Goring Accepted
The first four videos about the Goring Gambit Series talk about the “Accepted Variation”, a very important player that played this variation is the World Champion Mihail Tal , he played a very nice game against Russel in 1999
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[Event “Munich ol (Men) qual-A”]
[Site “Munich”]
[Date “1958.??.??”]
[Round “6”]
[White “Tal, Mihail”]
[Black “Russell, A.”]
[Result “1-0”]
[ECO “C44”]
[WhiteElo “2625”]
[PlyCount “51”]
[EventDate “1958.09.30”]
[EventType “team”]
[EventRounds “9”]
[EventCountry “GER”]
[Source “ChessBase”]
[SourceDate “1999.07.01”]
[WhiteTeam “Soviet Union”]
[BlackTeam “Ireland”]
[WhiteTeamCountry “URS”]
[BlackTeamCountry “IRL”]
1. e4 e5 2. Nf3 Nc6 3. d4 exd4 4. c3 dxc3 5. Nxc3 d6 6. Bc4 Be7 7. Qb3 Na5 8.
Bxf7+ Kf8 9. Qa4 Kxf7 10. Qxa5 Be6 11. O-O Kf8 12. Nd5 c6 13. Nc7 Bf7 14. Nd4
Qc8 15. Nxa8 Qxa8 16. Nf5 b6 17. Qc3 Bf6 18. Qg3 Ne7 19. Qxd6 Ke8 20. Bh6 Rg8
21. Rad1 Qc8 22. Bxg7 Nxf5 23. exf5 Be7 24. Rfe1 Be6 25. Rxe6 Rxg7 26. f6 1-0
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In the first video you can see the variation with Bb4, the idea is block the knight on c3, with will develop his pieces and will do a very big pressure against f7.
[fen csl=Rc3,Yf7 cal=Rb4e1,Gf1c4,Gc4f7]r1bqk1nr/pppp1ppp/2n5/8/1b2P3/2N2N2/PP3PPP/R1BQKB1R w KQkq – 0 1[/fen]
In the second video we go forward with the variation Bb4, and we focus the study about this position
[fen csl=Gf7,Rd6 cal=Gc4f7,Rd6e5,Yg8f6]r1bqk1nr/ppp2ppp/2np4/8/1bB1P3/2N2N2/PP3PPP/R1BQK2R w KQkq – 0 1[/fen]
In which black moves d6 to allow the Nf6, white must play to attack f7!
In the third video about the series, we see the push “d6” immediately , so the position to study will be this
[fen csl=Rd6,Gf7 cal=Rd6e5,Yf8e7,Yg8f6,Gf1c4,Gc4f7]r1bqkbnr/ppp2ppp/2np4/8/4P3/2N2N2/PP3PPP/R1BQKB1R w KQkq – 0 1[/fen]
black idea is castle quickly, again white will make pressure against f7
The fourth video talks about the variation Bc5
[fen]r1bqk1nr/pppp1ppp/2n5/2b5/4P3/2N2N2/PP3PPP/R1BQKB1R w KQkq – 0 1[/fen]
and about the variation “h6”
[fen csl=Rh6,Gf7 cal=Rh6g5,Gc4f7]r1bqkbnr/ppp2pp1/2np3p/8/2B1P3/2N2N2/PP3PPP/R1BQK2R w KQkq – 0 1[/fen]