Page 107 - Petrology of Sedimentary Rocks
P. 107

In  general,   immature   rocks   have   very   little   original   visible   pore   space,   and   little
       chemical   cement;   submature   (and  higher)   rocks   have  the  pores   either   open  or  filled   with
       chemical   cements   like  quartz   or  calcite.
              2.  Determine   the  sorting   of  the  rock.


                   a.     If  it  is  not  well   sorted   with  a  over   0.5$  (mm  diameter   ratio   over   2.0),  it
                          is  SUBMATURE.

                   b.     If  it  is  well   sorted   with   a  under   0.54,  go  to  (3).

       The  sorting   borderline   between   submature   and  mature   is  depicted   in  the  bottom   figure
       on  the   next   page,   which   represents   a  set  of  spheres   with   a  phi  standard   deviation   of
       exactly   0.5.   Compare   your   sample   with   the  figure   to  see  if  it  is  well   enough   sorted   to
       be  mature.
             3.    Determine   the  roundness   of  quartz   grains   of  sand  size.


                   a.     If  the  grains   are  subangular   to  very   angular   on  Powers’   scale   (p  under
                         3.0,  it  is  MATURE.

                   b.     If  the  grains   are  subround   to  well   rounded   (p  over   3.0)  it  is  SUPER-
                         MATURE.

       The   figure   below   represents   a  sand   with   roundness   of   3.0,   the   borderline   between
       mature   and  supermature.














                               TEXTURAL   MATURITY   FLOW  SHEET












                  IMMATURE               SUBMATURE              MATURE   SUPERMATURE

            Sorting   may  be  estimated   in  hand  specimen   with  a  phi  size  finder   mount   by  determining
       what   phi   diameter   has   one-sixth   of   the   grains   by   area   larger   than   itself;   then
       determining   the   phi  diameter   that   has  one-sixth   of  the  grains   (by  area)   smaller   than
       itself.   In  this  way,   one  has  estimated   the   16th  and  84th   percentiles   of  the  distribution.
       Subtract   these   two  diameters   and  divide   the  result   by  2  to  get  the  standard   deviation.
       For  example,   if  the  one-sixth   limits   are   I .2  and  2.84,   a  is  l.6/2   or  0.84.   In  thin  section,
       the   same   process   of   mentally   determining   the   diameters   at   the   16th   and   84th





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