Binghamton challenger 1999 (USA)
(50,000 $, hard, men)
Start | Round | S | 1 | 2 | 3 | 4 | 5 | H | A | ||
14.08. --:-- |
F | Sprengelm / Weir-Smit (2) | 2 | 5 | 6 | 6 | info | ||||
Kim / Lee | 1 | 7 | 4 | 2 | |||||||
12.08. --:-- |
SF | Kim / Lee | 2 | 6 | 6 | 6 | info | ||||
Grant / Peterson (1) | 1 | 7 | 3 | 4 | |||||||
12.08. --:-- |
SF | Sprengelm / Weir-Smit (2) | 2 | 7 | 7 | info | |||||
Sell / Van Emburgh | 0 | 6 | 6 | ||||||||
11.08. --:-- |
QF | Grant / Peterson (1) | 2 | 7 | 7 | info | |||||
Sela / Syed | 0 | 5 | 6 | ||||||||
11.08. --:-- |
QF | Kim / Lee | 2 | 7 | 7 | info | |||||
Manta / Niemeyer | 0 | 6 | 6 | ||||||||
11.08. --:-- |
QF | Sell / Van Emburgh | 2 | 6 | 6 | info | |||||
Massu / Whitehouse (4) | 0 | 3 | 3 | ||||||||
11.08. --:-- |
QF | Sprengelm / Weir-Smit (2) | 2 | 6 | 6 | info | |||||
Kokavec / Robichaud | 0 | 4 | 4 | ||||||||
10.08. --:-- |
R16 | Grant / Peterson (1) | 2 | 3 | 7 | 6 | info | ||||
Hanley / Healey | 1 | 6 | 6 | 4 | |||||||
10.08. --:-- |
R16 | Sela / Syed | 2 | 4 | 7 | 7 | info | ||||
Ivanov-Sm / Labadze | 1 | 6 | 5 | 6 | |||||||
10.08. --:-- |
R16 | Manta / Niemeyer | 2 | 6 | 6 | info | |||||
Behr / Spencer (3) | 0 | 4 | 4 | ||||||||
10.08. --:-- |
R16 | Kim / Lee | 2 | 6 | 6 | info | |||||
Blake / Russell | 0 | 4 | 4 | ||||||||
10.08. --:-- |
R16 | Sell / Van Emburgh | 2 | 6 | 6 | info | |||||
Motevassel / Pozzi | 0 | 3 | 3 | ||||||||
10.08. --:-- |
R16 | Massu / Whitehouse (4) | 2 | 6 | 6 | info | |||||
MacLagan / Nielsen | 0 | 3 | 4 | ||||||||
10.08. --:-- |
R16 | Kokavec / Robichaud | 2 | 7 | 6 | info | |||||
Hadad / Okun | 0 | 6 | 4 | ||||||||
10.08. --:-- |
R16 | Sprengelm / Weir-Smit (2) | 2 | 6 | 6 | info | |||||
De Chauna / Lisnard | 0 | 2 | 2 |
round of 16
Grant / Peterson [1]
Behr / Spencer [3]
quarterfinal
Grant / Peterson [1]
semifinal
Grant / Peterson [1]
final
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