Wayne Black vs Tim Henman — Indian Wells Masters Round of 64
Indian Wells Masters · ATP · Round of 64
2 — 0
Sets
Monday, 9 March 1998 ·
hard
Black W.6-3 6-4
Henman T.
Black W.
662
Henman T.
340
Wayne Black defeated Tim Henman 6-3 6-4 in the Indian Wells Masters round of 64 on hard. Wayne Black extended the head-to-head to 2–1.
Key context
- Match story: Wayne Black closed it out in two sets, 6–3 6–4.
Head to head
1–1| Date | Tournament | Round | Surface | Winner | Score |
|---|---|---|---|---|---|
| 1999-02-08 | Dubai | R32 | hard | Tim Henman | 6-0 6-4 |
| 1992-10-05 | Central Africa Masters 1 | R16 | hard | Wayne Black | 6-4 6-1 |
Statistics
This season1st serve in·Even
57%
61%
1st serve points won·Slight edge
65%
73%
2nd serve points won·Even
51%
52%
Service points won·Slight edge
59%
65%
Return points won·Even
39%
39%
Break points saved·Slight edge
55%
61%
Aces / match·Clear edge
3.0
5.4
Double faults / match·Clear edge
2.4
3.5
Averages over each player's completed matches in the selected window (100 vs 200). ATP, WTA & Challenger.
Conditions
Court speed
Medium-slowfavours grinders
SlowFast