Registered: Jun 2008
Location: Burlington, CT
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Exactly...

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mnthunter said "handler declares split tree for dog C".

Handlers A&B are happy as larks with their dogs doing their thing on the shut out tree.

Handler C did not procrastinate You can't call a dog treed before he is treed.Well, I guess you could and some of yall wonder why he didn't pitch him before the five and I think that's what has alot of people confused. He called C honest. and if C had his coon backed by B. (Read what he said) then the rest is pure conjecture about B. B was happy taking second tree and shutting C out remember? And Minus the one dog C that treed the coon and move second fiddle B up.HaHa LOL no wonder there is such confusion.

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He did have something to gain by declaring him split as proved out when B jump over and backed him . If he had just kep his mouth shut and walked in Then and only then can I see there be a question if B had treed it first. But it did not happen like that; He declared his C dog split.

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Just treeing a dog after the five, IS declaring a dog as split treed. Perhaps the handler of Dog C, had no idea what he had done.

Why do you continue to disregard the fact that all three of these dogs were close to one another, and therefore there's no way to determine who treed on what tree first??? Heck maybe A had first tree and then chose to get off by himslef and left his tree to B and C?!? I've seen dogs act jealous like that...

As far as you can prove, B wasn't happy to take a second to A, he actually got first on a split tree, and C was actually happy to play second fiddle. Worst than that, C took over 5 minutes to over B!

It's as if you have decided to make C and it's handler your personal hero, and therefore apply your conjecture in their favor!!!

David Schmidt

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According to the rules and what can be proven, Dog C must be minused... His handler had everything to lose.

Since you can't prove your conjecture on Dog B moving, you would have to assume as a judge, that B was always split.

Unless you just want to take it upon yourself to start making assumptions on how great C is and how poor B must have been...

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quote:

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Originally posted by deschmidt27
According to the rules and what can be proven, Dog C must be minused... His handler had everything to lose.

Since you can't prove your conjecture on Dog B moving, you would have to assume as a judge, that B was always split.

Unless you just want to take it upon yourself to start making assumptions on how great C is and how poor B must have been...

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Last edited by l.lyle on 03-02-2012 at 03:13 AM

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Yes, that's exactly what I'm saying. I know you love nothing more than to turn a thread into your own interpretation of reality, but the author of this thread said that you couldn't tell if they were split until you got there, as here's what Gary wrote...

quote:

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Originally posted by Gary E Redfern
If not obvious, how could you tell that maybe dogs A and B where not split first and dog C treed 2nd behind dog B?

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Registered: Jun 2003
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you can't just declare your dog is split. The judge still has some what of a job to do. Even if he is just a glorified score keeper. Handler C is who messed up. Should have known better

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Michael Ghorley

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quote:

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Originally posted by l.lyle
What you just described also applies when you can tell that dogs are split even a short distace apart .

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quote:

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Originally posted by deschmidt27
Yes, that's exactly what I'm saying. I know you love nothing more than to turn a thread into your own interpretation of reality, but the author of this thread said that you couldn't tell if they were split until you got there, as here's what Gary wrote...



...and here's how the thread starter, Mtn Hunter, replied:


And under these circumstances, only you think C had the coon. Every other rationale being on this thread believes B had the coon on a split tree, and C covered him, and the handler of C made the mistake of treeing after the five was up prior to getting in there and seeing what was going on. You still think C was the coondog, when as far as anyone can tell, he was a Johnny Come Lately.

David Schmidt

David Schmidt

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quote:

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Originally posted by deschmidt27
Actually no... that's a completely different situation! If Dogs A and B are treed over here and Dog C trees over there, and when we get there Dog B isn't here but is now there, that's as simple to score as everyone else finds this scenario.

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quote:

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Originally posted by Mtn Hunter
As per the March coonhound advisor, a handler can call his dog treed on a closed tree even when a split tree is not obvious. If dogs A and B are declared treed in that order and the tree is closed, then a handler declares a split tree for dog C. When you get there dogs A and C are treed seperately and Dog B is on Dog C's tree. How is dog B scored?

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Last edited by l.lyle on 03-02-2012 at 04:07 AM

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quote:

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Originally posted by Mtn Hunter
As per the March coonhound advisor, a handler can call his dog treed on a closed tree even when a split tree is not obvious. If dogs A and B are declared treed in that order and the tree is closed, then a handler declares a split tree for dog C. When you get there dogs A and C are treed seperately and Dog B is on Dog C's tree. How is dog B scored?

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Registered: Jun 2003
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It says that a split tree was not obvious. What you think that's saying then? To me its saying they are right there together and sounding treed in the same spot.

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Michael Ghorley

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Registered: May 2009
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quote:

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Originally posted by Mtn Hunter
As per the March coonhound advisor, a handler can call his dog treed on a closed tree even when a split tree is not obvious. If dogs A and B are declared treed in that order and the tree is closed, then a handler declares a split tree for dog C. When you get there dogs A and C are treed seperately and Dog B is on Dog C's tree. How is dog B scored?

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UNITED WE STAND DIVIDED WE FALL
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CH 'PR' Grady's Dark Woods Waylon -Bluetic

NITECH 'PR' Grady's Insane Tinker Bell (Tink) - Treeing walker --Okla. State Hunt open redg. winner

'PR' Grady's Barley - Treeing Walker

Last edited by Okie Dawg on 03-02-2012 at 05:27 AM

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quote:

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Originally posted by Jeff M.
A & B were declared treed the 5 is run and expired...C starts treeing his options are be declared treed for 125 in which he must not be on A or B's tree as it or they are dead he has to be by himself..So unless he is sure his dog is not treed with either dog his best bet would have been to do nothing... walk in.. if his dog is on either tree and they had a coon he would only get strike minus.. if it is slick or off game he gets next available tree and strike minus..if it a circle tree he gets his strike circle.. if C is split when they get there tree him then.. in the case of a 3 dog cast all dogs are treed and he can handle his dog doesn't have to wait the 5.. he put himself in this low percentage position by not treeing before the 5... if he is split with a coon he is a hero if he just covered the dogs he's a zero.. he had nothing to gain by declaring his dog treed.... if he is split wait until you get there and then tree him!!!

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UNITED WE STAND DIVIDED WE FALL
Grady Jarvis
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CH 'PR' Grady's Dark Woods Waylon -Bluetic

NITECH 'PR' Grady's Insane Tinker Bell (Tink) - Treeing walker --Okla. State Hunt open redg. winner

'PR' Grady's Barley - Treeing Walker

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quote:

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Originally posted by Okie Dawg
I would say A and B get 125 on tree points each and C is shut out on both trees. A and b might had been split to begin with. If the handler of C didn't hear A or B move then I would think he would have to acept the fact that A and B had the trees first and C just came in to one to late.

I think the big misconception is he calls it C's tree and it's not. Once you get there C just treed to late on dog B's tree.

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Last edited by l.lyle on 03-02-2012 at 06:02 AM

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quote:

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Originally posted by Jeff M.
I agree completely with deschmidt... hunting around here you can have alot of split tree's but there is no way to tell it until you get there.. I would score it.... Dog A 125 +/- Dog B 125 +/- Dog C 125- Since dog A & B were not obviously split the time was run as one tree.. then changed to a split tree upon cast arrival with B moving up... and since C didn't tree before the 5 and was not obviously split from A & B and was found to be treeing on either tree he gets minused.. Dog C 's handler shot himself in the foot by not treeing before the 5 and then treeing after the 5 without being 100% sure his dog was split and convincing the judge of this before the cast took one step towards the treed dogs!!.... the key words to this whole scenario is OBVIOUSLY SPLIT TREED!! which apparently they weren't...... and if all the dogs were were moving back and forth from tree to tree and no one in the cast noticed and called attention to it... so be it... it wasn't noticed before the cast arrived so it officially never happened... Jeff

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From what I got out of Allen's post on these situations you would declare the first tree ( where dog A is ) to be the ROOT tree and go from there. C has to be on a split tree from the root tree so using that A has 125 , B 75 minus for moving and C 125.
Allen said it may not be perfect but since you will never know for sure you have to use a consistent method of scoring it.

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Dale - I'm fairly confident that Allan was not referring to a situation where the dogs are so close you couldn't tell that they were split treed. Otherwise, every time you walked in and found a dog treed 20 feet away, you would have to minus him. Or the same with a loud dog that's just a bit deeper than the rest, but straight behind the tree.

Okie Dawg - yes, I would agree that a judge could start the 5 minutes on a dog that starts treeing without being called, but in this case you will be a the trees involved, before that 5 is up and it will be evident that he was already treed in the vicinity as the other dogs. If he hadn't been called, you simply give him next available and score it appropriately.

Most everyone recognizes "the fact" that if nobody ever mentioned that some dog was seen or heard moving, then you can't assume that was the case, when they're so close that a split is not obvious until you get up on them. l.lyle is willing to ASSUME that Dog B moved with absolutely no proof or mention of it, but is unwilling to assume that Dog C simply got there late. When the latter is all that you can assume, without proof to the contrary!

What's amazing is virtually all of us, quickly made the same accurate (based on what we know and don't assume, like Dog B maybe moving) judgement call, but a day later, someone else still can't get that straight! He must be a treat to draw out with, when accurate decisions must be made quickly, based on what can and can't be proven!!!

David Schmidt

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Let's start back over with the basics. Let's forget about C for right now. Same Scenario but C is NOT there. It's not obvious that A & B are split and it's not obvious that B moved. You find them split when cast arrived at the tree.

How do you score A and B tree points?

A 125
B 125

Right?

Now that A&B's tree points have been adjusted on the card, you can now bring C back into the equation. Since C was found to be treed on a closed tree, he is minused 125 for moving per the rule 4k.

woulda, coulda, shoulda, don't mean squat.

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Registered: Jun 2008
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Absolutley right... it seems to be that the only guessing and assumptions are l.lyle assuming that Dog B must have moved.

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QUOTE:
Why in the world Would a dog thats in for second place on the card and we have to "accept the fact" What fact? That A&B had the first tree, singular, not trees plural. So now it's C's fault who just calls his dog like what the dog does because he doesn't go to calling the B dog that Moved. Maybe this thing has got more to do with calling somebody elses dog than there own dog? Sounds like it to me. B calls B and shuts his mouth about C . C calls C and better keep his mouth shut about calling my tree swappingdog B. The miss conception is they walk all the way up to the tree and wonder where dog B is. Give me a break ! Is the whole cast deaf dum blind and crosseyed? Like you said, "A and b (Might have been) split tree to start with. Mighta maybe shoulda coulda Iffn , but she was on the card for 75 , fact period............

PLEASE, PLEASE, LORD!!! don't ever let me draw this dude.. I promise I'll change my ways!!!!!!

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quote:

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Originally posted by WEBBER
Let's start back over with the basics. Let's forget about C for right now. Same Scenario but C is NOT there. It's not obvious that A & B are split and it's not obvious that B moved. You find them split when cast arrived at the tree.

How do you score A and B tree points?

A 125
B 125

Right?

Now that A&B's tree points have been adjusted on the card, you can now bring C back into the equation. Since C was found to be treed on a closed tree, he is minused 125 for moving per the rule 4k.

woulda, coulda, shoulda, don't mean squat.

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UNITED WE STAND DIVIDED WE FALL
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CH 'PR' Grady's Dark Woods Waylon -Bluetic

NITECH 'PR' Grady's Insane Tinker Bell (Tink) - Treeing walker --Okla. State Hunt open redg. winner

'PR' Grady's Barley - Treeing Walker

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