1978 Wimbledon Championships – Women's doubles
1978 tennis event results / From Wikipedia, the free encyclopedia
Helen Cawley and JoAnne Russell were the defending champions, but lost in the quarterfinals to Françoise Dürr and Virginia Wade.
Quick Facts Women's doubles, Final ...
Women's doubles | |||||||||||||
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1978 Wimbledon Championships | |||||||||||||
Final | |||||||||||||
Champions | ![]() ![]() | ||||||||||||
Runners-up | ![]() ![]() | ||||||||||||
Score | 4–6, 9–8(12–10), 6–3 | ||||||||||||
Details | |||||||||||||
Draw | 48 (4 Q ) | ||||||||||||
Seeds | 8 | ||||||||||||
Events | |||||||||||||
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Main article: 1978 Wimbledon Championships
Kerry Reid and Wendy Turnbull defeated Mima Jaušovec and Virginia Ruzici in the final, 4–6, 9–8(12–10), 6–3 to win the ladies' doubles tennis title at the 1978 Wimbledon Championships.[1]