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How do you use a 30cm ruler to find DMS on a paper map? The locations I would like to find are the 'corner' points so I can generate an extent based on the four corners.

I have an old paper map (3 actually) for Northern Canada (late 1800's) that doesn't provide the Ellipsoid or the Datum. It provides a representative fraction (1:660,000 approx) and a scalebar (1" = 10 2/3 mile). The map shows grid lines spaced every 1 degree. No minutes or seconds are labelled.

I understand that NOT knowing the datum or ellipsoid will automatically introduce a margin of error into the calculations, but this is not a big deal for this exercise.

I determined the Lat/Lon of the intersecting grid lines, and from this question, was able to infer that it is closest to Lambert Conformal Conic (Statistics Canada, EPSG 3347).

Below is the index map showing all 3 maps with grid lines every 2 degrees: enter image description here

I will need to do this process for all three maps as those grid lines are spaced every 1 degree, and not 2 as in the index above.


Of course, I could geo-reference to a known spatial reference in a computerized GIS and digitize the extent, but what if your GIS is PC-less and you have traveled back in time and are now stuck...

If it is easier to provide an answer using say, an engineers ruler (1:100, 1:2500 etc) then feel free. It's just a 30cm ruler seems to be more readily available in a given situation.

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Do you also have a T-square? –  Kirk Kuykendall Jun 16 '11 at 0:07
    
@kirk No, but I can get one. Are you suggesting it might be necessary to use because of the orientation, spacing in the intervals of the grid lines (not all the same) and lack of grid lines for the desired locations (ie, the corners that have no grid lines beyond the edge of the map)? –  SaultDon Jun 16 '11 at 1:01
    
Given your maps are in the 1800's you do realize that you can rule out a whole load of datums. I would search what Canada was using then (can't remember) to limit your search. –  Dan Patterson Jun 16 '11 at 1:26
    
@dan Yes. My reference to the 3347 was just a quick visual. I am thinking it is based of the Clarke 1866 ellipsoid. I may call Geological Survey of Canada (NRCAN) to get some more info. They have a stamp on the outside of the map. This map dates around 1897-1899. –  SaultDon Jun 16 '11 at 2:33
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What about a calculator or at least trig tables? –  MerseyViking Jun 16 '11 at 9:54

2 Answers 2

up vote 5 down vote accepted

This isn't so old-fashioned: I remember having to solve exactly this problem back in the 80's when we didn't have scanners readily available and had to lift coordinates and elevations off large-format printed maps for geostatistical analysis.

In effect you can already read the longitude accurately along any line of longitude on the map. You want to interpolate these measurements to four specific points (the corners). Ditto for the latitude. Thus, this problem is a special case of interpolating between contours on any contour map. Therefore you don't need to know anything about the projection or datum to do it.

Because this is supposed to be done simply, we cannot easily exploit the fact we have full contours. It will suffice to identify a few discrete points along each contour and use them. This makes the problem equivalent to the following:

Given a collection of points on map, each labeled with a (smoothly varying) numeric value, to estimate the value at one other specified point on the map.

In order to solve this we need to establish a coordinate system for the map itself. The choice doesn't matter as long as the coordinate isolines are evenly spaced (they don't even have to be mutually perpendicular!) A simple way to accomplish this is to use the ruler to measure distances from the left edge (x) and bottom edge (y) of the map. (If you have a scanned image, just use the row and column indexes of the pixels.)

The interpolation can be accomplished by fitting a trend to the data.

We know, just by looking at the map (that is, by observing the locally regular spacings of the contours), that a linear estimator will work fairly well and a quadratic estimator will work even better. It's probably overkill (and too much work) to use any higher-order estimator. A quadratic estimator requires at least six control points. Use a collection of points clustered near the estimation point: this will assure high accuracy. Use more than the minimum: this provides useful cross-checks and can even yield error estimates.

This results in the following procedure, to be done for latitude and repeated for each corner point and then repeated all over again for longitude:

  • Mark off more than six points along relevant contour lines in the vicinity of a corner point. Use several different contour levels.

  • Measure (x,y) at the marked points and at the corner point.

  • Record (x,y,dependent value) at each marked point.

  • Compute the least squares fit of the data using the model:

    (lat or lon) = a + b*x + c*y + d*x*x + e*x*y + f*y*y + error
    
  • Apply the fitted model to the (x,y) value for the corner point.

People have been computing least-squares fits far longer than they have had mechanical calculators available. If you really don't have a computer or calculator available, settle for a linear trend and for the (easy) calculations consult any textbook on regression published before about 1970. Otherwise, you can do the fit with a graphic calculator, a spreadsheet, or (best and easiest) any full-featured statistical package. The latter will be able to provide you a prediction interval to assess the uncertainty in the estimates.

For example, I applied this procedure twice to find (lat, lon) at the upper left corner using the marked points (red for longitude, blue for latitude, yellow for the corner):

marked map

Using obvious variable names, I obtained the predicted values with two Stata 11 commands for each calculation:

regress lat x y c.x#c.y c.x#c.x c.y#c.y if lat!=0
predict lathat
regress lon x y c.x#c.y c.x#c.x c.y#c.y if lon!=0
predict lonhat

The estimated (lat, lon) of the corner point is (61.05, -136.80). The estimated error is surprisingly large (around 0.04 degree), about twice what I would expect from the resolution of the screen image. These contour lines might not be very accurately placed.

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Thanks whuber! I most certainly will not have a computer at all time, nor a calculator. I live in a cabin. Me any my horse will ride these trails on the map. Finding these 'extents' or any unknown coordinate in the future for that matter will be a nightly exercise so your answer is greatly appreciated. –  SaultDon Jun 16 '11 at 18:30
    
You're right about the contour lines... Notice how you can see from the naked eye, that the longitude lines, East to West, get closer together. Such a large area. –  SaultDon Jun 16 '11 at 19:03
    
@SaultDon In a cabin, I would be content to linearly interpolate the points of intersection of the contour lines with the edges of the map. Actually, I wouldn't even bother with that: there are plenty of better ways to follow a route on a map, anyway :-). People have been navigating from maps since long before least squares (or Euclidean geometry, for that matter) were invented. –  whuber Jun 16 '11 at 19:20
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+1 very nice answer whuber. –  MerseyViking Jun 18 '11 at 13:43
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@SaultDon For multiple regression with the TI-83, see web.centre.edu/lesley.wiglesworth/TI-83%20guide.pdf . –  whuber Jun 19 '11 at 16:30

Right, a bit of trig, some simple algebra, and a ruler should get you there... assuming it is a conic projection with the north pole at the centre.

First you need to determine the location of the north pole. To do that, you need to measure the distance along the bottom of your map of two points, A and B. To keep things positive, you can add a horizontal offset as in the image, but it is not essential.

Measure the angles a and b from the map using a protractor or Pythagoras (don't use the angles as they are written because the cone's meridian probably won't be the prime meridian), you can calculate the y-interceptions of the two lines with ya = tan(a) * A and yb = tan(b) * B Note angles a and b are the internal angles, that is they are less than 90 degrees. You also need the slopes of the lines, which can be had with ma = tan(180 - a)

With those four numbers, use the maths described here (or use the handy calculator at the bottom of the page), which will give you the position of the pole relative to your origin O. From here you can shift the origin so it is in line with the cone's meridian (the dotted line in the illustration), and also note the difference between your measured angles and the ones on the map, both of which should be identical and also equal the projection's meridian.

Finding the pole

To calculate longitude for any given point now, just measure its distance along the x axis from the map's meridian, call it p, and get the y-coordinate of i, call it q, and use atan(q/p)

To calculate latitude, note that lines of latitude are equidistant from each other, so the length of a line from the point of interest to the pole will be linearly proportional to the latitude of that point.

Caveat cartographer: I've not tried this on an actual map, just some scribbles in a notebook and a quick google, so YMMV.


A purely pen and ruler method has just come to mind: pick two lines of longitude that are either side of the corner you're interested in. Find where a line of latitude intersects the longitudinal lines, draw a line from one intersection to the next, and find the midpoint. Do the same for another line of latitude. Then draw a new longitudinal line joining those two midpoints. Then do the same with one of the halves that contain the corner. Rinse and repeat until your line is as close to the corner as you can get. Assuming your longitudinal lines are 1 degree apart, the fractional part of your new longitudinal line will be 2^-n * l where n is the number of bisections you did, and l is the integer number of n s from the known longitudinal line.

After that, calculating the latitude is the same as above, just simply measure the distance along your new line from corner to a line of latitude, and divide it by the length of 1 degree.

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On yet further thought, if it's a Lambert equal area projection, your lines of latitude won't be equidistant. But as I suspect in this case the maps are very close to one of the standard parallels and the distances involved are relatively small, it'll probably be close enough. –  MerseyViking Jun 16 '11 at 12:22
    
+1, I am going to try this tonight. I was initially finding the distance in mm between two lines of longitude (this distance would intersect my mystery point) to determine how many seconds in a mm and then adding or subtracting how many seconds from a longitude line that I measured from. But it's the latitude lines where that kind of brute approximation doesn't seem to work well (do I need to position my ruler at an angle somehow to get the lat?)... I will report the results this evening! –  SaultDon Jun 16 '11 at 16:34
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+1 The second approach is simple and practical. The first one is a great idea but I fear that in practice the (unavoidable) errors may be so great that it just won't work out. For one thing, the origin of the conic system is going to plot far off the map. It will be hard to pin it down exactly, and that error will propagate throughout all subsequent estimates. –  whuber Jun 19 '11 at 16:30
    
I tried the first suggestion using all that trig but because I am uncertain as to the projection, some of the(my) results were off (map may not be in LCC) but obtainable! The webpage calculator sped things up a bit and using a horizontal offset. The second method was straightforward and tedious (math isn't?), so much 'halving' if I want precision, but what can one expect? Sometimes this method gets tricky because of these particular maps. They have at some point in time, been taped back together along the fold creases thus causing overlaps (this changes the distance between some lat/lons)... –  SaultDon Jun 20 '11 at 18:23

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