# Re: MI Regional Centroids

```At 10:06 AM 07/07/1999 -0400, you wrote:
>Hello all,
>
>I'm trying to determine the centroid of population of our 2-County
>region.  Our population data is broken down into Traffic Analysis Zones
>(TAZ) of which there are roughly 500 in the region.  Is there any way to
>plot this centroid of population?  My goal is to track the movement of
>this centroid over the years.

Yes. Do a weighted average of the long/lat coordinates. For instance, query
your set of 500 centroids and multiply centroidx(obj) * population in each
one, then divide by the total population. Do the same thing with the
latitude and you will come up with a single weighted population (x,y) that
will be what you want.

For instance, I have a table with the USA by county. When I perform the
query...

Select sum(Population_70), Sum(Population_95) from USA_by_County into Selection
Browse * From Selection

...I get...

203293816       262748737

_____________________________________________

Then run the query...

Select sum(CentroidX(obj)*Population_70),
sum(CentroidY(obj)*Population_70),sum(CentroidX(obj)*Population_95),
sum(CentroidY(obj)*Population_95) from USA_by_County into Selection
Browse * From Selection

... and get....

-18201208482.75 7820955814.87   -24043161734.83 9931834862.45

_____________________________________________

Divide the 1970 x/y's by the 1970 population total and get x = -89.5315403
y = 38.4711939, which is near Mascoutah, IL. Good way to check the data,
since I used to go to High School there, and the only claim to fame the
town had was it was the center of the USA's population!

For 1995 you get x = -91.5062885, y = 37.7997435, that is 116.5 miles WSW
of the 1970 centroid.

HTH,

Steve Wallace
GIS & Market Information Manager
Florida Farm Bureau Insurance Companies

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