Question on scale

[paddywak]

I came across this question while working through some examples on scale and I'm still trying to get my head round it.

On a direct Mercator chart at 15 degrees South, a certain length represents 120nm on Earth. The same length on the chart will represent on Earth, at 10 degrees North, a distance of ?

a.) 124.2 nm
b.) 118.2 nm
c.) 121.3 nm
d.) 117.7 nm

Can anyone help?

If the change in longitude is 120nm or 2 degrees longitude at 15'S

Then Departure= ch in lg (minutes) X cos lat

So: Departure= 120nm X cos 15

Departure= 115.2 nm at latitude of 15'S

If the change in longitude is 120nm or 2 degrees longitude at 10'N

Then Departure= 120nm X cos 10

Departure= 118.2 nm at latitude of 10'N

So the distance at 10'N lat is greater than the distance at 15'S lat by 3nm

It's at this point where I became lost

[flip2]

Before I confuse you, are they saying the answer is 117.7nm or 118.2nm?

[paddywak]

Apparently the answer is c.)

I can do the scale from and to the Equator to another latitude

ie. If the scale is 1:1000,000 on a Mercator chart, what is the scale at 60N

Scale = scale at Equator x secant 60N

Scale = 1000,000 x cos 60N

Scale = 1000,000 x 0.5

Scale = 1:500,000

I know it must be to do with the change in secant but I just cant figure


The reverse is true for knowing the scale at a latitude ie. scale = 1:500,000 at 60N

so 500,000 / cos 60N = 1000,000

I dont even know if these questions will appear in the CPLH exams but it's just bugging me now

[flip2]

Ah, and therein lies the importance of the gross error check before starting! I lazily neglected to establish whether the answer should be bigger or smaller before starting

Unfortunately, I can't help you because the answer I get is 122.3, not 121.3.

Do post back if you find the answer

[paddywak]

I worked it like this after much faffing about

Denominator A = cos A
------------------ -------
Denominator B = cos B


Denominator A = cos10N 0.9848
----------------------------------------
Denominator B = cos15S 0.965925


Therefore Denominator A = cos10N 0.9848 * 120
------------------------
cos15S 0.96592

Therefore Denominator A = 118.1769304
----------------
0.965925

Therefore Denomintor A = 122.34

Scale is 1:122.34

Sometimes on days like today I doubt my capabilities of becoming a professional pilot. I have not been (up) since my return from Canada, I think maybe it's time I reminded myself why I want this so much. Soon as there is break in the wx I'm booking out.

Thank you for trying Flip, I think I have it now , it's just dissapointing when you dont arrive at the same answer as the one in the book, this is where distance learning falls down.

[flip2]

Ah, you have the same answer as me... 122.3
But the answer is 121.3, no?

[paddywak]

Yes the "correct" answer in the book is c.) 121.3 nm

But as 2 people got the same answer independently then I suspect it's a misprint or typo, and we are indeed correct

[flip2]

I'm glad you got it sorted

Edit: Full explanation below

[paddywak]

I like that better, I think I'll use that instead, cheers Flip

[flip2]

[paddywak]

Thanks for the morale boost.

Im trying to get them completed before the change over in April 2012

So I have the first 5 exams in Jan and the other 4 in March, I'm going down the CPL route as I dont have the funding for an IR.

But one step at at a time. If I dont pass then I will have to sit the new exams which are 13 for the CPLH so fingers crossed.

[LoachBoy]

Best tip I can offer is, by all means work through your answer, certainly take advantage of the knowledge-base on here..... but don't spend too long beating yourself up - you will bump into many 'typos', including on the CPL(H) exams.

I lost count of the hours I repeatedly quizzed myself over typos!

This question specifically - flying in the UK it is not really relevant, and our ground-school guru basically said it would only become of notable use if you are operating in the northern parts of Canada for example. (Or similar latitudes to the south).

And don't get me started on rhumb-line tracks...!!!

[paddywak]

Ive got to be honest I dont think I can see myself trying to work this type of question out holding a calculator, chart, rule, chinagraph and fly the 22

I still have a job doing it in the planning stage with a cup of tea and a warm room to sit in.

I just hope the actual exam doesn't throw in too many of this type of question.

The clock is ticking with other subjects to cover yet

on to the critical point, PNR etc.......head down cracking
on




A great reminder of why Im trying to learn all this stuff. He landed today in Worcester at the Grammer School cricket pitch at around 13:30 UTC and I just happened to be passing, was it you?

[LoachBoy]

Oooh we like that!

PNR is very useful though, and is dead easy to get your chops around.

That's the good thing about scales & rhumb line tracks - they make everything else seem sooo straight forward!!!

[flip2]

This was bugging me... slow day

So I did some research and 122.3 is indeed the correct answer from the question banks, not 121.3.

With that said, I'll explain how I got my answer and perhaps it will be of some assistance if a similar question arises.

Gross Error Check
As my initial post proved, always do one

A direct mercator has straight, parallel meridians.

Thinking just in terms of longitude (to keep it simple) the chart distance between 2 meridians is always the same on a direct mercator, irrespective of whether you are looking at the Equator, 20N, 45S, 60N etc. But we know that the earth distance between 2 meridians changes as you move away from the equator - it gets smaller as the meridians converge on the poles.

For this question, we are moving closer to the equator - from 15S to 10N. That means we should expect the earth distance to be bigger at 10N for the same chart distance.

ABBA
To answer the question, I used the ABBA forumla.
This formula presumes you have information on two points... point A and point B.
We will call 15S "A", and 10N "B"

The formula goes like this:
Scale at A x Cos B = Scale at B x Cos A

Let us consider the scale at A as being 1:1. It almost certainly isn't, but that doesn't matter. All we want to know is the relationship between Scale A and Scale B, not what their actual values are. I'll also point out that the North / South makes no difference to the question, so you can disregard that information.

1 x Cos 10 = ? x Cos 15
(1 x Cos 10) / Cos 15 = ?
1.0195 = ?

So this tells us that the relationship between the scale at A and the scale at B is 1:1.0195.
Plugging in "120nm" in place of "1"... 120nm x 1.0195 = 122.3nm

Edited for terrible spelling!