.

I hope.

[General Mutley]

Thankyou all. I can do this now - .
Cheers for your help :ok:

[General Mutley]

Can anyone recommend a good, plain english ATPL gen nav book. I see oxford do one as do CAT but would be interested in your thoughts.
Thankyou:)

[RichardH]

No picture needed and the answer based on your wording is actually 100.
The question is basically testing your definition knowledge.

The question states AVERAGE GC = 100 this means half way which is exactly the same as the rhumb line. Remember a rhumb line is constant bearing and is also the same as average great circle.

IF the question had stated INITIAL GC tk 100 then 104 would be correct.

[General Mutley]

Is there anyone out there who can explain this to me in really 'Simple' terms please. I know this must be easy once you know what is going on but at the moment I dont!!!lol

[RichardH]

Try this:-
On your North Atlantic Plotting Chart 27th.
Draw a straight line (great circle as it's a Lamberts) from
Tiree N56:30 W007 to Oystr N54 W055

Now measure the INITIAL GC at Tiree - should get 284
Now measure the FINAL GC at Oyster (might have to extend line)
should get 245
Maths average = 245 + 284 /2 = 264

Now measure the AVERAGE GC exactly half-way along track you should get 264. So your rhumb line/constant direction/average GC between Tiree & Oystr is 264.

Please PM me for any further assistance.

[wheelbrace]

RichardH summed it up beautifully; 100 degrees true.

The following link is an interesting history which I found useful for context; a few terrifying looking equations at the end which can be ignored!

http://www.cwru.edu/artsci/math/alexander/mathmag349-356.pdf

In order to picture a rhumb line, try this, courtesy of Wikipedia...

http://upload.wikimedia.org/wikipedia/commons/0/0c/Loxodrome-1.gif (http://upload.wikimedia.org/wikipedia/commons/0/0c/Loxodrome-1.gif)

[downwind24]

Quote 'The question states AVERAGE GC = 100 this means half way which is exactly the same as the rhumb line. Remember a rhumb line is constant bearing and is also the same as average great circle.' This is true however....

From the answers given on the original question the wording is incorrect? , there is a question on the DB that gives those answers but it lists co-ordinates , its been a few months so i cant recall. Il see if i can find it.

Richard what are you like with Met and Instruments :}

[General Mutley]

thanks for all your time :ok:

[wheelbrace]

I sympathise with your plight, but the best books for nav. are the one's issued for your groundschool.

Personally I found that Gen. Nav. and Meteorology were the type of subjects that you need to play with concepts and one day it just, well, clicks.

I am sure that you could waste a whole load of dosh on extra books, but plug away at the examples and it will come good, I promise!

[General Mutley]

I struggle with certain things. I tend to need things explained simply first then working fully through examples using layman terms!. Found the oxford cd made a massive difference for the met exam.

[wheelbrace]

I found the OAT Met. cd worked for me. If there is a Gen. Nav. equivalent, go with that, then.

Thrash through your groundschool examples, however, and it will come good; I would suggest that you hit the BGS database, but only do so when you understand the concepts. I found that helps you to eliminate the two nonsense answers immediately and gives a grace period for working out the two likely candidates with reason rather than hope.

[General Mutley]

I had this yesterday and now I can't fathom it! Please help if you can - thanks.

A rhumbline track between 45 degrees N 010 degrees, W and 48degrees 30 minutes N, 015 degrees W is approximately.....?

330, 315, 345 or 300.

Thanks

[Poose]

If I remember rightly...

Draw the line from point A to B, which becomes the hypotenuse of a right angled triangle.

Remember that going North to South and vice versa that: 1' = 1 nm
and that the East to West part of the triangle you have drawn is the DEPARTURE.

This is where they try and catch you out! http://www.arrse.co.uk/images/smiles/icon_hump.gif (javascript:emoticon('post', 'message', ':hump:'))
(Remember: Departure = Change of Longitude in minutes x cosine latitude)

These should give you the horizontal and vertical dimensions of the triangle, which you can then punch into the basic trigonometrical equations. (SOH, CAH, TOA)

I guarantee that one or two of these will crop up in your General Navigation exam. Just remember to use the Departure formula for the horizontal bit of the triangle!

Let me know if it works! :ok:

[Poose]

For General Navigation learn the formulas off by heart and be nifty with the whizz wheel.

I work better at remembering formulas than functions on the whizz wheel, so I went down that route. Other people are a demon with the CRP-5. You do have to be proficient with the whizz wheel for certain parts of the exam where there is no way of using an equation.

Get yourself on the Bristol Question Bank and use your school pass papers (Bristol were excellent.). I'd even suggest buying the General Navigation Manual off Bristol. Give them a shout. They also hold open days just for General Navigation and Meteorology. Well worth it if you think you need it.

In all honesty, the General Navigation exam is nowhere near as bad as what it first looks. You are actually only tested on a very small amount of the content. PM if you want anymore advice. I passed General Navigation and Meteorology at Oxford back in January with 87% and 78%, respectively. It's still fairly fresh! ;)

P.S Be good at drawing your diagrams of Rhumb Lines etc.

[Poose]

Mutley,

Remember. "East/West distance is Departure." This is the key to why you are getting stuck on both of these.

Thus the East/West elements of those question are equal to the Departure.

Departure = change of longitude x cosine latitude

(Remember that the change of longitude is in minutes in the departure equation above, not degrees as it is in the Convergency equation.)

Just rearrange the departure equation to find the change in longitude along the East/West bits... and Robert's your Father's Brother. PM if you want more help! :ok:

[Keith.Williams.]

This problem can be solved by approximating the situation to a right angled triangle.

The base of this triangle will be the line from 45N 010W to 45N 015W.

The distance from 45N 010W to 45N 015W can be found using the departure equation:

Departure in nm = Change in longitude (in minutes) x Cos Latitude.

From 010W to 015W is 15 degrees. Which is 15 x 60 = 300 minutes change of longitude.

So between 45N 010W and 45N 015W the
departure 300 x Cos 45 = = 212.132 nm.

So the base of the triangle represents a distance of 212.132 nm.

The vertical left side of the triangle is the line from 45N 015W to 48 30N 015W.

Moving up or down any meridian each minute of latitude is equal to 1 nm. The change of latitude from 45N to 48 30N is 3 degrees 30 minutes, which is 210 minutes. So the distance from 45N to 48 30 N is 210 nm.

So the vertical left side of the triangle represents a distance of 210 nm.

Use the two distances above to draw a right angle triangle joining the point 45N 010W, 45N 015W and 48 30N 015W.

The hypotenuse of this triangle is the rhumb line from 45N 010W to 48 30N 015W.

In this triangle the track from 45N 010W to 45N 015W is 270 degrees.

To calculate the Rhumb Line from 45N 010W to 48 30N 015W we must add the internal angle at 45N 010W.

From the triangle it can be seen that the Tangent of this angle = 210nm / 212.132 nm, which is 0.9899.

So the internal angle at 45N 010W is 44.71 degrees.

Adding this to 270 gives the rhumb line track of 314.7 degrees.

The closest answer to this in the question is 315 degrees.

I have made this explanation very detailed because the important thing is to get the method sorted out in your head. This will enable you to solve any new problems of this type when they appear in your exam.