1810 Delaware gubernatorial election

For related races, see 1810 United States gubernatorial elections.

1810 Delaware gubernatorial election

October 2, 1810

Nominee Joseph Haslet Daniel Rodney Party Democratic-Republican Federalist Popular vote 3,664 3,593 Percentage 50.49% 49.51%

Haslet:      60–70% Rodney:      50–60%      60–70%

Governor before election George Truitt Federalist Elected Governor Joseph Haslet Democratic-Republican

The 1810 Delaware gubernatorial election was held on October 2, 1810.

Incumbent Federalist Governor George Truitt was not eligible for re-election under the Delaware Constitution of 1792.

Democratic-Republican nominee Joseph Haslet defeated Federalist nominee Daniel Rodney with 50.49% of the vote.

General election

Results

1810 Delaware gubernatorial election^[1]^[2]^[3]^[4]

Party		Candidate	Votes	%	±%
	Democratic-Republican	Joseph Haslet	3,664	50.49%
	Federalist	Daniel Rodney	3,593	49.51%
Majority			71	0.98%
Turnout			7,257	100.00%
	Democratic-Republican gain from Federalist		Swing

References

1. "DE Governor, 1810". Our Campaigns. Retrieved December 8, 2020.
2. Congressional Quarterly 1998, p. 46.
3. Glashan 1979, pp. 50–51.
4. Dubin 2003, p. 26.

Bibliography

• Gubernatorial Elections, 1787-1997. Washington, D.C.: Congressional Quarterly Inc. 1998. ISBN 1-56802-396-0.
• Glashan, Roy R. (1979). American Governors and Gubernatorial Elections, 1775-1978. Meckler Books. ISBN 0-930466-17-9.
• Dubin, Michael J. (2003). United States Gubernatorial Elections, 1776-1860: The Official Results by State and County. Jefferson, North Carolina: McFarland. ISBN 978-0-7864-1439-0.