Tillage is the agricultural preparation of soil by mechanical agitation of various types, such as digging, stirring, and overturning. This presentation will discuss different tillage operations and equipment used for the land preparation.

1. BULACAN AGRICULTURAL STATE COLLEGE
Course code & title: AGEN 100 – Basic Farm Machineries and Farm Mechanization
Prepared by: Engr. Vincent S. Dangan, Professional Agricultural Engineer

2. • any physical soil manipulation which changes the structure of the soil.

4. 1. To develop a desirable soil structure for a seedbed.
2. To control weeds, cut roots and bury green materials.
3. To incorporate manure and chemical fertilizer.
4. To turn the soil into a soft “puddle”.
5. To form a hard layer which reduces water leaching.

5. 1. To develop a desirable soil structure for a seedbed.
2. To control weeds, cut roots and bury green materials.
3. To incorporate manure and chemical fertilizer.
4. To turn the soil into a soft “puddle”.
5. To form a hard layer which reduces water leaching.
wetland
conditions

6.  Primary Tillage

7. Definition
 initial cutting, breaking and usually inversion of the soil. Cutting the soil to
a depth of 6” to 36”. Often referred to as
Implements
Tractor drawn
moldboard plow
Disc plow Chisel plow
Subsoiler
Animal drawn
moldboard plow
plowing.

8. Animal drawn
moldboard plow
Share
Shoe
Body
Beam
Handle

9. Disc plow
Disc
Scraper
Rear furrow wheel

10.  Primary Tillage
 Secondary Tillage

11. Definition
 subsequent breaking, pulverization
and leveling of the soil making it ready
for planting. Preparing the soil to a
depth of 3” to 6”. Often referred to as
Implements
Disc harrow
Spiked-tooth harrowSpring-tooth harrow
harrowing.

12.  Primary Tillage
 Secondary Tillage
 General-purpose Tillage

13. Definition
 combined primary and secondary
tillage in one operation. Cutting the soil
to a depth of up to 6”. Often referred
to as
Implements
Rotavator
Floating tiller
rotavating.

14.  Theoretical Field Capacity

15. Formula
FCT =
SW
10
where: FCT = effective field capacity, ha/hr
S = speed of travel, km/hr
W = width of cut, m

16.  Theoretical Field Capacity
 Theoretical time
 Effective operating time

17. Definition
 It is the time that would be required at
the theoretical field capacity
Definition
 It is the time during which the machine
is actually performing its intended
function.

18.  Theoretical Field Capacity
 Theoretical time
 Effective operating time
 Effective Field Capacity

19. Formula
FCA =
SW
10
x
E𝑓
100
FCA = FCT x
E𝑓
100
FCA =
AF
EOT
where: FCA = effective field capacity, ha/hr
FCT = theoretical field capacity, ha/hr
Ef = field efficiency, %
AF = Area of the field, ha
EOT = Effective operating time, hr

20.  Theoretical Field Capacity
 Theoretical time
 Effective operating time
 Effective Field Capacity
 Field Efficiency

21. Formula
E𝑓 =
FCA
FCT
where: Ef = field efficiency, %
FCA = effective field capacity, ha/hr
FCT = theoretical field capacity, ha/hr

22.  Theoretical Field Capacity
 Theoretical time
 Effective operating time
 Effective Field Capacity
 Field Efficiency
 Draft of Plow

23. Ds = Specific draft x W x d
where:
Ds = draft
Specific draft = Kg/cm2 (see Table 1)
W = width of cut of the machine, cm
d = depth cut of the machine, cm
Formula
Da = Ds x Increase in draft
where:
Da = adjusted draft
Increase in draft = % (see Table 2)

24. Table 1. Specific draft of different soils
SOIL TYPE SPECIFIC DRAFT, SD
Kgf/cm2
0.21
0.21-0.42
0.35-0.49
0.42-0.56
0.70-0.77
Sandy soil
Sandy loam
Silty loam
Clay loam
Heavy clay

25. Table 2. Increase in draft due to speed
SPEED INCREASE IN DRAFT, %
MPH KPH
1
2
3
4
5
6
1.6
3.2
4.8
6.4
8.0
9.6
100
114
128
142
156
170

26.  Problem #1

27. Given:
A 5 x 20 cm double action disc harrow is operated by a
tractor having a speed of 5 km/hr. Calculate the effective
field capacity, assuming the field efficiency of 80 percent.
Solution:
W = 5 x 20 = 100 cm or 1m
S = 5 km/hr
Ef = 80 %
Find:
FCA = ?
FCA =
SW
10
x
E𝑓
100
FCA =
(1)(5)
10
x
80
100
FCA = 0.4 ha/hr Answer

28.  Problem #1
 Problem #2

29. Given:
A 3 x 30 cm plough is moving at a speed of 4 km/hr
calculate how much time it take to plough 500 x 500 m
field when the field efficiency is 70 %.
Solution:
W = 3 x 30 = 90 cm or 0.9 m
S = 4 km/hr
Ef = 70 %
Area of the field (AF)
= 500 m x 500 m
= 250,000 sqm
= 25 ha
Find:
Effective operating time (EOT) = ?
FCA =
SW
10
x
E𝑓
100
FCA =
(0.9)(4)
10
x
70
100
FCA = 0.25 ha/hr
EOT =
AF
FCA
EOT =
25 ha
0.25 ha/hr
EOT = 100 hr Answer

30.  Problem #1
 Problem #2
 Problem #3

31. Given:
Determine the hectares plowed per hour when a tractor is
operating at 6.4 kph and is pulling four 36-cm moldboard
bottoms at a depth of 20 cm. How many hectares can be
plowed in 10 hours if field efficiency is 78 percent? If the
soil is clay loam, what is the draft required to work the soil?
What is the adjusted draft?
Solution:
S = 6.4 km/hr
W = 4 x 36-cm = 144 cm or 1.44 m
d = 20 cm
Effective operating time = 10 hr
Ef = 78 %
Specific draft = 0.49 Kgf/cm2
Find:
Area of the field = ?
Draft = ?
Adjusted draft = ?
Area of the field
AF
EOT
=
SW
10
x
E𝑓
100
AF =
SW
10
x
E𝑓
100
x EOT
AF =
(6.4)(1.44)
10
x
78
100
x 10
AF = 7.2 ha
Draft
Ds = Specific draft x W x d
Ds = 0.49 x 144 x 20
Ds = 1,411 Kgf
Adjusted draft
Da = Ds x Increase in draft
Da = 1,411 Kgf x 1.42
Da = 2,004 Kgf
Answer
Answer
Answer

32. THANK YOU.
BULACAN AGRICULTURAL STATE COLLEGE
Course code & title: AGEN 100 – Basic Farm Machineries and Farm Mechanization
Prepared by: Engr. Vincent S. Dangan, Professional Agricultural Engineer