The disc angle of a disc plough is the angle between:

the disc axle and the horizontal

The threshing cylinder most suited to paddy, whose grain detaches easily and must not be broken, is the:

plain drum without any elements

Variable-rate application of fertilizer in precision agriculture requires, at minimum:

a positioning receiver and a prescription map linked to a rate controller

a combine yield monitor on the same machine

a tractor of at least four-wheel drive

a uniform application rate over the field

Farm Machinery: Tillage Forces and Hitching, Field Equipment, Performance Parameters and Cost Analysis

The second chapter of Section 2, Farm Machinery, of the GATE Agricultural Engineering (AG) paper, covering the sub-heading the syllabus calls Farm Machinery. It follows the order of a season: soil tillage and the forces a tillage tool meets; hitch systems and how an implement is hitched so that it runs level and straight; the functional requirements and working principles of equipment for sowing and planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing, whether manual, animal-drawn, tractor-drawn or powered by renewable energy; the performance parameters the syllabus lists by name — field capacity, efficiency, performance index, application rate and losses; the cost analysis of implements and tractors; and the equipment of precision agriculture. The numericals here — field capacity, draft and draft power, calibration, spray rate, cost per hour — are among the most dependable marks in the paper.

1. Soil tillage and the forces on a tillage tool

Primary tillage opens and inverts or loosens undisturbed soil — the mouldboard plough, disc plough, chisel plough and subsoiler. Secondary tillage breaks clods, levels and prepares the seedbed after it — disc and spike-tooth harrows, cultivators, planks and rollers; the rotary tiller can do either in one pass. Tillage is judged by the soil condition it leaves (clod size, inversion, depth) and by the energy it costs, which is why the same draft appears in both halves of this chapter.

The soil reaction on a tool resolves into three components. Draft is the horizontal component parallel to the direction of travel — the pull the power source must supply. The vertical component (suction) pulls the tool down into the soil, which is why a mouldboard plough needs vertical suction to hold its depth; the side (lateral) component is taken by the landside of a plough or the side forces of discs. Draft grows with the cross-section worked, so it is normalised as unit draft = draft/(furrow width × depth), in N/cm², and as specific draft per unit width. A 30 cm × 15 cm furrow cut against 3150 N has a unit draft of 3150/450 = 7 N/cm². The draft power is draft × speed: 10 kN at 5 km/h is 10 × 5/3.6 = 13.9 kW.

Primary tillage tools compared
Tool	How it works	Where it wins
Mouldboard plough	share cuts the furrow slice, mouldboard lifts, turns and inverts it	full inversion, burying weeds and residue
Disc plough	a concave rotating disc set at a disc angle to travel and a tilt angle to the vertical	hard, dry, stony or root-infested soil; rolls over obstructions
Chisel plough	narrow tines shatter the soil without inverting it	conservation tillage, residue kept on the surface
Subsoiler	a heavy tine run below the normal tillage depth	breaking a hardpan or plough sole

⚠️ Disc angle and tilt angle are measured from different references

The disc angle is between the plane of the disc edge and the direction of travel; the tilt angle is between the plane of the disc and the vertical. Increasing the disc angle widens the cut and reduces penetration; increasing the tilt angle also reduces penetration and helps inversion. Swapping the two definitions is the usual wrong option.

2. Hitch systems and hitching of tillage implements

An implement is pulled either from a drawbar (trailed), through the three-point linkage (mounted), or partly carried and partly trailed (semi-mounted). For a trailed implement the rule is geometric: the line of pull must pass through the implement’s centre of resistance — the point where the resultant soil force can be taken to act — and through the hitch point on the tractor. If it passes above the centre of resistance the implement noses into the soil and runs too deep; if below, it rides up and runs shallow. In plan, a line of pull offset from the centre of resistance makes the implement crab sideways, which is corrected by moving the hitch point laterally, not by steering.

Hitching also moves weight. A high hitch transfers more weight from the tractor’s front axle to its rear axle, improving traction but approaching the point where the front wheels lift and the tractor can overturn backwards — the geometry of weight transfer is worked in the tractor chapter. Mounted implements carried on the three-point linkage add weight transfer by their own weight and by the vertical soil force, which is one reason a mounted plough needs a lighter tractor than a trailed one of the same width.

3. Equipment for sowing, fertilizing, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing

Sowing and planting. A seed drill meters seed continuously in rows — by a fluted roller (rate set by the exposed length of the flutes) or a cup feed — and places it with furrow openers (shovel, shoe, disc) at a controlled depth, covering it after. A planter places individual seeds at a set spacing with an inclined plate, vertical plate, cell-type or pneumatic (vacuum) meter. Transplanters set seedlings (rice, vegetables). The metering device is driven from a ground wheel so the rate stays in step with travel, and a drill is calibrated by turning that wheel the number of times that corresponds to a known area: for a drill 2.0 m wide on a 1.2 m wheel, 1/25 ha (400 m²) needs 400/(π × 1.2 × 2.0) = 53.1 revolutions, and the seed collected is multiplied by 25 to give kg/ha.

Fertilizer application uses broadcasters (spinning disc), drills with a separate fertilizer box and star-wheel or auger meter, and band placement. Inter-cultivation — weeding and stirring between rows — uses manual wheel-hoes, animal-drawn blade harrows and tractor cultivators with sweeps or rotary tines. Sprayers atomise a liquid through hydraulic nozzles (hollow-cone for insecticides and fungicides that must cover foliage, flat-fan for herbicides that need an even band) or by air blast; types run from the hand-operated knapsack and foot sprayer to tractor boom sprayers and solar- or battery-powered knapsacks. The application rate follows from a unit-conversion identity: Q (L/ha) = 600 q/(w v), with q the total nozzle discharge in L/min, w the swath in m and v the speed in km/h. For q = 12 L/min, w = 10 m and v = 4.8 km/h, Q = 150 L/ha.

Mowing uses a reciprocating cutter bar (knife sections shearing against stationary guards, driven by a pitman from a crank) or rotary drum and disc mowers that cut by impact. A chaff cutter chops fodder between a rotating knife (flywheel or cylinder type) and a shear plate, fed by rollers whose speed sets the length of cut. Harvesting: reapers and reaper-binders cut and windrow or bind; the combine cuts (header with reel and cutter bar), threshes (cylinder and concave), separates (straw walkers or a rotary separator) and cleans (sieves and fan) in one pass. Threshers detach grain by impact and rubbing: the spike-tooth cylinder (wheat, sorghum), rasp-bar (the usual combine cylinder), wire-loop (paddy, whose grain shatters easily) and axial-flow designs. Losses are counted at each stage — header, threshing (unthreshed grain), separation (grain carried out with straw), cleaning (grain blown out) — and grain damage; threshing efficiency is 100 minus the percentage of unthreshed grain.

ℹ️ Four power sources, one set of functions

The syllabus asks for manual, animal, tractor and renewable-energy versions of each machine. The function — meter, place, cut, detach, separate — does not change with the power source; what changes is capacity, and therefore the field capacity and cost per hectare worked out below.

4. Performance parameters: field capacity, efficiency, performance index, application rate and losses

The theoretical field capacity is the rate of work if the machine used its full width all the time: TFC (ha/h) = w (m) × v (km/h)/10. The effective field capacity EFC is the area actually covered divided by the total time spent, including turning at headlands, filling seed or chemical, unclogging and adjustments. The field efficiency is their ratio, η_f = EFC/TFC, equivalently the theoretical time over the actual time. A 3 m cultivator at 4 km/h has TFC = 1.2 ha/h; at 80 per cent field efficiency its EFC is 0.96 ha/h.

Performance parameters and how they are computed
Parameter	Definition	Note
Theoretical field capacity	w·v/10 (ha/h)	w in m, v in km/h
Effective field capacity	area covered / total time	includes all non-productive time
Field efficiency	EFC/TFC × 100	rises with longer fields and fewer turns
Application rate	quantity applied per hectare, e.g. 600q/(wv) L/ha for a sprayer	set by calibration
Losses	grain or material lost at each stage, as a percentage of the total	header, threshing, separation, cleaning
Performance index	output weighted by quality of work, per unit power	read the definition the question gives

A performance index combines the rate of work, the quality of work and the power used into one figure of merit, so that two machines can be ranked: it rises with field capacity and with a quality measure (for a rotary tiller, the fraction of area actually tilled and the wheel slip avoided) and falls with the power consumed. Its exact form depends on the machine and the test procedure, so a GATE question on it states the formula; the skill examined is substituting the right quantities in consistent units.

⚠️ Divide by 10, not by 3.6

w (m) × v (km/h) gives thousands of m² per hour, and 10 000 m² is a hectare, so w·v/10 is already ha/h. Dividing by 3.6 converts km/h to m/s and belongs to power calculations, not to field capacity.

5. Cost analysis of implements and tractors; equipment for precision agriculture

The cost of owning and running a machine splits into fixed costs, which accrue whether it works or not, and variable (operating) costs, which accrue per hour of use. Fixed costs are depreciation, interest on the investment, and housing, insurance and taxes; variable costs are fuel, lubricants, repair and maintenance, and the operator’s wages. With the straight-line method, annual depreciation is D = (P − S)/L for purchase price P, salvage value S and life L years, and interest is charged on the average investment (P + S)/2. The cost per hour is the annual fixed cost divided by the hours used per year, plus the variable cost per hour — so fixed cost per hour falls as annual use rises, the whole economic argument for custom hiring and for sharing machines.

Worked: a tractor costs ₹6,00,000, has a salvage value of ₹60,000 after 10 years, and is used 1000 h a year. Depreciation is 5,40,000/10 = ₹54,000 a year; interest at 10 per cent on (6,00,000 + 60,000)/2 is ₹33,000; housing, insurance and taxes at 2 per cent of the price are ₹12,000. The fixed cost is ₹99,000 a year, or ₹99 per hour of use. The break-even use between owning and hiring is the annual area at which the costs match: with ₹45,000 of annual fixed cost, a variable cost of ₹600/ha and a hire charge of ₹1500/ha, it is 45,000/(1500 − 600) = 50 ha a year — below that, hiring is cheaper.

Precision agriculture manages variation within a field instead of treating it as uniform. Its equipment is a chain: GNSS/GPS receivers (with differential or RTK correction for centimetre accuracy) locate the machine; sensors measure soil (electrical conductivity, moisture, nutrients) and crop (canopy reflectance indices such as NDVI); yield monitors on combines map output; GIS software turns the data into prescription maps; and variable-rate technology (VRT) applicators vary seed, fertilizer or chemical rate on the go, while auto-steer and section control prevent overlaps and skips.

Key takeaways

Draft is the horizontal soil force along travel; unit draft divides it by the furrow cross-section, and draft power is draft × speed.
For a trailed implement the line of pull must pass through the centre of resistance and the hitch point.
TFC = w·v/10 ha/h; field efficiency = EFC/TFC = theoretical time/actual time.
Sprayer rate Q = 600q/(wv) L/ha; drill calibration turns the ground wheel for a known fraction of a hectare.
Fixed cost = depreciation (P − S)/L + interest on (P + S)/2 + housing, insurance and taxes; break-even area = fixed cost/(hire rate − variable cost per ha).

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

What is the theoretical field capacity of a 2.4 m wide harrow operated at 5 km/h, in ha/h?
Numerical answer — type the value.
Show answer
Answer: 1.2
TFC = w·v/10 = 2.4 × 5/10 = 1.2 ha/h, because 2.4 m × 5000 m/h = 12 000 m²/h = 1.2 ha/h. Dividing by 3.6 instead gives 3.33, which is a speed conversion, not an area rate.
A 3 m wide cultivator works at 4 km/h with a field efficiency of 80 per cent. How many hours does it need to cover 9.6 ha?
Numerical answer — type the value.
Show answer
Answer: 10
TFC = 3 × 4/10 = 1.2 ha/h; EFC = 0.8 × 1.2 = 0.96 ha/h; time = 9.6/0.96 = 10 h. Using TFC alone gives 8 h, which leaves out the turning and filling time that field efficiency accounts for.
A field that should take 5 h of effective work at the implement’s full width actually takes 6.25 h because of turning and adjustments. What is the field efficiency, in per cent?
Numerical answer — type the value.
Show answer
Answer: 80
Field efficiency = theoretical time/actual time = 5/6.25 = 0.80 = 80 per cent. Inverting the ratio gives 125 per cent, which is impossible for an efficiency, and 1.25/6.25 = 20 per cent is the lost-time fraction, not the efficiency.
A boom sprayer with a swath of 10 m travels at 4.8 km/h, and its nozzles together discharge 12 L/min. What is the application rate, in L/ha?
Numerical answer — type the value.
Show answer
Answer: 150
Area per minute = 10 m × 4800 m/60 min = 800 m²/min = 0.08 ha/min, so Q = 12/0.08 = 150 L/ha; the identity 600q/(wv) = 600 × 12/(10 × 4.8) gives the same. Leaving out the minutes-to-hours factor gives 2.5 L/ha, far too low.
A seed drill 2.0 m wide is driven by a ground wheel of 1.2 m diameter. How many revolutions of the wheel correspond to 1/25 ha during calibration (to one decimal place)?
Numerical answer — type the value.
Show answer
Answer: 53.1
1/25 ha = 400 m². One revolution covers π × 1.2 × 2.0 = 7.540 m², so N = 400/7.540 = 53.05 ≈ 53.1 revolutions. Using the radius in place of the diameter gives 106.1, twice the correct figure.
A mouldboard plough turns a furrow 30 cm wide and 15 cm deep against a draft of 3150 N. What is the unit draft, in N/cm²?
Numerical answer — type the value.
Show answer
Answer: 7
Unit draft = draft/(width × depth) = 3150/(30 × 15) = 3150/450 = 7 N/cm². Dividing by the width alone gives 105 N/cm, which is the specific draft per unit width, a different quantity.
An implement with a draft of 10 kN is pulled at 5 km/h. What is the draft power, in kW (to one decimal place)?
Numerical answer — type the value.
Show answer
Answer: 13.9
Power = force × speed = 10 kN × (5/3.6) m/s = 10 × 1.389 = 13.9 kW. Multiplying 10 × 5 directly gives 50, mixing kN with km/h without the 3.6 conversion.
A tractor costs ₹6,00,000, has a salvage value of ₹60,000 after a life of 10 years and is used 1000 h per year. Taking straight-line depreciation, interest at 10 per cent on the average investment, and housing, insurance and taxes at 2 per cent of the purchase price, what is the fixed cost per hour of use, in rupees?
Numerical answer — type the value.
Show answer
Answer: 99
Depreciation = (6,00,000 − 60,000)/10 = ₹54,000; interest = 0.10 × (6,00,000 + 60,000)/2 = ₹33,000; housing, insurance and taxes = 0.02 × 6,00,000 = ₹12,000. Total ₹99,000 a year over 1000 h = ₹99/h. Charging interest on the full price (₹60,000) instead of the average investment gives ₹126/h.
A machine has an annual fixed cost of ₹45,000 and a variable cost of ₹600 per hectare. Custom hiring the same work costs ₹1500 per hectare. Above what annual area, in hectares, does owning become cheaper than hiring?
Numerical answer — type the value.
Show answer
Answer: 50
At break-even 45,000 + 600A = 1500A, so A = 45,000/900 = 50 ha. Dividing the fixed cost by the hire charge alone (30 ha) forgets that the owner still pays the variable cost on every hectare.
The disc angle of a disc plough is the angle between:
1. the plane of the disc edge and the direction of travel
2. the plane of the disc and the vertical
3. the disc axle and the horizontal
4. two adjacent discs on the gang
Show answer
Answer: A — the plane of the disc edge and the direction of travel
The disc angle is measured in plan, between the plane of the disc’s cutting edge and the line of travel; the angle between the disc plane and the vertical is the tilt angle. Confusing the two is the standard trap, because both are adjusted to control penetration.
Which of the following are secondary tillage implements?
1. Disc harrow
2. Spike-tooth harrow
3. Subsoiler
4. Cultivator
Show answer
Answer: A — Disc harrow; B — Spike-tooth harrow; D — Cultivator
Harrows and cultivators work soil that has already been opened, breaking clods and preparing the seedbed, so they are secondary. The subsoiler breaks a hardpan below normal tillage depth in undisturbed soil — a primary operation, and a heavy one.
The threshing cylinder most suited to paddy, whose grain detaches easily and must not be broken, is the:
1. wire-loop type
2. spike-tooth type
3. hammer-mill type
4. plain drum without any elements
Show answer
Answer: A — wire-loop type
Wire loops strike the panicles and strip the grain with little rubbing, which suits a crop that threshes easily and cracks under heavy impact. Spike-tooth cylinders beat and tear, suiting wheat and sorghum where the straw is also broken up; a hammer mill is a grinding machine, not a thresher.
Variable-rate application of fertilizer in precision agriculture requires, at minimum:
1. a positioning receiver and a prescription map linked to a rate controller
2. a combine yield monitor on the same machine
3. a tractor of at least four-wheel drive
4. a uniform application rate over the field
Show answer
Answer: A — a positioning receiver and a prescription map linked to a rate controller
VRT needs to know where the machine is (GNSS) and what rate that spot should get (the prescription map), and a controller to change the rate on the go. A yield monitor helps make future maps but is not needed to apply one, and a uniform rate is exactly what VRT replaces.