## Monday, 8 September 2008

### About online's exp.

Let it be a continous function (continues in my online game -,-") f(x), time used to up lv, unit: hr/lv f'(x), time accel. : hr/lv^2 t(x), exp per time(hr), exp/hr t'(x), exp per time accel., exp/hr^2 exp([x])=exp to up lv *t(x) and f(x) are continue becoz exp per (something) fall even in lv but not fall just on the lv. exp(x)=f(x)t(x). exp'(x)=[f(x)t(x)]'=f'(x)t(x)+f(x)t'(x) so that we can count the time needed. Let's have an example: (forget about unit, m=million time *) (p.s., Δx is a small variable, about 1000~10000, too small so not to count here) Sub., x=81, (81lv) f(81)=28 f'(81)=0.785 t(81)=0.8m t'(81)=-0.01m+Δx exp(81)=28*0.8m=22.4m exp'(81)=[(0.785)(0.8)+(28)(0.01)]m=0.908the formula for diff. game is different so that we prefer you to get your own formula, in this game, we have
Exp(x)=(x/2-0.7)x^3=x^4/2 – 0.7x^3
Notice that Exp(x)’s two definition won’t get the same answers becoz they only is a approx. value.So that the time needed (function) to up lv totally→Exp(81) =21.1m (3 sig.fig.)
Time = Exp/speed = 21.1/Intergrate[1.61-0.01x dx]
=22.4 / (1.61x-0.005x^2)
sub x=81~82 = [1.61(82) -0.005(82)^2]- [1.61(81)-0.005(81)^2]=0.795
22.4/0.795 is about 28.17Comparing to the original time, In fact only about 10 minute is needed more than expect but still important XD The increasing rate might much more than that... This is the way of 外掛 to calculate time needed to up lv...
Of course, t’(x) may not as a ‘stright line’. Then the new T will not be x+t’/2.
But in this case, easily, we found that new time
=ft/(x+t’/2)
So easy comparing to a long calculating at the first……..---- the data is come from a famous online game in Taiwan =]~

Subscribe to:
Post Comments (Atom)

## No comments:

## Post a Comment