UKC Forums - score this please

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-- score this please (http://forums.ukcdogs.com/showthread.php?threadid=333810)

Posted by TREE BLITZIN on 02-12-2010 05:21 AM:

score this please

three dog cast all dogs are struck in position don't matter in this case dog a treed for 125 he was along way in there dog b treed for 75 and dog c for 50 they was deep and we just assumed they were on the same tree well when we get there dog a had went back to trailing and dogs b and c are split do they move up to 125 or stay at their original positions ??? i'll tell you what the other kc ruled after i hear some answers

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Posted by Jack Bingham on 02-12-2010 05:24 AM:

dog a -125
dog b 75
dog c 125

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Posted by RowdyWalker on 02-12-2010 05:34 AM:

I would say if b and c are split and both have coons you would score it

dog a 125-
dob b 125+
dog c 125+

with one set of strike points

Posted by TREE BLITZIN on 02-12-2010 05:47 AM:

jack how you going to move dog c up and not b when you don't know which tree dog a left if he was ever really treed you just going to move dog c up for treeing last??

Posted by ESTEP on 02-12-2010 05:55 AM:

Dog's b and c would be moved up for 125 each. Dog a would have taken his 125-.

Posted by Lowell Lynn on 02-12-2010 05:57 AM:

WHO IS TO SAY THAT DOG A WAS NOT ON A SPLIT TREE OF HIS OWN BEFORE HE LEFT? I WOULD SCORE IT THIS WAY. DOG A 125-, DOG B,125+ IF COON IS SEEN. DOG C 125+ IF COON IS SEEN.

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Posted by TREE BLITZIN on 02-12-2010 06:18 AM:

well this is how pkc scored it when we called it in. roy tramel said since you don't know what tree dog a left you have to each stay in the position you treed for which with how pkc works was a 25 each for dog b and c.. don't know how it be scored in ukc that why i ask incase this situation should arise again

Posted by Bill(Chew) on 02-12-2010 10:02 AM:

Dog A -125

Dog B +75

Dog C +75

You do not know which tree dog A left. The dog that treed 2nd gets what he deserves and the dog that split got a bad break.

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Posted by WattsFlatsRedbo on 02-12-2010 11:17 AM:

quote:
Originally posted by TREE BLITZIN
well this is how pkc scored it when we called it in. roy tramel said since you don't know what tree dog a left you have to each stay in the position you treed for which with how pkc works was a 25 each for dog b and c.. don't know how it be scored in ukc that why i ask incase this situation should arise again

Roy was right and its the same way in UKC. They would stay scored in the same positition that they were called treed in because theres no way of knowing what dogs were where. Did you guys get blasted with the snow storms that come through? How much did you guys get? Adam

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Posted by on 02-12-2010 12:36 PM:

quote:
Originally posted by WattsFlatsRedbo
Roy was right and its the same way in UKC. They would stay scored in the same positition that they were called treed in because theres no way of knowing what dogs were where. Did you guys get blasted with the snow storms that come through? How much did you guys get? Adam

I know it doesn’t sound correct but in UKC dog b is going to stay at 75 and c is going to move up. Please don’t go off on me wanting to see it justified in the rules because you can’t find it, but that is how they do it. It has been gone over on this board a thousand times.

Posted by treberta on 02-12-2010 01:01 PM:

quote:
Originally posted by Joey
I know it doesn’t sound correct but in UKC dog b is going to stay at 75 and c is going to move up. Please don’t go off on me wanting to see it justified in the rules because you can’t find it, but that is how they do it. It has been gone over on this board a thousand times.

Yep youre right. It doesn't make any sense whatsoever but youre right.

Posted by WEBBER on 02-12-2010 02:18 PM:

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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Posted by nccoonhunter197 on 02-12-2010 02:26 PM:

Wow, how many times has this situation been explained in the past few months? Not saying it isn't a good question. I had always assumed all dogs would move up and I was wrong. I am glad it was explained but it just doesn't seem right. Either leave them alone or move both dogs up in tree points. Simple fix.

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Posted by RowdyWalker on 02-12-2010 03:35 PM:

This seems to me to reward the dog that treed last. Anyone else agree or disagree?

Posted by on 02-12-2010 04:37 PM:

quote:
Originally posted by RowdyWalker
This seems to me to reward the dog that treed last. Anyone else agree or disagree?

Yes you are correct however there is no fair way to do it somebody is going to get or not get something they deserved. So the only thing to do is pick a way and rule it that way consistently.

Posted by TREE BLITZIN on 02-12-2010 08:48 PM:

i would say the way pkc scored it was the only fair way to score it who is to say dog b wasn't the one to be split from the get go i just can't see moving the dog up that treed last and not moving the other hound up also either you move them both up or as i believe you should do just leave them in at their original tree position

Posted by coon's age on 02-12-2010 09:02 PM:

YOU tree the dogs,125,75,50.

you,arrive to find first treed on trail,125 -. B&C ARE Split from each other,coon found each split tree,125+ each.

Posted by Big Bawler on 02-12-2010 09:07 PM:

jay where the heck you huntin in this crap?

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Posted by Okie Dawg on 02-12-2010 09:19 PM:

Re: YOU tree the dogs,125,75,50.

quote:
Originally posted by coon's age
you,arrive to find first treed on trail,125 -. B&C ARE Split from each other,coon found each split tree,125+ each.

Well that is what I would think. Who is to say that dog a wasn't on a total differant tree than either of the other two.

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Posted by on 02-12-2010 10:02 PM:

Re: YOU tree the dogs,125,75,50.

quote:
Originally posted by coon's age
you,arrive to find first treed on trail,125 -. B&C ARE Split from each other,coon found each split tree,125+ each.

It dosent matter how unfair it sounds or what we think is right or wrong, but that is not how you score it. See above post.

Posted by coon's age on 02-13-2010 02:16 AM:

Re: Re: YOU tree the dogs,125,75,50.

quote:
Originally posted by Joey
It doesnt matter how unfair it sounds or what we think is right or wrong, but that is not how you score it. See above post.

Seems to me that calling it as I see it when arriving at the dogs is fair.A is not treed after being treed first and now is trailing after arriving near where he was treed AFTER THE FIVE IS UP ,but B&C are treed on two separate trees right near by and coon seen in both,minus dog A,for leaving his first tree position,plus both B&C's split trees as separate trees.125-,dog A,125+,for dogB,125+,for dog C.
And this was a UKC hunt I presume and not a PKC hunt.Since their situation on ruling and scoring are different in many ways.

Posted by Big Bawler on 02-25-2010 09:39 PM:

btt this is a good one

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Posted by wesley wickline on 02-26-2010 01:47 AM:

dogs a,b,c,d are treed in that order. upon arriveing b, c are off trailing. this is a registered cast. a snd d are split. a sees his coon d has a possum.

a=125+
b= 75-
c=50-
d=125-

would this be correct?

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Posted by JustinM on 02-26-2010 03:19 AM:

quote:
Originally posted by wesley wickline
dogs a,b,c,d are treed in that order. upon arriveing b, c are off trailing. this is a registered cast. a snd d are split. a sees his coon d has a possum.

a=125+
b= 75-
c=50-
d=125-

would this be correct?

yup

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Posted by gfults on 02-26-2010 10:28 AM:

Re: YOU tree the dogs,125,75,50.

quote:
Originally posted by coon's age
you,arrive to find first treed on trail,125 -. B&C ARE Split from each other,coon found each split tree,125+ each.

You dont know for sure where dog A was treed. They were assumed to have all been together. Dog A 125- dog B 75+ dog C 125 +. This has been explained on here several times.

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