What is the max base value of a ladies FS?

[Dan]

I have always wanted to determine the maximum base value for a ladies long program without a quad or triple axel. If such a number existed, skaters and coaches could use this as a benchmark to help determine how good their choices for their program are. Of course, this isn't the only benchmark for this but it is one that never gets talked about. However, I start to do the math and my head explodes. Can anyone help me answer this question from a shear math point of view?

To simplify things I am going to use what most of the ladies are doing today. Not the superstars of the sport but most ladies. Thus, I assume for this question that only loops and toes can be the second or third jump of a combination. And again I'm going to assume for this calculation that a triple axel or quads are not allowed.

The ladies do 12 elements in their free skate which consist of 7 jumping passes, 3 spins and 2 sequences. The spins and sequences part of this seems easy. As I can't find in the scale of values table a spin higher than 3.5 base value or sequence higher than 3.9 base value that means the maximum base value of these five elements is (3 * 3.5) + (2 * 3.9) = 10.5 + 7.8 = 18.3

Now where I get stuck is finding the max base value that results from arranging the six types of jumps (axel, toe, loop, salchow, flip, lutz) into seven jumping passes under Zayak. Zayak says that you can only attempt a triple or quad jump twice and AT LEAST one of these attempts must be in combination. Both can be in combination if you want. I could just play with combinations in a trial and error way and see what comes up. However, it seems like a mathematician would have a better way to determine a max than trial and error.

I suspect that coaches and skaters just do trial and error with the math along with what they know they can do consistently to determine the content of their jumping passes. However, if they knew what the max was, they could get a feel for how well they have chosen this content. To use an example, let's make up a number for the max base value of the jumping. I am randomly going to choose 52. I made this up since single jumps are around 6, two jump combinations with a double are around 8 and three jump combinations are around 10 base value. Then with 18.3 for the spins and sequences the max base value of most of the skaters today would be 52 + 18.3 = 70.3. Then a skater like Medeveda who doesn't have a triple axel or quad could see that her base value of 62.33 was pretty good but could garner 8 more points of base value if she changed what she did. Does anyone else see the "value" (pun intended) in knowing what the max base value could be?

Now, my favorite question for you...Does more than one mix of jumps give you this highest base value?

 

[caelum]

I'm a mathematician! (well almost, haven't quite finished the PhD yet) and "trial-and-error" is (probably) going to be best solution!

As far as mathematically, this problem is very similar to linear integer programming (and could be formulated as one if we wanted). In general, these types of problems are NP-hard, which is a fancy way of saying no efficient methods (polynomial time) exists for finding a solution (and most complexity theorists strongly suspect one doesn't exist, though this is unconfirmed). This may or may not be that difficult depending on the specific constraints. I haven't worked out the mathematics of it, my area of research is not in optimization and its been forever since I've touched linear programming, but I suspect this problem is actually quite difficult in an NP sense. However, the constraints of this problem (e.g. jumps have to sum to at most 11) make the solution space small enough in this problem that a computer could be easily programmed to check all possible outcomes and report back the correct answer. I might do it in the next week if I find the time. Honestly though, the best way to construct the answer would probably be heuristically - i.e. we know two triple lutz's are going to be part of the solution and build it from there.

 

[Dan]

I'm a mathematician! (well almost, haven't quite finished the PhD yet) and "trial-and-error" is (probably) going to be best solution!

As far as mathematically, this problem is very similar to linear integer programming (and could be formulated as one if we wanted). In general, these types of problems are NP-hard, which is a fancy way of saying no efficient methods (polynomial time) exists for finding a solution (and most complexity theorists strongly suspect one doesn't exist, though this is unconfirmed). This may or may not be that difficult depending on the specific constraints. I haven't worked out the mathematics of it, my area of research is not in optimization and its been forever since I've touched linear programming, but I suspect this problem is actually quite difficult in an NP sense. However, the constraints of this problem (e.g. jumps have to sum to at most 11) make the solution space small enough in this problem that a computer could be easily programmed to check all possible outcomes and report back the correct answer. I might do it in the next week if I find the time. Honestly though, the best way to construct the answer would probably be heuristically - i.e. we know two triple lutz's are going to be part of the solution and build it from there.

Thank you for your answer. I was an electrical engineer and had minor in math but it was long before Michelle Kwan was born so I'm not sure we speak the same language. What does NP-hard mean? I think by linear programming you mean that you write a program that will look at all possible combinations which is just a more sophisticated heuristic, trial and error method, isn't it? From experience we know that the better ladies are routinely getting 62 or more in base value in the FS and that almost always involves 3Lz + 3T so that should bound the problem a little bit more. If you are a PhD candidate, shouldn't you have some underclassmen as students of some kind? Assign the problem to them!!!! What is your PhD theses? I'm am always intrigued by the complex, detailed areas people investigate these days.

 

[caelum]

Thank you for your answer. I was an electrical engineer and had minor in math but it was long before Michelle Kwan was born so I'm not sure we speak the same language. What does NP-hard mean? I think by linear programming you mean that you write a program that will look at all possible combinations which is just a more sophisticated heuristic, trial and error method, isn't it? From experience we know that the better ladies are routinely getting 62 or more in base value in the FS and that almost always involves 3Lz + 3T so that should bound the problem a little bit more. If you are a PhD candidate, shouldn't you have some underclassmen as students of some kind? Assign the problem to them!!!! What is your PhD theses? I'm am always intrigued by the complex, detailed areas people investigate these days.

Sometimes I take for granted mathematical language! Linear programming is just a type of problem where we are trying to find the maximum of sum quantity (in this case the maximum possible sum value of jumps) alongside some constraints that are linear (for example, the # of jumps is less than or equal to 11) Linear programming problems are relatively simple to solve actually with efficient algorithms and don't really involve much trial-and-error, so that's not what I mean by linear programming. The difference here is we are restricted to discrete values, which makes it like integer linear programming. What I mean is, the allowed values for our jumps are the discrete set 6.0, 5.3, 5.1, 4.4, 4.3, 3.3 etc, I can't use 5.6 since that's not an allowed value. This difference complicates the problem enormously and the usual algorithms go out the window. While this specific case might be simpler, linear integer programming is, in general, NP-hard. NP-hard is something specific, but what NP means is that no algorithm currently exists that can solve NP problems in polynomial time. Polynomial time means that if our input size into the computer is "n" (where n is a number) then we can solve the problem if the number of steps required to solve the problem is of order n^k where k is a non-negative integer. NP problems are problems that this is not true for. We call these "inefficient" because NP problems generally aren't solvable quickly (and we suspect that these problems cannot be though this hasn't been proven, this is the famous P=NP problem you may have heard of). I will also point out the known NP algorithms for solving restricted integer programming problems are terrible in the sense of computational speed. I will point out there are certain restrictions that allow integer programming to be solved efficient, but I think there are certain restrictions that would prevent this for this problem (it's a really technical detail), and even so it would be a slow algorithm. But my general point was, that the problem is likely to be difficult from a computational perspective and any available algorithm won't sufficiently speed-up the process relative to just an informed trial-and-error solution.


I would think a "3Lz + 3Lz" would be the best lutz combo, but that's almost impossible to perform since who has lutz's in both directions and could pull that off? Perhaps the best "realistic" combination is "3Lz + 3Lo" since you could then do two lutz's and two loops.

Anyway, I know someone who is an expert in optimization in the statistics department who might give me some information about this problem.

 

[Dan]

I would think a "3Lz + 3Lz" would be the best lutz combo, but that's almost impossible to perform since who has lutz's in both directions and could pull that off? Perhaps the best "realistic" combination is "3Lz + 3Lo" since you could then do two lutz's and two loops.

Anyway, I know someone who is an expert in optimization in the statistics department who might give me some information about this problem.

Wow, that is an excellent explanation that I do understand much better. Thank you so much. The base values for all the jumps can be found on page two of this document. http://static.isu.org/media/207718/1944-sptc-sov-communication-2015-2016.pdf

I think you should use your obvious good communication skills to persuade someone to assign this to a class of students. The one with the best answer/problem solving method gets to help a skater chose the jumps for their program in the 2016-2017 season. Can you imagine what some of the old guard in skating would say to mathematicians being involved in creating a program? I think Frank Carroll's head would explode, Tarasova would lose her fur coat and Mr. Nicks would simply be rendered without voice for the remainder of the decade. Of imagine what this would mean to skaters that obsess about perfection? They could say their program was X number of points from the biggest possible base value as determined by students at such and such university.

 

[Dan]

Now where I get stuck is finding the max base value that results from arranging the six types of jumps (axel, toe, loop, salchow, flip, lutz) into seven jumping passes under Zayak. Zayak says that you can only attempt a triple or quad jump twice and AT LEAST one of these attempts must be in combination. Both can be in combination if you want. I could just play with combinations in a trial and error way and see what comes up. However, it seems like a mathematician would have a better way to determine a max than trial and error.

I think that I have the answer. These seven jumping passes allow for all triple jumps to be repeated twice in combination according to Zayak. The jumps that aren't triples will have to be the most difficult doubles to pull off in combination with the other jumps. Right now I think that is the 2Lo. I don't know anyone that does a 2S as the second jump of a two jump combination but if they did, that would increase the result I have here by an additional 2.1 points.

3lz+3t = 6 + 4.3 = 10.3
3lz+3t = 6 + 4.3 = 10.3
3f+3lo = 5.3 + 5.1 = 10.4
3f+3lo = 5.3 + 5.1 = 10.4
3S+2lo = 4.4 + 1.8 = 6.2
3S+2lo = 4.4 + 1.8 = 6.2
2a+2lo = 3.3 + 1.8 = 5.1

This totals 58.9. If we add that to the 18.3 for the spins and sequences then the max base value in the FS that we could expect from our ladies under the assumptions that I have made is 77.2. Does anyone agree? Does anyone want to do a similar calculation assuming the skater can do a 3A? Is this a useless exercise?

 

[ruffledgrouse]

I think that I have the answer. These seven jumping passes allow for all triple jumps to be repeated twice in combination according to Zayak. The jumps that aren't triples will have to be the most difficult doubles to pull off in combination with the other jumps. Right now I think that is the 2Lo. I don't know anyone that does a 2S as the second jump of a two jump combination but if they did, that would increase the result I have here by an additional 2.1 points.

3lz+3t = 6 + 4.3 = 10.3
3lz+3t = 6 + 4.3 = 10.3
3f+3lo = 5.3 + 5.1 = 10.4
3f+3lo = 5.3 + 5.1 = 10.4
3S+2lo = 4.4 + 1.8 = 6.2
3S+2lo = 4.4 + 1.8 = 6.2
2a+2lo = 3.3 + 1.8 = 5.1

This totals 58.9. If we add that to the 18.3 for the spins and sequences then the max base value in the FS that we could expect from our ladies under the assumptions that I have made is 77.2. Does anyone agree? Does anyone want to do a similar calculation assuming the skater can do a 3A? Is this a useless exercise?

It's two in the morning so I might be misunderstanding your post, but you're not allowed to have seven jump combinations in you're free skate. IIRC, you're allowed 1 3-jump combo, 2 2-jump combos, and 5 solo jumps. Again, sorry if I totally misread this thread.

 

[gkelly]

It's two in the morning so I might be misunderstanding your post, but you're not allowed to have seven jump combinations in you're free skate. IIRC, you're allowed 1 3-jump combo, 2 2-jump combos, and 5 solo jumps. Again, sorry if I totally misread this thread.

Four solo jumps (and 3 combos as you detail) in a ladies freeskate -- 7 jump elements possible, 11 total jumps.

One of those jumps must be an axel-type jump (single, double, or triple).

Only 2 triples may be repeated and must be in combination at least one of those times to receive full credit.

Putting as many of those jumps as possible after the halfway mark increases their base value.

Does including all of those limits and requirements make the problem less linear?

We're not looking for the highest single element, but the highest total jump layout.

 

[rollerblade]

Oh, I'm not a mathematician but I wanna play. It's actually very simple! Approach it like you're doing a large jigsaw puzzle - sort them into segments.

Note: Satoko actually has the highest BV of the ladies this season, followed by Elena, then Evgenia. Very small differences, due to the ladies' choice of jumps to repeat.
Satoko repeats 3Z/3T/2A, Elena repeats 3Z/3L/2A, Evgenia repeats 3F/3T/2T. (Technically Elena's repeats are worth more, but she backloaded the "wrong" jumps.)

-----------------------------------------------------------------------------------------------------------------------------------------------------------------

Let's start with Spins:
The highest scoring spin is worth 3.5 points each ==> CoSp3p4 and FCoSp3p4 (Combination spin, with change foot and 3 change of positions. The "F" stands for flying.)
The next highest scoring spin is worth 3.2 points each ==> there are several choices here, too lazy to list them all

They cannot repeat the same spins, right?
If that's the correct assumptions, then 3.5 + 3.5 + 3.2 = 10.2 points from all Spins combined.
(But based on what both Russian ladies did, it was a 3.5 + 3.5 + 2.7 = 9.7) <---- I need clarification about spin rules.


Next up, Step and Choreographic Sequences:
This one is simple. One of each.
3.9 + 2.0 = 5.9 points for Sequences.


Now the most complicated, Jumps:
3A = 8.5 / 3Z = 6.0 / 3F = 5.3 / 3L = 5.1 / 3S = 4.4 / 3T = 4.3
2A = 3.3 / 2Z = 2.1 / 2F = 1.9 / 2L = 1.8 / 2S = 1.3 / 2T = 1.3

Since the OP rejected 3A (but why?!) for his hypothetical max BV, let's just go down the list. Of course you'd want to repeat the highest scoring jumps, so repeating 3Z and 3F is a must.

With 11 jumps total, the just list becomes:
3Z / 3Z / 3F / 3F / 3L / 3S / 3T / 2A / 2A / 2Z / 2F ---> a BV of 47.0 (ignore 10% bonus)

However, 2Z & 2F are kinda hard to put in a combo. When Liza did her 3S+2A combo, it was marked as 3S+2A+SEQ.... which was worth less points than the sum of 3S and 2A. For this reason, I'm modifying the jump list to:
3Z / 3Z / 3F / 3F / 3L / 3S / 3T / 2A / 2A / 2L / 2T ---> a BV of 46.1 (ignore 10% bonus)

Tuning #1:
3F+3L ................ 10.4
3Z+2L ................ 7.8
3S+3T+2T .......... 10.0
3Z .................... 6.0
3F .................... 5.3
2A .................... 3.3
2A .................... 3.3

The combos can be cherry-picked, I just made combos that seem the most reasonable. There's no brownie points for performing implausible combos like 3Z-3Z, unless GOE comes into play... but we're not here to discuss GOE.

Tuning #2:
All that's left to do is BACKLOAD! Simply reverse the list above.

2A .................... 3.3
2A .................... 3.3
3F .................... 5.83 x
3Z .................... 6.6 x
3S+3T+2T .......... 11.0 x
3Z+2L ................ 8.58 x
3F+3L ................ 11.44 x

And we get 50.05 points for all Jumps.


FINALLY! Time for easy peasy 1st grade math, add 'em up!

If my assumptions on spins are correct, we get 10.2 + 5.9 + 50.05 = 66.15
Going by the spins of Russian ladies, we get 9.7 + 5.9 + 50.05 = 65.65

- Just for comparison: Satoko's BV is 62.96, Elena's 62.50, Evgenia's 62.33. Not that far off, really!
- GOE's is where Evgenia gets her advantage.
- 3A matters. Mao's Sochi FS BV was 66.34. If she gets no UR/e deductions, using today's value, it would have been 69.38. And that's not even "maxing-out" the potential BV.

TLDR: I can't believe I just wrote my analysis for barely a 3 point increase. Let me know if my math is off.

 

[gkelly]

Also, for any given skater the optimum jump layout would not necessarily be the maximum base value, because you also have to take into account the likelihood of 1) best possible GOE and 2) actually rotating all the jumps as planned. And also meeting judges' expectations for unity and proportionality in the layout of the program -- backloading will get you more jump points but might lose some points in the Choreography component.

Different individuals have different best jumps/jump combination skills, so what's optimum for one skater might not work for another. And they're all subject to fatigue as the program progresses and lactic acid builds up, which is why there is a bonus for doing jumps later in the program and GOEs tend to be better for the earlier successful jumps.

Oh, and double jumps including double axel can only be done twice each, but they don't have to be in combination and you can repeat as many different doubles as you want.

Some possibilities:

3Lz+1Lo+3F
3S+3Lo
2A+3T
3Lz
3F
2A
maxed out on difficult jumps -- the remaining jump could be 2Lz

3Lz+3T
3F+3Lo
2A+2Lo+2Lo
3Lz
3F
3Lo
3S

3Lz+3Lo
3S+3T+3Lo
2A+2Lo+2Lo
3Lz
3F
2A
2Lz

 

[petitebrie]

Well, only two types of triple jumps can be repeated. If you start out with two 3Lz+3T combos, you cannot repeat any other triple throughout your program. You’re also limited to three combinations, leaving four solo jumps. (As far as I know. There are probably some scoring rules I'm missing here, so please correct me if I'm wrong!)

With that in mind, I'll give it a go:

The two most valuable jumps to feasibly repeat are Lutz and flip. You could do a solo 3Lz, a 3Lz+2Lo (making a combination without ‘occupying’ a triple), a solo 3F and a 3F+3T, then add a 3S, 3Lo, 3S and the two permitted 2Axels. Combinations are counted as jump value + jump value, so moving jumps around wouldn't make much of a difference (e.g. 3Lz+3T, 3F+2Lo, 3Lo+1Lo+3S and two solo 2A). For example (in no particular order):

1. 3Lz+2Lo (6 + 1.8 = 7.8)
2. 3F+3T (5.3 + 4.3 = 9.6)
3. 2A+1Lo+3S (3.3 + 0.5 + 4.4 = 8.2)
4. 3Lz (6)
5. 3F (5.3)
6. 2A (3.3)
7. 3Lo (5.1)

Theoretically you’d max out your values if all jumps were done in the second half, but that probably isn’t physically feasible. So I’ll forget the 10% bonus for now, even though this is something skaters do use to bring up their BV, putting as many as five or six jumping passes in the second half.

This makes 45.3 points for jumps. Adding your 18.3 estimate for non-jump elements gives 63.6, which actually isn’t far away from the highest BVs you see in the wild (Evgenia Medvedeva and Elena Radionova both had BVs over 62 at Euros this season, much due to the 10% second half bonus). Mao Asada has a 3A and repeats 3F and 3Lo, which should give her one of the highest theoretical BVs. High GOE is what gives some ladies skaters over 70 in TES.

What if you’re some otherworldly prodigy who gets +3 GOE on all elements? I’m not going to pull up the scale of values to check manually, but if someone has access to the scoring sheet judges use, it would seem easy enough to just plot in the aforementioned technical elements, assign GOE and see what comes out.

(ETA: three new posts while I was writing this! This is fun. )

 

[rollerblade]

Ya sorry, I just saw I could have used a stonger combo, and fixed. It only added half a point, but still.... points are points!

Elena did two solo 2A in Europeans. It's why I had them there.

 

[rollerblade]

3Lz+3T
3F+3Lo
2A+2Lo+2Lo
3Lz
3F
3Lo
3S

3Lz+3Lo
3S+3T+3Lo
2A+2Lo+2Lo
3Lz
3Lo
3F
2A

Can they do that? You had THREE 3Lo in the 2nd scenario. And in both cases, THREE triples were repeated on top of the repeated 2Lo. I didn't dig up the rulebook, but deduced from GPF, EC, 4CC.... that you're only allowed to repeate TWO triples and ONE double. If they are allowed, then I have to edit my post again.

 

[gkelly]

Oh, I'm not a mathematician but I wanna play. It's actually very simple! Approach it like you're doing a large jigsaw puzzle - sort them into segments.

Yeah.

Let's start with Spins:
The highest scoring spin is worth 3.5 points each ==> CoSp3p4 and FCoSp3p4 (Combination spin, with change foot and 3 change of positions. The "F" stands for flying.)
The next highest scoring spin is worth 3.2 points each ==> there are several choices here, too lazy to list them all

They cannot repeat the same spins, right?[/quote]

Right.

There are three spins: one must be a combo spin, one must be a spin in one position, and the third must fly (could be either a combo or one position). In theory they could all fly. But it's more valuable to change feet and you can't do two FCCoSp.

Each spin must have a different code.

And you can only get credit for each level feature, including each difficult position, once in the program. Which features you do don't show up in the codes, but it does affect what combination of spins a given skater is best off planning depending which features she's best at.

If that's the correct assumptions, then 3.5 + 3.5 + 3.2 = 10.2 points from all Spins combined.

Yes.

(But based on what both Russian ladies did, it was a 3.5 + 3.5 + 2.7 = 9.7) <---- I need clarification about spin rules.

Did they all do LSp4 (regular layback with no fly or change of foot) as the spin in one position?

 

[rollerblade]

All 3 ladies went for LSp4 as their 3rd spin, yes. It's worth 2.70.

Mao did a FCSp4 as her 3rd spin in Sochi, it was worth 3.20. But I don't know if the spin rules have changed since 2014.

 

[gkelly]

The requirements haven't change since then. I think some of the base values may have. FCSp4 is still worth 3.0.

 

[petitebrie]

Summarising new information, then. HIghest value for StSq4 + ChSq + three required spins is 16.1, not 18.3 as OP first suggested.

A revision to my jump content suggestion:

3Lz+3Lo = 11.1
3F+3T =9.6
3S+2Lo+2Lo = 8
3Lz = 6
3F = 5.3
2A = 3.3
2A = 3.3

= 46.6 BV

Interestingly, a lot of people are talking about 3+3 combinations, but they’re really only an advantage in the SP. In the scenario above you could rearrange the jumps into 3Lz+2Lo, 3F+2Lo, solo 3S, 3Lz, 3F, 2A and a 2A+3T and retain the same BV. AFAIK it would only garner more points in the FS if you put the 3+3 combos in the second half.

One rationale might be that practicing the 3+3 for the SP should give you a higher success rate on that combination, in which case it makes sense to put it in the FS as well. A case for not doing 3+3: if you make a mistake you'll suffer less from the negative GOE. But doing two 2Lo AND a 3Lo makes you susceptible to Zayaking if you pop the 3Lo.

Backloading is obviously inevitable. For general modelling purposes I just took half of the total sum above and add 10%, giving 48.93. Add 16.1 = 65.03.

A 3A is worth 5.2(!) points more than a 2A, so to replace a 2A with a 3A just add 5.2 to the score = 70.23. And my example is finally over 70!

 

[Dan]

I think part of the discussion here is that we all understand the Zayak rule differently. What made me create this thread was my new understanding of the Zayak. I thought like some of you did that only 11 jumps were possible in the seven allowed jumping passes. Gkelly posted with this understanding. 4 jumps without combo, 2 two-jump combos and 1 three jump combo. This is not the case. This is just what most skaters are doing right now. Some other rule besides Zayak limits you to seven jumping passes but it doesn't specify limitation in those jumping passes like this. However, the Zayak rule states that you can only attempt a triple twice in one program to receive their full value. In the FS, you can do as many doubles as you want. I think the short is different starting this year regarding doubles. Let's stick with the FS. You can attempt the same triple jump twice and AT LEAST one of those has to be in combination. This doesn't prevent you from doing both attempts in combination. The key word here is AT LEAST. Also, you could do it a third time if you wanted but if you do you receive 70% of the base value. So I am sticking with my previous max base value of 77.2 based on these 7 jumping passes. You could increase them slightly by adding 2Lo to the end of each pass bumping up the total base value by (7 * 1.3) = 9.1 but now we have gone from unlikely passes to highly unlikely passes.
3lz+3t = 6 + 4.3 = 10.3
3lz+3t = 6 + 4.3 = 10.3
3f+3lo = 5.3 + 5.1 = 10.4
3f+3lo = 5.3 + 5.1 = 10.4
3S+2lo = 4.4 + 1.8 = 6.2
3S+2lo = 4.4 + 1.8 = 6.2
2a+2lo = 3.3 + 1.8 = 5.1

 

[Dan]

Since the OP rejected 3A (but why?!)

I wanted to select from jumps that MOST of the ladies are doing today. Mao Asada and one other lady are the only ones attempting a 3A. I think I have seen one lady that attempts a quad salchow but other than these I don't think the ladies are there yet consistently. I also didn't consider jump combinations that seemed unlikely such as a 3Lz+3Lz combo for the same reason.

 

[Dan]

They cannot repeat the same spins, right?
If that's the correct assumptions, then 3.5 + 3.5 + 3.2 = 10.2 points from all Spins combined.
(But based on what both Russian ladies did, it was a 3.5 + 3.5 + 2.7 = 9.7) <---- I need clarification about spin rules.

I assumed that you could repeat a spin. I thought just changing the positions would work if you kept the flying and the change of foot, etc. Thus my 77.2 max BV calculation based on 3.5 + 3.5 + 3.5 = 10.5 I couldn't find anything about repeating spins so like rollerblade, I would also appreciate anyone that can clarify this for us. I love your enthusiasm for this question!!!!!