UKC Forums - scoring question

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Posted by tpettit on 05-09-2017 07:56 PM:

scoring question

4 dog cast. A,b,c,d are treed in that order. We walk into tree find a alone and c and d together about 30 yards away. The question is with dog b leaving, how do you score dogs c and d? All were treed as if they were together. So does dogs c and d move up or is it a tuff break because we didn't know where dog b was? Would like opinions and an official answer. Thanks

P.s. my answer is I thought c and d should have stayed as they were treed in because b wasn't there and we didn't declare/couldn't tell dogs were split until we got there. This was a youth hunt, I just walked along. No questions were asked. The judge moved the split,dogs up to 125 and 75.

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Posted by Tom anderson on 05-09-2017 08:08 PM:

Not saying I am write but if you where a two dog cast and you get in there and they are split you give them both 125 so I would say it would be same way here

Posted by msinc on 05-09-2017 08:12 PM:

The judge was correct. Split tree speaks for itself...it sounds like C and D were split treed all along and just sounded like they were at the same tree with it only being 30 yards away. B of course gets minus because he left. The fact that B left has nothing to do with the scoring of the others.
As to the question "how do you score???" you don't say whether or not the trees were slick, dens or had a coon. Also, dogs C and D did not "move up", they received a first and second because they were split on a different tree. Dogs don't "move up" because one of them leaves or gets scratched, if they are all on one tree they retain points as called.

Posted by tpettit on 05-09-2017 08:18 PM:

Both trees had coons. I was just wondering if c and d move up? They all treed as one tree, not knowing there was a split tree in the mix. I asked because we didn't know wether dog b was with dog a or with dogs c and d.

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Posted by Nick Jennings on 05-09-2017 08:26 PM:

Root Tree

Jim? Allen?

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Posted by joey on 05-09-2017 08:36 PM:

It's what they called the root tree rule. C and D move up. In ukc way of thinking everything is considered treed with the first dog until proven other wise. If dog A would have left and B was split from c and D then dog B would have stayed and c and D move up. Crazy I know but that's how it's done.

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Posted by shadinc on 05-09-2017 08:47 PM:

So B would get 75 minus?

Posted by Tim MACHA on 05-09-2017 08:49 PM:

quote:
Originally posted by joey
It's what they called the root tree rule. C and D move up. In ukc way of thinking everything is considered treed with the first dog until proven other wise. If dog A would have left and B was split from c and D then dog B would have stayed and c and D move up. Crazy I know but that's how it's done.

Exactly!

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Posted by tpettit on 05-09-2017 09:12 PM:

I've never heard that " root tree rule". Thanks for informing me. I still don't agree with it but if that's the way it is, then so be it. I was thinking they couldn't move up because or,jot knowing where b was for sure.

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Posted by tim griffin on 05-09-2017 09:15 PM:

Dog
A-125+
B-75-
C-125+
D-75+

That's how there tree points should look

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Posted by Donnie Stevens on 05-09-2017 10:16 PM:

Who come up with the "tree root rule" ...
Todd or Allen ?

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Posted by joey on 05-09-2017 10:35 PM:

quote:
Originally posted by Donnie Stevens
Who come up with the "tree root rule" ...
Todd or Allen ?

I believe that was Todd

Posted by Donnie Stevens on 05-09-2017 11:23 PM:

Maybe that's why they hired Allen lol

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Posted by MARSHALL AYERS on 05-09-2017 11:52 PM:

I know for a fact it's totally different in Pkc. Same situation in Pkc and b leaves every stays the same because you can't prove where b left from

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Posted by indiana1 on 05-10-2017 12:47 AM:

There is a thread on it in the advisor section. (Root tree)

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Posted by indiana1 on 05-10-2017 01:23 AM:

Allen w/ UKC Posted:::::::::
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.

Posted by Allen / UKC on 05-10-2017 02:10 PM:

Re: scoring question

quote:
Originally posted by tpettit
4 dog cast. A,b,c,d are treed in that order. We walk into tree find a alone and c and d together about 30 yards away. The question is with dog b leaving, how do you score dogs c and d? All were treed as if they were together. So does dogs c and d move up or is it a tuff break because we didn't know where dog b was? Would like opinions and an official answer. Thanks

P.s. my answer is I thought c and d should have stayed as they were treed in because b wasn't there and we didn't declare/couldn't tell dogs were split until we got there. This was a youth hunt, I just walked along. No questions were asked. The judge moved the split,dogs up to 125 and 75.

C and D move up to 125 and 75. B is minused 75 for leaving A's tree.

Determining whether or not dogs are split, is the decision of the judge. Not the handler. The only exception to this is in the event where a tree is closed and a handler declares a dog treed, which results in awarding the dog 125 tree points as a split tree.

Otherwise, it must be obvious to the judge that a dog(s) is split. If not obvious, tree position points are awarded as being on the same tree, until or unless it becomes obvious at some point later.

In this case it was not obvious at the time the dogs were called, therefore, they all went on the card as being with the first dog declared treed. It did become obvious later (upon arrival) that C and D were split from A. It was never obvious that B was anywhere else other than where it had originally been treed, which was with Dog A. Hence the scoring as suggested.

Posted by tpettit on 05-10-2017 02:54 PM:

Thank you Allen.

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