On Wed, 12 Sep 2012 08:27:47 +0000, Spencer101 wrote:

tomcat82;1605414 Wrote:
Can't upload files at work, but let me try to explain it...

In baseball, innings pitched are calculated by the amount of batters
the pitcher has retired in one inning. He needs to retire three (3)
batters to end the inning. So one retired batter amounts 0.1 innings
pitched and 3 retired batters to 1.0.

So when a pitcher pitches 6.1 innings in Game 1 and 5.2 innings in Game
2, he has pitched a total of 12.0 innings in 2 games.

Does that help? Normally 6.1+5.2 = 11.3, but in this case it should be
12.0 and so onÂ…

The result should look like:

---A-------------------- B 1 12.0 (B1+B2)----- 6.1
2-----------------------5.2


Does *=ROUNDUP(SUM(B1:B2),0)* in A1 (with 6.1 in B1 and 5.2 in B2) do
the trick?

It might as long as you only add 2 games together... If you add 3 that
are 1.2, 1.2 and 1.2, you approach yields 4, but the answer should be 5.

I'd do something a little more complicated, along the lines of splitting
the screwy numbering scheme that baseball uses into 2 more columns.

B1 = int(a1)
C1 = (a1-b1)*10

B{n} = sum(b1:b{n-1})
C{n} = int(sum(C1:C{n-1})/3)
D{n} = mod(sum(c1:c{n-1}),3)/10

innings pitched = B{n}+C{n}+D{n}

Can probably be simplified a bit, but it's a start.