Throwing wpf and armor weight

[Apedude]

I'm having trouble understanding how heavy armor i can wear and still utilize my wpf.
I tried calculating it with the formula in the Game mechanic thread, and i hope someone can tell me what i came up with.
My effective weight right now is 9.8. i am going to have 6 Power Throw and i have 105 Throwing profficiency right now.
Here is what i came up with.

(105 - 13x6) x (1 - 9,8 / 100)
(105 - 78) x (1 - 0,098)
27 x 0,902 = 24,354

Those 24,354, I have no idea what they are. Are they left  over wpf, so that i can wear heavier armor?

[Vibe]

Your effective weight is 10 (rounded), that means your effective Throwing WPF will be 95. You need 78 wpf to fully utilize power throw, meaning you have 17 "spare" wpf. Someone correct me if wrong.

[Apedude]

Damn....
Now i get it. 1% of each effective weight. Thats 1,05 x 9.8 =10,29wpf lost because of my armor.
It was so simple after all.
Thanks for enlightening me Vibe!

[Okkam]

Your effective weight is 10 (rounded), that means your effective Throwing WPF will be 95. You need 78 wpf to fully utilize power throw, meaning you have 17 "spare" wpf. Someone correct me if wrong.

14*6=84

You need 84 wpf in throwing for 6PT. This way you'll have 6 effective PT
TBH you need 120+ wpf, because with 1 effective wpf you'll miss even at point-blanc range

[Vibe]

14*6=84

You need 84 wpf in throwing for 6PT. This way you'll have 6 effective PT
TBH you need 120+ wpf, because with 1 effective wpf you'll miss even at point-blanc range

Okkam, according to the calculator it's 13 WPF per PT for throwing and 14 WPF per PD for archery.

[Okkam]

Seriously? Well, you are right and my post is more dumb then usually

[Lactose_the_intolerant]

http://forum.c-rpg.net/index.php/topic,29149.0.html

tadaaaaaaa