8 2025-12-15  16.12.2025
SCHUCHLENZ Arthur - TEIML Martina
   

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Paar: 8    SCHUCHLENZ Arthur - TEIML Martina

Score: 79,0    36,6 %    Platz: 11    
                       
Rnd  Tisch  Als  Geg  Board  Kont      Aus  Ergebnis  Score    %    Runde    %      Ges      %        Name d. Gegners

 1     2    N-S   12    4    4P   W +1 HA       -650    0,0    0                                      HEIL - KRAJNY
                        5    4H   O +1 KA       -450    3,0   38                                            
                        6    2P   S -2 HD       -100    2,0   25      5,0   21      5,0   20,8              

 2     4    N-S   4     13   5K   O  = TA       -600    1,0   13                                      HEIL - MEIER
                        14   4H   O +1 PK       -450    3,0   38                                            
                        15   4H   S -1 H2       -100    2,0   25      6,0   25     11,0   22,9              

 3     5    N-S   2     19   3K   W +1 TB       -130    0,0    0                                      SCHINDLAUER - SCHOBER
                        20   3P   O +2 K5       -200    8,0  100                                            
                        21   3H   N -2 K6       -200    2,0   25     10,0   42     21,0   29,2              

 4     3    N-S   11    16   3H   S +2 PA    200        0,0    0                                      DRAPELA - SCHUCHLENZ
                        17   4P   O  = TK       -420    2,0   25                                            
                        18   4H   N  = K10   620        4,0   50      6,0   25     27,0   28,1              

 5     6    O-W   6     1    5Hx  O  = P6       -650    8,0  100                                      FENZ - ROPPOSCH
                        2    2P   S +2 KB    170        2,0   25                                            
                        3    4P   S  = P2    420        5,0   63     15,0   63     42,0   35,0              

 6     4    O-W   3     25   4P   S -2 H6       -100    6,0   75                                      KOGLER - KOTTULINSKY
                        26   2H   N  = T2    110        8,0  100                                            
                        27   3N   N  = T10   400        8,0  100     22,0   92     64,0   44,4              

 7     6    N-S   1     7    4H   S  = H4    620        2,0   25                                      MALDET - MALDET
                        8    3N   W +2 H9       -460    0,0    0                                            
                        9    2P   O  = KA       -110    2,0   25      4,0   17     68,0   40,5              

 8     1    O-W   10    22   3P   N -1 H10       -50    6,0   75                                      MORETTI - WENDLER
                        23   3P   N  = KA    140        1,0   13                                            
                        24   3Hx  O -3 PA    500        2,0   25      9,0   38     77,0   40,1              

 9     5    O-W   5     10   3N   N +2 PD    660        0,0    0                                      LACKNER - ZIMMERMANN
                        11   2K   W +1 H4       -110    1,0   13                                            
                        12   3T   O -3 P10   150        1,0   13      2,0    8     79,0   36,6              


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