Estimating tractor power needs

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Tractors throughout Australia tend to be over-powered for the tasks they typically perform. This is likely because the most common approach farmers acquiring tractors have taken to date has been to opt for those with the highest ‘horsepower-per-dollar’ they can afford. This outdated tactic, however, may be a false economy, since smaller tractors may be as capable of handling the major tasks farmers require, and will save those who use them considerable amounts of fuel.

The following guidelines present a method by which you can determine the real power requirements of your farm’s specific field operations. Nonetheless, it is imperative that you obtain additional advice from experts and local industry leaders so you can adapt the method to your specific needs and situation.

Step 1: Identify your ‘priority critical field operation’

Complete a thorough review of your tractor’s requirements in relation to its intended usage.

Begin by listing all the tasks you’ll need the new machine to perform, such as: fertiliser/chemical application, tillage, loader work, PTO-driven operations, etc. Consider that an occasional high-horsepower task may not justify dragging extra tonnage around on a daily basis. It may be more cost-effective and fuel-efficient to subcontract out such tasks.

Review the list of tasks that the machine will perform and identify the most critical operation that will require an implement with the highest draft force.[1] You may wish to use Table 1 to help you estimate the draft force from various implements.

Step 2: Estimate the time you’ll have available to complete this priority task

Determine the time period available (or the period expected) to complete your priority critical field operation.

As an illustration, your requirements may be for tillage work to be completed over the span of a week, working Monday through Friday, eight hours a day. This equates to having 40 hours available for the job.

Step 3: Find the work rate (hectares per hour)

Once you have determined the time available, you can calculate the required hectares-per-hour (ha/h) rate using your farm’s dimensions. Continuing from our previous example, assume that the size of the field for tillage is 120 hectares. This means that the task must operate at the following rate:

Step 4: Determine the width of the implement required

The next step is to use the work rate of your operation (i.e. how many hectares you need to cover per hour) and the expected working speed of the job to determine the width of the implement required. This is given by the following equation:

The ‘work factor’ 11.8 is a dimensionless number that represents the number of hours required to cover an entire hectare when using a metre-wide implement, at 1 km/hr, assuming losses of 18 percent from overlapping, turning and other field inefficiencies[1]. Following from the previous example, let’s establish that we’ll be using an offset disc harrow to conduct primary tillage at 8 km/h. The required width of the offset disc plough can therefore be determined as follows:

 

Alternatively…

If you already know the size of the implement that will be used with your machine, you may wish to determine the work rate your equipment will allow for. You may do this using the following equation:

If the resulting work rate is too low for your requirements, consider obtaining a wider implement

Step 5: Determine soil resistance

When conducting your priority field operation, identify the type of soil your field will have (i.e. clay, loamy or sandy). Use this information and Table 1 to find the resistance that will be offered by the soil per unit of width and depth (i.e. its draft force). Then calculate your total soil resistance by multiplying this measure by your expected working depth and the full length of the implement.

Soil resistance = implement width x working depth x resistance per width

Expanding from our previous example, let’s assume that the soil type is sandy, that our offset disc harrow is for primary tillage and that the working depth is 10 centimetres. Table 1 tells us that this particular implement for this soil type will present a draft force of 598 Newtons per metre of implement width, per centimetre of depth. Hence, for our total width of 4.425 metres and the 10-centimetre working depth, we have:

Step 6: Determine power required at the drawbar

Our next step is to equate this pulling force into the power required at the drawbar from the machine. This is obtained by using the following equation:

The number 3,600 is a required conversion factor that must be used if the working speed is provided in km/hour. Alternatively, the equation can be simplified if the working speed is known in metres per second, as so:

Following, once more, from our previous example, we determine that the drawbar power required is:

Table 1: Draft and power requirements for tillage and seeding implements Adapted by NSW Farmers from (Williams, 2007).

Implement

Unit

speed km/h

Draft force for soil type (N/unit/cm depth)

Clay

Loamy

Sandy

Major tillage tools

  Subsoiler/manure injector

Narrow point

tools

8

517

361

233

12-inch winged point

tools

8

669

468

301

Moldboard plough

metre

7

1,281

896

580

Chisel plough

2-inch straight point

tools

8

201

172

131

3-inch shovel/14-inch sweep

tools

9

243

207

159

4-inch twisted shovel

tools

9

280

238

182

Sweep plough

Primary tillage

metre

8

781

666

511

Secondary tillage

metre

8

517

437

333

Disk harrow, tandem

Primary tillage

metre

9

672

592

529

Secondary tillage

metre

9

408

356

316

Disk harrow, offset

Primary tillage

metre

8

770

672

598

Secondary tillage

metre

8

471

414

368

Disk gang, single

Primary tillage

metre

9

224

195

172

Secondary tillage

metre

9

155

132

121

Coulters

Smooth or ripple

tools

8

95

84

74

Bubble or flute

tools

8

114

100

89

Field cultivator

Primary tillage

tools

8

88

75

58

Secondary tillage

tools

8

61

53

40

Row crop cultivator

S-tine

rows

8

226

191

147

C-shank

rows

8

419

356

271

No-till

rows

8

730

620

475

Rod weeder

metre

7

362

310

236

Disk-bedder

rows

8

366

315

285

Minor tillage tools

Rotary hoe

metre

11

305

305

305

Coil tine harrow

metre

8

115

115

115

Spike tooth harrow

metre

8

299

299

299

Spring tooth harrow

metre

8

1,046

1,046

1,046

Roller packer

metre

8

345

345

345

Roller harrow

metre

8

1,551

1,551

1,551

Land plane

metre

8

4,585

4,585

4,585

Seeding implements

Row crop planter, prepared seedbed

Mounted – seeding only

rows

8

242

242

242

Drawn – seeding only

rows

8

438

438

438

Drawn – seed, fertiliser, herbicides

rows

8

767

767

767

Row crop planter, no-till

Seed, fertiliser, herbicides –
1 fluted coulter/row 

rows

8

898

898

898

Row crop planter, zone-till

Seed, fertiliser, herbicides –
3 fluted coulter/row

rows

8

1,809

1,809

1,809

Grain drill w/press wheels

< 6.5 feet drill width

rows

8

198

198

198

6.5 to 10 feet drill width

rows

8

147

147

147

> 10 feet drill width

rows

8

54

54

54

Grain drill, no-till

1 fluted coulter/row

rows

8

378

378

378

Hoe drill

Primary tillage

metre

8

3,620

3,620

3,620

Secondary tillage

metre

8

1,724

1,724

1,724

Pneumatic drill

metre

10

2,155

2,155

2,155

Step 7: Determine the PTO power required

The final step is to determine the power that your machine should have at the power-take-off point (PTO) so that it can achieve the required power at the drawbar. This is calculated using a rule-of-thumb multiplying factor, which takes into account the type of soil condition you will experience. These multiplying factors are shown in Table 2 (below).

Table 2: PTO power multiplication factors for different soil conditions. From (Williams, 2007).

Soil condition

Multiply drawbar kW by

Firm, untilled soil

1.5

Previously tilled soil

1.8

Soft or sandy soil

2.1

For our running example, we will assume that conditions are for untilled soil. This means that our final PTO power requirement is given by:

Step 8: Adjust for further considerations

Before finalising your goal power, consider the following:

  • The outcome produced by the previous steps is representative of the minimum power requirements for the set conditions that were inputted into the calculations. Different soil conditions, inefficiencies and various other set-up variables (hydraulics, driving methods, etc.) will play a role, and can substantially influence the real power that will be available and that will be required for a given task.
  • To operate correctly, certain implements may require minimum PTO power and engine speeds. Consult your implement’s specifications to identify such cases.
  • Continuously using a machine that is underpowered for the tasks it undertakes will damage it and decrease its lifespan.
  • When taking advice or considering power requirements, it’s important to remember that most engines are at their most fuel-efficient when working at 80 percent of their rated ‘max’ power output. Operation at or around 80 percent of full power will provide both optimal fuel efficiency and the promotion of a long, productive service life. Likewise, running an engine at light loads (30–50% of rated capacity) may not be the most fuel-efficient method for completing the task. Consider using a lower-powered tractor where possible.

Given this, we recommend that you increase the goal power calculated in the previous steps by 10 to 15 percent.

However, as noted previously, experts say that tractors throughout Australia tend to be oversized for the tasks they perform. Moreover, running an engine at light loads (30–50% of rated capacity) may not be the most fuel-efficient method for completing a task. It is therefore crucial that you consider the realistic scenarios you expect to perform with your tractor, and use a lower-powered tractor wherever possible.

PTO versus engine power

Be careful with power metrics as they may be expressed differently for different machines! For 2WD and MFWD tractors, rated power is usually stated at the power take-off (PTO). For 4WD tractors, however, rated power is typically quoted at the engine (engine power). PTO power is approximately equal to 85 percent of engine power (Michelin North America, Inc., 2001).

Further information

Calculator tool

The NSW Farmers’ Association tractor power calculation tool is available to assist you in following the guidelines instanced in this paper, and is available in an easy spreadsheet format. You may download it by visiting the following link:

http://www.aginnovators.org.au/sites/default/files/AgInnovators - Tractor Power Calculator.xlsx

References

Michelin North America, Inc., 2001. Optimum Tire Performance Worksheet For Tractors, s.l.: s.n.

Williams, P. E. S. a. E. J., 2007. What Size Farm Tractor Do I Need?. [Online]
 

[1] You may calculate this factor yourself for differing efficiency values using the following equation:

e.g.: For 18% field efficiency losses, 

[1] Draft is a measure of the force imposed by the implement on the machine as it travels through the soil.

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