Site data updated: 2026-05-28 18:14 UTC
David Wheaton def. Taylor Dent
6-7(1) 7-5 6-1
Match Information
| Date | Aug 17, 1998 |
|---|---|
| Tournament | Indianapolis |
| Category | ATP 250 |
| Surface | Hard |
| Round | R64 |
| Winner | David Wheaton |
| Score | 6-7(1) 7-5 6-1 |
| Loser | Taylor Dent |
| Duration | 2h 41m |
| Stat | David Wheaton | Taylor Dent |
|---|---|---|
| Aces | 6 | 4 |
| Double faults | 9 | 10 |
| First serve % | 49 | 42 |
| First serve points won % | 72 | 65 |
| Second serve points won % | 54 | 47 |
| Break points saved % | 70 | 62 |
| Break points converted % | 38 | 30 |
| Service points won % | 63 | 55 |
| Return points won % | 45 | 37 |
| Total points won % | 54 | 46 |
Pre-match Comparison
No prior-path match data is available for this tournament stage.