slugging %

What is slugging % and how is it calculated?

Slugging percentage is used to see if someone is a power hitter. The way you calculate it is to take a player’s total bases (Hits + 2B + (3B times 2) + (HR times 3)) and divide that by their total at bats. If a player hits a lot of doubles, triples, or home runs, the slugging percentage will be much higher. This is why power hitters have high slugging percentages.

Thanks!

Except shouldn’t that be 2Bx2, 3Bx3, HRx4?

But the general idea is right. It’s essentially the average number of bases gained per at bat by the player. So every HR by McGwire makes up for 4 of those whiffs. Still, even his sa is only in the .600s. Like ba, at the beginning of a season someone will go on a tear and have a sa of over 1.000, but then of course they have to come back to reality and someone like MM takes over the final lead.

The precise formula is as follows:

S+(D2)+(T3)+(H*$)=TB

TB/AB

When:
S = Singles
D = Doubles
T = Triples
H = Home Runs
TB = Total Bases
AB = At Bats


Brian O’Neill
CMC International Records
rockuniverse.com/cmc/cmc.html

ICQ 35294890
AIM Scrabble1
Yahoo Messenger Brian_ONeill

sigh

The dollar sign above should be a 4, as I’m sure you figured out…

I wish this BB had an “edit” feature for these 2:45 AM postings…

>>(Hits + 2B + (3B times 2) + (HR times 3)) and divide that by their total at bats >>llardball

>>Except shouldn’t that be 2Bx2, 3Bx3, HRx4? >>Dave Swaney

No, see below.

>>The precise formula is as follows:
S+(D2)+(T3)+(H*4)=TB
TB/AB>>Satan

The 2 formulas are slightly different ways at arriving at exactly the same thing.

Llardball’s formula starts with “Hits”, which includes singles, doubles, triples and home runs. Since all of those hits have been counted once, if you add doubles in again you have attributed 2 bases to them. If you add (triples X 2) in you have attributed 3 bases to each triple. If you add in (HRs X 3), then you have attributed four bases for each HR.

Satan’s formula starts with “singles” instead of hits. Therefore, doubles, triples and HRs have not yet been counted once. Therefore, it is necessary to multiply doubles by 2, triples by 3 and HRs by 4.

The results are identical.