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How are ties broken in F class matches?
So in this scenario:
Shooter 1. 200 +11x, 200 +9x, 200 +9x
Shooter 2. 200 +10x, 200+9x, 200 + 10x
Guys I've read and studied this until I was blue. By far the most confusing damn rule ever written in the history of man. I put in one sentence for ya. Read it 50 times and see if I got it right.
Hi Mike. Could you explain the part "Highest relay wins", not sure I understand that one. Not saying it is wrong, I was not sure how to interpret it.
Thanks
They wrote it in a backwards explanation for some reason. Go to C-1 and read it closely. Fewest misses means high score wins.

In this scenario, the way I would interpret the rules, match 1 went to shooter 1, match 3 to shooter 2, so the decision for the aggregate would need to be done by rule 15 (c)-4 for match 2, for which the scorecards would need to be available and evaluated going from shot 20 backwards.
If the tie is not broken by this then 15.13 would kick in, i.e. either a complete or partial re-shoot between the competitors (15.13 (a)) or a coin-toss, or similar (15.13 (b)).
Note the wording, "the way I would interpret the rules".
To decide the winner of the Match 2, you would start with shot 20 on that scorecard and work backwards on that scorecard. For the aggregate as mgdietrich stated you go to the last shot of the aggregate match which is shot number 60 (shot 20 on Match 3 scorecard) and work backwards...
To decide the winner of the Match 2, you would start with shot 20 on that scorecard and work backwards on that scorecard. For the aggregate as mgdietrich stated you go to the last shot of the aggregate match which is shot number 60 (shot 20 on Match 3 scorecard) and work backwards...
 

