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Score This!

4 dog cast.

Strike
A 100
B 75
C 50
D 25

Tree
A 125
B 50
C 25
D 75

All dogs were thought to be on the same tree. When calling them treed.

While going into the tree its determined that dogs are split.

C & D are handled on tree.

Cast proceeds 50 yards deeper to Dog B to handle. Cast decides to score Dog B's tree while there. Dog B is moved up to 125 tree. Tree is slick.


Dog A has its head down barking on the ground in between (C & D's tree) and Dog B's tree. So he is -125, but not handled yet.


Cast goes to Dogs C & D to score the tree. Dog A comes in and starts treeing and is handled. Coon is seen.

Note: Dog A never showed treed while we were in there except when we were scoring C & D, but was definitely treeing before. Assume he was on the same tree as C & D.

A -225
B -200
C ?
D ?

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Dog A would be minused his strike.

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A screwed it up for everyone. No one should have moved up cause you don't know which tree dog a left.by the way a's strike is minused cause a coon is seen.

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guess you could make an argument to move B up to 75 tree and c up to 50 but not 125 for b cause like I said you don't know what tree dog a left. s
so I say A -225 B-150 C+100 D +100

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I'd leave C&D where they are on the card and plus'em.

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quote:

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Originally posted by JiM
I'd leave C&D where they are on the card and plus'em.

---

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John Queen

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Dog A minus strike point all other dogs stay as is no dogs move up in tree points that's the way I have always seen it scored besides dog B he would move up to 125 tree points

what if dog b had a coon? for all we know dog a could have been first on dog b tree since it was a split and dog a was barking on ground between trees when cast arrives he gets his minus for not being on tree all others move up a split is a split is a split and dog a minus track for comming into closed tree coon seen dog a 225- dog b 200- dog c 125+ dog d 150+ now if dog b had coon 200+ rest stay the same tell me i'm wrong

A -- 225 minus--1st strike & off the tree (100+125)

B --200 minus --3rd treed-- SPLIT & slick & holding 2nd strike (75+125)

C--100 plus--3rd strike & last treed & moves up next available because of B split. (50+25+25)

D--100 plus--last strike & 2nd tree with A holding 1st tree (25+75)

That's my guess.

quote:

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Originally posted by StrawberryMt
guess you could make an argument to move B up to 75 tree and c up to 50 but not 125 for b cause like I said you don't know what tree dog a left. s
so I say A -225 B-150 C+100 D +100

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This scenario makes my head hurt but i remember in the past a ruling on a similar situation that used a ROOT tree system. Jim would probably remember.

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B stays at 75
C and D move up to 125 and 75.

quote:

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Originally posted by elvis
B stays at 75
C and D move up to 125 and 75.

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Why would B not move to 125 when it is split by itself?

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I went with the root tree being the tree C & D were on since A who was holding first tree was off the tree and D was holding 2nd tree. With A wandering around it would be a guess as to the root tree .
A question : when A is minused for being off the tree do the dogs move up or is 1st on that tree just -125 for A and the rest scored as called ?

What you assume before you get there holds no bearing.

If you never saw A on a tree when you walked in, it gets minused its tree points, and all other dogs move up accordingly. If dog A comes in after the 5 it is also minused it's strike for coming in after judge, and coon being seen.

D-150+
C-125+
B-200-
A-225-

You can't assume they are all together, and what tree is what from the trucks. Take the "WHAT IF" out of it. When you walked in there were 2 dogs on 1 tree, 1 split, and one that isn't treed. Move dogs up and score accordingly.

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Alright Allen,time to jump in!

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quote:

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Originally posted by Tully

You can't assume they are all together, and what tree is what from the trucks. Take the "WHAT IF" out of it.

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John Queen

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Dog A gets -125 for leaving a tree, Dog B gets -125 tree and 50 strike minus for the slick. C & D gets moved to +125 & +75 on the tree points and strike plused. I score it on what I see and is apparent.

You can't say dog C & D didn't split tree when called and the same goes for B. You can say A left a tree and deserves 125- and the others were split on different trees. Which deserves two first tree points. A could have been with C and D or possibly B, all speculation. If A would have been treed in between the other two trees with a coon no minus and 125 plus on three seperate trees. I like to score situations as I see them, not hear or speculation.

p.s. dog A gets a strike minus when comming to the tree when coon is seen.

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I would leave C&D on the card the way they are and plused expect Dog C could move to third tree. Dog B recieves 150 minus for the slick and could be moved to second tree.

No dog can move up to first tree without knowing where dog A was treed.

Had a situation just like this occur on a hunt in Missoui about three months ago. this was scored liek this by another judge, it was my dog that was not able to move up on a split tree where she was treed another 100 yards down the creek and I thought it was scored correctly then and it cost me a NT Ch. loosing the hunt 725+ to 700+. I would have moved from second tree to first tree.


A is minused tree points and strike if comes into tree with the coon in it.

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quote:

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Originally posted by mleck
I would leave C&D on the card the way they are and plused expect Dog C could move to third tree. Dog B recieves 150 minus for the slick and could be moved to second tree.

No dog can move up to first tree without knowing where dog A was treed.

Had a situation just like this occur on a hunt in Missoui about three months ago. this was scored liek this by another judge, it was my dog that was not able to move up on a split tree where she was treed another 100 yards down the creek and I thought it was scored correctly then and it cost me a NT Ch. loosing the hunt 725+ to 700+. I would have moved from second tree to first tree.


A is minused tree points and strike if comes into tree with the coon in it.

---

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just keep everything as it was recorded and pluss or minus as where they were. either way someone will be pissed. lol

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dog a is called treed but is not treed unless he shows tree

How soon ya'll rule gurus forget!



Originally posted by Allen / UKC
Seems like I may have some of you very confused. Possibly because of my terminolgy or lack thereof. Please allow me to make this very clear. First, I am well aware of page 99 in the Advisor. What Todd Kellam wrote there has not changed. Not one bit.

My question is this; how is the situation on page 99 of the Advisor and the scenario posted by Mr. Hagood even related? Please read both of them again. The difference in the two is that in Darrell’s scenario dog C is “obviously” split from A.

Again, I guess I wasn't clear enough. We're talking about situations where you have dogs declared treed, then, when you get to the tree you find one or more has left and the rest are on separate trees. (split wasn’t "obvious" until you got there and "saw" it.) You don't really know which tree the running dog(s) actually left, therefore, making it a confusing or difficult situation to score.

What I am "trying" to get across is that in situations where it was never "obvious" that any of the dogs were split (until you got to the tree and "saw" it), and to have a consistent way of scoring these difficult situations, use dog A's tree as the "root". Why? Because according to Rule 11 a judge never declares a dog split unless it is obvious or until at a point when it does become obvious. In some instances it doesn’t become obvious until you get there and see it; as in Darrell’s scenario and in the examples shown below.

If we stick with that theory or rule of thumb that we score dogs as being on the same tree “unless or until” it becomes obvious then we have a consistent way of scoring these difficult situations. May not always be exactly what happened but neither are any other ways you might use to score them. At least we have a method to use that’s consistent and probably as fair as any other options if you really think about it. And, if you haven’t learned that the breaks don’t always go your way then you haven’t been huntin’ long enough.

Example 1 -
Dogs A, B, C and D are declared treed in that order. Upon arriving dog C is gone and dog D is on a separate tree.

Tree Points:
Dog A = 125
Dog B = 75
Dog C = 50-
Dog D = 125

Example 2 -
Dogs A, B, C and D are declared treed in that order. Upon arriving dogs A and D are treed together. Dog B is on a separate tree and Dog C is gone.

Tree Points:
Dog A = 125
Dog B = 125
Dog C = 75- (moved up one tree position because B is on separate tree)
Dog D = 50 (moved up one tree position because dog B is on separate tree)

Example 3 -
Dogs A, B, C and D are declared treed in that order. Upon arriving dog A is gone. B, C and D are treed together.

Tree points:
Dog A = 125-
Dog B = 75
Dog C = 50
Dog D = 25
( this example is similar to TK’s scenario on page 99 of the Advisor – difference is three dogs stayed instead of just the one.)

Example 4 -
Dogs A, B, C and D are declared treed in that order. Upon arriving dog A is gone. B and C are treeing together and D is on a separate tree.

Tree points:
Dog A = 125-
Dog B = 75
Dog C = 50
Dog D = 125

Example 5 -
Dogs A, B, C and D are declared treed in that order. Upon arriving A is on one tree. B and C are both gone. D is on separate tree.

Tree points:
Dog A = 125
Dog B = 75-
Dog C = 50-
Dog D = 125

Example 6 -
Dogs A, B, C and D are declared treed in that order. Upon arriving A is gone. B and D are treed together and C is on a separate tree.

Tree Points:
Dog A = 125-
Dog B = 75
Dog C = 125
Dog D = 50 (moved up one position because C is on separate tree.)

Example 7 –
Dogs A, B, C and D are declared treed in that order. Upon arriving dogs A and B are gone. C is on one tree and D is on a separate tree.

Tree points:
Dog A = 125-
Dog B = 75-
Dog C = 50
Dog D = 125
(this one will likely bring on the biggest argument, but again, it is staying consistent with the theory of using the root tree. Why? Because it was never obvious that the dogs were split until the cast arrived and “saw” D on a separate tree. We don’t assume A and B were anywhere else other than with C because it wasn’t obvious.)

Hopefully, these examples will help a little to get the idea and eliminate any confusion. Especially, in regards to page 99 of the Advisor as the two situations are not really related.

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quote:

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Originally posted by Dale Young
A -- 225 minus--1st strike & off the tree (100+125)

B --200 minus --3rd treed-- SPLIT & slick & holding 2nd strike (75+125)

C--100 plus--3rd strike & last treed & moves up next available because of B split. (50+25+25)

D--100 plus--last strike & 2nd tree with A holding 1st tree (25+75)

That's my guess.

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