Rube Foster
(1879-1930)

American Hall of Fame baseball player (1879-1930) – Rube Foster was born in Calvert (City in Robertson County, Texas, United States) on September 17th, 1879 and died in Kankakee (City in Kankakee County, Illinois, United States) on December 9th, 1930 at the age of 51. Today Rube Foster would be 144 years old.

Age

how old was Rube Foster when he died?

51

Rube Foster died in 1930 at the age of 51.


Biographical data

Birthday September 17th, 1879 (Wednesday)
Place of Birth Calvert
City in Robertson County, Texas, United States
Death Date December 9th, 1930 (Tuesday)
Death place Kankakee
City in Kankakee County, Illinois, United States
Birth sign (Zodiac) Virgo (The Maiden) ♍
Chinese Zodiac Rabbit 兔

Other personalities born on September 17

Also born in 1879

Rube Foster

Birthday

When was Rube Foster born?

Rube Foster was born on September 17th, 1879.

Death

When did Rube Foster die?

Rube Foster died on December 9th, 1930 at the age of 51 in Kankakee (City in Kankakee County, Illinois, United States). Today Rube Foster would be 144 years old.

Place of Birth

Where was Rube Foster born?

Calvert (City in Robertson County, Texas, United States).

Death place

Where did Rube Foster die?

Kankakee (City in Kankakee County, Illinois, United States).

Zodiac

What is the zodiac sign of Rube Foster?

Rube Foster was born in the zodiac sign Virgo (The Maiden).

Chinese Zodiac

Which Chinese zodiac sign does Rube Foster have?

Rube Foster was born in 1879 in the year of the Chinese zodiac Rabbit.

Rube Foster: Future birthdays

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Died on December 9th, 1930:
Anniversaries of death

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American Hall of Fame baseball player (1879-1930)