Pass Efficiency

Discussion in 'The Tiger's Den' started by Ch0sn0ne, Sep 4, 2006.

  1. Ch0sn0ne

    Ch0sn0ne At the Track

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    PASS EFFICIENCY Team Cl G Att Cmp Int Pct. Yds TD Eff.
    ------------------------------------------------------------------
    1. Erik Ainge.......... UT JR 1 17 11 1 64.7 291 4 274.4
    2. J. Russell.......... LS JR 1 17 13 0 76.5 253 3 259.7

    Not complaining, because it doesn't matter. Just trying to figure this out.

    How does Ainge have a higher efficiency than Russell?

    Lower Completion %, threw an INT, but more TD's and Yrds.

    Anyone know the formula?
     
  2. red55

    red55 curmudgeon Staff Member

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    I found this:

    NCAA Pass-Efficiency Formula

    To determine a pass-efficiency rating, multiply a passer's yards per attempt by 8.4; add the number obtained by multiplying pass completions per attempt by 100; add the number obtained by multiplying touchdowns per attempt by 330; and subtract the number obtained by multiplying interceptions per attempt by 200.
    The formula is as follows:

    ER = TY/PA*8.4+PC/PA*100+TD/PA*330-I/PA*200,

    where TY=total yards; PC=pass completions; PA=pass attempts; TD=touchdowns; I=Interceptions; ER=Efficiency Rating.
     
  3. COTiger

    COTiger 2010 Bowl Pick 'Em Champ

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    Damn, I got a headache just reading it :lol:
     
  4. Ch0sn0ne

    Ch0sn0ne At the Track

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    Well, I did the math using this formula. JR came out to 257.09, Ainge came out to 316.52.

    Amazingly this formula only penalizes Ainge 10 points for the pick. Doesn't make much sense.

    Either that or my math sucks.
     

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