Talk:Accuracy Formula (UFO2000)

There seems to be some kind of problem with the latex math formula generating. Either that or a problem with me. I'd appreciate some feedback about this, since it would look nice to have actual pretty looking formulas. And there are some very ugly formulas coming.

I guess MediaWiki isn't set up here. But, couldn't those formulas be simplified somewhat?

 $hp &=& acc \times \frac{max.health - current.health}{\frac{max.health}{2}}$
$hp &=& \frac{2 \times acc \times (max.health - current.health)}{max.health}$

 $mp &=& acc \times \frac{100 - current.morale}{\frac{100}{2}}$
$mp &=& \frac{acc \times (100 - current.morale)}{50}$


Or, even down to non-TeX format:

 hp = acc * (max.health - current.health) / (max.health / 2)
hp = 2 * acc * (max.health - current.health) / max.health

 mp = acc * (100 - current.morale) / (100 / 2)
mp = acc * (100 - current.morale) / 50


Well, these can. My problem is the ones that come afterwards. Like (2/(1/wacc^2+1/sacc^2)) <- this looks real ugly. And I was wondering about maybe delving into the statistics of hitting, but without decent formula display, I really shouldn't. Gaussians and things like that. No good.

$\frac{2}{\frac{1}{wacc^2} + \frac{1}{sacc^2}}$
$\frac{2}{\frac{sacc^2}{wacc^2 \times sacc^2} + \frac{wacc^2}{wacc^2 \times sacc^2}}$
$\frac{2}{\frac{wacc^2 + sacc^2}{wacc^2 \times sacc^2}}$
$\frac{2 \times wacc^2 \times sacc^2}{wacc^2 + sacc^2}$

2 / (1 / wacc2 + 1 / sacc2)
2 / (sacc2 / (wacc2 * sacc2) + wacc2 / (wacc2 * sacc2))
2 / ((wacc2 + sacc2) / (wacc2 * sacc2))
2 * wacc2 * sacc2 / (wacc2 + sacc2)


Don't ask me about gaussians, though. I've been out of college too long.

Ist the formula itself incorrect.?

The lines

hp = acc * (max.health - current.health) / (max.health / 2)

and

hp = 2 * acc * (max.health - current.health) / max.health

are mathematically equal, but I'm really not sure if the formula itself is right.

One Example:

Accuracy=80

max.Health=50

current.health=25

hp = 80 * (50 - 25) / (50 / 2) = 80 * (25 / 25) = 80 (!)

This can not be right. If I understood it correctly, the hp can be the half of the total accuracy in maximum. In this simple case I did use, the lost accuracy iss 100%! Not mentioned, what would happen, if the morale is also low... Negative accuracy can't be possible.

Isn't it right this way?

hp = acc * (max.health - current.health) / max.health /2

For the example I used:

hp = 80 * (50 - 25) / 50 / 2 = 80 * 25 / 50 / 2 = 20

For the mp it's the same:

80 accuracy

50 morale

mp = acc * (100 - current.morale) / 50

mp = 80 * (100 - 50) / 50 = 80 * 50 / 50 = 80

I guess this is the proper way:

mp = acc * (100 - current.morale) / 200

mp = 80 * (100 - 50) / 200 = 80 * 50 / 200 = 20

Am I wrong, or is there a mistake in this article :-) ?

BTW: I think this formula style is OK:

hp = acc * (max.health - current.health) / max.health / 2

Looks like a mistranslation. The LaTex is fine.

As you noted, the intent is to have 50% of the penalty come from health and 50% from morale. However, omitting the parentheses is technically accurate but misleading, because division is not commutative in general.

The clear way to write the parenthesized versions should be:

hp = (acc * (max.health - current.health) / max.health) / 2

mp = (acc * (100 - current.morale) / 100) / 2)

i.e.

hp = acc * (max.health - current.health) / (2*max.health)

mp = acc * (100 - current.morale) / 200

-- Zaimoni, 14:28 Sept 20 2006 CDT