Map location question Zero point for NAD 83

I was looking over some maps and noticed the border numbers corresponding to North American Datum 27 (NAD 27) or NAD 83. It turns out that its easy to find the zero point for NAD 27 as its easily Googable.

but its not clear to me where the zero point for NAD 83 is. I’ve read some mumbo jumbo about how NAD 83 is based on the center of mass of the earth and its a refinement of NAD 27, but they don’t clearly say where exactly the numbers go down to 0.

Is that point the same as NAD 27 (Meades Ranch), or is it somewhere else? Or is it close enough to the original NAD 27 marker to be good.

Where is this for NAD 83?

The ellipsoid is meant to fit the global mean sea level, or MSL by the least squares method.

Let me see if I can make my question more clear.
In Southwest Colorado is Stewart Peak. Its NAD 83 coordinates are 4210xxx and 331xx (problem: I don’t know how many zeros go in place of the 'x’s)

As I go south the first number gets smaller 4209…4208…etc. and as I go west, the second number gets smaller 330, 329 etc.

Where on the map do these numbers become 000 and 000?

Its difficult for me to scroll through bunches of maps trying to find that point, so maybe I’m looking for an easy way to find it?

Some of the mumbo jumbo from the Wikipedia page

“The North American Datum of 1983 is based on a newer defined spheroid (GRS 80); it is an Earth-centered (or “geocentric”) datum having no initial point or initial direction.”

For finding where the numbers go to zero, it depends on what numbers you are referring to when:

If these numbers are latitude / longitude (the basic numbers of the NAD) then the numbers go to zero on the equator and the prime meridian (the one that goes through Greenwich, UK), in the Atlantic Ocean just south & west of the African coast.

It should be noted the NAD 27 system does NOT have the numbers go down to zero at the Meades Ranch Triangulation Station. This benchmark is at latitude 39°13’26.686" N and longitude 98°32’30.506" W. This point is the United States standard horizontal datum because all land measurements in the nation start at this point. To figure the lat/long of any other point, you measure the distance and direction from Meades Ranch, then calculate the lat/long difference and add or subtract from the Meades Ranch lat/long numbers.

If the numbers on the map border are in feet or meters, they are State Plane Coordinate (SPCS) numbers:

These numbers are a Cartesian system, NOT a Spherical system. That is the SPCS is trying to warp the round surface of the earth onto a flat piece of paper. As such the SPCS is only approximately accurate, and only for a limited area of the earth. Therefore the US has 124 geographic zones with different SPCS. Small states like Maryland have one zone for the entire state, large states like Texas has 5 different zones.

I believe (but do not know) all the SPCS zone are set up to have positive numbers, that is all locations in each zone are north and east. So the point where the numbers go to zero in each zone are either at the south-west corner, or they are off from the south-west corner.

I picture of the the SPCS zone in the contiguous US:

These are State Plane Coordinates, pretty sure you need to add 3 zeros to the end of each numbers. You can confirm this by looking at the map scale, and measure between the numbers along the map border. If the distance scales to 1,000 feet, you need to add 3 zeros.


for each map the numbers go to zero for the southwest corner of the state they are in? Is that what is meant by “State Plane Coordinates?”

Oh for latitude and longitude? I did forget how weird those are. That is the Universal Transverse Mercator zone 13 and the numbers are “False Easting” and “False Northing” from that UTM of 13N

I always use software to translate them to be honest, but the NAD 1983 UTM Zone 13N will have the equator start at zero meters and the central meridian will be 500,000 meters.

Sorry I forgot just how backwards, old and painful those were.

Those are UTM coordinates and are meters North of the equator. UTM is a projected coordinate system. NAD 83 is a datum.

Well, more correctly, the UTM coordinates are meters north from the equator and meters east or west of that particular UTM zone’s central meridian with any given zone’s meridian being defined as 500,000 meters east.

Thank you, I knew I was screwed up but the edit window didn’t allow me to fix it, so I just got rid of the bad info.

For meters East they broke up the earth into 60 zones starting at the international date line. They established the midpoint of each zone as 500,000 meters and then built is east and west from there.

Should have looked at the map linked in the 3rd post of this thread:

This map does not have “…border numbers corresponding to North American Datum 27 (NAD 27) or NAD 83.” In the lower left of this map is says:

“Produced by the USDA Forest Service.
North American Datum of 1983 (NAD 83).
Projection and 1,000-meter tics: Universal Transverse Mercator, zone 13

So you still add 3 zeros to the border numbers, but they are 1,000 meters apart, NOT feet.

The “the numbers go down to 0” at the equator, and at the center of the time “regular” time zone, the one established by map makers (not the time zone established by the US Congress, that mostly follows the state borders.)

Ok, I think I’m getting the hang of this.

I follow that the 0 point for the NAD 83 zones is the equator for the latitude.

But for longitude its more complicated. They set the midpoint of the zone at 500,000 meters which gets smaller as you go west and larger as you go east (until you go into the next zone).
An interesting feature of this is that for zone 13 (for example), in the BigHorn mountains of Wyoming, the longitude never collapses to zero.
Check out these two maps that are next to each other.
It goes down to around 263,000 until it hits the next adjacent zone which starts at about 737,000 (in zone 12)

and this adjacent map

Since the central line of zone 13 is set at 500,000 and the zone in this area is NOT 500,000 meters wide, it will never get down to zero.

However, I’m going to assume that very close to the equator where the zone is much wider, it will get closer to zero on the western edge. I’m also going to guess that even at its widest point along the equator, zone 13 is still not 500,000 meters wide, so there will never be a true 000,000 longitude. (Now I wonder how small those numbers get at the equator)
Anyone want to check my work and see if I understand correctly?

Look up the Transverse Mercator method, and yest it is more complicated to change the basis to Lat/Long.

from EPSG Guidance Note #7-2, reverse this.

For the calculation of easting and northing from latitude and longitude, first calculate constants for the projection:
n = f / (2-f)
B = [a/(1+n)] (1 + n^2/4 + n^4/64)
h1 = n/2 – (2/3)n^2 + (5/16)n^3 + (41/180)n^4
h2 = (13/48)n^2 – (3/5)n^3 + (557/1440)n^4
h3 = (61/240)n^3 – (103/140)n^4
h4 = (49561/161280)n^4

Then the meridional arc distance from equator to the projection origin (Mo) is computed from:
If LatO = 0 then Mo = 0
else if LatO ? 90°N ? ?/2 radians
   Mo = B (?/2)
else if LatO ? 90°S ? -?/2 radians
   Mo = B (-?/2)	
 Qo = asinh(tan LatO) – [e atanh(e sin LatO)]
 ?o = atan(sinh Qo)
 ?O0 = asin (sin ?o)
Note: The previous two steps are taken from the generic calculation flow given below for latitude Lat, but here for LatO may be simplified to ?O0 = ?o = atan(sinh Qo).
 ?O1 = h1 sin(2?Oo)
 ?O2 = h2 sin(4?Oo) 
 ?O3 = h3 sin(6?Oo) 
 ?O4 = h4 sin(8?Oo) 
 ?O = ?O0+ ?O1+ ?O2+ ?O3+ ?O4
 Mo = B ?O

Note: if the projection grid origin is very close to the pole (within 2" or 50m), the tangent function in the equation for Qo is unstable and may fail. Mo may instead be calculated as:
Mo = a[(1 – e^2/4 – 3e^4/64 – 5e^6/256 –....)LatO – (3e^2/8 + 3e^4/32 + 45e^6/1024+....)sin2LatO 
+ (15e^4/256 + 45e^6/1024 +.....)sin4LatO – (35e^6/3072 + ....)sin6LatO  + .....]
with LatO in radians.

Q = asinh(tan Lat) – [e atanh(e sin Lat)]
? = atan(sinh Q)
?0 = atanh [cos ? sin(Lon – LonO)]
?0 = asin (sin ?  cosh ?0)	
?1 = h1 sin(2?0) cosh(2?0)
?1 = h1 cos(2?0) sinh(2?0)
?2 = h2 sin(4?0) cosh(4?0)
?2 = h2 cos(4?0) sinh(4?0)
?3 = h3 sin(6?0) cosh(6?0)
?3 = h3 cos(6?0) sinh(6?0)
?4 = h4 sin(8?0) cosh(8?0)
?4 = h4 cos(8?0) sinh(8?0)
? = ?0 + ?1 + ?2 + ?3 + ?4
? = ?0 + ?1 + ?2 + ?3 + ?4
Easting, E =  FE + ko B ?
Northing, N =  FN + ko (B ? – Mo)

For the reverse formulas to convert Easting and Northing projected coordinates to latitude and longitude first calculate constants of the projection where n is as for the forward conversion, as are B and Mo:
h1' = n/2 – (2/3)n^2 + (37/96)n^3 – (1/360)n^4
h2' = (1/48)n^2 + (1/15)n^3 – (437/1440)n^4
h3' = (17/480)n^3 – (37/840)n^4
h4' = (4397/161280)n^4

?' = (E –  FE) / (B ko)				
?' = [(N – FN) + ko Mo] / (B ko)
?1' = h1' sin(2?') cosh(2?')
?1' = h1' cos(2?') sinh(2?')
?2' = h2' sin(4?') cosh(4?')
?2' = h2' cos(4?') sinh(4?')
?3' = h3' sin(6?') cosh(6?')
?3' = h3' cos(6?') sinh(6?')
?4' = h4' sin(8?') cosh(8?')
?4' = h4' cos(8?') sinh(8?')
?0' = ?' – (?1' + ?2' + ?3' + ?4')
?0' = ?' – (?1' + ?2' + ?3' + ?4')
?' = asin(sin ?0' / cosh ?0')
Q' = asinh(tan ?')
Q" = Q' + [e atanh(e tanh Q')] = Q' + [e atanh(e tanh Q")] which should be iterated until the change in Q" is insignificant. Then
Lat = atan(sinh Q")
Lon = LonO + asin(tanh(?0') / cos ?')

I thought the big idea was that NAD coordinates were pinned to North America, as opposed to, say, GPS coordinates:

You’re getting closer, but some big misconceptions here.

You meant to write “I follow that the 0 [in the Y direction] point for the UTM zones is the equator”.

When you are talking in UTM coordinates, you must say “X” and “Y” (or “easting” and “northing”), NOT “latitude” and “longitude.” Latitude and longitude refer to a NON-projected coordinate system – no zones, like UTM.

And…in ANY UTM zone, the X (easting) never “collapses to 0.” The 0,0 XY point is deliberately outside the zone, for ANY zone.

Okay…NOW, we can start to talking about NAD 83. That’s a datum (and/or the spheroid – model of the earth – corresponding to that datum). That’s about deciding WHERE every point is on the earth, relative to each other, as well as choosing the exact spot on Earth for your 0,0 origin (where the equator crosses the Greenwich meridian).

THAT is a few meters to a few hundred meters away (in X and Y directions) from the previous popular datum/model (NAD 27) – exactly how many meters depends on where you on Earth.

GPS units, for example, by default assume you want to use the NAD 83 datum. Modern maps showing a UTM grid will use that datum. Some maps from the 1990s show BOTH a UTM NAD 83 grid, AND a UTM NAD 27 grid – it makes it easy to see the typically-several-tens-of-meters difference between them.

I thought consumer GPS units typically defaulted to WGS 84?

Most modern GPS chipsets output NEMA messages. NMEA $GPGGA messages use the WGS84 the datum…except for Garmin, or at least you can’t assume Garmin is.

Technically that’s true, but for most practical purposes, NAD83 and WGS84 are essentially the same:

I see various links that say there’s a difference of somewhere between 1 and 2 meters; close enough for the consumer.

I get a huge laugh out of this statement though…

… and I don’t really know why. I guess I like the image of some poor cartographer having to figure for drift and hoping that he or she can just retire before the number becomes significant.

Ha! That’s actually really interesting — I had no idea.