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general navigation questions

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general navigation questions

Old 11th Oct 2011, 06:23
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Join Date: Jan 2011
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general navigation questions

Q1. sun rise at 50N072E at 0254 on 25th jan. what time will the sun rise at 50N007E on that day?
0714UTC-ans{please explain how to solve this}

Q2. The angle b/w true circle track and true rhumbline track joining points A{60S165W} B{60S 177E}, at the place of departure A,is
7.8degree-ans

Q3.what is UTC time of sunrise in CANADA{49N 123 30W} on 6th december?
1552UTC-ANs

Q4.A great circle track joins position A{59 S 141 W} and B{61S148W}.What is the difference b/w great circle track at A and B?
It increases by 6degree---------ans

Q5.As value for ellipticity of earth is 1/297. Earths semi major axis ,as measured at the equator equals 6378.4km. what is semi minor axis of the earth at the axis of the poles?
6356.9-----ans

Please guide me in solving these problems
thanks
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Old 11th Oct 2011, 08:29
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Hi

1) Use the sunrise table, and find the LMT of [email protected] 50N (interpolate it for the 25th of jan)
Once u have the LMT, Convert it to UTC by considering ur Long (i.e. 7E/ 15)
Since long is E, UTC is least(UTC<LMT), now uve got ur ans.
(i cannot determine this ans for, since the sunrise tables dat i have might be for a diff year)

2) Here u have to find the angle between the R.L and G.C at A, This is nothing but ur conversion angle, i.e. Convergency/2.

Con= D.long*Sin Lat
= 18*sin 60
=15.58
Therefore C.A.=15.54/2=7.79

3) Find the LMT the of sunrise on the 6th of dec from the sunrise tables.
Once u have the LMT, Convert it to UTC by considering ur Long (i.e. 123 30W/ 15)
Since long is W, UTC is best(UTC>LMT), now uve got ur ans.

4) I think u have to use the meridoinal parts tables to determine the ans here. (correct me if wrong).

5)Its just a simple formula based problem.
Compression ratio= (Equatorial diameter-polar diameter)/ Equatorial diameter
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Old 11th Oct 2011, 10:02
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thanks abhi

still have problem with regards to 4th and 5th Q......

As in 4th}what is meridonial parts of table where can i find those

and in 5th}
Compression ratio= (Equatorial diameter-polar diameter)/ Equatorial diameter
so 1/297=polaraxis/6378.4
polar axis comes out as=21.47.........the ans only comes when i subtract 6378.4 and 21.47............i.e.6356.9...........please give your views on this...........
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Old 11th Oct 2011, 10:12
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gen. navigation Q

Q1. The sun moves from east to west at a speed of 15degree longitude an hr.what ground speed will give you the opportunity to observe the sun due south at all times 60 00N?
450KT-------ans

Q2.What is the initial great circle direction from 45N 14 12 W to 45N 12 48E?
a}90degree{M}{acc. to me ans should be this one..???????}
b}80.4degree{T}--ans

Q3. The great circle track X-Y meansured at X is 319degree,and Y 325degree then
a}northern hemisphere,rhumbline track is 322degree
b}southern hemisphere,rhumbline track is 322degree-----ans
{i knw this question need to solved using drawing ....but just give a hint in terms of g/c track and r/l track of x and y both.....i`ll try to figure out myself..}
thanks...
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Old 11th Oct 2011, 11:19
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Hi

4)I think u can google it or locate it in any ATP Nav books...
Meridoinal parts are used when we deal wid 2 points which are neither on the same lat nor on the same long.
There would be a trigonometrical formulated table to find the track and dist between the two points.

5) The ans u get (i.e. 21.47) is== Equatorial D- Polar D
hence
Polar D= Equatorial D-21.47
Work it on a piece of paper,and it will make sense.

1) Sun moves from west to east isnt it?? Ans is pretty simple.
Sun moves at 15 deg/ hr. And he wants us to cover the same dist as the sun.
dat is dep= d long cos lat
==180*60*cos60
=5400 nms.

And, Sun travels at 15 deg/hr so 180 deg of long in how many hrs??
=> 180/15= 12 hrs

Hence the req ground speed is D/T=> 5400/12= 450kts.

2) The emphasis here is given on (T). coz anything measured on a chart is a true bearing. i.e. G.C. QUJ or G.C. QTE

3) G.C. trks are given.
We know dat Initial G.C trk-Final G.C. Trk = Convergency
And C.A= Con/2
SO hence its 3 degs.
And to determine the hemisphere its best to draw them and see.
And the ans seems to be correct according to me.
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Old 12th Oct 2011, 05:25
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some gen nav Q.

Q1. The total length of 53N parallels of latitude on a direct Mercator chart is 133cm.what is approximate scale at latitude 30S?
1:25 000 000-----ANs{ I did this by first finding scale of equator from 53degree N parallel...then using it to calculate scale at 30S...............but not able to get the correct ans....please guide with this Q.}

thanks

Last edited by AVIATROZ; 12th Oct 2011 at 07:56.
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Old 12th Oct 2011, 07:33
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Hi

2) I think u made a mistake in the D long.
They are in the same hemisphere, so they have to be subtracted.
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Old 12th Oct 2011, 07:56
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hey thanks abhi

thank abhi for correcting me on that................that was a dumb mistake from my side....i apologise for that..............
Thanks again
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Old 12th Oct 2011, 08:18
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few more gen nav Q.

Q1.At 0020 UTC an a/c is crossing the 310degree radial at 40NM of a VOR/DME station.At 0035UTC the radial is 40degree and DME distance is 40NM.magnetic variation is zero. The true track and ground speed are.
085degree and 226kts........ans{ i am not able to get how to solve this....plz guide me on that}

Q2. On a mercator chart 47cm represents 247nm at 35S.the scale of the chart is:
a} 1:1 700 000 at the equator
b} 1:1200 000 at the equator........ans
{i solved this question and getting an ans 1:1186879.........which is close to both the above option....please advice.......first i got scale at 35S then calculated the scale at equator.}
thanks
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Old 12th Oct 2011, 12:08
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Hi

Q1. The total length of 53N parallels of latitude on a direct Mercator chart is 133cm.what is approximate scale at latitude 30S?
1:25 000 000-----ANs{ I did this by first finding scale of equator from 53degree N parallel...then using it to calculate scale at 30S...............but not able to get the correct ans....please guide with this Q.}

thanks

wat i would do is to find the scale @ 53N first.
When i get dat i would use the ABBA formula to find the scale @ 30S

i.e. sc= 133/ 360*60*cos53*185300
sc=1/18110921

Now using the ABBA formula

18110921*cos30= X*cos53

X= 26062025~25000000.
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Old 12th Oct 2011, 12:27
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Hi

1) The problem should be solved by drawing a sketch

once u draw it u can calculate the length of the trk by using the pythagoras theorem
i.e. X^2= 40^2+40^2
X=56.56

We now the dist now. the time is given in the ques i.e. 15 mins.

hence Speed is 226kts.
Im not sure abt the trk. it should be 090 degs, but u since its 085, it must be a hdg correcting for 5 deg right drift.

Last edited by abhimakam; 12th Oct 2011 at 16:39.
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Old 12th Oct 2011, 12:31
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Hi

2)Sc at 35s= 973810

Sc at 0 deg = (973810*cos 0)/ cos 35
=>188803~=1200000

Hope it helps
Abhi.

Last edited by abhimakam; 12th Oct 2011 at 16:39.
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Old 13th Oct 2011, 08:55
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An aircraft departs Hong Kong 2219N 11412E at 2234 Standard Time on the 17 April. If the flight time is 11 hours 17 minutes, the Standard Time and date of arrival at Honolulu (Hawaii) 2125N 15755W will be:
A) 1151 17 April
B) 1551 17 April
C) 1151 18 April
D) 0551 18 April


ans is b...
so how do we decide here in this question what the standard times are rounded off to the lower hour or the upper hour?


An aircraft has hard iron magnetism only, and this hard iron magnetism is represented by a red pole in relative bearing 070 from the compass. On what heading will the westerly deviation be maximum?
A) Heading 110.
B) Heading 340.
C) Heading 020.
D) Heading 200.

ans is c...how do we do this

Last edited by Shabez; 13th Oct 2011 at 13:34.
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Old 13th Oct 2011, 21:45
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On a Lamberts chart a straight line track crosses a meridian with a direction of 043 T and a second meridian with a direction of 055 T. If the constant of the cone is 0.75 the difference in longitude between the meridians is:
A) 9
B) 6
C) 16
D) 12

ans is c

conv= dlong* constant of cone ...so how do we get the ans as c?
Constant of cone = 0.75

Convergence = 055 - 043 = 12 degrees.


As you said Convergence = dlong x constant of cone.

Rearranging this gives dlong = convergence / constant of cone

Inserting the data gives dlong = 12 degrees / 0.75 = 16 degrees.
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Old 14th Oct 2011, 10:19
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At 0020 UTC an aircraft is crossing the 310 radial at 40 NM of a VOR/DME station. At 0035 UTC the radial is 040 and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are:
A) 088 - 232 kt.
B) 080 - 226 kt.
C) 090 - 232 kt.
D) 085 - 226 kt.
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Old 14th Oct 2011, 15:40
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At 0020 UTC an aircraft is crossing the 310 radial at 40 NM of a VOR/DME station. At 0035 UTC the radial is 040 and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are:
A) 088 - 232 kt.
B) 080 - 226 kt.
C) 090 - 232 kt.
D) 085 - 226 kt.
Draw a vertical line to represent the 360/000 radial

At the bottom of this line draw a cross to represent the VOR

From the VOR draw a straight line upwards and to the left to represent the 310 radial. The angle between this line and the 360/000 line will be 360 - 310 = 50 degrees.

From the VOR draw a straight line upwards and to the right to represent the 040 radial. The angle between this and the 360/000 line will be 000 + 040 = 40 degrees.

On the 310 radial line draw a cross to represent the initial position 40 nm from the VOR.

On the 040 radial line draw a cross to represent the second position 40 nm from the VOR.

Draw a straight line joining the two crosses.

You now have a triangle with corners at the VOR, position 1 and position 2.

The internal angle at the VOR is 50 + 40 = 90 degrees.

The side from the VOR to position 1 and the side from the VOR to position 2 are each 40 nm long. So the internal angles at positions 1 and 2 are equal.

To calculate the size of these angles subtract 90 degrees (the internal angle at the VOR) from 180 degrees (the total internal angle in a triangle) then divide by 2. This gives an angle of 45 degrees.

We now have a triangle with internal angles of 90, 45 and 45, and two side of 40 nmn length.

Using pythag the distance from position 1 to position 2 is the square root of ( 40 nm squared + 40 nm squared), which is 56.5685 nm. This distance is flown in 15 minutes (between 0020 and 0035). 15 minutes is 1/4 hour so multiplying this distance by 4 gives a ground speed of 226.74 knots.

At position 1 the internal angle is 45 degrees so the external angle between the 310 radial and the track is 180 45 = 135 degrees.

Adding this to the 310 radial give 135 + 310 = 445 degrees true track. This is greater than 360 so subtract 360 to give the true track of 85 degrees.

So the true track is 085 degrees and the ground speed is 226 knots.

Now try this one so see if you have learned anything.

At 0020 UTC an aircraft is crossing the 300 radial at 50 NM of a VOR/DME station. At 0040 UTC the radial is 050 and DME distance is 50 NM. Magnetic variation is zero. The true track and ground speed are:
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Old 14th Oct 2011, 20:12
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thanks ...never seen this type of q before

ans is 212.132 kts and 85 deg T
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Old 15th Oct 2011, 22:49
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Yes,
ans is 212.132 kts and 85 deg T
is the correct answer.

An aircraft departs Hong Kong 2219N 11412E at 2234 Standard Time on the 17 April. If the flight time is 11 hours 17 minutes, the Standard Time and date of arrival at Honolulu (Hawaii) 2125N 15755W will be:
A) 1151 17 April
B) 1551 17 April
C) 1151 18 April
D) 0551 18 April

ans is b...
so how do we decide here in this question what the standard times are rounded off to the lower hour or the upper hour?
I suspect that this question has been taken from the notes of a UK FTO. The original version of these notes would include extracts from the Air Almanac. You are expected to use these extracts when answering the questions.

From the Air Almanac exatract:

Standard Time in Hong Kong is GMt (UTC) + 8 hours.
Standard Tine in Hawaii is GMT (UTC) - 10 hours.

This means that Standard time in Hawaii is 18 hours behind that in Hong Kong. So Hawaii Standard Time = Hong Kong - 18 hours.

This means that when the aircraft took of at 2234 ST in Hong Kong the ST in Hawaii was 2234 - 18 = 0434.

Adding the 11 hours 17 minutes flight time gives ST for time of arraival in Hawaii as 0434 + 1117 = 1551.

Because the times in these calcualtions did not become less than zero nor more than 2400 the date did not change.

So the correct answer is 1551 on 17th April.

Had the calculations produced a time of less than zero then you would need to subtract one day from the date and add 24 hours to the time. If the calculations produced a time greater than 2400 then you would need to add one day to the date and subtract 2400 from the time.

If your were using an unofficial (pirate) copy of the notes they may not have included the Air Almanac extracts. In this case you would probably be using the Arc OF Time method to calculate the time difference.

If you were using the arc of time method (1 hour time change per 15 degrees longitude) then you would have come up with a time difference of something like 18 hours 8.5 minutes. This would not have matched any of the options. Is this why you were concerned with rounding up or down of the numbers?
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Old 17th Oct 2011, 09:38
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guys can u plz help

guys can u plz help
On a Lambert conformal chart the distance between two parallels of latitude having a difference of latitude = 2 , is measured to be 112 millimetres. The distance between two meridians, spaced 2 longitude, is, according to the chart 70 NM. What is the scale of the chart, in the middle of the square described?
A) 1 : 1 056 000
B) 1 : 756 000
C) 1 : 1 984 000
D) 1 : 1 233 000
ans c
how exactly thats has come could you please tell me
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Old 17th Oct 2011, 09:55
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scale varies as sec lat
so
112 mm = 11.2 cm
diff is 2 deg lat = 60 *2 = 120 nm on earth

scale denom = 120*185200 /11.2 = ANS
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