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-   -   Mercator scale question (https://www.pprune.org/professional-pilot-training-includes-ground-studies/147793-mercator-scale-question.html)

Foz2 9th October 2004 18:36

Mercator scale question
 
Hi,

Having trouble with this one,

'On a mercator chart the distance between two meridians 1deg apart is 7.45cm. What is the scale at 48degN?'

Now, I came out with the answer 1:491,543 = 1:500,000

However the answer given is 1:1000,000?

I worked it out by: 1deg = 60nm so 60nm =7.45cm

60nm = 11112000cm so, divide that by 7.45 and you get 1:1491543.

I just cant see how it could be 1:1000,000?

If anyone could shed some light that would be great.

Thanks

Foz

OneIn60rule 9th October 2004 18:55

Well
 
Only way to get 1:1.000.000 is by using

Dep= 60xCos48= 40 NM which is74080 meters or 7408000 cm.


CL/ED = 7.45/ 7408000 = 0.000001005669

or if you prefer= 994362.4161 which is a darn sight close to 1.000.000 and therefore it must be 1:1.000.000

Aim High 9th October 2004 19:16

Aim High
 
'On a mercator chart the distance between two meridians 1deg apart is 7.45cm. What is the scale at 48degN?'

The distance between the two meridians is at 48 N is found by departure formula:

Dep (in NM) = dlong in min x cos mlat
Dep (in NM) = (1 x 60 x cos 48)
Dep (in NM) = 40.147836381

1 / d = CL / ED
1 / d = 7.45 cm / 7435379.29785963 cm

d = 991383.9 (approximately 1,000,000)



The scale at the Equator could also be found if requested
Since on a std Mercator scale expands as a function of secant lat

secant lat = 1 / cos lat

New d / Old d = Old secant lat / New secant lat

Must rush dinner on table (lol)

Foz2 9th October 2004 22:39

Thanks very much guys,

All sorted.


Cheers

Foz:ok:


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