Seabury Ford
(1801-1855)

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.

Age

how old was Seabury Ford when he died?

53

Seabury Ford died in 1855 at the age of 53.


Biographical data

Birthday October 15th, 1801 (Thursday)
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

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Also born in 1801

Seabury Ford

Birthday

When was Seabury Ford born?

Seabury Ford was born on October 15th, 1801.

Death

When did Seabury Ford die?

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.

Place of Birth

Where was Seabury Ford born?

Cheshire (Town in New Haven County, Connecticut, United States).

Death place

Where did Seabury Ford die?

Burton (Town in Ohio, USA).

Zodiac

What is the zodiac sign of Seabury Ford?

Seabury Ford was born in the zodiac sign Libra (The Scales).

Chinese Zodiac

Which Chinese zodiac sign does Seabury Ford have?

Seabury Ford was born in 1801 in the year of the Chinese zodiac Rooster.

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Died on May 8th, 1855:
Anniversaries of death

American politician (1801-1855)

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