Do Land Surveys Account for the Curvature of the Earth?

I assume they do not.

If you have a parcel of land with an irregular surface - or one on a hill - the surface area will be larger than the amount of airspace above the land. It would be overly complex to figure out just what the surface area is. And certainly it’s hard to imagine reflecting the curvature of the Earth for small areas.

But what about large-scale surveys? If you took a measurement of the land area of the entire US, it would be bigger as a result of its being a curved surface than it would otherwise be. Is this reflected in official measurements? If it is, then the official size would be greater than the sum of the parts which don’t reflect it.

I am not a surveyor but hopefully this will do until one comes along.

A parcel is defined by horizontal measurements between points so, no, elevation is not taken into account and a hilly parcel would have more surface area than a flat parcel with the same legal area. Caveat emptor.

Surveys are referenced to a geodetic datum that models the Earth as an ellipsoid and long enough measurements differ from what you would calculate assuming a flat Earth in quantifiable ways.

Individual parcels don’t take elevation into account, but yes, large-scale land surveys do.

The Canadian federal government surveyed the prairies in western Canada in the 19th century, using a large scale grid system. Every so often, the survey lines running north-south are corrected to take curvature into account. North-south grid roads accordingly have a jog every so often, called a correction line. And although Saskatchewan looks like a rectangle, if you zoom in on the Saskatchewan-Manitoba boundary, you’ll see that it has regular little jogs to correct the boundary line as it goes north.

IANAS either, but… from what I’ve read -

It’s horizontal flat measurements, so no, your land is not bigger if there’s a hill or cliff in the middle of it.

Most localized mapping is done using “universal transverse Mercator” coordinates. IIRC this means:

  • consider the area you are locally surveying, let’s say some area of interest a few dozen miles in each general direction… (I.e. a farm, county, mining claims area, etc.)
    -map it as a Mercator projection, with the center of the map being the nearest line of longitude (north-south) rather than the equator.
    -when you get too close to the next line (i.e. from 3-degreesW to 4-degrees west) use that new line instead. For a localized area like a county or township, everything should fit in one fla plane with minimal distortion for earth curvature.

In fact, the spherical earth means that “parallel” northbound roads converge… or that if you lay out land in square mile sections, sooner or later they will not line up as you go north. You can see this effect, I think, in the boundary between Manitoba and Saskatchewan - in big maps, it looks like a straight line. Zoom in, and it jogs regularly to compensate for the shrinking earth.

EDIT - Darn, Piper beat me to that…

(Also look at the map of Toronto’s north boundary. Along Steeles Ave east and west of Yonge street, most major township north-south streets (about 1.25 miles apart) take a jog toward the Yonge St centerline. One story says this was a surveyor error, another explanation says the township road separation gets less as the globe shrinks going north. )

The same thing happens in the Upper Midwest. My brother was on a township board and wanted to do something about people driving the gravel roads that were the section lines (one mile apart). Every so often the road would jog left or right to correct for curvature, and some people (meaning teenagers, or drunks, or drunk teenagers) would speed down the road and drive straight ahead into a farm field; this happened usually at night. They solved the problem by putting T-intersection signs where the jogs happened.

Not quite correct…the jogs are not for curvature–they are due to the incredibly inaccurate surveys that were done 150 years ago. The original surveyors were supposed to mark off “Sections” of land, one mile by one mile. They did this in groups of 36 (i.e 6 rows and 6 columns of 1-mile squares.) Theoretically, every line in the grid is parallel to the others, but in reality they are mistaken by many dozens of feet. So the surveyors corrected by simply jogging over to the right or left by as much as necessary, when they found out they were off course.

Well, actually, your land is bigger if it’s hilly. If you and your neighbor each own an “identical size” lot of land, but his lot is totally flat and yours (silly example) slopes at a 45 degree angle*–your lot is bigger (contains more square feet)-but on the map both lots will be drawn the same size, and the number of square feet will be calculated as the same size.

*(to visualize this: imagine that the lots of land are the rooftops of 2 adacent buildings, with different roof lines …one has a flat roof, one has a steeply sloped roof)

Platting, lot surveys, etc. are done with horizontal measurements. However, when you are doing large-scale cadastral surveys, curvature does come into play. In the US, the Bureau of Land Management has equations for dealing with the curvature that are published in the Manual of Surveying Instructions.

I have a bit of experience that is loosely related.

The scenario: Humpback whale researchers go to Alaska, stake out a cliff top overlooking one of the bays, and observe humpback whales and boats in the bay. The object of the research is to investigate whale-boat interactions. In particular, they want to know if the boats are scaring the whales away or otherwise disturbing them. The study will track whale movements around these bays, with and without boats nearby, and attempt to tell if the whales change there movements when boats are around.

Researchers observe whales and boats with a theodolite (basically, like a surveyors transit), which gives them an angle from north and a declination (angle below the horizontal, since they are looking down on the bay from a cliff top). They record these, enter data into a computer, and the program computes the actual x-y coordinates of the whale. A later program will analyze the whale and boat sightings to determine, specifically, the closest the whales and boats ever approach one another. I wrote those computer programs.

At first glance, using a “flat earth” model, the computations are trivial, as any trigonometry student would learn in the first week of the course. Over the rather short distances involved, one would think this would be just fine. But the lead investigator wanted to compute distances across the surface of the bay using a curved earth model – even though he acknowledged that the adjustments would be petty. Why? Because other researchers were doing that, and would snipe at his published results if he didn’t too, that’s why.

We both went to the library and dug through textbooks to find a relevant-looking formula for the computation. I looked in trig text books, and didn’t find any. He looked in surveying books, and didn’t find any that seemed quite to match the situation, although there were various formulas for vaguely similar computations.

I ended up having to devise the formula myself. It was moderately messy, and involved figures of such massively differing orders of magnitude as the radius of the earth and the height of the cliff. This was 30-some years ago. It’s just possible that I might still have the formula laying around somewhere, if I go digging through all my old stuff, although I doubt it.

One might wonder what kind of accuracy one could get. Note that the radius of the earth itself varies from place to place on the earth, and the variation alone is of the roughly the same order as the altitude of any cliff top from which observations were made. I had no data on the radius of the earth at those particular places, only a general figure from various general sources. (Equatorial radius, 6378 m; polar radius, 6357 m.)

So I ran the computations with a whole bunch of hypothetical test figures, varying the earth radius over a range of several hundred meters, and varying the cliff-top altitude over hundred or so meters. The results were just the way one would have liked: The formula was highly insensitive to rather large (but reasonable) variations in earth radius, but highly sensitive to small changes in cliff altitude. IIRC I also tested variations in the angle of declination, also with desirable results. So I considered our computations to be highly reliable.

To anyone out there who is up on your trigonometry, you could try to reconstruct the formula from what I’ve given here. I think I’ve described the scenario completely enough.

Let’s be sure we’re all talking about the same phenomenon here.

The OP seems to be mixing talking about both topography and planetary curvature. And using the word “curvature” for both things. Or maybe his title & his post don’t match real well. Or I’m dense. Or something …

At any rate, *topography *relates to the fact that an irregular surface has more surface area than the same footprint of a smooth surface. This would be true whether the underlying surface was a true plane or a section of a very large sphere.

Whereas *curvature *properly relates to the difference between the Earth’s surface imagined or simplified as a true plane versus the reality of it being a section of a darn big more-or-less sphere.

Looks tome like different folks are talking about different aspects of the problem.

Fans of land-surveying would be interested in this book:
The Great Arc: The Dramatic Tale of How India was Mapped and Everest was Named



They couldn’t have been accurate, if they didn’t account for the curvature of the Earth. Over an area the size of, say, Ohio, it really does make a significant difference, which necessitates the jogging of the lines.

The Public Land Survey System absolutely took curvature into account, at least ideally:

Those places where the roads that follow the survey lines don’t line up, those are the points where the surveyors reset the starting point of their surveying. One spot where this is clearly visible on a map is in Chicago on the north and south sides of North Ave roughly west of Western Ave.

Not quite correct. To quote the Wiki:

That said, errors in surveying are common, especially in Ohio and other points toward the east of the country.

Alas, even the Master has had to correct himself on this very point. The PLSS generally put correction lines every four townships (24 miles). There’s one south of Chicago, and additional weirdness where two surveys done at different times meet along the Indian Boundary. But North Avenue is not a correction line.

Actually, and interesting - I had always assume the major township roads laid out around Toronto in the early 1800’s followed regular patters - 1.25 miles on a side.

Looking at Google earth, the layout first appears to be trapezoidal, and then the GE measuring stick appears to show the roads are not evenly distributed, ranging from 1.2 to 1.4 miles apart. The jogs north of Steeles appear to be because the Markham township roads are closer than the Toronto ones. Weird.

I assumed the Scarborough township roads were half a mile apart, they also exhibit a certain amount of irregularity.

I assume this is from design or laziness. Even the worst tools of the day were not that inaccurate.
Sorry, the point is that land surveys just care about the horizontal measurements (“plan view”), not hills or slopes. If your land is a hill, that’s your bonus or curse depending on circumstance. Before computers and such, it would be too complex and awkward and meaningless to account for slope. After all, if your land is a corduroy of parallel ditches, does that mean it’s 2 acres not 1?

Found an interesting article in the Globe & Mail about the Saskatchewan-Manitoba boundary, confirming that the jogs are necessary to account for the curvature of the earth as the boundary goes north.

It also provides info on the corrections needed in the township lines.

"Why the zigzag between Sask. and Man.?

Opinion: Why the zigzag between Sask. and Man.? - The Globe and Mail

Simple back-of-envelope math calculation, with major simplifications, and assuming I remember my trig from 1970…
So let’s assume we are at 45-degrees. Simplify calculations by assuming a cone with slope of 45-degrees instead of a sphere…
Earth equator circumference approx. 40,000 km.
At 45 degrees, 40,000/sqrt(2)= 28284 circumference
Go north 10 km. radius of cone now 10/sqrt(2) less, circumference (2pi)*10/sqrt(2) less, or about 44.43km less circumference at that latitude.

Divide by 360 degrees.
At 45 degrees, each degree of longitude is 28284/360= 78.6km (about 46 miles) apart.
Go north 10km, 6 miles, and east-west you lose 44.43km/360=0.123 km or 0.7 miles between “parallel” longitude lines 46 miles apart.

If you are doing 1-mile square sections, then at 45 lat roughly every 10 miles north you go, you lose one 1-mile section in 45 east-west.

Of course, the further north you go from the equator, the faster the loss. At 60-degrees, the rate is 1/2 not 1/sqrt(2) (i.e. 0.5 not 0.7), it’s essentially cos(latitude). near the north pole it’s about 1 for 1.

Here’s a USGS map of an area of east central Wisconsin…,-87.77397&z=13&t=T

The survey squares on the north and south sides of the line in the middle of the map. The main survey north-south baseline for Wisconsin’s PLSS (the 4th Principal Meridian) starts out not too far from where Wisconsin, Illinois, and Iowa meet at the Mississippi. As a result, the compounded offset corrections grow as you move east and thus you get the clearly offset township lines near Lake Michigan.

About 1/8 of a km is not equal to about 3/4ths of a mile. You slipped a decimal place. 0.123km is ~= 0.07mi. Which means your conclusion is off by some similar factor.

But other than that, your trig seems good to me.

Doh! Thanks, I always seem to make an egregious error doing math in my head.

So - go north 10 miles at about 45 Lat, and you lose about .1 mile in 46 miles east-west.