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Q: Dividing the earths surface into equidistant cells ( Answered,   6 Comments )
Subject: Dividing the earths surface into equidistant cells
Category: Science > Math
Asked by: jim_p-ga
List Price: $50.00
Posted: 31 Aug 2005 12:49 PDT
Expires: 30 Sep 2005 12:49 PDT
Question ID: 562752
I'm looking for a formula to divide up the earths surface into
equidistant cells such that given a lat/long pair I can retrieve its
cell and all neighboring cells.

So, for example, if I have something like the following:

| 1,1 | 1,2 | 1,3 | 1,4 |
| 2,1 | 2,2 | 2,3 | 2,4 |
| 3,1 | 3,2 | 3,3 | 3,4 |
| 4,1 | 4,2 | 4,3 | 4,4 |
| 5,1 | 5,2 | 5,3 | 5,4 |

... and I specify a lat/long [that falls in (4,2)], I need the formula
to return (3,1), (3,2), (3,3), (4,1), (4,2), (4,3), (5,1), (5,2),

I need to be able to specify a distance (i.e. 10mi, 20km, ...), with all
cells being equal in end-to-end distance (either from top-left to
bottom-right, or from top-to-bottom/left-to-right), and accounting for
the curvature of the earth.

I'm looking for a formula, algorithm, or program to do this
(preferably a Java program).

Request for Question Clarification by maniac-ga on 31 Aug 2005 17:47 PDT
Hello Jim_p,

Can you explain if the solution can impose some constraints on the
region of the earth that it operates in (e.g., within N 60 degree
parallel and S 60 degree parallel)?

Is there a preferred method you wish to use to divide the world up
into "equidistant" cells?
(e.g., township & range)

By "equidistant", I assume you mean that the distance north / south
and east / west is uniform on the cells (with perhaps some slight

The general problem for dividing up cells is illustrated at
The specific illustration I refer to shows how the US aligned the
township and range relative to a know point. For example, in Kansas
the reference point was in this case the N 40 degree parallel and the
"6th principal meridian". There is other good material in this
reference, referring to the difficulty the surveyors had in performing
the initial survey.


Clarification of Question by jim_p-ga on 31 Aug 2005 18:53 PDT
Definately S 60 ... N 60, I suppose. What are the constraints?

You're correct on what I meant by equidistant.

I'm not sure I understand what you mean by preferred method though.
Range I think. Driving distance. I need to be able to define multiple
grids with variable distances (i.e. A grid with cells that are 10mi
wide, A grid with cells 25mi wide, a grid with cells 10km wide,
etc...) Is that what you were asking?


Request for Question Clarification by maniac-ga on 31 Aug 2005 20:02 PDT
Hello Jim,

To clarify why a constraint (e.g., 60 N / 60 S) is helpful, look at a
Mercator map illustrated at
In this map, the areas near the poles are distorted. If you chopped
this map into squares, the squares north of the equator represent an
area on earth with a wider area to the south than the north. The
difference is not much for the center region of the map but the
difference is quite significant near the poles (top and bottom of the

This type of method (chopping up a map projection) would be a straight
forward solution and there are plenty of sites with source code - for
has a form to convert between lat/long and meters (or an Excel
spreadsheet at a small fee for bulk conversions). This could be
adapted to determine the cells (and adjoining ones) in a straight
forward manner. Note that the square cells (on the map) are "almost"
square (on the earth), with the distortion relatively minor until you
get near the poles.

Would a solution like this be suitable?

Clarification of Question by jim_p-ga on 31 Aug 2005 20:34 PDT
Sorry, I misunderstood your earlier post. The constraint is that the
formula wouldn't work in areas N of 60 or S of 60. I suppose that
would be fine, if it keeps things simple.

What's important is that the cells are all equal in size and that I
can alter this size (in miles and kilometers) to increase or decrease
the granularity of the grid.



Clarification of Question by jim_p-ga on 31 Aug 2005 20:37 PDT
Out of curiousity, would N of 70 instead of 60 complicate things much?
Not that it's all that important, but I'd be nice to include a bit
more of Canada and Alaska.


Request for Question Clarification by maniac-ga on 01 Sep 2005 17:56 PDT
Hello Jim,

Hmm. I read the comments and I believe I understand the application
  (quickly looking up nearby locations from a lat/long with a specific radius)
and not necessarily dividing up the world into squares. The problem
with squares was described by myself (distortion, different sizes) and
the commenter myoarin (alignment of cells) - if they are not
necessary, that should simplify the solution quite a bit.

Based on what you stated (and some searching on the web), it may be
possible to solve that problem more directly using latitude and
longitude. Let me outline a solution:
 - each degree of latitude covers about 69 miles. So on the north /
south axis, you have a direct conversion between the difference of
latitude and distance.
 - each degree of longitude covers about 69 miles at the equator but
zero miles at the poles. The distance per longitude can be calcuated
based on the latitude as illustrated at
[scroll to "Part 2", (7)]

Using the 100 W longitude and 45 N latitude as an example, the 20 mile
x 20 mile area centered would be roughly:
 - 100 degrees + / - (10 miles / (cosine(45 degrees) x 69
miles/degree)) or 100W +/- 0.205
 - 45 degress + / - (10 miles / 69 miles/degree) or 45N +/- 0.145
[using decimal degrees, you can convert to degrees / minutes / seconds if needed]

Using these ranges of values, you could then search your data [I
assume in a database] for values within those two ranges to do the
range based lookup. Would this be satisfactory or do you really need
the squares (which won't quite be uniform in size if they line up or
won't line up well if they are of uniform size)?


Clarification of Question by jim_p-ga on 01 Sep 2005 21:30 PDT
No, actually, that's the problem, I can't use ranges. Imagine if you
weren't allowed to use database 'less than' or 'greater than'
operators. How would you solve this problem then?

One approach is to calculate the distance of every object in the
database, but that's just not feasible. Another approach is to
catagorize your data, which is essentially what I'm trying to do.

If I had 3 fixed distances (5mi, 15mi, 35mi), I could create a grid
with cells that are 5mi x 5mi, another grid that's 15mi by 15mi,
etc... Then, I could create columns in my database for 5MI_CELL,
15MI_CELL, and 35MI_CELL, and use the formula to add the cell labels
for each entry. Now if I want to find locations within a 15mi radius
of a lat/long, I just search for all entries where 15MI_CELL='(3,1)'
or 15MI_CELL='(3,2)' or 15MI_CELL='(3,3)'... or whatever the labeling
scheme is...

Make any sense? Is there a better way to do this?


Request for Question Clarification by maniac-ga on 02 Sep 2005 14:51 PDT
Hello Jim,

It boggles the mind a little bit that your database does not allow you
to extract values based on range selections. However, I can think of
at least three ways to work around that.

[1] Fetch more data / reduce as a second step. Encode the data with
the lat / long as whole numbers as well as fractions. Extract all the
values that match the whole numbered lat / long of the central point
(plus up to three adjoining lat / long values if you are within 10
miles of a whole lat / long position - for a 10 mile radius). For the
values fetched, do the distance algorithm (such as suggested by
myoarin) to throw out the values that are not within the radius.
[though I can understand if the fetching of "too many" values is not

[2] If you need to cover a large part of the world using cells, it
would probably be best to use a grid system that is standardized. For
example, the "Military Grid Reference System" (MGRS) described in a
tutorial at
allows you to specify a location on the earth to any resolution (e.g.,
1000 or 100 meters). You could encode the locations with the MGRS
location and source code for conversions between lat / long and MGRS
(and several others) is available at
It is pretty straight forward to compute the MGRS values of adjoining
areas - though if you used 1000 meter resolution and had a 15 mile
radius, there would over a thousand MGRS values to look up. Using
multiple ranges (e.g., 1000 and 10000 meter) for the encoding would be
one way to reduce that problem.

[3] Get a better database. If the effort needed to work around the
limitations of the database is more than the costs of switching to a
new database, it may be work it. I can certainly refer you to several
good ones [though if your hadware is limited - I can see this would be
a problem as well]

Please let me know if one of those solutions would be acceptable or if
not, additional constraints on the solution.


Clarification of Question by jim_p-ga on 06 Sep 2005 12:04 PDT
If you think of directories or search engines that typically build
indexes on keywords and are optimized for keyword lookups, then this
problem makes a little more sense.

I had thought about your suggestion, basically using an existing grid
system like UTM or MGRS [2], and then reducing as a second step [1],
but reducing as a second step is inefficient because I'd have to apply
distance calculations on every entry returned, and that's going to
slow things down significantly. So, the grid system being used would
have to be fairly granular to minimize the number of invalid entries
returned.  Then, I figured that if grid systems like UTM and MGRS
exisit, then it must be possible to encode a custom grid system. Hence
my asking here. I would prefer a grid system with cells that are the
size of the search area radius, but if that's not possible, then if
the grid system is fairly granular, I guess I could use something
close to my search area radius and reduce as a second step.

So, maybe MGRS would work. Could you elaborate on it some more. How do
I define my grid size? What is the max size for grid cells and what is
the min size that I could have for grid cells under MGRS?



Request for Question Clarification by maniac-ga on 06 Sep 2005 16:49 PDT
Hello Jim,

Well, MGRS - as explained briefly at
(and in some more detail in the tutorial already referenced) - can
represent locations in a variety of resolutions. As you add more
digits - you get better resolution like the following:
 -  2 digits - 10 km resolution
 -  4 digits - 1000 meter (1 km)
 -  6 digits - 100 meter
 -  8 digits - 10 meter
 - 10 digits - 1 meter resolution

So, you have a wide range of resolutions available for the fixed sized
cells. Of course, at higher resolutions, you have a LOT more possible
values to extract from the database within the selected range of a

You say the distance calculation is "expensive" - perhaps on a hand
held device (or another with limited computational resources). If that
is the case, there are work arounds like:
 - eliminate the square root calcuation from the distance calculation
  (you are "within the range" if dx*dx + dy*dy < range*range)
 - if you have several values on the same "row" or "column" (I use
those terms loosely), determine the maximum "within range" cell -
throw out all the ones above that point (and keep the ones below)
 - use fixed point arithmetic (if floating point is expensive) if you
can do double precision integer arithmetic
A combination of these methods could be used as well.

Let me know if you want to follow up on one of these more fully.

Request for Question Clarification by maniac-ga on 09 Sep 2005 14:23 PDT
Hello Jim,

Let me recap a previous clarification request [removed with the
answer] and then address it.
[reference to Google Local removed]

Note on the results page how there are fixed proximities (Search
within: 1mi, 5mi, 15mi, 45mi).

Suppose I wanted to implement something similar to this. Then my cell
sizes would need to be, approximately:

1  mi = ~1600 m
5  mi = ~8 km
15 mi = ~24 km
45 mi = ~70 km

Using MGRS, could I define my grid resolutions to these values? If so,
then this is probably closer to the answer I'm looking for.

I took a look at the source code at
(a large package doing a variety of coordinate conversions)

and after unpacking, looked at the MGRS source code in
it looks like once the UTM cell is identified (the first part of the
MGRS value), the "northing" and "easting" values are then manipulated
to compute the offset (modulo the resolution) to compute the remaining
digits. There are some hard coded values in the code (e.g., 100000.0
in Make_MGRS_String) but the code could be modified to use a different
scaling factor.

For example, if you used 5 mile squares (8046.72 meters each), you
could use that value to divide the easting / northing values with a
fixed precision (e.g., 2 digits) and compute a reference like this
  UTM cell (e.g., 18SWD)
  offset East (two digits)
  offset North (two digits)
Or something like
with references for that cell and adjacent ones laid out like this
  18SWD0116, 18SWD0216, 18SWD0316
  18SWD0115, 18SWD0115, 18SWD0315
  18SWD0114, 18SWD0214, 18SWD0314
[north to south is top to bottom, west to east is left to right]

Of course, if you are near the top / bottom and/or left / right of an
UTM cell, computing the adjacent cells would have to take that into
account (and pull cells from the adjoining UTM cell).


Clarification of Question by jim_p-ga on 12 Sep 2005 23:46 PDT
Ok, I think that's what I'm looking for. Why are the offsets only two digits?

Subject: Re: Dividing the earths surface into equidistant cells
Answered By: maniac-ga on 13 Sep 2005 18:43 PDT
Hello Jim,

OK. Let's summarize the solution as the answer and explain how I got
to this solution.

Q: How to cover the globe with "equal sized rectangles" (size is
selectable) and find the adjacent cells?

A: As has been commented before, the "uniform squares" are not quite
rectangles and if squares are placed on the earth as "stamps", they
would not line up right from row to row. The solution used by the US
military (and many others) is called the "Military Grid Reference
System" (MGRS) and illustrated at several facilities like
or search for others using phrases like
  convert lat long squares
  convert lat long grid
  grid squares military
  utm mgrs

Now the UTM cells themselves are not of uniform size (wider at the end
closest to the equator) but MGRS gets better resolution (to a meter)
by subdividing the UTM cells into uniform squares. So the tiling
problem is eliminated with a UTM cell and minimized between cells
because the row of cells always starts at a 15 degree boundary and
restarts at the next boundary.

There is source code for converting between lat /long and MGRS (or
other grid systems) at several locations. I mentioned
in part because it is freely available, has some level of support, and
appears to be VERY comprehensive. For example, as others have
commented, the earth is not quite round and GEOTRANS can be used to
take that into account if needed.

There are other source programs or tools available, a search using something like
  MGRS source code
  MGRS source code (desired language)
will provide leads to other sources if this one is not suitable.

You also wanted to choose one of several sizes (e.g., 1 mile, 5 miles,
10 miles) for the size of the cells. In MGRS, the sizes are a power of
10 (e.g, 1 meter, 10 meters, 100 meters) but by changing the code in a
few locations (e.g., changing the fixed value of 100000.0 in the code
or passing the number of meters in the cell size) the coordinate
transformation from lat / long to a "Jim's MGRS" can be done in a
straight forward manner (as described previously - dividing the
northing / easting offset by the cell size).

You asked - why two digits are enough - actually they are not quite
enough for a 5 mile grid. I had a slight mistake in my calculations.
At the equator, a degree covers a distance of about 69 miles. Using
the google calculator
  convert 15*69 miles to meters
I get
  15 * 69 miles = 1 665 671.04 meters
for the maximum distance in a UTM cell or perhaps more directly
I get
  (15 * 69) / 5 = 207
for the number of 5 mile squares across a UTM cell at the equator so
you need three digits at 5 mile resolution (and four digits for 1 mile
resolution). You could get by with fewer digits if you used a
different radix - for example 207 in hexidecimal would fit within two
digits. [a side note - 2 digits would be enough for 1035 cells (1 mile
squares) if you used radix 36 - 10 digits + 26 alpha]

A suggestion, I would stick with a fixed number of digits with
different resolutions to simplify the code unless you have serious
storage problems. If you have serious storage problems, I suggest a
bigger radix (as described above) or store a binary value to minimize
the storage needs.

With this answer, I covered the original question and I believe I
covered most (if not all) of the issues raised in the clarifications.

Please make a clarification request if some part of the answer is
unclear or if you want me to expand on some of the other concepts
(e.g., reducing computational time of the distance formula) I raised
in the question clarification requests.

Good luck with your work.

Subject: Re: Dividing the earths surface into equidistant cells
From: myoarin-ga on 01 Sep 2005 07:31 PDT
Hello Jim,
After all the clarification, I get the impression that you would like
to find a way to cover a globe with equal-sized squares or rectangles 
- much as though one were pasting stamps on it, each one of which
being a map of, say, 100x100 miles.
If this is what you want, starting with a row of squares at the
equator, obviously the next rows will need less squares, and many
fewer as they approach lat. 60 N & S.  That means that there cannot be
a regular lattice of squares; in each successive row, the vertical
lines won't line up.  That is to say, the formula to identify adjacent
squares will be tricky.

Regardless whether I am right or wrong with my impression, I have
difficulty understanding how this could be useful.
Perhaps you could explain what you want to do with the results if the
problem described can be solved.  Maybe there is another way to
achieve your ultimate goal.
Subject: Re: Dividing the earths surface into equidistant cells
From: jim_p-ga on 01 Sep 2005 10:49 PDT
If you're trying to paste stamps onto a globe then this might be true,
but if you're projecting the globe onto a plane then I don't think
this is an issue. If it were, then I would think that coordinate
systems like Universal Transverse Mercator (UTM) wouldn't be possible
(which, incidently is defined between 80 degrees south and 84 degrees
north). But I'm no mathematician or cartographer, so I don't know
what's possible and what's not, hence my posting the question here.

The ultimate goal is to quickly build bounding areas around lat/long
points for a fixed set of radii. It's an optimization for lookup
systems that inefficiently handle range queries.

Subject: Re: Dividing the earths surface into equidistant cells
From: myoarin-ga on 01 Sep 2005 12:54 PDT
Thanks, Jim, you can tell that I am even less of a mathematician or
cartographer than you are, but I hope your comment can help some one
who is to come up with an answer.
I will be very interested.
Subject: Re: Dividing the earths surface into equidistant cells
From: myoarin-ga on 02 Sep 2005 03:08 PDT
Hi Jim and Maniac,
I must be pretty dense, 'cause I still have trouble understanding the project.
But maybe another perhaps wrong example will help.

I get the impression that you would like something like a radar at the
starting point, with variable distance, its beam defining a radius of
5, 15, 35 miles and identifying the secondary points within the chosen
range.  I assume that starting point and secondary points are
identified by Lat. and Long. Then the distances and directions should
be calculated.

This should be possible.  This question demonstrates that the distance
can be calculated:

The direction must also be possible; that is what great circle
navigation is all about.  It should be possible to have a program that
calculated the distances and directions to a set of secondary points,
rejecting those beyond the range, and producing a file of those within
the range.
If you need a file of secondary points between ranges, >5, <10 miles,
this should also be possible.  (But not by me, ;)

HOping this helps, Myoarin
Subject: Re: Dividing the earths surface into equidistant cells
From: myoarin-ga on 02 Sep 2005 03:35 PDT
Here is a comment to the identical question linked above:

" Subject: Re: dublin spire
From: frehan-ga on 01 Sep 2005 14:32 PDT 	 

The answer is inaccurate as the distance calculator uses the
approximation that the Earth is a perfect sphere. In fact it is oblate
with a slightly larger diameter at the equator than pole-to-pole.
A more accurate answer can be obtained by using a distance calculator
that uses the WGS-84 global mapping reference and that answer differs
by some 8km.
For details of the geometry see"

The imperfection of the earth's shape won't make much difference for
the ranges mentioned, but maybe the WGS-84 reference system could be
Subject: Re: Dividing the earths surface into equidistant cells
From: maniac-ga on 08 Sep 2005 06:24 PDT
Hello Jim,

Let me clarify the situation briefly. At this point, you have an
"answer" from hedgie (and not from myself). If you find that answer
inadequate, I suggest you either:
 - request a clarification - in particular, emphasize how the answer
is inadequate (e.g,. your original question asked specifically for
"... a formula, algorithm, or program to do this (preferably a Java
 - if you are not satisifed by hedgie's response to the request for
clarification, you can ask hedgie or the editors to withdraw the
answer so another researcher can answer it more completely.

If you find the answer adequate to the original question but want more
information specifically on MGRS, you could post another question.

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