ATPL Navigation Question

[Mattoosular]

I recently did my ATPL Nav exam, and uh.. failed. There was a question in there that I was very confused about which was something like this:

Standard parallels at 25N and 20N (Parallel of origin therefore at 22.5N). If a straight line (S.L) is drawn in a south/west or north/east direction, the great circle at 22.5N will be:
a) left of S.L
b) right of S.L
c) on the S.L

I immediately dismissed 'c)' since the GC is concave to the parallel of origin. But then I was confused as to where the GC would be left, or right (as this would depending on direction of travel?). Does this therefore mean that the answer is in fact, C)?

Also on another question, Does GPS operate on the principle of "time difference" from transmitter to receiver, or on the time taken to get to the satellite.

Thanks for your input :)

[southernskyz]

Have you thought about doing your atpl subjects by correspondance?
I did it that way, except for flight planning and loading.
Mr.Secombe at bankstown airport, runs the atpl course all year round with classroom or self study courses available.
Great service and all correspondance queries are answered immediately through email.

[ThreeWheels]

My understanding of GPS is that the signal is sent as 'the time' at the instant it transmits.
The time of arrival then gives the distance between the satellite and receiver.
So if I understand what you are asking correctly they are both right but the first is the 'most correct'.

[Hasselhof]

Basically, yes, C is correct. The straight line track approximates the great circle track which is concave to the parallel of origin when it is on either side of that point. When it is smack bang on the parallel of origin, the straight line track is the same as the great circle track.

With the GPS, you aren't sending any info to the satellite. The GPS unit detects the transmissions from the satellite and works the time difference between what is should be seeing "now" and what it is actually receiving that was sent "then".

Using this information it "draws" for itself a spherical arc, any point on the surface of which could be your location.

Once it has calculated multiple arcs from the transmissions from multiple satellites, and knowing the current position of those satellites, it is able to calculate your position using the intersection of those spherical arcs.

[alphacentauri]

Mattoosular,

In answer to your gps question. All of the gps satellites send out the same coded signal at the same time at the speed of light. The gps reciever also knows what time it is and knows what part of the coded signal it should have recieved at what time. Obviously the reciever recieves the coded signal slightly late (due to the time taken to travel from the satellite to the reciever). Because we know the speed of light and the reciever also knows the time delay. It can accurately measure a distance to that satellite. If it does it with three independant satellites then it can accurately triangulate your position. So your answer is, the principle of gps is based on time taken for the signal to travel from satellite to reciever.

To try and answer you mapping question....it does depend on your projection. If you are working on a mercator projection and you draw a straight line and a great circle line. The great circle line will always be concave to the EQUATOR. (Note standard parallels and parallel of origin are irrelevant).If you are working on a conformal conic projection then (by definition of conformal conic) a great circle line is a straight line. The standard parallels are again irrelevant and at 22.5N your answer is c

Hope this helps you

Alpha

[FGD135]

Mattoosular,

Re the great circle question:

The projection in this case must be a conical (e.g Lambert's Conformal). I know this because it has two standard parallels. Many aeronautical charts, the world over (most, I suspect) are based on this projection, so it is well worth knowing its properties.

The property of this projection that is relevent to your question is that a great circle approximates a straight line. So, that would make "C" the correct answer. Note that "approximates" in our context can be taken to mean "equates to".

I have never heard of a "parallel of origin" in relation to a conical projection. I don't believe (and cannot see how) any such thing exists for the conical. All projections have an "origin" but something that could be called a "parallel of origin" could only exist for the Mercator projection.

I think this "parallel of origin" thing has thrown you.

Re the GPS question:

GPS uses the time differences from all the received satellites to determine the ranges to those satellites. Once the receiver knows the ranges to the satellites, it can compute its own position (since it knows the instantaneous position of all the satellites).

Note that the receiver does not transmit anything. It is not a transmitter of anything. It is a receiver only.

All the satellites transmit a coded "pulse". Each satellite transmits that pulse at precisely the same instant. This is the key to the whole thing.

These pulses arrive at the receiver at slightly differing times. Those differences are converted into ranges (distances) by the receiver. The receiver then knows the distance to each satellite and, (computed separately), the position of each satellite. Calculating its own position is then trivial.

[Mattoosular]

Wow - I wish I had joined this site earlier... thanks for your help :)

[Mattoosular]

I had never heard of "parallels of origin" either. Apparently for a Lambert's conical chart, the 'parallels of origin' are midway between the standard parallels. And although great circles are approximately a straight line on a Lambert's, they are actually 'concave to the parallel of origin'. If this is drawn in the exam question I was given, the great circle almost appears as an "S" shape when it passes through the origin.

It seems like unnecessary detail...

[knox]

"I have never heard of a "parallel of origin" in relation to a conical projection. I don't believe (and cannot see how) any such thing exists for the conical. All projections have an "origin" but something that could be called a "parallel of origin" could only exist for the Mercator projection."

2 standard parallels and half way between is the parallel of origin.
Standard parallels show scale correctly and the parallel of origin is where the scale is smallest.




Knox.

[FGD135]

2 standard parallels and half way between is the parallel of origin.


Ok, so such a thing as the "parallel of origin" does seem to exist, but it has absolutely no practical purpose. Unlike the standard parallels, it is not used anywhere in the mathematics of the projection.

Instead of saying "scale is smallest at the parallel of origin" you could just say "scale is smallest along that parallel that is mid-way between the standard parallels"!

[knox]

Ok, so such a thing as the "parallel of origin" does seem to exist, but it has absolutely no practical purpose. Unlike the standard parallels, it is not used anywhere in the mathematics of the projection.

Instead of saying "scale is smallest at the parallel of origin" you could just say "scale is smallest along that parallel that is mid-way between the standard parallels"!

I'd most likely think that it would be used in the "mathematics of the projection", but not sure how and don't really care.

Really don't understand why you would want to make the explanation any more wordy than what it already is... just seems to complicate things.

But at the end of the day, not really practical for hands on flying. Just know how to read the map. :ok:

Some good info here Jar professional pilot studies - Google Books (http://books.google.co.nz/books?id=KY-MBUeQoZEC&pg=SA10-PA12&lpg=SA10-PA12&dq=parallel+of+origin+lamberts+chart&source=bl&ots=cM1lTi2TWQ&sig=o9bF5w7bhZSka3KNd_gcFbHt1R0&hl=en&ei=ZUTkTN6uGIn6sAOKztFm&sa=X&oi=book_result&ct=result&resnum=3&ved=0CC0Q6AEwAg#v=onepage&q=parallel%20of%20origin%20lamberts%20chart&f=false)

Knox.

[VIMD]

Mmm... to comment or not. I might be digging up an old thread here, but cannot help commenting and hopefully helping others that come across this thread.

Contrary to the belief of some posters here, a Lambert Conformal Conical Projection does indeed have a Parallel of Origin, as well as two Standard Parallels.
Scale is correct at the Standard Parallels, whereas Convergency is correct at the Parallel of Origin, which is approximately midway between the Standard Parallels (actually slightly closer to the Upper Standard Parallel). Convergency relates to the change in great circle track between given meridians, so the Parallel of Origin is actually very important. If you know the Parallel of Origin, you can calculate the angular difference between meridians and the change in the straight line track on the chart. For Conformal Conical charts, reference is also made to n, which is the constant of the cone, and this is the same as sine of Parallel of Origin (Sin PoO).
On the Earth, the change in great circle track can be estimated from Earth Convergency, and as the meridians on the Earth are not straight lines as on the chart, the angle of convergency between two meridians on the Earth will change as a function of sine latitude.
Earth convergency changes with latitude between two meridians whereas chart convergency between meridians is constant, and depends on Sin PoO. The straight line can therefore only accurately depict the great circle where chart and earth convergency is the same - at the parallel of origin. Anywhere else on the chart, the straight line approximates the great circle, but the great circle is concave to the parallel of origin. An exception to this will be the meridians which cross the parallel of origin at 90°, and will exhibit no concavity towards the parallel of origin.

Bit late at night, but hope this clears up some confusion.

PS: Scale is correct at the standard parallels, but will be contracted maximum at the parallel of origin, so you could say that scale expands away from the parallel of origin.

[MakeItHappenCaptain]

Slight correction to the GPS/GNSS explanations here.

Three satellites would work if the receiver had the accuracy of an atomic clock (loses less than 1x10-9 secs per day or roughly 1 second every 30 million years) and was correctly synchronised to the satellites.

Definitely not.

At the speed of light (300,000,000m per second) you can see how much distance error a poofteenth (technical term) of a second would produce.

Therefore a fourth satellite is required to differentially solve for the timing error and give an accurate position.