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1948 edition of the Davis Cup From Wikipedia, the free encyclopedia
The 1948 Davis Cup was the 37th edition of the most important tournament between national teams in men's tennis. 25 teams entered the Europe Zone, and four teams entered the America Zone. Pakistan and Turkey made their first appearances in the competition.
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Duration | 22 April – 6 September 1948 |
Edition | 37th |
Teams | 29 |
Champion | |
Winning nation | ![]() |
← 1947 1949 → |
Australia defeated Mexico in the America Zone final, and Czechoslovakia defeated Sweden in the Europe Zone final. Australia defeated Czechoslovakia in the Inter-Zonal play-off, but fell to defending champions the United States in the Challenge Round. The championship was played at the West Side Tennis Club in Forest Hills, New York, United States on 4–6 September.[1][2][3]
Mexico vs. Australia
![]() Mexico 1 |
Centro Deportivo Chapultepec, Mexico City, Mexico[4] 6–8 August 1948 Clay |
![]() Australia 4 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Czechoslovakia vs. Sweden
![]() Czechoslovakia 4 |
Prague, Czechoslovakia[5] 27–29 July 1948 Clay |
![]() Sweden 1 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Australia vs. Czechoslovakia
![]() Australia 3 |
Longwood Cricket Club, Boston, MA, United States[6] 19–22 August 1948 Grass |
![]() Czechoslovakia 2 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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United States vs. Australia
![]() United States 5 |
West Side Tennis Club, Forest Hills, NY, United States[3] 4–6 September 1948 Grass |
![]() Australia 0 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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