Steven Dutch, Professor Emeritus, Natural and Applied Sciences, University
of Wisconsin - Green Bay

Information on the UTM system

**New! Javascript-Convert Between Geographic and UTM
Coordinates**

Spreadsheet For UTM Conversion

Help! My Data Doesn't Look Like A UTM Grid!

I get enough inquiries on this subject that I decided to create a page for it.

Caution! Unlike latitude and longitude, there is no physical frame of reference for UTM grids. Latitude is determined by the earth's polar axis. Longitude is determined by the earth's rotation. If you can see the stars and have a sextant and a good clock set to Greenwich time, you can find your latitude and longitude. But there is no way to determine your UTM coordinates except by calculation.

UTM grids, on the other hand, are created by laying a square grid on the earth. This means that different maps will have different grids depending on the datum used (model of the shape of the earth). I saw US military maps of Germany shift their UTM grids by about 300 meters when a more modern datum was used for the maps. Also, old World War II era maps of Europe apparently used a single grid for all of Europe and grids in some areas are wildly tilted with respect to latitude and longitude.

Previously, the two basic references for converting UTM and geographic coordinates
were
U.S. Geological
Survey Professional Paper 1395 and U. S. Army
Technical Manual TM 5-241-8 (complete citations below). Each has advantages and
disadvantages. These have been entirely superseded by Charles F. F. Karney (2010)
*Transverse Mercator with an accuracy of a few nanometers*.
(arXiv:1002.1417)

**(dd + mm/60 +ss/3600) to Decimal degrees (dd.ff)**

dd = whole degrees, mm = minutes, ss = seconds

dd.ff = dd + mm/60 + ss/3600

Example: 30 degrees 15 minutes 22 seconds = 30 + 15/60 + 22/3600 = 30.2561

**Decimal degrees (dd.ff) to (dd + mm/60 +ss/3600)**

For the reverse conversion, we want to convert dd.ff to dd mm ss. Here ff = the fractional part of a decimal degree.

mm = 60*ff

ss = 60*(fractional part of mm)

Use only the whole number part of mm in the final result.

30.2561 degrees = 30 degrees

.2561*60 = 15.366 minutes

.366 minutes = 22 seconds, so the final result is 30 degrees 15 minutes 22 seconds

**Decimal degrees (dd.ff) to Radians**

Radians = (dd.ff)*pi/180

**Radians to Decimal degrees (dd.ff)**

(dd.ff) = Radians*180/pi

**Degrees, Minutes and Seconds to Distance**

A degree of longitude at the equator is 111.2 kilometers. A minute is 1853
meters. A second is 30.9 meters. For other latitudes multiply by cos(lat).
Distances for degrees, minutes and seconds** **in latitude are very similar
and differ very slightly with latitude. (Before satellites, observing those
differences was a principal method for determining the exact shape of the
earth.)

Okay, take a deep breath. This will get *very* complicated, but the
math, although tedious, is only algebra and trigonometry. It would sure be
nice if someone wrote a spreadsheet to do this. Or
better yet, a Javascript page.

P = point under consideration F = foot of perpendicular from P to the central meridian. The latitude of F is called the footprint latitude.O = origin (on equator) OZ = central meridian LP = parallel of latitude of P ZP = meridian of P OL = k _{0}S = meridional arc from equatorLF = ordinate of curvature OF = N = grid northing FP = E = grid distance from central meridian GN = grid north C = convergence of meridians = angle between true and grid north |

Another thing you need to know is the datum being used:

Datum |
Equatorial Radius, meters (a) |
Polar Radius, meters (b) |
Flattening (a-b)/a |
Use |

NAD83/WGS84 | 6,378,137 | 6,356,752.3142 | 1/298.257223563 | Global |

GRS 80 | 6,378,137 | 6,356,752.3141 | 1/298.257222101 | US |

WGS72 | 6,378,135 | 6,356,750.5 | 1/298.26 | NASA, DOD |

Australian 1965 | 6,378,160 | 6,356,774.7 | 1/298.25 | Australia |

Krasovsky 1940 | 6,378,245 | 6,356,863.0 | 1/298.3 | Soviet Union |

International (1924) -Hayford (1909) | 6,378,388 | 6,356,911.9 | 1/297 | Global except as listed |

Clake 1880 | 6,378,249.1 | 6,356,514.9 | 1/293.46 | France, Africa |

Clarke 1866 | 6,378,206.4 | 6,356,583.8 | 1/294.98 | North America |

Airy 1830 | 6,377,563.4 | 6,356,256.9 | 1/299.32 | Great Britain |

Bessel 1841 | 6,377,397.2 | 6,356,079.0 | 1/299.15 | Central Europe, Chile, Indonesia |

Everest 1830 | 6,377,276.3 | 6,356,075.4 | 1/300.80 | South Asia |

Don't interpret the chart to mean there is radical disagreement about the shape of the earth. The earth isn't perfectly round, it isn't even a perfect ellipsoid, and slightly different shapes work better for some regions than for the earth as a whole. The top three are based on worldwide data and are truly global. Also, you are very unlikely to find UTM grids based on any of the earlier projections.

The most modern datums (jars my Latinist sensibilities since the plural of *datum*
in Latin is *data*, but that has a different meaning to us) are NAD83 and
WGS84. These are based in turn on GRS80. Differences between the three systems
derive mostly from redetermination of station locations rather than differences
in the datum. Unless you are locating a first-order station to sub-millimeter
accuracy (in which case you are way beyond the scope of this page) you can
probably regard them as essentially identical.

NIMA Technical Report 8350.2 states:

The WGS 84 Ellipsoid is for all practical purposes identical to the GRS 80 ellipsoid. They use the same value for the semi-major axis and have the same orientation with respect to the center of mass and the coordinate system origin. .... The difference between the GRS 80 and WGS 84 values for f creates a difference of 0.1 mm in the derived semi-minor [polar] axes of the two ellipsoids.

Based on these definitions, geodetic positions determined with respect to NAD 83 or WGS 84 have uncertainties of about one meter in each component. For mapping, charting and navigation, the two systems are indistinguishable at scales of 1:5,000 or smaller and with accuracies of about 2 m.

Earlier versions of this page were written based on Army and USGS references. It has been updated according to the formulas of Karney (2010), which are accurate to within nanometers. The Army and USGS references are now obsolete. Refer to the Appendix for the Army and USGS formulas.

First step, project the ellipsoid onto a sphere and compute the *
conformal latitude *(lat'). In the formulas below, sinh and tanh are *
hyperbolic functions*, which behave much like regular trigonometric
functions. They are pretty docile and do not bite if not provoked.
"Argsinh," etc. are analogous to arcsin, etc and are *inverse hyperbolic
functions.*
The hyperbolic and inverse hyperbolic functions are standard in pretty much
all programming languages. The quantity e = SQRT(f(2-f), where f is the
flattening of the ellipsoid.

argsinh tan lat′ = argsinh tan lat − e argtanh(e sin lat).

We can solve for tan lat': tan lat' = sinh(argsinh tan lat − e argtanh(e sin lat))

It's convenient to work with tangents, so define tau = tan lat and tau' = tan lat'

Define sigma = sinh(e argtanh(e tau/SQRT(1 + tau^2)))

tau' = tau SQRT(1 + sigma^2) - sigma SQRT(1 + tau^2)

Now we can define xi' = arctan(tau'/cos long) and eta' = argsinh(sin (long/SQRT(tau'^2 + (cos long)^2)))

Somewhat confusingly, xi refers to the north south direction

The next step requires some series computations. We need to define:

A = a/( 1 + n ) (1 + (¼) n^2 + (1/64)n^4 + (1/256)n^6 + (25/16384)n^8 + (49/65536)n^10 + ...), defines the scale of the ellipse. 2piA= circumference of meridian.

α1 = (1/2)n − (2/3)n^2 + (5/16)n^3 +
(41/180)n^4 - (127/288)n^5 + (7891/37800)n^6 + (72161/387072)n^7 -
(18975107/50803200)n^8 + (60193001/290304000)n^9 +
(134592031/1026432000)n^10 + ...

α2 = (13/48)n^2 - (3/5)n^3 +
(557/1440)n^4 + (281/630)n^5 - (1983433/1935360)n^6 +
(13769/28800)n^7 + (148003883/174182400)n^8 -
(705286231/465696000)n^9 + (1703267974087/3218890752000)n^10 + ...

α3 = (61/240)n^3 - (103/140)n^4 +
(15061/26880)n^5 + (167603/181440)n^6 - (67102379/29030400)n^7 +
(79682431/79833600)n^8 + (6304945039/2128896000)n^9 -
(6601904925257/1307674368000)n^10 + ...

α4 = (49561/161280)n^4 - (179/168)n^5 +
(6601661/7257600)n^6 + (97445/49896)n^7 - (40176129013/7664025600)n^8
+ (138471097/66528000)n^9 + (48087451385201/5230697472000)n^10 + ...

α5 = (34729/80640)n^5 -
(3418889/1995840)n^6 + (14644087/9123840)n^7 +
(2605413599/622702080)n^8 - (31015475399/2583060480)n^9 +
(5820486440369/1307674368000)n^10 + ...

α6 = (212378941/319334400)n^6 -
(30705481/10378368)n^7 + (175214326799/58118860800)n^8 +
(870492877/96096000)n^9 - (1328004581729009/47823519744000)n^10 + ...

α7 = (1522256789/1383782400)n^7 -
(16759934899/3113510400)n^8 + (1315149374443/221405184000)n^9 +
(71809987837451/3629463552000)n^10 + ...

α8 = (1424729850961/743921418240)n^8 -
(256783708069/25204608000)n^9 + (2468749292989891/203249958912000)n^10
+ ...

α9 = (21091646195357/6080126976000)n^9 -
(67196182138355857/3379030566912000)n^10 + ...

α10 =
(77911515623232821/12014330904576000)n^10 + ...(24)/a)n^d/

Now we calculate:

xi = xi' + SUM(αj sin 2j xi' cosh 2j eta') and eta = eta' + SUM(αj cos 2j xi' sinh 2j eta'), or:

xi = xi' + α1 sin 2xi' cosh 2eta' + α2 sin 4xi' cosh 4eta' + α3 sin 6xi' cosh 6eta' + α4 sin8xi' cosh 8eta' etc.

xi = xi' + α1 sin 2xi' cosh 2eta' + α2 sin 4xi' cosh 4eta' + α3 sin 6xi' cosh 6eta' + α4 sin8xi' cosh 8eta' etc.

finally scale xi and eta to obtain easting = k0Aeta and northing = k0Axi.

Since these formulas are derived from elliptic functions, they are both more rigorous and easier to troubleshoot than the Army and USGS references.

Earlier versions of this page were written based on Army and USGS references. It has been updated according to the formulas of Karney (2010), which are accurate to within nanometers. The Army and USGS references are obsolete. Refer to the Appendix for the Army and USGS formulas.

The process involves inverting the procedure above.

xi' = xi + SUM(βj sin 2j xi cosh 2j eta) and eta' = eta + SUM(βj cos 2j xi sinh 2j eta).

Since the sums involve xi and eta instead of x' and eta', the coefficients are not the same:

β1 = (1/2)n - (2/3)n^2 + (37/96)n^3 -
(1/360)n^4 - (81/512)n^5 + (96199/604800)n^6 - (5406467/38707200)n^7
+ (7944359/67737600)n^8 - (7378753979/97542144000)n^9 +
(25123531261/804722688000)n^10 + ...

β2 = (1/48)n^2 + (1/15)n^3 -
(437/1440)n^4 + (46/105)n^5 - (1118711/3870720)n^6 +
(51841/1209600)n^7 + (24749483/348364800)n^8 -
(115295683/1397088000)n^9 + (5487737251099/51502252032000)n^10 + ...

β3 = (17/480)n^3 - (37/840)n^4 -
(209/4480)n^5 + (5569/90720)n^6 + (9261899/58060800)n^7 -
(6457463/17740800)n^8 + (2473691167/9289728000)n^9 -
(852549456029/20922789888000)n^10 + ...

β4 = (4397/161280)n^4 - (11/504)n^5 -
(830251/7257600)n^6 + (466511/2494800)n^7 + (324154477/7664025600)n^8
- (937932223/3891888000)n^9 - (89112264211/5230697472000)n^10 + ...

β5 = (4583/161280)n^5 -
(108847/3991680)n^6 - (8005831/63866880)n^7 + (22894433/124540416)n^8
+ (112731569449/557941063680)n^9 - (5391039814733/10461394944000)n^10
+ ...

β6 = (20648693/638668800)n^6 -
(16363163/518918400)n^7 - (2204645983/12915302400)n^8 +
(4543317553/18162144000)n^9 + (54894890298749/167382319104000)n^10 +
...

β7 = (219941297/5535129600)n^7 -
(497323811/12454041600)n^8 - (79431132943/332107776000)n^9 +
(4346429528407/12703122432000)n^10 + ...

β8 = (191773887257/3719607091200)n^8 -
(17822319343/336825216000)n^9 - (497155444501631/1422749712384000)n^10
+ ...

β9 = (11025641854267/158083301376000)n^9
- (492293158444691/6758061133824000)n^10 + ...

β10 =
(7028504530429621/72085985427456000)n^10 + ...

We can then obtain tau' = sin xi'/SQRT(sinh(eta')^2 + cos(xi')^2) and long = arctan(sinh eta'/cos xi')

We can invert the forward formulas to obtain: tau SQRT(1 + sigma^2) - sigma SQRT(1 + tau^2) - tau' = 0.

There are two problems: tau is the unknown and sigma is a function of tau. So we have to resort to approximation methods. Newton's Method works nicely.

f = tau SQRT(1 + sigma^2) - sigma SQRT(1 + tau^2) - tau' = 0.

Start with tau' as the initial guess for tau. Compute sigma = sinh(e argtanh(e tau/SQRT(1 + tau^2)))

Now we need the defivative of f.

df/dtau =(SQRT((1 + sigma^2)(1 +
tau^2)) - sigma tau)(1 - e^2)SQRT(1+tau^2)/(1 + (1 - e^2)tau^2)

Estimate f using tau' as the initial guess, calculate sigma and df/dtau.
Estimate a new value for tau

tau = tau - (df/dtau)/tau, compute a new
value for f and keep going until f converges to zero and tau stops changing.
Three iterations were sufficient for me.

Just to check your formulas, Karney generated a huge number of test values. Here are the first ten

Latitude | Longitude | Easting | Northing |

70.57927709 | 45.59941973 | 1548706.792 | 8451449.199 |

10.01889371 | 23.31332382 | 2624150.741 | 1204434.042 |

19.47989559 | 75.66204923 | 9855841.233 | 6145496.115 |

21.07246482 | 29.82868439 | 3206390.692 | 2650745.4 |

5.458957393 | 36.38523737 | 4328154.084 | 749647.6237 |

70.1754537 | 22.86535023 | 847598.2665 | 7947180.962 |

61.96560497 | 58.93137085 | 2727657.338 | 8283916.696 |

11.11604988 | 20.90106919 | 2331001.752 | 1313608.225 |

32.21054315 | 60.70584911 | 6035557.239 | 5791770.792 |

79.1874509 | 61.53238249 | 1064553.126 | 9417273.737 |

Note that all longitudes and eastings are with respect to a central meridian
of zero degrees and zero easting. *Yes, these formulas are accurate to more
than 75 degrees from the central meridian*. However, beyond 80 degrees, and
at low latitudes, they can blow up in Excel.

**In response to innumerable e-mails, you cannot use UTM grid
coordinates without knowing your zone. There are sixty points on the earth's
surface that have the same numerical UTM coordinates, 120 if you consider that
northing is duplicated in both hemispheres.**

This program was recently updated using the formulas of Charles Karney
(2010). His paper claims accuracy of a few nanometers. Excel has 15-digit
accuracy, so the spreadsheet doesn't achieve that level, but is accurate to
about 0.01mm. Most importantly, Karney's paper provides links to a site that has
a huge set of test data. The spreadsheet has been tested against the data in
both directions. Before linking to the program, note (**especially the last item**):

- It is an Excel spreadsheet. If you have Excel on your computer, it will (or should) open when you click the link. Most major spreadsheet programs can read spreadsheets in other formats.
- A spreadsheet is not an applet or program. In particular, you can't manually enter data into a cell and preserve any formulas that are there. That's why some aspects of data entry are clunkier than they might otherwise be with, say, a Visual Basic program.
- There are three computation pages, one for single conversions, the other two for batch conversions. Other pages contain information on datums, test data and the specific conversion formulas. To use the batch conversions you need to be somewhat proficient in spreadsheets as you will have to copy data and cell formulas.
- For our mutual peace of mind, run a virus check. The spreadsheet has a couple of simple macros.
- You may copy the program for your own non-commercial use and for non-commercial distribution to others, but not for commercial use. Please give appropriate credit when citing your calculations. You may also modify it as needed for your personal use. In Internet Explorer, right click on the link and select Save Target As... to save the spreadsheet to your own disk.
- There is now a Javascript Page which I hope is free of bugs. Since it's a program, it should handle complex what-ifs a lot better than a spreadsheet can. All the permissions for the spreadsheet apply here, too. Persistent bugs caused me to delete links to the older spreadsheet.
- These pages are somewhat OBE (overtaken by events). http://www.earthpoint.us/Convert.aspx can convert between a variety of coordinate systems and has many links to other useful pages. The National Geodetic Survey has a site at http://www.ngs.noaa.gov/TOOLS/spc.shtml with that converts latitude and longitude to State Plane Coordinates.

JavaScript Page to Convert Between Geographic and UTM Coordinates

**Yay! Most of these are on-line now.**

Snyder, J. P., 1987; Map Projections - A Working Manual. U.S. Geological
Survey Professional Paper 1395, 383 p. **If you are at all serious about maps
you need this book. **On-line at
http://pubs.er.usgs.gov/publication/pp1395

Army, Department of, 1973; Universal Transverse Mercator Grid, U. S. Army
Technical Manual TM 5-241-8, 64 p. Superseded by DMATM 8358.2
The Universal Grids: Universal Transverse Mercator (UTM) and Universal Polar
Stereographic (UPS).

On-line at
http://earth-info.nga.mil/GandG/publications/tm8358.2/TM8358_2.pdf

NIMA Technical Report 8350.2, "Department of Defense World Geodetic System 1984, Its Definition and Relationships with Local Geodetic Systems," Second Edition, 1 September 1991 and its supplements. The report is available from the NIMA Combat Support Center and its stock number is DMATR83502WGS84. Non-DoD requesters may obtain the report as a public sale item from the U.S. Geological Survey, Box 25286, Denver Federal Center, Denver, Colorado 80225 or by phone at 1-800-USA-MAPS. On-line at http://earth-info.nga.mil/GandG/publications/tr8350.2/wgs84fin.pdf

Karney, Charles F.F., 2010, Transverse Mercator with an accuracy of a few nanometers, http://arxiv.org/abs/1002.1417v3. Provides accuracy to within nanometers (a few atoms). Supersedes Snyder and Army, above. His test data and numerous other programs are at http://geographiclib.sourceforge.net/.

Return to Professor Dutch's Home Page

*Created 12 September 2003, Last Update
14 September 2018
*