THE SOIL

CONSERVATIONIST'S

DILEMMA

GENERAL STATEMENT: A corn- soybean rotation on highly erodible land is not sustainable - because of excessive erosion.

- Adding a greater percentage of corn to the rotation may help to control erosion but is probably even less sustainable because of the increased nitrogen and insecticide requirements.

WHAT ABOUT NO-TILL?

The best current technology for controlling erosion in a corn soybean rotation is no-till planting - but that is still not enough to provide adequate erosion control on most highly erodible fields.

Using the Northern Illinois "C" factor chart for No-till soybeans following corn with 80% residue cover (drilled or wide rows.) C= .06
For No-till corn following soybeans- Assume some carryover residue to provide 60% ground cover. Estimated C= .10 The total for both years is .16 and the average "C" for the rotation is .08

In Jo Daviess County the best group 1 highly erodible soils average:

8% slope, 175 ft. slope length (LS 1.31) K=.37, R=180

Using the Universal Soil Loss Equation:

180 X 1.31 X .37 X .08"C" = 6.98 tons/Ac./Yr. or nearly 2 tons per acre higher than the 5 ton/acre/year tolerable soil loss level for deep soils.

WHAT ABOUT CONTOURING?

If the new "P" factors for contouring are used it is very difficult to achieve more than a 0.8 "P"

6.98 tons X .8 = 5.6 tons/Ac. Yr.- Still too high. (Ridge tillage with greater than 6" ridge height or contour terraces would be possible ways to reduce the C factor and achieve adequate erosion control.)


WHAT IS THE ANSWER

The 10% solution.

Add a minimum of 10% small grain and hay to the rotation.

- Two acres out of twenty. - 1 acre would be small grain and the other acre would be full year established grass legume hay.

The new "C" factor would be:

17 Ac. corn- soybean rotation: C=.08 X17=1.36

1 Ac. no-till corn after hay 80% cover C= .02

1 Ac. small grain- disk tillage- 30% residue C= .07

1 Ac. grass, legume hay- Full year established C= .004

Total 1.454

1.454 divided by 20 Ac. = The new "C" is .073

Soil loss- 180 X 1.31 X .37 x .073 =6.4 tons/ac./yr.

multiply by the contouring factor or.8 = 5.1 tons/ac./yr.

multiply by the 10% contour buffer strip factor X .9= 4.6 tons/Ac./Yr -which is well within the 5 ton tolerable soil loss level

EXAMPLE OF CONTOUR BUFFER STRIP LAYOUT

A 40 acre square field has 1320 ft. per side Use 33 ft. wide field borders on both ends and two 33 ft. wide contour buffer strips for a total of 4 buffer strips.

Approximately 1320 ft. X 4 buffer strips X 33 ft. = 174240 ft. Divide by 43560 ft. per Ac.= 4 Ac. or 10% of the field.