viernes, 6 de febrero de 2015

SERENA AND VENUS WILLIAMS IN ARGENTINA. EEUU VS ARGENTINA


Argentina’s hopes of reaching the Fed Cup by BNP Paribas World Group play-offs for the second year running – and just the second time since 2009 – appear to have been dashed before they have even begun. 

 Or so a glance at this weekend's draw might have them believe. Friday’s draw for the clash between Maria-Jose Gaidano’s Argentina and USA in Buenos Aires pitted world No. 121 Paula Ormaechea and No. 197 Maria Irigoyen against the mighty Venus and Serena Williams, respectively. 
The American sisters, who now have 56 different Grand Slam singles, doubles and mixed titles between them after Serena’s recent Australian Open triumph, present a formidable challenge for the Argentines. In the host nation’s favour, at least, is the surface. Ormaechea certainly gave Venus a run for her money in the pair’s only previous meeting – which the American ultimately won in three sets on the clay of Roland Garros in 2012. 
Irigoyen and Serena have never played before, but the Argentine will hope the 196 places between them in the rankings will be negated by the world No. 1s relative unfamiliarity with clay, having not contested a match on the surface since losing to Garbine Muguruza in the second round at last year’s Roland Garros. 
Serena and Venus are competing in the same Fed Cup team for the first time since they defeated Sweden in the 2013 World Group play-offs. They'll hope their influence can pull them out of the doldrums of World Group II. USA is featuring in the second tier of the competition for the first time since 2012, and only the second time since the formation of World Group II after they were stunned from 2-1 up by France in last year’s play-offs. More to follow... 

R1: Paula Ormaechea (ARG) v Venus Williams (USA) 
R2: Maria Irigoyen (ARG) v Serena Williams (USA) 
R3: Paula Ormaechea (ARG) v Serena Williams (USA) 
R4: Maria Irigoyen (ARG) v Venus Williams (USA) 
R5: Tatiana Bua/Nadia Podoroska (ARG) v Taylor Townsend/Coco Vandeweghe (USA) 

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