American politician (1801-1855) – Seabury Ford was born in Cheshire (Town in New Haven County, Connecticut, United States) on October 15th, 1801 and died in Burton (Town in Ohio, USA) on May 8th, 1855 at the age of 53. Today Seabury Ford would be 222 years old.
Seabury Ford died in 1855 at the age of 53.
Birthday | October 15th, 1801 (Thursday) |
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Place of Birth | Cheshire Town in New Haven County, Connecticut, United States |
Death Date | May 8th, 1855 (Tuesday) |
Death place | Burton Town in Ohio, USA |
Birth sign (Zodiac) | Libra (The Scales) ♎ |
Chinese Zodiac | Rooster 雞 |
Seabury Ford was born on October 15th, 1801.
Seabury Ford died on May 8th, 1855 at the age of 53 in Burton (Town in Ohio, USA). Today Seabury Ford would be 222 years old.
Cheshire (Town in New Haven County, Connecticut, United States).
Burton (Town in Ohio, USA).
Seabury Ford was born in the zodiac sign Libra (The Scales).
Seabury Ford was born in 1801 in the year of the Chinese zodiac Rooster.