Golf Club Atlas
GolfClubAtlas.com => Golf Course Architecture Discussion Group => Topic started by: John Kavanaugh on July 25, 2007, 02:40:29 PM
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Given reasonable boundries every hole has a finite set of options and or recoveries. Please begin the process of developing a mathmatical formula for calculating the number of options for any given hole.
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Given reasonable boundries every hole has a finite set of options and or recoveries. Please begin the process of developing a mathmatical formula for calculating the number of options for any given hole.
Looks like your understanding of math is about as good as your understanding of spelling.
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Car posting...let's not kill a decent thread.
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X to the Nth power + Y to the Nth power = Z to the Nth power, where N is greater than 2.
I think some folks have been working on this for awhile.
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X to the Nth power + Y to the Nth power = Z to the Nth power, where N is greater than 2.
I think some folks have been working on this for awhile.
And X, Y, and Z are natural numbers.
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Actual theorem states it can't be solved for N > 2, according to my recollection. I think it was also solved a few years ago - at least, casual perusal of the bookstore shelves (and my own geek shelves) would imply so - haven't gotten beyond the prologues myself.
There's no way the formalize Jax's request, imho. The continuous nature of nature would defy formulas, again, imho.
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(1) Number of different clubs you could hit off the tee to reach the green in the same amount you could best expect with your longest clubs.
(2) Number of different clubs you could hit off the tee in an effort to play the hole for a one putt par.
(3) Number of second shot clubs that coincide with the goal of (1).
(4) Number of second shot clubs that coincide with the goal of (2).
(5) Number of different clubs you would plan to use for your chip/pitch shot in hopes of making a par
(6) Number of different directions viable to play these clubs off the tee or the fairway in hopes of attainng your goal score.
((1*3) + (2*4)) * 5 = total options for a par 4.
Split fairway holes would factor in 6 if there really is an option...
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Actual theorem states it can't be solved for N > 2, according to my recollection. I think it was also solved a few years ago - at least, casual perusal of the bookstore shelves (and my own geek shelves) would imply so - haven't gotten beyond the prologues myself.
There's no way the formalize Jax's request, imho. The continuous nature of nature would defy formulas, again, imho.
Fermat's conjecture (http://en.wikipedia.org/wiki/Fermat's_last_theorem)
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You know, I'm not sure if JES is on the right track, though I think he may be, at least in getting close to a ball park figure for the number of options a hole provides (even in the context of JK's 'reasonable' parameters). But:
What I find even more interesting is the NATURE of those options, as JES sees them, i.e. not so much in terms of direction (e.g. playing left or right of a centre-line bunker) as in terms of club selection, and that in the very concrete context of trying to get the ball in the hole in the fewest number of strokes.
It makes me think that good players really DO think differently about options (and design?) than mediocre ones...or at least, than THIS mediocre one, whose thoughts immediately went to the "A" "B" "C" "D" kind of drawings with which the old designers would illustrate a hole's various options, depending on skill level
Peter
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The proof that this is possible lays in the fact that an analytical mind can give you the correct answer.
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You know Peter, thinking about it like that I would probably only use #'s (3) and (4) as multiples if they were also divided by (5).
In other words, if I am 200 yards away and would really only want to hit a pitch shot from between 40 and 70 yards (but I might use three different clubs from 40-70 yards) I have to factor that in.
In the context of JK's question, a 100 yard 9 iron followed by another 100 yard 9 iron would not seem to qualify as a reasonable attempt to get the ball in the hole as quick as we can.
Take away that caveat and this effort is useless...how many shots would it take me to go around the world the wrong way?
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Most if not all of these conjectures, theorems and problems have already been solved by the guys at The Institute in Morgan Hill, but they are not allowed to tell us the answers.....
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Isn't it a serum, not a theorem, that reveals all the options, for some?
Go Barry! ;)
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e=mc^2... no wait, if the hole is a 90 degree dogleg then a^2 + b^2 = c^2, a being the length of the drive, b being the length of the approach and c being the length of the drive if you drive the green. Yea, that's the ticket! Do I win? When can I book my hit and run at Vic National? ;D
Cheers,
Brad
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JK,
Three- left, right, or straight
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Car posting...let's not kill a decent thread.
Decent? Yes I suppose since there has been no vulgarities here, it must be decent.
Decent? pshaw!
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"Given reasonable boundries every hole has a finite set of options and or recoveries. Please begin the process of developing a mathmatical formula for calculating the number of options for any given hole."
No, absolutely not. Reducing golf to mathematical formulae is for those who see no mystery in golf. For those who want to reduce golf to a pat logic. For those who make fortunes off toilets, like Joshua Crane.
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An interesting fact is that the number of bunkers is inversely related to the number of options with the one exception that one bunker provides five more options than none.
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JK,
Three- left, right, or straight
True, to narrow it down I would suggest that every 40 ft of fairway width equals one option on holes less than 400 yds. For holes greater than 400 yds every 60 ft equals one option.
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I pondered on this for a while when I sat througha video replay CLE seminar today. (Those are always great for contemplation and mind wandering)
I did not come up with a precise formula but I did discover a postulate so to speak:
The best option is the one that leaves the fewest options for the next stroke.
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An interesting fact is that the number of bunkers is inversely related to the number of options with the one exception that one bunker provides five more options than none.
Just keep thinking, Barney, that's what you're good at.
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Jaka's question depends entirely on what "reasonable" means. You could probably define it narrowly enough to come up with an equation. I doubt it would mean much though, for the reason George Pazin pointed up.
In Isaac Asimov's Foundation books, a scientist comes up with mathematical formulas that predict, seemingly 100% successfully, all future history. So options on a golf course should be easy as pie, yes?
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Most if not all of these conjectures, theorems and problems have already been solved by the guys at The Institute in Morgan Hill, but they are not allowed to tell us the answers.....
Or allow more then 36 golfers a day on the course which of course is identified as "Institute = golfers <= 36/day"
Reminds me of Lt. Scheisskopf's wife in Catch-22 who had a Mathematics degree but couldn't count to 28.
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Reminds me of those motion studies from the turn of the last century. They thought they could maximize efficiency on the factory floor by mathematizing the workers' movements. You ran the raw data through some sort of weighting formula and, presto changeo, you make Model T's for 30% less labor.
The parallels with golf scream out. You could give players similar data for each hole and, presto changeo, you get 30% less mucking around and more rounds per day.
Bob
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What do you think is the first information that Tiger processes when looking at how to play a hole? What is the first information that you process?
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You know Peter, thinking about it like that I would probably only use #'s (3) and (4) as multiples if they were also divided by (5).
In other words, if I am 200 yards away and would really only want to hit a pitch shot from between 40 and 70 yards (but I might use three different clubs from 40-70 yards) I have to factor that in.
In the context of JK's question, a 100 yard 9 iron followed by another 100 yard 9 iron would not seem to qualify as a reasonable attempt to get the ball in the hole as quick as we can.
Take away that caveat and this effort is useless...how many shots would it take me to go around the world the wrong way?
Quite a few once you reach either the Atlanic or Pacific, unless you are the guy Gary Player was talking about who is using steroids ;D