2019 Indian Open
Tournament information | |
---|---|
Dates | 27 February – 3 March 2019 |
City | Kochi |
Country | India |
Organisation(s) | WPBSA |
Format | Ranking event |
Total prize fund | £323,000 |
Winner's share | £50,000 |
Highest break |
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Defending champion |
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Final | |
Champion |
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Runner-up |
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Score | – |
← 2017 |
The 2019 Indian Open is a professional ranking snooker tournament. It was due to take place between 18 and 22 September 2018 at the Grand Hyatt Kochi Bolgatty in Kochi, India but was postponed due to the 2018 Kerala floods.[1] The re-scheduled Indian Open will now be played in Kochi from 27 February to 3 March 2019.[2] It will be the fifteenth ranking event of the 2018/2019 season.
Qualifying took place on 15 and 16 August 2018 in Preston, England.
Prize fund
The breakdown of prize money for this year is shown below:
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|
The "rolling 147 prize" for a maximum break: £
Main draw
Last 64 Best of 7 frames | Last 32 Best of 7 frames | Last 16 Best of 7 frames | Quarter-finals Best of 7 frames | Semi-finals Best of 9 frames | Final Best of 9 frames | |||||||||||||||||
Final
Final: Best of 9 frames. Referee: Grand Hyatt Kochi Bolgatty, Kochi, India, 3 March 2019. | ||
– | ||
– | ||
Highest break | ||
Century breaks | ||
50+ breaks |
Qualifying
These matches were held between 15 and 16 August 2018 at the Preston Guild Hall in Preston, England. All matches were best of 7 frames.
- Notes
Century breaks
Qualifying stage centuries
Total: 10
- 135
Hamza Akbar - 122
Rod Lawler - 118
Joe Swail - 115
Lyu Haotian - 114
Ross Muir - 110
Oliver Lines - 102
Gerard Greene - 102
Michael Holt - 101
Stuart Bingham - 101
Elliot Slessor
References
This article is issued from
Wikipedia.
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