1978 Wimbledon Championships – Women's doubles explained

Score:4–6, 9–8(12–10), 6–3
Draw:48 (4)
Seeds:8
Type:grand slam

See main article: 1978 Wimbledon Championships. Helen Cawley and JoAnne Russell were the defending champions, but lost in the quarterfinals to Françoise Dürr and Virginia Wade.

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]

Seeds

See also: 1 and 1. Billie Jean King / Martina Navratilova (quarterfinals)

See also: 2 and 4. Evonne Cawley / Betty Stöve (third round, withdrew)

See also: 3 and 3. Françoise Dürr / Virginia Wade (semifinals)

See also: 4 and 2. Kerry Reid / Wendy Turnbull (champions)

See also: 5 and 3. Helen Cawley / JoAnne Russell (quarterfinals)

See also: 6 and 1. Sue Barker / Mona Guerrant (semifinals)

See also: 7 and 4. Mima Jaušovec / Virginia Ruzici (final)

See also: 8 and 2. Ilana Kloss / Marise Kruger (quarterfinals)

Draw

Top half

Section 2

Bottom half

Section 4

External links

Notes and References

  1. Book: Barrett, John. Wimbledon: The Official History. 2014. Vision Sports Publishing. 9-781909-534230. 4th.