Soil Acre Furrow Slice

Soil Acre Furrow Slice

By Donald G. McGahan

Two values that are handy to calculate when managing land are the pore space in the soil and the mass of soil in any given volume of soil. The volume of the pores is important because the total pore space, together with the pore size distribution, influences water retention and soil atmosphere exchange with the above ground atmosphere. The mass of soil is critical to determine exchange capacities and nutrient buffering capacities.

To calculate many things on the basis of an acre furrow slice of soil we must have more information about the soils bulk density, or alternately we can estimate the bulk density. Since soils have varying densities we are not quite sure about the bulk density unless it is physically measured.

To calculate the mass of soil by using an estimate of the bulk density of the soil we proceed as follows, and if the actual bulk density is known the calculation remains the same. Simply substitute the measured bulk density for the assumed bulk density.

afs = acre furrow slice; or the volume of soil to the nominal depth of plowing over one acre (ac).

The nominal plowing depth can be something like 0.5 ft, or 15 cm, depending on the way the question is asked.

For this segment I will address the total volume of the acre furrow slice then the mass.

Since the bulk density is not given I will assume a bulk density of 1.475 Mg / cubic m or 1.475 g / cubic cm. Bulk density is nearly always reported in metric terms. The exception to this is with engineers. Producers in the USA also often use lb so we must be able to calculate and convert.

Remember that a megagram (Mg) is 1,000,000 g.

1 cubic m = 1.31 cubic yd. Where m = meter.

You could proof this, right?

1 lb = 454 g, or 0.454 kg, or 0.000454 Mg.

1 ac. = 43,560 sq. ft.

You memorized this previously, right?

Going to the nominal plowing depth, 0.5 ft., for an acre furrow slice (afs):

(43,560 sq. ft. / ac.) x (0.5 ft / fs) = 21,780 cu. ft. / afs

[where furrow slice is fs and this example fs = 0.5 ft – two dimensional]

1 cu. yd. = 3 ft. x 3 ft. x 3 ft., or 27 cu. ft.

(21,780 cu. ft. / afs) x (1 cu. yd. / 27 cu. ft) = 806.667 cu. yd. / afs

[cubic yard (cu. yd.) is a common volumetric unit used when unconsolidated materials are transported. That that once was soil is included in these unconsolidated materials. Any volumetric unit can be used and calculations by scientist frequently use metric whereas engineers commonly use cubic feet.]

Now we use the assumed density to calculate the weight of an afs.

(1.475 Mg / cubic m) x (1 cu. m / 1.31 cu. yd.) x ( 1 lb. / 0.000454 Mg) = 2480 lb. / cu. yd.

Therefore the mass of an afs is:

(2480 lb. / cu. yd.) x (806.6667 cu. yd. / afs) = 2,000,516 lb.

Obviously if you assume (but alway better if you have an emperical measurement) a different density you get a different mass.