1999 Nottingham Open – Doubles
Justin Gimelstob and Byron Talbot were the defending champions, but Talbot did not partner this year. Gimelstob partnered Patrick Galbraith.
Doubles | |
---|---|
1999 Nottingham Open | |
Champions | |
Runners-up | |
Final score | 5–7, 7–5, 6–3 |
Galbraith and Gimelstob won the title, defeating Marius Barnard and Brent Haygarth 5–7, 7–5, 6–3 in the final.
Seeds
David Adams / John-Laffnie de Jager (Semifinals) Andrew Kratzmann / Fabrice Santoro (Semifinals) Patrick Galbraith / Justin Gimelstob (Champions) Piet Norval / Kevin Ullyett (Quarterfinals)
Draw
Key
- Q = Qualifier
- WC = Wild Card
- LL = Lucky Loser
- Alt = Alternate
- SE = Special Exempt
- PR = Protected Ranking
- ITF = ITF entry
- JE = Junior Exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
Draw
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | 6 | 6 | |||||||||||||||||||||||||
4 | 0 | 1 | 4 | 7 | 7 | ||||||||||||||||||||||
3 | 4 | 6 | 5 | 5 | |||||||||||||||||||||||
6 | 6 | 1 | 3 | 2 | |||||||||||||||||||||||
4 | 4 | 7 | 6 | 6 | 6 | ||||||||||||||||||||||
WC | 6 | 5 | 2 | 4 | 3 | 6 | |||||||||||||||||||||
6 | 1 | 3 | 6 | 7 | |||||||||||||||||||||||
3 | 6 | 6 | 7 | 5 | 3 | ||||||||||||||||||||||
6 | 6 | 3 | 5 | 7 | 6 | ||||||||||||||||||||||
WC | 4 | 4 | 4 | 7 | 4 | ||||||||||||||||||||||
Q | 6 | 1 | 3 | 6 | 6 | 6 | |||||||||||||||||||||
3 | 7 | 6 | 3 | 6 | 7 | ||||||||||||||||||||||
WC | 2 | 4 | 2 | 1 | 6 | ||||||||||||||||||||||
6 | 6 | 6 | 7 | 6 | |||||||||||||||||||||||
6 | 4 | 4 | 2 | 7 | 6 | 7 | |||||||||||||||||||||
2 | 1 | 6 | 6 |
gollark: i.e. if feeding in input A gives output X, and input B gives output Y, then feeding in A+B gives X+Y.
gollark: But linear/passive circuits *do* obey the "principle of superposition".
gollark: Not all components meaningfully have "resistance".
gollark: I'm sure you can make it work somehow.
gollark: Well, if you can do NAND, you have achieved logic.
References
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