Agriculture Reference
In-Depth Information
200 kg x 0.3 gram/kg = 60 grams of Calcium needed per m
3
.
Calculating Weight Per Depth of Light-Weight Soils
Suppose we were working with an acre or hectare of loose-textured high-organic
matter soil that, rather than weighing 92 lbs/ft
3
or 1333 kg/m
3
,
weighed only 60
lbs/ft
3
or 870 kg/m
3
?
Obviously a 6 inch or 15 cm depth of this soil would not weigh 2 million lbs/acre or
2 million kg/hectare, but for the sake of simplicity, we choose to work with 2 million
lbs or kg. This poses two questions:
“How deep is 2 million lbs or kg of this
soil?”
and
“How much does the top 6” or 15 cm actually weigh?”
The answer to the second question is easy enough. We figured out above that
there were 1500 cubic meters per hectare to a depth of 15 cm. Multiply 870 kg/m
3
x
1500m
3
= 1 305 000 kg/ha to a depth of 15 cm.
We also calculated above that an acre 6” deep is 21 780 ft
3
;
60 lbs/ft
3
x 21 780 ft
3
= 1 306 800 lbs/ac to a depth of 6” (the difference between
the numbers for lbs/ac and kg/ha here are due to rounding off earlier).
We see that the top 6” or 15 cm of this soil weighs about 700 000 lbs/ac or kg/ha
(35%)less than our “standard” of 2 million. If we decided to amend only the top 6”
or 15 cm of the soil, we would take our amendment weight needed for 2 million
weight units and multiply by 65%. For example, 600# Ca x 0.65 = 390# Ca needed
to bring the top 15cm or 6” of this light-weight soil into balance.
The other question was
“How deep is 2 million lbs/ac or kg/ha of this soil?”
If 15 cm or 6” of the soil weighs 65% as much as our standard soil, then 2 million
lbs or kgs of this soil would cover a depth equal to the standard depth divided by
0.65:
15 cm / 0.65 = 23 cm deep, or 6” / 0.65 = 9.23” deep. If we add the required 600
lbs/ac or kg/ha Calcium, that will be enough to balance the Ca in the top 23cm or
9.23 inches of this soil.
