VeeshanVault

Looking at my ffXI regen rates and how it speeds up over time..and it kinda intrigues me....I'm decent at math but it's not really my strong suit either.

What calculation would you run where you could figure out how many ticks it requires a character with X mana to regen to full?

The regen rate is funny....

10 the first tick
11 the second
12 the third
13 the fourth

I mean it obviously has a very easy to understand pattern....but still....I was thinking to myself...what equation would that require to find my answer?

I mean I obviously know how many ticks it takes....I can do a running total on my calculator or just count the ticks as they occur on my screen.

But at 3am...I admit I stumped myself. I know it can't be hard but I can't figure it out for some reason.

Who can do it? I know someone here will show my ass up and solve it in like 30 secs. Who will be first?

I'm not very good at math. In high school I used to be alright at it, but things sure changed when I got to college....

I roomed blind with two other guys.... Chris and Jim. Jim was a big time pothead, Chris was from a farm. We all got along pretty well at first but then Chris and I started to get kind of sick of Jim.

One time Jim got really sick and was throwing up for like 24 hours.... i was making fun of him and telling him to quit puking and finally we put him in a cab and sent him to University Health Services. That's like the student infirmary/hospital.

Jim's appendix had burst, they sent him to the real hospital and the doctors there said that it was one of the worst cases they'd seen.... Jim ended up having to leave school for a month, and when he came back Chris and I both liked him again.

I think Jim lives in like Colorado now.... I haven't really talked to him since he transferred schools after freshman year.

I can't figure out your equation, sorry.

It was like 25 cent wing night at this bar and I love wings so I went down with my drinking friends who I hang out with every night because where I come from I am known as the party animal.

While I was there I met the most amazing woman I have ever seen. Only 30 lbs overweight with glorious, bleach blonde hair, she was vision of mediocrity the likes I have have only seen on Jerry Springer and sometimes Povich when he has a really classy guest on. I'm serious man she was at least a 5 or a 6.

She let me hold her hair while she vomited into the trash can so I reached under grabbed her boob. She was so turned on that we started kissing right there. It tasted pretty gross but I was in the moment. Then she slapped me and screamed something about being a daddy to her kids. I think I'm in love. I want to buy her a ring but the payments on my riced out Neon are pretty high (like $110 a month) and after that and my coors light money I hardly have enough to pay for my gym membership so I can pump my body up on the excercise bicycle and scope out all the honies.

I haven't really played FFXI.

*wipes a tear away*

Beautiful... just beautiful.

rofl

this has my vote for topic of the week

oh jesus

rofl

So what your trying to say is not only can you not answer my questions.

But you can't come up with a new flame either?
Weak.

Well, the answer isn't trivial (i don't think).

The sum of an arithmetic series, S, is:

S = (.5*n) * (a1 + an)

Where n is the number of terms, a1 is the first in the sequence and an is the last (nth) in the sequence.

So in the case of 10,11,12,13,14,15 the formula for how much you regen'd is S=(.5*6) * (10 + 15) or S=3*25 or S=75.

Your problem is that you have the sum already (how much mana you want to regen) and you have the first term of the sequence. For example, if you want to regen 500 mana, and the first tick you regen 10, the formula becomes:

500 = (.5 * n) * (10 * an)

You have two unknowns. You want to know n, but at the same time you have to know what an would be. You would need a second equation that relates n and an in order to solve this, and i'm not sure what that equation would be.

Someone else take over =P

Pilsburry wrote:So what your trying to say is not only can you not answer my questions.

But you can't come up with a new flame either?
Weak.

What I'm saying is not only that I don't want to bother trying to answer your question.

But I'm not trying to come up with a new flame either. If it ain't broke, don't fix it.

Emulating your grammatical style and blathering logorrhea.... it's fun.... so far 3 people have liked it to the 1 that didn't. Here's some math I'll figure out for you.... that's 75% of the people who have responded!

Yep. Still funny!

Integrate.

Take the integral of 9+t from 0 to t.

x = .5x^2 +9 t

Throw out the erroneous root.

It's not 100% accurate due to the fact that the mana regenerated per tick is always a whole number where integrating will connect say, 11 to 12 along the line y=9+t with a straight line instead of a perfectly horizontal line.

However, we can remedy this situation with the fact that the slope of the line is 1, therefore by adding .5 to the integral for every t, you can obtain the "exact" answer.

Final equation is:

x = .5t^2 + 9.5t

Where x is your total mana pool and t is the number of ticks.

edit: I can't spell pool

and I am a geek.

Guys, I am not even going to try to follow the logic above, but this problem isn't that complicated.

tick 1: 10
tick 2: 11
tick 3: 12
tick 4: 13
...

ok, take the "10" off of the top.

We end up with

Sigma (1 to n) of (n+10)

which simplifies to Sigma (1 to n) of n + n*10


The sum of an arithmetical progression is = (2a(1) + d(n-1))n/2.

Where the arithmetical progression of first term = a(1), common difference = d, number of terms = n and last term = a(1) + d(n-1) exists.
The sum of the arithmetical progression

In this case, d = 1 a(1) = 1 therefore

= 0.5*(n^2) + 0.5*n

so the full equation is 0.5*(n^2) + 10.5*n


Remembering that we "threw out the first tick", we accomodate that with...

=0.5*(n-1)^2 + 10.5*(n-1) + 10
=0.5*(n-1)^2 + 10.5*n - 0.5

For the first tick, we get:
=0.5*(0)^2 + 10.5*(1) - 0.5
=0 + 10.5 - 0.5
=10

(which matches 10 - correct)


For the second tick, we get:
=0.5*(1)^2 + 10.5*(2) - 0.5
=0.5 + 21 - 0.5
=21

(which matches 10+11 - correct)


For the third tick, we get:
=0.5*(2)^2 + 10.5*(3) - 0.5
=2 + 31.5 - 0.5
=33

(which matches 10+11+12 - correct)



For the fourth tick, we get:
=0.5*(3)^2 + 10.5*(4) - 0.5
=4.5 + 42 - 0.5
=46

(which matches 10+11+12+13 - correct)



For the fifth tick, we get:
=0.5*(4)^2 + 10.5*(5) - 0.5
=8 + 52.5 - 0.5
=60

(which matches 10+11+12+13+14 - correct)


For the sixth tick, we get:
=0.5*(5)^2 + 10.5*(6) - 0.5
=12.5 + 63 - 0.5
=75

(which matches 10+11+12+13+14+15 - correct)



So to know your mana regen'ed, M, after n ticks, the formula is:

M = 0.5 * (n-1)^2 + 10.5*(n) - 0.5
M = 0.5 * ( n^2 - 2*n + 1 + 21*n - 1)

M = 0.5 * (n^2 + 19 * n)
or
M = 0.5* n^2 + 9.5*n

(which is what Waikiki wrote, but I wanted to write a clear proof )


To solve this for n, so you know how many ticks to regen M mana, you solve as follows:

M = 0.5* n^2 + 9.5*n
2*M = n^2 + 19*n
2*M + 90.25 = n^2 + 19*n + 90.25
2*M + 90.25 = (n + 9.5)^2

n = sqrt(2*M + 90.25) - 9.5

so to test this...

how long to regen 21 mana?

n = sqrt(2*21 + 90.25) - 9.5
= sqrt(132.25) - 9.5
= 11.5 - 9.5
= 2 ticks

how long to regen 75 mana?

n = sqrt(2*75 + 90.25) - 9.5
= sqrt(240.25) - 9.5
= 15.5 - 9.5
= 6 ticks


bingo...

So, in short:

if M is mana regened
and n is number of ticks, then

M = 0.5* n^2 + 9.5*n
or
n = sqrt(2*M + 90.25) - 9.5

that or you can say..
You regen wierdly.

WTF Pils??! Don't I win a cookie or something?

Thank god someone posted the solution so I don't have to take 5 min doing the problem.

Ya you get a cookie, just took me a while to get back to this forum.

Couldn't find a cookie in the back shop?? WTF?

I sent you some pie instead and a bonus.

I was walking down the street the other day and this bar was having their 25 cent wing nite so I looked in my pockets and only had like 75 cents... so I walked into the bar and ordered 3 wings and they were like WTF no way and I was like WTF why not so I left.

Bah, I had this stuff covered from this thread awhile ago!

Winnow wrote:

Topper wrote:...but i refuse to believe he is a complete dumbass and think that anyone would really care about another guy scoring.

I dunno. I think this stuff can be highly interesting. For example:

I was at the bank the other day. I went up to the teller and it was a nice looking lady.(lesbian?...dunno!) I conducted my business and at the end she smiled at me and said, "have a good weekend" (not have a nice day mind you!). I told her the same and started walking away but OMG!!!11! I forgot and took the pen that was at the counter. I turned back and said "Opps, tee hee! Here's your pen back" She smiled and said, "Don't worry. Keep it" I said, "Thanks!!1111!1' , waved, and started walking away but I stumbled a bit, looking a little clumbsy. I turned back to see if she noticed and she was staring right at me!11!!11!!. She giggled a little (but not the "you're a jackass loser" giggle) and I went on my way.

I'll tell you if more sparks fly the next time I go to the bank!

http://www.veeshanvault.org/forums/view ... light=bank