Here are the answers to the questions: a) Drawbar power available = 75% of 96 HP = 72 HP Speed = Drawbar power x 375 / Load = 72 HP x 375 / 5000 lbs = 4.32 mph b) Width of plow = 3 ft = 36 in Area of plow = Width x Depth = 36 x 8 = 288 in^2 Load per inch = Soil resistance x Area = 20 psi x 288 in^2 = 5760 lbs Total load = Load per inch x Width = 5760 x 36 = 207,360 lbs Speed = 5 mph Power required = Load x Speed / 375 = 207,360 x 5 / 375 = 110 HP

1. MAINTAINANCE OF FARM
MACHINERY
En Mohd Fauzie Jusoh
Lecturer
Agrotechnology Programme
Faculty of Agro-Based Industry
Universiti Malaysia Kelantan (Jeli Campus)
Locked Bag No.100, 17600 Jeli, Kelantan.
014-2903025/fauzie.j@umk.edu.my/

2. FARM RECORD

3. Logbook
• With greater reliance on agricultural machinery in
crop production, the ownership cost of tractors
and various farm implements are becoming a
bigger component of the total cost of production.
• As such, care must be taken not only in the
proper selection and purchase of tractors and
farm implements but also in their efficiently use.
• Logbook : Daily record/log book of each tractor.

4. Logbook
Consist of :
Date
Name of operator and attendant
Whether a trailer is hitched
Job description
Starting time and hour meter reading
Finishing time and hour meter reading
Total hours worked
Distance travelled (km)
Load carried (kg)
Fuel and oil consumption
Parts replacement and repairs

5. Maintenance schedule
• Breakdowns of machinery at work not only cause
direct repair cost but also losses in revenue due
to poorer or lesser yield
• As such a farm manager should set up a proper
maintenance schedule for all tractors and farm
implements under his care
• Different machine manufacturer will propose
slightly different maintenance schedule
• A general maintenance schedule will be given in
the hard copy

6. Maintenance schedule
Machines that are well looked after will have:
Longer useful lives
Resale values
Experience less breakdown
Lower fuel and oil consumption

7. MACHINE CAPACITY

8. Machine Capacity
The machine capacity is its rate of machine performance
The performance capacity will be measured in terms of
hectare per hour (ha/hr), tons per hour, liter per minute,
etc
Understanding how to estimate the capacities of machines
will help farm managers making a plan in selecting and
buying machine for future use
To know machine capacities for selection of power units
and an equipment that can complete field operations on
time
Important to avoid the added expenses of larger that
necessary machines

9. Factors that influence field efficiency
of the machine
Speed – average rate of travel in km/hr
Width – distance in feet or meters across the
processing portion of the machine
Efficiency
Ratio of effective capacity of a machine to its
theoretical capacity.
an indicator to determine how much working
time is spent versus turning, filling hoppers and
other jobs

10. Theoretical Field Capacity (TFC)
• Is the maximum possible capacity obtainable
at a given speed, assuming the machine is
using its full width
• TFC depend on factor : speed and width
• TFC[Area/Time] = Speed (m/hr) x Width (m)
= m2 /hr or ha/hr

11. Theoretical field capacity (TFC)
• Ex : Suppose a tractor with a 4.25 meter disk travels 8
km/hr. What is the theoretical field capacity in ha/hr?
 The theoretical field capacity (TFC) would be :
= speed x width
= 8000 m/hr x 4.25 m
= 34,000 m 2/hr
 Convert to hectare --- 1 hectare = 10,000 m 2
 TFC = 34,000 m2 /hr
10,000 m2
= 3.4 ha/hr

12. Theoretical Field Capacity (TFC)
Theoretical Field Capacity cannot be sustained
for long periods of time because field
operations will be interrupted by turns, filling
hoppers and breakdowns
The Effective Field Capacity (EFC) is always less
than the Theoretical Field Capacity (TFC)

13. Effective Field Capacity (EFC)
• The number of acres/hectares actually covered
over a long period of time
• EFC brings in the factor of efficiency. This capacity
determination represent the real life or actual
capacity obtainable over a period of time
• EFC [Area/Time] = Total Area (ha)
Total Time (hr)

14. Effective Field Capacity
• Ex : If a 4.25 meter disc actually covers 28 ha
while operating for 10 hours with no
breakdowns, its EFC would be :-
EFC = Total hectares (ha)
Total hours (hr)
= 28 ha
10 hr
= 2.8 ha/hr

15. Field Efficiency @ Field Capacity
• EFC for one day’s experience may not give a
true picture of the EFC for the season. It will
be different depending on the period use.
• The different can be determine by calculating
the Field Efficiency
• Field Efficiency (FE) = EFC x 100%
TFC

16. FACTORS THAT AFFECT FIELD
EFFICIENCY
• Not using the full capacity of the tractor
• Refilling procedures
• Transporting procedures
• Repairing of machine in the field / machine
breakdown
• Irregular resting time
• Cleanliness of the implements
• Rest of operation
• Adjustment of implement

17. Example 1
Suppose a tractor with a 4.25 meter disk travels 8
km/hr. If the disc of this tractor is used for 2
weeks period and cover 195 hectares, calculate
the Effective field capacity and field capacity. This
tractor work 8 hour/day.
Total day = 14
Total working hour/day = 8
Total hours in field = 14 x 8 = 112
Total hectares covered = 195

18.  EFC = Total hectares
Total hours
= 195
112
= 1.74 ha/hr
 Field efficiency = 1.74 x 100%
3.4
= 51%

19. Example 2
• Assume a 6 rows 30 inches row corn head
equipment combined is working at a speed of
3.2 km/hr with long row well organizes
unloading pattern and no break-down. Field
study indicated field efficiency of 70%
achieved. Find effective field capacity.

20.  Width = 30 in x 6 rows
= 180 in
@ = 4.572 m
 TFC = width x speed
= 4.572 m x 3200 m/hr
= 14,630 m2/hr
1 hectare = 10,000 m2
 TFC = 14,630 m2/hr
10,000 m2
= 1.46 ha/hr

21. FE = EFC x 100
TFC
 EFC = FE x TFC
100
= 0.7 x 1.46 ha/hr
= 1.022 ha/hr

22. Example 3
• The theoretical field capacity (TFC) of a disc
harrow is 2.0 ha/hr. If its field efficiency (FE)
is 70%, calculate :
i. The effective field capacity (EFC)
ii. The time taken to plow a 18 ha field, in minutes

23. TFC = 3.0 ha/hr, FE = 65%
a. EFC = FE x TFC
100
= 0.65 x 3.0 ha/hr
= 1.95 ha/hr
b. Time taken to plow a 12 ha field, in minutes
1.95 ha = 1 hr
12 ha = x
1.95x = 12
x = 6.15 hr x 60
= 369 minute

24. Example 4
• One farmer has a working hour of 8 hrs per
day using a TS90 Ford tractor to plow 25
hectare of corn farm. If the theoretical field
capacity (TFC) of a moldboard plow is 2.5
ha/hr and the overall field efficiency is 70%
then can he finish plowing in 2 days time?

25. • TFC = 2.5 ha/hr, FE = 70%, Area = 25,
• Working Hours = 8
 EFC = FE x TFC
100
= 0.7 x 2.5 ha/hr
= 1.75 ha/hr
 1.75 ha = 1 hr
25 ha = x hr
1.75x = 25 ha
x = 14.29 hr (required time to finish plowing)

26. Working hour for 2 day = 8 x 2
= 16 hr
The farmer can finish plowing in 2 day
because he only need 14.29 hr to do the job
while working hour in 2 day is 16 hr

27. MACHINE POWER REQUIREMENT

28. Introduction
To match power units to the size and type of
machines, so all field operations can be carried
out on time with a minimum cost.
If tractor is oversized for implements, the costs
will be excessive for the work done.
If the implements selected are too large for the
tractor, the quality or quantity of work may be
lessened or the tractor will be overloaded usually
causing expensive breakdowns

29. Factors to consider when selecting a
power unit
1. Engine type
2. Power ratings
3. Soil resistance to machines
4. Tractor size
5. Matching implements
6. Sizing for critical work

30. 1. Engine type
The combustion process in the cylinder
converts the energy contained in fuel to a
rotating power source
This rotating power source can be further
converted into 3 forms :
 Drawbar pull
 PTO output
 Hydraulic System Output

31. 2. Power rating
Power is a measure of the rate at which work is
being done
The English power unit is defined as 550 foot-
pounds of work per second
The metric power unit is measured in kilowatts
(kW)
1 kW = 1.34 horsepower
1 hp = 550 ft.lb/sec
Work = Force x Distance

32. Power rating
If a load require a force of 20 pounds to move it
vertically a distance of 3 feet, the amount of work
done is :
Work = Force x Distance
= 20 lb x 3 feet
= 60 ft.lb
The amount of work done is 60 ft.lb with no
reference to time

33. • If a 1000 lb force is moved 33 feet in one
minute, the rate of doing work is one
horsepower, because one horsepower equals
33,000 ft.lb per minute
• The equivalent rate of work in one second to
equal 1 horsepower is :
1 HP = 33,000 ft.lb = 550 ft.lb per second
60 seconds

34. • When working with field machinery, we
usually think of miles per hour and pounds of
draft. For these conditions the formula for
horsepower is :
HP = Force ,lb x Speed , mph
375

35. Metric equivalent
• Metric unit for power is kilowatt (kW)
• Force is measured in newtons or
kilonewtons
1 HP = 0.746 kW
1 kW = 1.34 HP
1 N = 0.225 lb
1 kN = 224.8 lb force

36. Metric equivalent
• Drawbar power is the measure of pulling
power of the engine by way of tracks, wheels
or tyres at a uniform speed
• Formula for Drawbar (kilowatt) is :
Drawbar (kW) = Force (kilonewtons) x Speed
(km/hr)
3.6
Draft is the horizontal component of pull,
parallel to the line of motion

37. Metric equivalent
• Ex: If the draft of a trailing implement such as a
disc harrow is measured at 11.1 kilonewtons
and is pulled at a speed of 8 km/hr, what is the
drawbar in kilowatt?
Drawbar (kW) = Force (kilonewtons) x Speed
(km/hr)
3.6
= 11.1 kN x 8 km/hr
3.6
= 24.7 kW

38. Metric equivalent
• Ex: A tractor is pulling a plow with a total draft of
22.2 kilonewtons. How fast can the plow be
pulled if the tractor has 50 drawbar (kilowatts)?
Speed = Drawbar x 3.6
Force
= 50 kW x 3.6
22.2 kN
= 8.1 km/hr

39. Metric equivalent
• Ex : Given 65 kW tractor, speed 8 km/hr, field
cultivator draft is 4 kN per meter of width when
used in a given field. What width of cultivator
could be pulled?
Draft = Power (kW) x 3.6
Speed (km/hr)
= 65 kW x 3.6
8 km/hr
= 29.25 kN

40. Metric equivalent
Width = Total draft
Draft per meter
= 29.25 kN
4 kN/meter
= 7.3 meters

41. Metric equivalent
• 1 dbhp = 45000 kgm/min
• Drawbar horsepower = Force, lbs x speed, mph
375
• This formula can be used to determine how fast an
implement could be pulled with a given size of tractor
• Speed = Drawbar horsepower x 375
Draft,lbs
• This formula can also be used to determine how large an
implement can be pulled but an extra step is involved
• Size of the implement have to be related to the amount of
soil resistance

42. Metric equivalent
• Ex : If the draft of a trailing implement like a disc
harrow is measured at 2500 pounds and is pulled
at a speed of 5 mph. What is the drawbar
horsepower?
Drawbar horsepower = Force, lbs x speed, mph
375
= 2500 x 5
375
= 33.3 HP

43. 3. Soil resistance to machines
• With a given tractor, there is a set of
amount of power available. This available
power is used for :
Moving the tractor over the ground
Pulling the implement over the ground
Powering the implement for useful work
• This reduces the available usable drawbar
power

44. 4. Determining tractor size needed
• There are various kind of power
Brake
PTO
Drawbar
• Tractor power is measured in horsepower
(USA) or in kilowatts (kW) – metric equivalents

45. E.g:
A tractor is mounted with a plow that is 5 feet
wide. The resistance from the soil is 2 500 lb
per foot of plow. Calculate the total draft.
Total draft = Soil Resistance x Width

46. 5. Matching tractors and implements
• When matching a tractor and implement, 3
important factors must be considered :
The tractor must not be overloaded or early
failure of components will occur
The implement must be pulled at the proper
speed or optimum performance cannot be
obtained
The soil conditions and their effects on machine
performance must be considered

47. Questions a
• The brake horsepower (BHP) of a tractor is 96
HP. The power available at the drawbar (DHP)
is only 75% of the BHP. The tractor is pulling
5000 lbs of harvested crops through a field.
What is the maximum speed in miles per hour
that the tractor can travel ?

48. Questions b
• A tractor has a plow that is 3 feet wide.
Plowing is carried out at a depth of 8 inches.
The resistance from the soil is 20 psi (pounds
per square inch). The tractor must be able to
plow the field at an average speed of 5 miles
per hour (MPH). How much horsepower is
required?

49. Questions c
• The tractor has an indicated power (IHP) of 90
HP where else its brake power (BHP) is 6.6 Hp
less than IHP. The average pulling load cause
by the moldboard plow is 4500 lbs. Calculate
the average forward travelling speed if the
available power at the drawbar (DHP) is 90%
of BHP.

50. Questions d
• The brake horsepower (BHP) of a tractor is
100 HP. The power available at the drawbar
(DHP) is only 60% of the BHP. The tractor is
pulling 4000 lbs of harvested crops through a
field. What is the maximum speed the tractor
can travel, in miles per hour (MPH) ?

51. Questions e
• Calculate the drawbar horsepower required
to pull an 80 inch tiller at a depth of 5 inch in
a soil with a resistance of 30 psi and at a
forward speed of 3miles per hour. If the
approximate PTO horsepower equals to
125% of the drawbar horsepower, what is
the approximate PTO horsepower needed?

52. Q&A
TQ