Site data updated: 2026-05-28 18:14 UTC
Byron Black#35 def. B. Ellwood#297
6-2 6-2
Match Information
| Date | Aug 17, 1998 |
|---|---|
| Tournament | Indianapolis |
| Category | ATP 250 |
| Surface | Hard |
| Round | R64 |
| Winner | Byron Black |
| Score | 6-2 6-2 |
| Loser | B. Ellwood |
| Duration | 1h 6m |
| Stat | Byron Black | B. Ellwood |
|---|---|---|
| Aces | 3 | 2 |
| Double faults | 5 | 0 |
| First serve % | 50 | 50 |
| First serve points won % | 79 | 58 |
| Second serve points won % | 55 | 46 |
| Break points saved % | — | 50 |
| Break points converted % | 50 | — |
| Service points won % | 67 | 52 |
| Return points won % | 48 | 33 |
| Total points won % | 58 | 42 |
Pre-match Comparison
| Field | Byron Black | B. Ellwood |
|---|---|---|
| Country | ZIM | AUS |
| Birth date | Oct 6, 1969 | Mar 12, 1976 |
| Age | 28 | 22 |
| Height | 175 | 178 |
| Handedness | Right-Handed | Right-Handed |
| Backhand style | Two-Handed Backhand | — |
| Ranking (entering) | 35 | 297 |
| Ranking points (entering) | 1,125 | 125 |
| Elo (entering) | 1,758 | 1,471 |
| Career W-L | 239–229 | 4–14 |
| Career earnings | $5,159,775 | $388,461 |
No prior-path match data is available for this tournament stage.