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Old Jul 11th, 2005, 02:04 PM   #29
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Jeca's points (1417.25):

BEST 17
Dubai (28.02.2005) - 242
Berlin (02.05.2005) - 191
Linz (25.10.2004) - 162
Filderstadt (04.10.2004) - 137.25
Birmingham (06.06.2005) - 107
Wimbledon (20.06.2005) - 92
Beijing (20.09.2004) - 63
New Haven (23.08.2004) - 63

Amelia Island (04.04.2005) - 50
Australian Open (17.01.2005) - 48
Sydney (10.01.2005) - 48
US Open (30.08.2004) - 40
Auckland (03.01.2005) - 36
Rome (09.05.2005) - 35
Philadelphia (01.11.2004) - 35
Montreal (02.08.2004) - 35

Paris (07.02.2005) - 33

non-rank
Los Angeles (19.07.2004) - 29
San Diego (26.07.2004) - 26
Roland Garros (23.05.2005) - 2
Eastbourne (13.06.2005) - 1
Charleston (11.04.2005) - 1
Miami (21.03.2005) - 1
Indian Wells (07.03.2005) - 1
Tokyo (31.01.2005) - 1
Guangzhou (27.09.2004) - 1
Bali (13.09.2004) - 1
Athens (16.08.2004) - 1
Stanford (12.07.2004) - 1


Jelena must defend 564.25 (2004 season) pts.



Ana's points (1390.75):

Roland Garros (23.05.2005) - 364
Warsaw (25.04.2005) - 174
Miami (21.03.2005) - 171
Zurich (18.10.2004) - 112.75
Canberra (10.01.2005) - 108.5
Australian Open (17.01.2005) - 94
Wimbledon (20.06.2005) - 84
Rome (09.05.2005) - 73
Luxembourg (25.10.2004) - 51
Doha (21.02.2005) - 33
Antwerp (14.02.2005) - 19.75
Dubai (28.02.2005) - 15.75
US Open (30.08.2004) - 2
Berlin (02.05.2005) - 1

Ana must defend 165.75
(2004 season) pts.




Conclusion:
Ana have "maximum tour per year" restriction, but her (points & rank) situation is better for 2005 championship tour race at this moment than Jelena, no doubt.
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and I'm not even fan of Ana Ivanovic... just an objective observer who realize how perfect she is

AMAZING BELGRADE MIRACLE: In just three years (2008-2011) 5 different players born in same city reach world #1 in rankings and several GS titles and finals (WTA singles: Ana Ivanovic and Jelena Jankovic, ATP singles: Novak Djokovic, ATP doubles: Nenad Zimonjic and Daniel Nestor... with notable performances of Bojana Jovanovski, Janko Tipsarevic and Viktor Troicki which is also all three born in Belgrade)
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