Hubbard House (Crescent City, Florida)

From Wikipedia, the free encyclopedia
  (Redirected from Hubbard House (Florida))
Jump to: navigation, search
Hubbard House
Hubbard House (Crescent City, Florida) is located in Florida
Location: Crescent City, Florida  United States
Coordinates: 29°26′2″N 81°30′22″W / 29.43389°N 81.50611°W / 29.43389; -81.50611Coordinates: 29°26′2″N 81°30′22″W / 29.43389°N 81.50611°W / 29.43389; -81.50611
Built: 1877
Architect: Henry G. Hubbard
Architectural style: Frame Vernacular with Queen Anne and Shingle-style elements
NRHP Reference#: 73000601
Added to NRHP: August 14, 1973

The Hubbard House (also known as San Sui) is a historic site in Crescent City, Florida, United States. It is located at 600 North Park Street. On August 14, 1973, it was added to the U.S. National Register of Historic Places.

[edit] References

[edit] Gallery

Personal tools
Namespaces
Variants
Actions
Navigation
Interaction
Toolbox
Print/export