Can Takahashi Close The Gap On Patrick Chan?

[SkateFiguring]

Here is the question: Can Patrick Chan win with multiple falls over a perfect Daisuke?

I've said before, it depends on the falls. I can see maximum a random fall and a fall on a fully rotated jump and everything else as perfect as Dai. 5 points is about the difference in their LP BV which would be what such falls would wipe out. Actually, the GPF LP is an example of that, with one point to spare for a random fall which however would likely affect the PCS as past examples have shown. The perfect Dai must also include a successful 4T+3T in the SP as he did at the Nats, which Patrick does quite regularly.

All in all, I can see Patrick possibly winning with one fall, no worse than on a fully rotated quad. Anything more or additional flaws elsewhere would almost certainly put him behind a perfect Dai with his current BV.

[skatinginbc]

Here is the question: Can Patrick Chan win with multiple falls over a perfect Daisuke?

Actually, the more practical question would be: Can Daisuke produce two perfect programs?

[Boeing787]

Actually, the more practical question would be: Can Daisuke produce two perfect programs?

It's going to be very hard. Last time he did so (with programs of such level of difficulty) was back in 2008 4cc. Even for Patrick it's not an easy task, simply because programs today are so complex and difficult.

[Mathman]

[url=http://www.youtube.com/watch?v=cyu8GYBRIjE&t=1m36s]

3A+1Lo+3F

Under CoP scoring Plushenko's first three jumping passes (4T+3T+3Lo, 4T, 3A+half-loop+3F) have a base value of 43.4.

The current scoring has already been tremendously successful in encouraging quads.

Do you mean in comparison to the 6.0 ordinal system, or in comparison to the CoP of two years ago?

[ImaginaryPogue]

Which they did and rightfully so.

I don't know. The win they deserved wasn't the one they got in the end.

[Mathman]

Well, that's what the debate is about. A fall on a quad his deemed different than a fall on a double axel, which is different than a fall while stroking. I think that's a fair and accurate take on figure skating. Others do not.

I think this is kind of a 6.0-ish way of looking at things.

In the IJS you don't start with the base value of your program and then go up or down from there. You start at 0 and get points for each thing that you do.

Under the "0 points for a failed element proposal" if you fall on a quad attempt, you lose nothing. You also gain nothing.

[ImaginaryPogue]

My issue, Mathman, is that no one's willing to follow the "0 points for a failed element" proposal to it's logical conclusion. If they were, it would be a different debate.

[DianaSelene]

Under CoP scoring Plushenko's first three jumping passes (4T+3T+3Lo, 4T, 3A+half-loop+3F) have a base value of 43.4.



Do you mean in comparison to the 6.0 ordinal system, or in comparison to the CoP of two years ago?

Too bad he can't do that today. He'd beat Chan in no time.

[DianaSelene]

I don't know. The win they deserved wasn't the one they got in the end.

It is too bad the scandal ruined their victory.

[Boeing787]

Too bad he can't do that today. He'd beat Chan in no time.

Really? COP limits the number of jumps a skater can do. After these jumps, there are not many jumps he can do, and his SS and TR marks won't be as high as Patrick's. He might still beat Patrick, but not "in no time", not that he'd have the gold in his bag.

eta: under COP, he can only do one three-jump combination. so the 3A+1/2Lo+3F has to go. Oops, his chance of beating Patrick gets even slimer.

[Mathman]

My issue, Mathman, is that no one's willing to follow the "0 points for a failed element" proposal to it's logical conclusion. If they were, it would be a different debate.

Let's go back to this

How about this:

No credit system

4T (solid, +1 GOE)
2A (fall)
3T (fall)
3S (fall)

Total: 11.3 points

4T (fall)
2A (0 GOE)
3T (0 GOE)
3S (0 GOE)

Total: 11.7 points

Is that what you feel would be an accurate representation of what you want to see.

Yes, I would be satisfied that the person who did three easy imps and fell on his quad should beat the skater who did a successful quad but fell three times later in the program on easy elements. I would be surprised if anyone felt otherwise, under any system.

But if the question is really what I want to see, I would want to see those two guys competing for the bronze in a four-man field, while the winner did 4T, 3A, 3Lz+3T (28.1) and the silver medalist did 3A+3T, 3Lz, 3Lo (22.9).

Having said that, however, I don't think this proposal will ever fly. The stumbling block that beats it is, what to do in the case of a hands-down sort of half-fall.

[seniorita]

Too bad he can't do that today.

On practice in Euros he did 3sal -a sth hop -3axel, I can send you my video I think he could do the old combo but he doesnt do the flip anymore. I used to like his flip, it was not a huge as Joubert's but it looked effortless. And I dont know if it worths this combo now.

I m reading this thread, and I have a feeling we will jinx Takahashi and he will end behind other skaters, and not only Chan :(

[ImaginaryPogue]

yes, i would be satisfied that the person who did three easy jumps and fell on his quad should beat the skater who did a successful quad but fell three times later in the program on easy elements. i would be surprised if anyone felt otherwise, under any system.

... Surprise!!!!!

[Mathman]

Really? COP limits the number of jumps a skater can do. After these jumps, there are not many jumps he can do, and his SS and TR marks won't be as high as Patrick's. He might still beat Patrick, but not "in no time", not that he'd have the gold in his bag.

eta: under COP, he can only do one three-jump combination. so the 3A+1/2Lo+3F has to go. Oops, his chance of beating Patrick gets even slimer.

That is a good point. The CoP is quite unfriendly to jump combinations. As Hurrah has been arguing on this thread, the only thing you gain by doing a triple triple of any kind is that you get to score an extra double Axel at the end.

Even allowing Plushenko two three-jump combinations, he could do only something like 4T+3T+3Lo, 4T, 3A+half-loop+3F, 3A+2T, 3Lz, 3S, 2A, 2A -- which I think is pretty close to his actual jump card for that program. (Lysacek could beat him by doing all of his jumps in the second half. )

... Surprise!!!!!

I am indeed! Your guy attempted four jumps and fell three times! He fell on a double Axel, for goodness sake. Who is he, Brezina at 2011 Skate America?

[ImaginaryPogue]

Actually, I read your post wrong. I agree with you. My error.

I'll just point out that the partial credit system, however, with the -1 deduction, means that instead of 0.4 point separating them, the three fall skater has 14.2 points. The one-fall skater has 19.0. I think that's better.