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Jinkster
14th Oct 2003, 16:55
Given that the value of ellipticity of the Earth is 1/297 and that the semi-major axis of the Earth, measured at the axis of the Equator is 6378.4km, what is the semi-major axis of the Earth measured at the axis of the Poles?

Answer = 6356.9km

Can anyone please tell me what this question is talking about?

Is it still valid?

Jinkster :ok:

GonvilleBromhead
14th Oct 2003, 17:51
This takes me back about a year. We had this same question in the exams and to be honest none of knew really what it was on about, but we all seemed to cobble together the answer, more by lateral thinking than anything else.

Forgive me if you know how the answer is derived, if not, it goes something like this.

6378.4 / 297 = 21.47

6378.4 - 21.47 = 6356.9

We all kinda figured it has something to do with the ratio of the width of the earth (left to right along the equator) to the "height" of the earth, measured "top to bottom" i.e. between the poles. i.e. as the earth is not a perfect sphere, but more like a squashed sphere (imagine squashing a squash ball between your finger and thumb, it would get "wider" left to right, but also "flatter" top to bottom).

We could have all been completely wrong and none too scientific, but it seemed we all got the same answer.

Came in dead handy too when I was trying to nail the ILS on my IR...........hmmm...

Jinkster
14th Oct 2003, 18:34
GonvilleBromhead,

Not sure how long ago you did the exam but was the figures exactly the same if they were then I will probably just learn the answer - I dont like doing this!!

GonvilleBromhead
14th Oct 2003, 19:11
About a year and a half or so ago so I must admit I can't remember the exact numbers but with this type of question the answer must have been the same, and the figures look vaguely familiar.

I remember this question standing out though because as I say, in the post exam post mortem, we all chatted about this one as we hadn't remembered anything in the notes about it to a man, and yet we all came to the same conclusion with the above thought process.

Basically, it's just about the earth being fatter than taller, by a factor of 1 in 297. So if they give you the equatorial value, then the polar value must be smaller by a factor of 1 in 297, nothing more complicated than that (I think !).

If anyone else can confirm this ?

jockeymon
14th Oct 2003, 20:54
Just doing my ATPL ground school now and as has been said above the earth is an elipse (fatter at the equator than across th poles) an elipse is measured by two "radius" measurements the semi-major (the biggest "radius" on the elipse in this case the center of the earth to any point on the equator) and the semi-minor axis (the smallest "radius" or in this case the center of the earth to the pole).Take the semi-major axis to be 1 if the semi-minor axis is 1/297th smaller than it will be 296/297 or 0.996632 etc.etc.multiply the 6400 or whatever it is by this and you get the semi-minor axis.I`ve just had pointed out to me that the earth should be refered to as an oblate spheroid as well hope it helps!!

oxford blue
14th Oct 2003, 22:52
I can confirm that this question has appeared in the ATPL Gen Nav exam and that, as far as I know, it is still in the Common Question Bank and has every possibility of turning up again.

Gonville and Jockey have the right approach. The geoid shape of the Earth can be approximated closely to an ellipsoid. The lengths of the Polar Diameter and the equatorial Diameter are in the ratio 296/297. Therefore if you know one, you can work out the other. The question gives you all the information; it is not really your navigation knowledge which is being tested, but your ability to re-arrange simple algebraic formulae. This is quite usual in Gen Nav, which is more than a simple test of knowledge - it is a test of numeracy as well.

Polar semi-diameter = 6378.4 x 296/297

Alternatively, you should have been taught that the Polar diameter is 43 km shorter that the Equatorial. In which case, the semi-diameter will be shorter by half that figure. Half of 43 is 21.5 km.

6378.4 - 21.5 = 6356.9

Jinkster, you're going to have to accept that there is an awful lot of simple calculation in Gen Nav - you can't possibly learn all the answers to all the questions by heart. None of it is above GCSE level - it's simple re-arrangement of formulae, and nothing more complicated than sine, cos and tangent - but you've got to be proficient, quick, and accurate.

stator vane
16th Oct 2003, 04:30
that is the bastards who write the ****ing exams and those who approve them for use.

i am so glad that i have finished all that ****!!!

i still hope to find some way in which this entire testing process can some how be brought back into touch with the real world of flight operations!!!

it should be, that a current captain operating to the max scheduling allowed by the regulations, should be able to make a bloody 95 percent on a current ground school test any time of the day!!!


but it will take the determined effort of many pilots to get that done.

good luck with your tests!!!! you will need it.

GonvilleBromhead
16th Oct 2003, 16:54
Agreed completely with Oxford Blue. Basically all this is asking you is "do you know the earth is an elipse ???". Assuming you do, it then becomes basic maths.

As with a lot of JAA questions, they will phrase them in such a way as to make you think "what the hell is this about" or they will incorporate lots of "big numbers" into the question to make it look daunting.

But when you look at them objectively, you soon realise they are actually not that bad, and it's just the examiner who is trying to phase you with "big numbers" or jargon.

Best get used to this fact, it does happen alot. Once you realise this is all that's going on, you're half way there. Otherwise, you will be trying to "learn" answers until you are blue in the face, and as stated by OB, that is not, nor should it be, feasible.

Good luck.