# Thread: Posts Per Day Mathematics

1. Posts Per Day: 0.69

2. Wolfy! Your Pness is in question.

I'm at 12.01ish atm, fun stuffs. Used to be much higher though.

3. post per day = total posts / days of membership

a = b^x with 10.00 = 10.06^x
x = log(b)a approximately = 0.99741

10.00 = ( 10.06 )^0.99741

10.00 = 10.06 * ( x/y )^0.99741

with x >= 1 and y >= 1 and x = new posts per day and y = new days since today

-> [10.00 / 10.06]^0.99741 = x/y

-> x/y approximately = 0,9941

means 9,941 posts in the next 10,000 days or 9.941 posts in the next ten days

-> 10.06 * 9.941 / 10 = approximately 10.00

of course you may vary x/y = 0,9941 by posts / day as you like:

For example:

9.941 posts / 10 days

-> 9.941 * x = 10 -> x = 1,005935

so: 1,005935 * ( 9.941 / 10 ) = 10 / 10,05935

so 10 posts in the next 10 days 1 hour 25 minutes 27 seconds 840 milliseconds

4. Milliseconds.... that's a nice touch.

5. Originally Posted by fidelia
ENTPs will do too. It's just that they might have flitted off to do something else. Can't be sure that they'll be around for this kind of information. Pretty dependable though when they are. Good work.
ENTPs are scary, esp entropie *flits off*

6. Originally Posted by entropie
post per day = total posts / days of membership

a = b^x with 10.00 = 10.06^x
x = log(b)a approximately = 0.99741

10.00 = ( 10.06 )^0.99741

10.00 = 10.06 * ( x/y )^0.99741

with x >= 1 and y >= 1 and x = new posts per day and y = new days since today

-> [10.00 / 10.06]^0.99741 = x/y

-> x/y approximately = 0,9941

means 9,941 posts in the next 10,000 days or 9.941 posts in the next ten days

-> 10.06 * 9.941 / 10 = approximately 10.00

of course you may vary x/y = 0,9941 by posts / day as you like:

For example:

9.941 posts / 10 days

-> 9.941 * x = 10 -> x = 1,005935

so: 1,005935 * ( 9.941 / 10 ) = 10 / 10,05935

so 10 posts in the next 10 days 1 hour 25 minutes 27 seconds 840 milliseconds
Dude, what?

Taken from Wolfy's statistics page:
Total Posts: 7,270
Posts Per Day: 10.07 (Which is unusable due to it being rounded to two decimals)
Join Date: 06-30-2008

06-30-2008 -> 06-22-2010 = 722 days

Doesn't take a genius to figure out what post difference he has between his current posts per day and the exact 10.00 posts per day. You should be able to see it instantly, but for funs I'll actualy post how to reach it. :P

7270-(722*10)=50 posts above the 10.00 mark.

Who needs a calculator for that.

So if he posts less than ten posts per day and accumelates the posts he is short of ten a day, until he accumelates 50 posts, he will be back at 10.00 with exact precision. Or just don't post for 5 days.

You say 10 posts in the next 10 days he will be at 10.00? Whilest in fact he will be 40 posts shy of exact 10.00 if he does that.

My method is both easier to calculate and much more exact. Hmpf! One mistake in your method is assuming his 10.06 was a precise factor, whilest in fact it is a rounded number, and currently is 10,069252077562326869806094182825 (Was probably closer to 10.06 at the time of your post though.)

edit: Ofcourse, I didn't take into account the post difference of Wolfy since your post, but you still should have calculated the exact coeffecient first. :P

7. Ya I figured it doesnt work, but well first I wanted to do a laplace transformation

8. Originally Posted by entropie
Ya I figured it doesnt work, but well first I wanted to do a laplace transformation
Well, I did like the structure of your method, but it's always best to look for the more obvious and easy ways, and not just overcomplicating things to make it look good.

9. obvious solutions are boring

Mistake is here: 10.00 = total post + x / days of membership + y

Tho were to go from here ?

I brought it to: 10.00 = tp ( 1 + x/tp ) / dom ( 1 + y/dom)

But I am running out of ideas from there.

think there is no workaround but to laplace transform it, like:

tpd + x = dom + y

with g(0) = 10.00

and f(t) = tpd + x(t) / dom + y(t)

-> g(0) = (1st derivation) f(t)

-> Transformation into imaginal area:
0 = 1 - s * g(t)

-> U(s) = 1 / -s

something, I dunno. Have to look into my controltechniques exam -.-

10. with laplace I end up with:

G(s) = 10 / [(tp + x) / dom + y]

So if he has 7,270 posts in 722 days its:

-> 7220 + 10 y / 7,270 + x = 1

So if he doesnt post for 50 days you get y = 50 and x = 0

That works

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