Seymour
City in Jackson County, Indiana
Seymour is a city in Jackson County, Indiana, with 21,569 residents (2020 Census). It sits at 604 ft (184 m) above sea level. The nearest town is Rockford, 2.7 miles N.
- 21,569residents (2020 Census)
- 604 ftelevation · 184 m
- 13.3 sq miland area · 34.4 km²
- 1,624 / sq mipopulation density
- 47274ZIP code
About Seymour
Seymour is a city in Jackson County, Indiana, United States. The population was 21,569 at the 2020 census, making it the largest city in Jackson County. It was platted in 1852 by landowner Meedy Shields and named after a railroad engineer.
Seymour developed as a railroad town after the north–south Jeffersonville Railroad and the east–west Ohio and Mississippi Railroad intersected in the early 1850s, leading to the city to be nicknamed the "Crossroads of Southern Indiana." In October 1866, the local Reno Gang carried out what is considered to be the first peacetime train robbery in the United States—an episode that brought vigilante reprisals and lasting notoriety. During World War II the U.S. Army Air Forces operated Freeman Army Airfield southwest of the city, a training base later known for the 1945 Freeman Field mutiny, a milestone in the desegregation of the U.S. military.
The modern economy is anchored by manufacturing and distribution led by the automotive-parts maker Aisin, which opened a Seymour plant in the 1980s. Beginning in the late 1980s, the city drew substantial immigration from San Sebastián Coatán, Guatemala. Its Hispanic population grew rapidly in the early 21st century, reshaping the community and drawing state scrutiny of the city's policing policies in 2024. Seymour is the hometown of rock musician John Mellencamp and holds an annual Oktoberfest reflecting its 19th-century German heritage.
Text: Wikipedia · CC BY-SA 4.0
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Living in Seymour
- Median household income
- $64,426
- Median home value
- $162,800
- Median rent
- $1,011/mo
- Median age
- 33.3
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Map of Seymour
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Open in Google Maps ›Seymour, IN population
Seymour had 21,569 residents in the 2020 Census.
| 2020 Census | 21,569 |
|---|---|
| Density | 1,624 per sq mi (627 per km²) |
Source: U.S. Census Bureau, Decennial Census.
Seymour at a glance
| State | Indiana (IN) |
|---|---|
| County | Jackson County |
| Type | City |
| Population | 21,569 (2020) |
| Land area | 13.28 sq mi (34.4 km²) |
| Water area | 0.00 sq mi (0.01 km²) |
| Elevation | 604 ft (184 m) |
| Coordinates | 38.9463° N, 85.8885° W |
| ZIP code | 47274 |
| Official website | seymourcity.com |
| Census code (GEOID) | 1868832 |
Sources: U.S. Census Bureau (Gazetteer, ZCTA), USGS GNIS, Wikidata.
Towns near Seymour
- Rockford2.7 mi N
- Cortland4.4 mi NW
- Reddington6.9 mi NE
- Jonesville7.8 mi N
- Hayden8.2 mi E
- Brownstown9.7 mi SW
- Azalia10.4 mi N
- Crothersville10.8 mi S
- Waynesville11.5 mi N
- Country Squire Lakes12.6 mi NE
- Scipio12.8 mi NE
- Mineral Springs12.9 mi N
Straight-line distance between town centers.
Frequently asked questions about Seymour
What county is Seymour in?
Seymour is in Jackson County, Indiana.
What is the population of Seymour, IN?
21,569 people lived in Seymour at the 2020 Census.
What is the ZIP code of Seymour, IN?
The ZIP code of Seymour is 47274.
What is the elevation of Seymour?
604 feet (184 m) above sea level.
What towns are near Seymour?
Rockford (2.7 mi), Cortland (4.4 mi), Reddington (6.9 mi), Jonesville (7.8 mi), Hayden (8.2 mi).
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