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SoSh
UKC Forum Member

Registered: Apr 2005
Location: Richland,Mo
Posts: 236

tim, are you saying that you never talked to darrel's 'friend' ?

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Old Post 02-15-2008 08:04 PM
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Allen / UKC
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Registered: Nov 2004
Location: Michigan
Posts: 9282

I think what Tim is saying is that he was at the World but it wouldn't have been him cause that's not the answer he would have given.

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Old Post 02-15-2008 08:08 PM
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gfults
UKC Forum Member

Registered: Mar 2006
Location: Petersburg, Tn. aka Redneck USA
Posts: 1184

quote:
Originally posted by Allen / UKC
I think what Tim is saying is that he was at the World but it wouldn't have been him cause that's not the answer he would have given.


Now that sounds like a correct interpretation!!!lol

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Old Post 02-15-2008 08:12 PM
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Allen / UKC
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Registered: Nov 2004
Location: Michigan
Posts: 9282

quote:
Originally posted by gfults
Now that sounds like a correct interpretation!!!lol



How do you know he was there? What if he wasn't and it was someone else? What if he did talk to the guy and just forgot, or worse yet, he's lying about it? Until we can get a camera on 'ole Tim and listen to the recording then I refuse to accept that whole messed up idea. No way, no how! lol!

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Old Post 02-15-2008 08:37 PM
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gfults
UKC Forum Member

Registered: Mar 2006
Location: Petersburg, Tn. aka Redneck USA
Posts: 1184

There u go again! I'm trying to be nice and u come on here and start disagreeing with me again. I've gotta get u out of that country you live in. The Confederate States of America could really do ur education alot of good!!

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Old Post 02-15-2008 09:10 PM
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Allen / UKC
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Registered: Nov 2004
Location: Michigan
Posts: 9282

That's it...........I knew there was something. First I thought it might just be the water in your neck of the woods but now I see otherwise. The war is over my friend and we are now all living in the same happy place where everyone gets along. It's called the UNITED States of America! lol!

Have a good weekend. I'm out.

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Old Post 02-15-2008 09:34 PM
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gfults
UKC Forum Member

Registered: Mar 2006
Location: Petersburg, Tn. aka Redneck USA
Posts: 1184

When did the war end?? Nobody told me!! Uhoh. I guess I need to go hide those 2 I got this morning!! See ya'll later

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Old Post 02-15-2008 09:39 PM
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larrypoe
Banned

Registered: Jan 2006
Location: bronaugh,MO
Posts: 2595

quote:
Originally posted by gfults
When did the war end?? Nobody told me!! Uhoh. I guess I need to go hide those 2 I got this morning!! See ya'll later



Now thats funny I dont care WHO you are.

Back to the original question.


Everybody has got stuck on the one situation where poor ol dog b gets the shaft and slowpoke dog d wins the lottery.

They have also assumed there is a coon in both trees. Now remember we have to rescore these trees BEFORE we even start shining. We have no idea whats up there or if anything is up there.

Now score them:

dog a 125- (for leaving)
dog b 75- (for leaving)
dog c 50 (still on tree)
dog d 25 moved up to 125 because there is now a split tree.


Now we start shining and both trees are slick as snot. I mean you couldnt hide a tick up there. Now who got the shaft?

Lets say we start shining and d has a coon and c is slick as snot.

Maybe one or the other has a possum and thats why a and b left dodge in the first place before we got there.

Lets change it up and c and d leaves. A and b are split. Since we now dont have to know where a split was we might as well assume they left there own trees and get 125- for it instead of the 25 and 50 they were carded for. I mean a and b had there own trees so c and d had to have there own as well right??

Lets say we never move a dog up if one has left. c and d still treed but a and b scoot out. c is slick and d has a coon. D gets 25+ and whatever strike he held for sticking under a coon while the rest of the cast run around like idiots. C scoots by with 50- and whatever strike he held for his slick because we dont know for sure who has the split and who is under the origenal tree (if anybody)

I see alot more points stolen by babblers and dogs who bump trash then I ever see stolen by the rare chance a dog gets to move up and mine has to stay were they were carded on a tree.

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Last edited by larrypoe on 02-16-2008 at 09:10 PM

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Old Post 02-15-2008 10:53 PM
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Darrell
UKC Forum Member

Registered: Jun 2003
Location: Central Arkansas
Posts: 1652

quote:
Originally posted by gfults
There u go again! I'm trying to be nice and u come on here and start disagreeing with me again. I've gotta get u out of that country you live in. The Confederate States of America could really do ur education alot of good!!


A man after my own heart...

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Old Post 02-16-2008 05:13 PM
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Bill(Chew)
UKC Forum Member

Registered: Jun 2003
Location: Washington, NC
Posts: 3315

Many years ago I was told that on a UKC hunt there is ONLY ONE coon in the woods until the dogs split tree.

Keeping that in mind makes scoring much eaiser. If dog A strikes to the right and dog B & C strike to the left it is still considered ONE track. When the dogs tree it is considered ONE tree unless obviously seperate. No one is argueing over an obvious split tree.
Allen's senerios work because of the ONE COON in the woods until proven different.

Yes, dog C may have been the one who treed by it's self but if you don't KNOW that he did then you must treat it as if it was treed with dogs A & B. That maybe incorrect but thats the ruling!

In my experence most of the time when one dog is split from the others he was the last to tree anyway.

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Last edited by Bill(Chew) on 02-16-2008 at 09:07 PM

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Old Post 02-16-2008 09:04 PM
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elvis
UKC Forum Member

Registered: Jun 2003
Location: Dog House
Posts: 4112

quote:
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.


gfults, I would gladly discuss this topic with you if you care to call or give me a phone number where you can be reached at. Might be able to eliminate any confusion. Thanks.



im sure everyone will be tickled i brought this one back up.

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Old Post 04-04-2009 08:53 PM
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CooperCreek
UKC Forum Member

Registered: Aug 2004
Location: NE/MO
Posts: 579

Bump....

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Old Post 07-14-2009 03:27 AM
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elvis
UKC Forum Member

Registered: Jun 2003
Location: Dog House
Posts: 4112

quote:
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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Old Post 11-26-2009 05:02 AM
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WEBBER
UKC Forum Member

Registered: Sep 2005
Location: Orangeburg, SC
Posts: 196

bttt

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