Sandusky Township, Crawford County, Ohio

From Wikipedia, the free encyclopedia
Jump to: navigation, search
Sandusky Township, Ohio
Township
Location of Sandusky Township in Crawford County.
Location of Sandusky Township in Crawford County.
Coordinates: 40°51′5″N 82°49′22″W / 40.85139°N 82.82278°W / 40.85139; -82.82278Coordinates: 40°51′5″N 82°49′22″W / 40.85139°N 82.82278°W / 40.85139; -82.82278
Country United States
State Ohio
County Crawford
Area
 • Total 18.0 sq mi (46.5 km2)
 • Land 17.9 sq mi (46.5 km2)
 • Water 0.0 sq mi (0.0 km2)
Elevation[1] 1,030 ft (314 m)
Population (2000)
 • Total 475
 • Density 26.5/sq mi (10.2/km2)
Time zone Eastern (EST) (UTC-5)
 • Summer (DST) EDT (UTC-4)
FIPS code 39-70366[2]
GNIS feature ID 1085944[1]

Sandusky Township is one of the sixteen townships of Crawford County, Ohio, United States. The 2000 census found 475 people in the township.[3]

Geography[edit]

Located in the eastern part of the county, it borders the following townships:

No municipalities are located in Sandusky Township.

Name and history[edit]

Statewide, other Sandusky Townships are located in Richland and Sandusky counties.

Government[edit]

The township is governed by a three-member board of trustees, who are elected in November of odd-numbered years to a four-year term beginning on the following January 1. Two are elected in the year after the presidential election and one is elected in the year before it. There is also an elected township fiscal officer,[4] who serves a four-year term beginning on April 1 of the year after the election, which is held in November of the year before the presidential election. Vacancies in the fiscal officership or on the board of trustees are filled by the remaining trustees.

References[edit]

External links[edit]