Re: Point System ideas
Posted: January 23rd, 2012, 8:08 pm
Don't worry, I know how to write understandable and testable specifications. I've done it for more projects than I care to think about
I know the point/day has ZERO support, but this statement shows that it is the simplest system programatically-no factor to multiply by. 1 day is 1 point, 1 week is 7 points, 2 months is about 60 points. If I'm going for a lonely cache and can see it's worth a 91 points, I know its either been out for 91 days if unfound or found once, 182 days if found twice, 273 days if found 3 times...no factor needed. A day still seems like the simplest, straight up unit that doesn't need a factor...just subtract the the 2 dates and be done with it.Unless we go for something overly complicated and/or stay with DGP's method, when the database does its calculation, it's going to subtract the publication date from the current date to find the number of days the cache has been active, then multiply it by a factor to get the total points available.
Guys, don't worry. However we do it, the math is really simple. For some number P (e.g. P=100) points per year:jcanyoneer wrote:I know the point/day has ZERO support, but this statement shows that it is the simplest system programatically-no factor to multiply by. 1 day is 1 point, 1 week is 7 points, 2 months is about 60 points. If I'm going for a lonely cache and can see it's worth a 91 points, I know its either been out for 91 days if unfound or found once, 182 days if found twice, 273 days if found 3 times...no factor needed. A day still seems like the simplest, straight up unit that doesn't need a factor...just subtract the the 2 dates and be done with it.Unless we go for something overly complicated and/or stay with DGP's method, when the database does its calculation, it's going to subtract the publication date from the current date to find the number of days the cache has been active, then multiply it by a factor to get the total points available.
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Total Points = P * (years + <day in year>/<days in this year>)Sorry Corfman Clan to make things seem more complicated. It's what I do!Corfman Clan wrote:Kripes, don't make it so complicated. It's not and there is no reason to make it so.
jcanyoneer, I see a few advantages with 1 point per day, especially with the "higher resolution" on cache and cacher scores. I liked watching my score move, as well as my hides and finds, and couldn't wait for the next update to see what the new scores were on my caches (hey, I'm a new hider). But personally, I like keeping it around 100/yr, 10/mon, 2/wk. We don't often think of cache age in days (or even weeks) unless we're chasing FTFs... well, maybe that would be a reason to change it.jcanyoneer wrote:I know the point/day has ZERO support, but this statement shows that it is the simplest system programatically-no factor to multiply by. 1 day is 1 point, 1 week is 7 points, 2 months is about 60 points. If I'm going for a lonely cache and can see it's worth a 91 points, I know its either been out for 91 days if unfound or found once, 182 days if found twice, 273 days if found 3 times...no factor needed. A day still seems like the simplest, straight up unit that doesn't need a factor...just subtract the the 2 dates and be done with it.
Is this right?Guys, don't worry. However we do it, the math is really simple. For some number P (e.g. P=100) points per year:
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Total Points = P * (years + <day in year>/<days in this year>)
Thanks. It may be possible to fit a curve to whatever we'd like it to be. The danger though is that the more complicated it is, the less likely it will be widely accepted and applicable, and the more room for politicking and contention.jcanyoneer wrote:I like Rocketsciguy's graphs and tables, but this does seem to degrade points for true Lonely caches also...I think that is undesirable. If anything, I would think people would prefer to have the lonelier caches have Bonus points or a scaler or something to make them worth more, not less. I like the different thinking though, any more ideas?
And I'm not sure I picked the best example case for a power trail, because the ET Highway has just had its grand reopening. I picked out a 5-year-old LPC micro at a mall nearby and it has a DGP score of 2 CPs or 2.23 if you include the fraction. Big caveat is obviously that DGP is out of date. Using DGP's nearest caches tool, I find there are 43 caches within "<1 mile" (meaning within the same DD° MM' grid by DGP rules) with a total score of 241. If I exclude the puzzles, there are 36 caches with a total of 101 CPs. Not all of these are PnG LPCs of course, a couple are very well concealed, at least one requires climbing a tree, at least one has been archived (4 CPs) and another ought to be (12 CPs). But otherwise, it's very likely that you could collect 75ish CPs by driving around the shopping district and adjoining neighborhoods for a couple hours. Maybe this is a better example of the kinds of caches that can be diminished on LCP.jcanyoneer wrote:Your example also shows that the TP = points/finds formula seems to take care of power trails already.
Jealousy arises within me. No cane here, just a handful of rugrats clinging to my legs, and a commitment not to buy any power toys or outdoor gear until my student loans are paid off.jcanyoneer wrote:I figure, if someone wants to get a 100 points by finding 1500 caches, let 'em! It's not my thing, but maybe that's all they are capable of finding (no offroad vehicle, uses a cane, no rappelling gear, no access to a boat...etc)
Yes, it's correct, you just applied the formula wrongjcanyoneer wrote:Is this right?Guys, don't worry. However we do it, the math is really simple. For some number P (e.g. P=100) points per year:
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Total Points = P * (years + <day in year>/<days in this year>)
A cache that has been out for 5 months...
TP= 100 * (0+24/24) = 0
I assume years is the number of years the cache has been out?
The <day in year> is today?
The <days in this year> in the number of days so far this year?
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TP = 100(0 + 150 / 365) = 41.096To me, that's really the heart of the issue. The scoring system needs to be simple and understandable. It may be fun to come up with interesting ways to weight how a cache is scored, but in the end, if the method appears to be magic to most people, its value is greatly diminished.rocketsciguy wrote:Thanks. It may be possible to fit a curve to whatever we'd like it to be. The danger though is that the more complicated it is, the less likely it will be widely accepted and applicable, and the more room for politicking and contention.jcanyoneer wrote:I like Rocketsciguy's graphs and tables, but this does seem to degrade points for true Lonely caches also...I think that is undesirable. If anything, I would think people would prefer to have the lonelier caches have Bonus points or a scaler or something to make them worth more, not less. I like the different thinking though, any more ideas?...
I see. So, in the (0 + 150 / 365) 0 + 150 is actually a formula too?From your example:
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TP = 100(0 + 150 / 365) = 41.096
I can't tell if you're just trying to bug me or whatjcanyoneer wrote:I see. So, in the (0 + 150 / 365) 0 + 150 is actually a formula too?
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TP = 100(0 + (150 / 365)) = 41.096Code: Select all
TP = 100(years + <fraction of year>)Yes, and that will continue to be the case.firennice wrote:currently an archived cache does not count at all...
No problem. Just to be clear, I'll plug in another example.jcanyoneer wrote:Sorry-not trying to bug ya-I was mainly questioning where the 150 came from. I know that it is the number of days that a cache has been out. I was meerly saying that once you know the number of days a cache has been out, you are done using a 1 point/day system. No factors to multiple or divide by. The cache is just worth the number of days it's been out.
sorry again...I am not a good communicator!
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TP = 100(7 + 44/366) = 712.022Code: Select all
TP = 100(2 + 358/365) = 298.082