**scale factor at different latitudes**

July 27 2008 |
3 comments

I need to decide what is the better scale to create a chart of a certain area in order to cover the route of a pipeline, and I need to calculate the scale factor at different latitudes. I am having hard time in finding a solution in this matter:

given the following area in lat and long:

Lat 31°N – 51°N

Long 12°W – 36°E

I think I would use the Mercator projection with a single Latitude of 41°N

as origin. and the Scale factor at Latitude of Origin = 1.00000. I

need to work out what the scale factor is at 51°N and 31°N with a Lat

of O of 41°N ?

If this is acceptable I would like to split the route up in to

Longitude 12° chunks this would mean eastings would be in the range 0

to @1,000,000 for each section. If I set a false Northing of

5,000,000 then there will be no ambiguity between eastings and

northings.

What do you guys think about this solution? Can you help me with this issue?

Thank you

### Mapping Center Answer:

It’s a bit hard to answer your question since I think you might be a little confused about what scale factor is. You see scale factor the relation (ratio) between the denominators of the representative fraction for the actual and principal scales at different places on the map, defined as SF = ACTUAL SCALE / PRINCIPLE SCALE. SF is scale factor, actual scale is the scale that you measure at any point on the map (it will differ from one location to another), and principle scale is essentially the same thing as the value in the representative fraction.

The scale factor is not the same across the entire map -- it will only be 1.0 at a point of tangency or along a line or lines of tangency. On any map the scale factor will vary across the map as a result of the map projection process - no map projection can maintain the correct scale throughout.

You suggest using the Mercator projection which is a world projection centered on the Equator -- you can either choose to have it tangent along the Equator -- so scale factor would be one along that line only, or you can choose to have it be secant at two latitudes symmetrical around the equator. In either case, you would not be using a projection that reduces scale distortion within your area of interest (because the line of tangency is either 31 degrees to the couth along the equator, or you have one lie of tangency that you define someone within your area of interest) -- if you could use a projection that had two lines of tangency in your area of interest, then you would be better off.

Perhaps you could consider using a conic projection instead -- this will give you the opportunity to define two lines of tangency within your mid-latitude area of interest. Since you are mapping a pipeline, I think that you probably are more interested in preserving the shape of features than distances or directions or areas -- you pretty much have to choose which of these you want to preserve as you cannot keep all of them in any map projection. A projection that helps keep shape distortion to a minimum is called a conformal projection. So now you are down to a conic conformal projection. A good choice for your area of interest might be the Europe Lambert Conformal Conic projection (you will find this in the Projected Coordinate Systems under Continental then Europe. Once you choose this, you can modify it to make two changes:

1. Set the standard parallels to be within your study area -- to do this use the 2:6 rule -- divide the range of latitude into 6 intervals (you have 31-51 degrees or a range of 20 total degrees -- each part would then be 3.3 degrees of latitude. Then go down two intervals from the top latitude of you’re a study area (51) and set one standard parallel there (51 - 6.6 = about 44) and go up two intervals from the bottom latitude (31 + 6.6 = about 38).

2. Set the central meridian to the middle of your study area -- your extent is from 12W to 36E -- a total of 48 degrees, so you want to set the central meridian to 12E.

This would probably be my choice if I had to use only one projection -- which of course is better if you want to see the whole area at once. But it does mean that there is more distortion than if you used a series of smaller areas, each of which you could define the projection parameters for -- as you see above, you can do this easily and as a result reduce the scale variation somewhat within the area.

However, you are working with a fairly large area, so really, the of scale distortion that you will be able to mitigate by using smaller areas will only be significant if you use a LOT of smaller areas -- just to show the area you are talking about on my screen requires a map scale (not the same as scale factor!) of over 1:20,000,000. You would have to be working with large scale maps with a map scale of about 1:20,000 -- that’s a huge difference.

Anyway, maybe the above info will help you to narrow down your choice somewhat.

**did you ever find a solution?**posted by Mark Amend on Apr 24 2009 9:35PM

**Re: Did you ever find a solution?**posted by Melita Kennedy on May 11 2009 6:42PM

k = ( sqrt(1.0 - e2*sin(lat)*sin(lat))/cos(lat) ) * ( cos(lat1)/sqrt(1.0 - e2*sin(lat1)*sin(lat1)) )

e2 = eccentricity squared of the spheroid (0.00669437999 for WGS84)

lat = latitude of point

lat1 = standard parallel

This is from John P. Snyder's Map Projections: A Working Manual, pages 44 (equation 7-8) and 47. This book is available online at http://pubs.er.usgs.gov/djvu/PP/PP_1395.pdf

Melita

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More information on Mercatorposted by Melita Kennedy on Jul 28 2008 4:19PMin the strict sense. The Y coordinate origin is always at the equator.

Instead, you can set a standard parallel which is then replicated

to both sides of the equator. If the standard parallel is non-zero,

that turns Mercator into a secant case where the latitude of the

standard parallel is a line of no distortion.

We calculated the point scale factor for Mercator using WGS84 and

a standard parallel of +/-41.

At 31N, k = 0.880957383726

At 51N, k = 1.198546560327

Moving the standard parallel to +/-43 equals out the scale a bit more:

At 31N, k = 0.853794016432

At 51N, k = 1.161590674561

You could also try to fit an oblique Mercator projection to the

general trend of the pipeline. Hotine oblique Mercator will always

make sure that true/geodetic north at the center point (using azimuth

or two point 'center') is rectified to grid north. If you using RSO

(rectified skew orthomorphic), you can set any angle to be grid north.

Melita @ ESRI