Someone is paying attention

I <3 Tornadoes! only if they were a tad more affordable.

Quite a good read, even if not all the historical details are the same as I have read earlier. Calling the M20 a modern B class is stretching things a bit in my opinion but it is more or less a philosophical question. Good to see the ITA take initative!
The F-16 definately is on the radar in the catsailing community. Next thing is to become a well known class in the sailing community at large.

To do this we need to post on other sailing forums like SA and www.yachtsandyachting.com

But if we do post in the outside world, we need to be clear, consise, helpfull and accurate. No point going out and pissing everyone off.

I'm finding that F16 is named more often in the broader catamaran sailing scene, especially publications.

Indeed it looks like we are becoming a generally accepted class. Now we only have to make sure we also become an established class the world over. We are halve way there.

The work is not completed yet, but if we keep our noses clean and stick at it then there is no reason why we will not succeed at that as well.

Onwards ! Everybody.

Wouter

This is great to see.

From our prospective, there was a significant amount of internet chatter in the past about the F16. After the Alter Cup last year though, we began to really see the F16 start to become recognized as a real option. There are now in excess of 50 boats in the US, and even in areas where there are no F16s, the F16 class is now regularly listed with the F18, 20's etc as an option for a boat style/group/class. Apparently this is happening inernationaly as well.

No class or style of boat will fit everyone, and the F16 is definitely not an exception. This class of boat is definitely my favorite of the many designs I have sailed and raced in my lifetime. The feedback we get from virtually everyone who gets on one continues to be extremely positive. As such, to some extent the class sells itself becuase of the boat.

While we have not reached critical mass, with 4 builders (and more at least watching it), and boats reasonably spread across the globe, if we can continue the press and promotion of events I feel the class has a great future.

Regards,

Matt

Not too sure what to make of this comment -

---
A-Catamaran, F-18, B Class, etc. builders are not licensed
Anyone can build a boat... and change it without any notice
In the Olympics, €100000 custom boats would be inevitable
---

Are they contrasting the Tornado with less restrictive rules of the other classes (though I'm not sure that's true)? What are they suggesting would lead to €100000 custom boats?

Mark.

There are some more statements in that ITA section that may be considered strenious.

I wouldn't know how to spend €100.000 on a compliant F18 or F16 and make it show up in the performance beyond say a boat made for €25.000.

Alot of people think that money is the only influence factor for performance (more = faster) and that something that is custom made is automatically better. This is quite often based on misunderstanding the situation. One can only improve a product by having it custom made when the builder knows exactly how the product behaves with respect to even very small changes. This may be true for clothing, but this is certainly not true for sailboat design. Sailboat design still is to a large extend a black art, mixed with large amounts of personal preference and emperical knowlegde. It is impracticle, if not impossible, to achieve significant performance differences by custom designing and custom building a racing boat like the F18.

Look at how much money and time is spend on sail development and how little the steps of improvement really are ?

Wouter

Spending GBP100.000 should be no problem. Just invest in enough computer modelling and CFD analyzes. It dont have to make the boat faster, those writing the checks just need to believe it makes the boat faster.

To build boats for sale, you need to be a licensed builder. Otherwise, amateurs are (still) allowed to build one boat a year for personal use. I think they are arguing that

open

rules are bad for cost in the games while the tornado rules are

closed enough

to give everybody a fair chance?

What ? You can't spend 100.000 on exotic tornado snuffers and sail development if you tried ?

There is no reason why you can spend the entire fortune of Bill Gates on a single beach catamaran, but I really don't believe that that will lead to any significant (= measureable) performance gain over say reasonably well designed a 25.000 boat when limiting yourself to say a ruleset like the F18's.

Too many people believe that this is a magic thing. That one only has to put a computer to work on it and that machine will figure out for you what is best. I do alot of computer simulation on noise disturbed data (= real life data) and it simply is not that simple or even that accurate. Alot of people also believe that they are skilled designers when they know how to operate a finite element software package but in reality they are simply not. A very large portion of the designing is to be found in intepreting the computer generated results and quite often you'll see numerical glitches. It take understanding of the algorithms used to make sound intepretations on accuracy and dependability of the generated data. Simply generating a plot with some colourings in it is not accuracy or dependability. Even plots with margins of error are often better at producing a false sense of accuracy and dependability.

An to make things worse. The behaviour of a hull when sailed over a disturbed watersurface is too complex too model accurately in all its details. That means that stochastic analysis (random variables) must be applied to even be able to

run the numbers

on a computer. Stochastic analysis is what you do to predict the outcomes in a Casino and we all know that we can only assign probability levels to those outcomes and thus never be sure. In my experience, probability analysis is something that is widely misunderstood even by well educated engineers. And I must admit it is a very tricky subject that requires you utmost focus to not overlook details that can change the whole model or your results.

To give an example :

Proof which situation has the highest probability of being truthful or else proof that all situations are just as likely the happen :

-1- it is easiest to throw at least the number 6 ones when throwing the dice 6 times
-2- it is easiest to throw at least the number 6 twice when throwing the dice 12 times
-3- it is easiest to throw at least the number 6 three times when throwing the dice 18 times

And then the best part of stochastic processes it that you need to be able to produce a component far more accurately then the difference that was designed into the boat from a prior design. Else the natural variance in production overpowers your designed in difference. Then all that work is just pointless as you will have to produce 10's of boats and test them all to see which boat is actually reflecting what you designed. Much like how the top Laser sailors go through scores of boats and mast to find the best combination. That is OD design for you !

As a matter of fact just producing a [censored] load of models and find the best one by testing them all may actually be the most inexpensive option as you won't need highly accurate (expensive) production tooling and you can still sell the lesser products to part time racers and recreational sailor, thus getting a pretty sizeable return on investment.

Indeed, one can easily spend 100.000 on a golden doornob with platinum and diamond decorations, but in the end of the day it is just a doornob whose function is implemented just as well by a 1 dollar plastic handle we all buy at wallmart.

Wouter

I did not say I tought it made sense <img src=

alt=

/> We both remember the brit T alu-mast development project which was used as leverage to get the carbon mast ballot trough. But let us not tear apart what the ITA is trying to do.

I am not quite sure why the sailing community is so fixated on having limits and extensive restrictions to the equipment. This is definitely not an Olympic contraint. Look at all the money and design that goes into the Bobsled event. To carry the argument even further, I think we should make all the runners wear the same shoes, that would equal things up as well.

100K is easy to spend on a boat if you really want. Unless the event supplies the equipment and swaps out every race, there will be variation that is boat and not sailor related. Part of sailing though is fitting the sailor to the boat; Weight, sailing slyle, etc, etc.

The real question is how does this fit into this thread.... Sorry.

Matt

situation -1- Probability is : 1 - (5/6)^6 = 0,67
situation -2- Probability is : 1 - (5/6)^12 - 12*(1/6)*(5/6)^11 = 0,62
situation -3- Probability is : 1 - (5/6)^18 - 18*(1/6)*(5/6)^17 - 18!/2!*(1/6)^2*(5/6)^16 = 0,60

I had to use the symbol

!

to denote the mathematical operation of faculty, which means :

18! = 18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

As such the expression 18!/2! denotes

(18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1) / (2*1) = 18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3

Now some people will say, hey look the difference between situation -1- and -3- is quite large as it is comparing 0.67 to 0.60, we could have just found the answer by just simulating it (actually throwing the dice and counting the number of times the statements are satisfied !)

In reality the difference is really not that large and actually simulating the results would require 1000's throws to have the variance of the averaged experimental results be smaller then the probabilty difference of 0.67 - 0.60 = 0.07. It requires a good number more experiments to have a sifficiently high accuracy level assigned to your conclusion to make the conclusions dependable that one situation is more likely to happen then another WHILE ASSUMING LABORATORY LIKE CONDITIONS (tightly controlled) during the whole time you are simulating.

All that for a simple problem as stated above.

Now try the same statistical simulating strick in an environment where the conditions can change each 10 minutes and where the overal problem is a 1000 times more complex as we have in sail boat racing. Now try to discover which change to the design has actuall improved the performance of the boat. And now we know why the aging Hobie Tiger is still just as fast in capable hands as the nacra Infusion when both are using similar suits of sails.

Back to the original topic now.

Wouter

I understand what the ITA is trying to do.. Place itself at the fore-front of cat sailing. Making its case for the T as the cat sailing world's olympic representative. So I'm not going to knock them in their efforts.

Personally I agree costs stated by the ITA and reasoning behind the dropping of the T are furfies. But this isn't a game of reality its a game of perception. At least they are trying to get cats back in.

Also, I don't believe there is anything new in the past 2 decades that have been introduced into cat/skiff building.. Nomex/carbon was state of the art back in 1985 and still is today. Sure they could go to boron/carbon but the benefit comes at a cost. Not in money but in brittleness.. To be honest the T isn't the lightest boat around so why bother? Nomex/carbon is a total waste in the steamroller weight classes. But that is only my opinion *smiles* Hell even the A class can be built down to weight without going to nomex/carbon..

Just thought it was cool to see the ITA acknowledge our little class!!

We call it

factorial

in English.

My favorite combinatorial problem: How many ways are there to put N identical balls in M distinct buckets? It's cool problem because you won't get it through brute force, but when you look at it the right way, you can just write down the answer, with only a basic background in combinatorics.

Another bit of brain candy for math lovers: How many zeros are on the end of

100!

? This requires only very basic (pre algebra) math, and insight.

--Glenn

My first guesstimate is 21...

the theory I've used has worked for 10! and 20!, but can't check higher numbers with my calculator.

Thanks.

Despite having done very large portions of my education in English I still fail to remember some of the correct names. Even if you become a fluent speaker, you will never become a native speaker.

Wouter

Well done! Want to explain why?

--Glenn

I'll go for 24 zeros.

Each of the numbers ending in a zero create a zero on the end so that's 9 zero's
the 100 adds 2 more (so total is 11)
BUT 50 adds 2 zeros (2x50) so that is a total of 12 (we already countered one of them)

Each number that ends in 5 also creates zero's (so that's 10 zeros)

BUT 25 and 75 create 2 zeros (2x25 and 6x75) so that's 2 more

Total 24 zeros....

Don't understand why it's 21.......

Lost my first try at typing down the proof due to the forum time-out feature that destroys your message. *@!#$@!*&@#

So here goes again.

- first step :

Each number ending on the digit 0 adds a 0 to the total multiplication. We have 10 such numbers, where the number 100 adds two zero's.

10, 20, 30, 40, 50 , 60, 70, 80, 90 and, 100 => 11 trailing zero's added.

For the remainder of this proof we'll continue with new series of 89 numbers were the above numbers have been removed.

- Second step

Note how the trailing digit of any product of two numbers is totally determined by the trailing digit of each individual number. Example 17*23 = 391 => ends on 1 as 7*3 = 21 and also ends on 1.

By virtue of this result we know that a number ending on the digit 5 will result in a number ending on the digit 0 when it is multiplied by an even number. Example : 25*6 = 120 as 5*6 = 30

As a result we have 10 more numbers that in combination with any even number will multiply into another number with one or more trailing zeros. These numbers are :

5, 15, 25, 35, 45, 55, 65, 75, 85 and 95

If we pick our even numbers carefully then we can garantee that the resulting numbers only have a single trailing zero (and not 2 or more) and that the second last digit is out of the range 1, 2, 3, 4, 6, 7, 8 and 9. This is a handy feature as shall be shown later.

Lets say we multiply these with 4, 6, 2, 8, 12, 14, 16, 18, 22 and 24 (10 and 20 are excluded already as per above) and we find the products :

20, 90, 50, 280, 540, 770, 1040, 1350, 1870 and 2280

Note how all the results have a forelast digit that is part of the set 1, 2, 3, 4, 6, 7, 8 and 9 where the results 50 and 1350 are the only exceptions. We need to remember these numbers for later.

These 10 numbers add 10 trailing zero's to any multiplication they are part off. The may even add more depending with what other numbers the are multiplied with. The only way to be sure is to remove the 10 zero's and keep the following truncated numbers in our remaining multiplication set.

2, 9, 5, 28, 54, 77, 104, 135, 187 and 228.

We take the numbers 5 and 135 from this set and multiply them again with 26 and 28 from the full dataset resulting in :

130 and 3780

Both these numbers have a second digit that now too is of the set 1, 2, 3, 4, 6, 7, 8 and 9.

Again we remove the trailing zero's and add this to our total arriving now at 11+12 = 23 trailing zero's.

Of course we also keep these two truncated numbers 13 and 378 in our remaining dataset from which where we have now also removed the numbers 2, 4, 6, 8, 12, 14, 16, 18, 22, 24, 26 and 28 of course.

thus our subset now looks like
2, 9, 13, 28, 54, 77, 104, 378, 187 and 228. + all other numbers not removed so far.

- Third step

We are now left with a dataseries where all the numbers that end on 5 or 0 have been removed. So all remaining numbers have a trailing digit that is enclosed in the set

1, 2, 3, 4, 6, 7, 8 and 9

Mutual multiplication by numbers ending on a digit from this set is said to be

mathematically closed

.

This means that any multiplication between such numbers results in a new number that is also ending on a digit out of this set (and therefor not ending on the digit 5 or 0). Of course this behaviour is recurring.

The overal result is that the multiplication of remaining series of numbers will result in a number that will also end on 1, 2, 3, 4, 6, 7, 8 or 9 and therefor never end with either a 5 or a 0. Therefor this series will never add another trailing zero to the multiplication expressed by 100! irrespectibally of how the numbers are multiplied to eachother. We can therefor scrap his whole series of remaining numbers from our notepad and just focus on the numbers that we had taken out earlier.

- Forth step

That leaves us with the following data sub result :

10, 20, 30, 40, 50 , 60, 70, 80, 90 and, 100 resulting from the numbers ending on 0

+ 12 more times the number 10 resulting from the numbers ending on 5.

This can be rewritten into 23 times the number 10 and the new dataset 1, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Of course the number 5 can be multiplied by 32 out of the old discarded set to produce 160, which breaks down into 10*16 leaving us with :

24 times the number 10 and the dataset 1, 1, 2, 3, 4, 6, 7, 8, 9, 16.

All the numbers of the last data set have a trailing digit from the set 1, 2, 3, 4, 6, 7, 8 and 9. and can in turn never, under multiplication, produce a number with a trailing digit of 5 or 0 again.

And thus we have proven that the number of trailing digits is exactly 24. No less and no more.

That is assuming I have made no error somewhere.

I will have to check my own reasoning in a couple of days to spot oversights and errors. Right now I will most likely only read over them.

Maybe there is a simpler proof, but if there is then I need to let things calm down a bit before I can spot that one. Personally I don't think that the proof is of made of

very basic math

the steps themselves may not be particulary complex but composing a fully

mathematically closed

proof is certainly alot more complex and beyond the level of novices. But that assumes that the above proof is the definate proof, There may always be a very elegant pathway that we (I) haven't discovered yet.

Wouter

Yes, but how many buns make 5....?

I really have to get working on my report now, but I couldn't resist to write a small excel program to numerically work out the number of trailing zero's. Of course excel can't handle sufficiently large numbers to describe the number 100! accurately. But you can proof the number of trailing zero's by multiplying each neightbouring pair of numbers and factoring out any multiples of 10 before executing the next series of neighbouring multiplications.

You stop when all (remaining) neighbouring pairs both have a trailing digit out of the series 1,2,3,4,6,7,8 or 9 where the left over series can never again produce another trailing zero (another 10 that can be factored out). This happens after defactorizing the 2nd series.

Totalling the number of factored out 10's results in 24, ergo the number 100! has 24 trailing zero's.

See the table attached : (POS = number is now Part Of Set with trailing digit 1, 2, 3, 4, 6, 7, 8 or 9)

Or just use my method.

Your method may have produced the (correct) number of 24 trailing zero's but it is not mathematically sound. Basically, you followed a pathway that is not

thoroughly logical

and that is most certainly not

closed

. The first meaning that you have not taken the full scope of the problem into account. Example ; who is to say that you have covered all eventualities in your

proof

, maybe you missed a special number or special combination ? The second means you have failed to proof that there can AT MOST be 24 trailing zero's.

I'll provide a few more details.

You wrote :

In this method you use the number 2 twice; ones in multiplying 2*50 and once in multiplying 2*25. That is obviously incorrect.

You also say : Each number that ends in 5 also creates zero's (so that's 10 zeros), without showing how this is the case. For example, a number that ends in 5 ONLY creates a trailing zero when multiplied by an even number. Your statement still allows it to create trailing zero's when multiplied with any given number.

You forget to show that such a multiplication can produce more then only 1 trailing zero. Example by using the next available even number = 4 we find : 25 * 4 = 100 and that is adding two zero's to the total instead of just 1. So how do you know that there aren't more then 24 trailing zero's ?

Then you leave out the proof that none of the other possible combinations can produce trailing zero's or can produce numbers that in combination with yet another number can produce zero's.

Even with some leniency regarding your phrasing of the proof (using the number 2 twice), you can at most proof that there are At LEAST 24 trailing zero's. Of course that was never the hard part of the proof. The hard part is to proof that there are AT MOST 24 trailing zero's as then you'll have to find a structure in the remaining series of numbers that prevents this series from ever creating more trailing zero's. Only by combining these two parts can the proof be completed.

Wouter

It was logically thought thru.

I looked at the problem. Understood what was causing the zero's to be produced, calcuated how to work out which numbers produced the zeros and then described the method in simple terms.

Simple is best IMO.

Simon,

We are talking mathematics here. Some words have slightly different meaning there then they do in the broader worlds.

Your approach was indeed logical, but it wasn't

thorougly logical

as in exhaustive.

Your method overlooks part of the possibilities/eventualities and basically implicetly assumes that they can not create more zero's. Assuming something (either explicetly or implicitely) does not equal to proof.

My point was whether you also understood why the other numbers could not produce zero's. That part was completely left out in coverage. Without it the proof is incomplete and not a proof at all. That is the fun about mathematics and the part that makes some proofs so darn difficult.

Point in case: proving Fermat's principle (look it up); we have know that it is true for about 250 years now, but no-one has yet been able to find a

closed

proof for it. Some mathematician thought he had just recently after working on it for some 15 years. Turns out a small part of his very elegant proof was not

thorough

, preventing the proof from being

fully closed

and collapsing his whole proof.

Only when you have the option of choosing between two or more pathways that are equal in

thoroughness

and

closedness

. A simple pathway that doesn't contain both or either one is absolutely worthless, irrespectibally whether it is simple or not.

Sorry,

I did remove the

lucky

part however as I do not think that was a fair statement

Wouter

Wouter,

OK I did a little digging around the web to try and find out how to show more cleanly what I did. I actually searched for

number of zeros in 100 factorial

via google.

Maybe I had done this calc in the past as part of my studies into Maths as part of my A levels and then deciding weather or not to take maths into a degree (I did not, I studied Computing, Management and software design).

I did essentially what is described here.

This is also worth a read

interesting

to see what happens with 1,000,000! <img src=

alt=

/>

More here

and here

Stop IT! My head hurts! <img src=

alt=

/>

I'd rather pack decubitus ulcers. <img src=

alt=

/>