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F class ties

From NRA High Power rule book, rule 15 - Decision of ties.

First part of the rule (15 - a and 15 - b) talks about decisions by X, and there are rules if shooting was done at multiple distances (e.g. Palma aggregate).

In a 'regular' F-class match, at a single distance, if shooters have the same score and the same number of X, the rule says (15 - c):

(c) In the event the tie is not broken, the scores at each range, in the above order will be ranked:
(1) By the fewest misses.
(2) By the fewest hits of lowest value.
(3) By the fewest hits of next lowest value.
(4) In slow fire individual stages, by the value of the hits in inverse order, counting singly from last to first (X’s being hits of highest value).

C-1 probably does not decide a lot of ties but in really tough conditions it could.

C-2 and C-3 would mean that if two shooters tied, but one shot an 8 and another only shot 9's (for the shots that were not 10 or X that is), the shooter who shot an 8 loses.

C-4 is what I have colloquially heard referred to as 'Creedmoring'. Starting from the last shot of the match, compare the two scorecards and working backwards, when the two competitors do not have the same score for the shot, the one with the higher wins.

For example, if both shooters ended with an X for shot 20, but shooter 1 has an X for shot 19 and shooter 2 has a 10 for shot 19, shooter 1 wins. Working back like that until there is a difference.

If this still does not break the tie there is rule 15.13 - Unbreakable Ties:

15.13 Unbreakable Ties — In any case where a tie cannot be ranked under the foregoing provisions of this section, the Match Director will direct that the tie be decided and prizes awarded under one of the following plans as appears necessary or advisable:
(a) By firing of a complete or partial score under the original match conditions or at longest range of the match.
(b) By drawing of lots for merchandise, medal or trophy awards, and combining any cash awards to which those tied may be entitled and equal division of such cash among those tied.

15.13 (b) essentially being the good old coin toss
 
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So in this scenario:

Shooter 1. 200 +11x, 200 +9x, 200 +9x
Shooter 2. 200 +10x, 200+9x, 200 + 10x
 
So in this scenario:

Shooter 1. 200 +11x, 200 +9x, 200 +9x
Shooter 2. 200 +10x, 200+9x, 200 + 10x

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".
 
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
 
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.
 
They wrote it in a backwards explanation for some reason. Go to C-1 and read it closely. Fewest misses means high score wins.

I read fewest misses to mean misses in the literal sense, as in 'M' or '0' on the scorecard. With these two shooters having shot no misses I thought it would go to 15-C-4. :confused:
 
I think you would start from the last shot of the last match and who ever shot a 10 first, would lose. So in the 200-9 vs 200-10 match, who shot the highest value on the last shot, then work backwards until someone has a 10 vs X.
 
Mgdietrich is correct, unless of course, different distances were shot at the match, then the highest X count of the longest match would be the defining factor.
 
You guys are right. This comes from shooting a match for too many years that broke ties by the first point dropped. My apologies
 
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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".

Highpower-FClass,


For whatever it might be worth, I think your interpretation/reading of the Rules is correct.

John
 
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...

Looks right to me...
 
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...

Makes sense
 

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