D-34 
/ 
0..13482000000000.,0.,0.,661000000000.,2978000000000.,9843000000000. ,0.,247400000 
0000. ,0. ,0.,0., 114000000000. ,2360000000000. ,0. ,49482000000000. ,3462000000000., 3540 
000000000.,4177000000000.,6012000000000.,32291000000000.,0. 
“07/16/1997”,"07/31/1997”,0.,50.,360000000000.,0.,0.,0.,0.,360000000000.,0.,12370000000 
00.,0.,0.,7000000000.,24000000000.,1206000000000.,0.,2391000000000.,0., 13300000000 
0.,159000000000.,282000000000.,1817000000000.,0.,9237000000000.,0., 1620000000000. 
, 1644000000000.,942000000000.,5031000000000.,0.,15388000000000.,0., 1453000000000 
.,2710000000000.,2005000000000.,9220000000000.,0.,6503000000000.,0.,0.,5500000000 
0..1231000000000.,5217000000000.,0.,13491000000000., 0 ., 0 ., 9000000000., 17930000000 
00..11689000000000.,0.,3160000000000.,0.,0.,0.,0.,3160000000000.,0.,51767000000000., 
0.,3206000000000.,4584000000000.,6277000000000.,37700000000000.,0. 
EXAMPLE MASS REPORT FILE (.mas) 
The mass file report contains the mass of a parameter computed for each cell in the 
interpolated (.est) file then summed in one of two ways. The default method (below 
example) is to sum the mass by segment and total for all segments in the bathymetry. 
The second method follows the format of the volume report and computes the mass 
bv concentration rang e for each segment. 
The mass that is computed and summed is saved to a mass report file (‘file¬ 
name’.mas). It is assumed the input data are measured in [units]/[liter], such as mg/1 
or ug/1 or counts/liter. In the mass report, the resulting mass estimates are computed 
by multiplying the [estimated concentration in the cell (often in mg/1)] * [the volume 
of the cell in m A 3 (for instance, 1000m east-west x 1000m north-south x lm deep)] 
* [1000 l/m A 3 to convert from m A 3 to liters]. Hence, if the input data were in mg/1 
and then the concentration is estimated to be 6mg/l in a cell, the resulting mass will 
be 6*10 A 9 mg for a 1km x lkm x 1 m cell. As a second example, if the input data 
were in mg/m A 3, which is equivalent to ug/1, then the reported mass values would be 
in micrograms to account for the volume being reported in m A 3 rather than liters. If 
the input data are counts (such as organism counts) per liter, then the mass report 
units would be total counts. If the input data are counts (such as organism counts) 
per cubic meter, then the total counts in the mass report must be divided by 1000 to 
account for the conversion from cubic meters to liters between the input data and the 
interpolated counts. The mass (or counts) for each cell is then summed for a total 
mass (or count) in the segment and also a grand sum of mass (or count) for the total 
for all segments under analysis. For instance, if the input data for CHLA were meas¬ 
ured as ug/1 and the resulting mass in Segment CB20H was reported after 
interpolation as 13,000,000,000,000, that represents 1.3 A 13 ug CHLA for Segment 
CB20H—i.e 1.3 A 13ug / 1.237 A 12 liters in CB2OH=10.5 ug/1 average. As a second 
example, if the input data were for mg biomass of organisms per cubic meter and the 
resulting mass in Segment CB20H was reported as 132,627,709,873,200, that repre¬ 
sents 1.326 A 14 / 1000 mg for Segment CB20H, since an adjustment for the input 
data must be made for the per cubic meter to per liter basis. A quick check can be 
made by multiplying the average input data value by the volume of a segment to 
determine if the results are within reason. For instance, if there were approximately 
appendix d 
User Guide and Documentation for the Chesapeake Bay Interpolator 
