The purchase of a tractor and associated equipment is a substantial investment. The result of improper size can be costly – – a tractor too small can result in long hours in the field, excessive delays and premature replacement.
A tractor too large can result in excessive operating and overhead costs. It is important to know how to determine the size and number of tractors needed for a farm operation. The ideal equipment should get the work completed on time at the lowest possible cost. The size of the largest tractor should be based on getting critical, high-horsepower jobs done within a specified time period.
FOUR WHEEL DRIVE, FRONT WHEEL ASSIST AND TWO WHEEL DRIVE TRACTORS
Tractors can be divided into 3 categories: 2-wheel drive, front-wheel assist or unequal 4-wheel drive, and equal 4-wheel drive tractors. Each one of these tractors has different tire configurations and different ballast requirements.
Two-Wheel Drive Tractors (2WD) are most commonly used in dry or upland farming situations and for transportation. They range in size from 5 HP – 200 HP and need 80% of the weight distributed over the rear axle to maximize traction. The biggest advantages of this type of tractor over other 4-wheel tractors are smaller turning circle, simplicity of design, fewer mechanical parts and lower purchase price. However, a 2WD tractor does not work at all well in wet, hilly and muddy conditions.
Front Wheel Assist (FWA) is commonly known as 4WD or unequal 4-wheel drive. It is the most popular 4-wheel tractor in many parts of the United States and worldwide. These tractors range in size from 5 HP – 240 HP and are capable of delivering between 50-55% of the rated power at the drawbar. Typically, between 75% and 85% of the rated engine HP is delivered to a rear PTO (Power Take-Off) on any diesel tractor. On a FWA tractor the front drive tires are smaller than the rear tires. These tractors require 40% of the weight distributed over the front axle and 60% over the rear axle. The major advantage in using this type of tractor is that it can deliver 10% more power to the ground at all 4 tires for the same fuel consumption, and thus has much better traction and flotation capabilities than 2-wheel tractors of the same size. FWA tractors normally cost about 15-35% more than the same horsepower two wheel drive tractor.
Equal Four-Wheel Drive (4WD) tractors have all four tires of equal size and range in size from 35 HP – 600 HP. This tractor type has the greatest power to weight ratio and can deliver between 55-60% of power at the drawbar. It is challenging to maneuver and often the size and expense makes these tractors impractical.
Several terms are used by equipment manufacturers to describe the capacity of their tractors. The basic definitions are:
Horsepower (HP) – – A measure of the rate with which work is done. By definition one horsepower is the amount of energy required to move 33,000 pounds a distance of one foot in a time span of one minute or likewise, to move 1 pound 33,000 feet in one minute. It is the measure of a machine’s ability to move a load.
Brake Horsepower – – The maximum power the engine can deliver without alterations. This figure is particularly useful in sizing stationary engines.
Power-Take-Off-Horsepower (PTO) – – The power as determined at the power-take-off shaft.
Draft (Drawbar) Horsepower – – The power transmitted by the tractor to the implement. The Nebraska Tractor Tests indicate that maximum drawbar horsepower will average approximately 85 percent of the maximum PTO horsepower for most tractors.
Of the various kinds of horsepower, maximum PTO horsepower is the one most commonly used in designating the size of a tractor. On tractors that do not have a PTO shaft, brake horsepower, or maximum drawbar hp ratings may be used.
Determining Minimum Horsepower Requirements
A suggested procedure for determining the minimum horsepower needed is:
Step 1. Determine the most critical field operation requiring implements with a high draft.
Step 2. From past experiences, estimate how many days are available to complete this critical field operation. If you plan to run a double shift be realistic about maintenance of the machine and the operator’s personal time.
Step 3. Calculate the capacity needed in acres per hour in order to get the job done within the time allotted.
Step 4. Determine the size of implement needed.
Step 5. Select a tractor of proper size to pull the implement. To do this:
1. Determine draft of implement (Table 2.),
2. Determine drawbar horsepower needed to pull implement, and
3. Determine PTO horsepower needed.
Obviously the most important decision is to determine the size of the largest tractor. Normally the size of the largest tractor should be based on getting the critical, high-horsepower jobs done within a specified time period. It follows, then, that as much use as possible should be made of the same tractor for other operations.
Keep in mind, also, that large tractors should be matched with large, heavy duty equipment to withstand the heavy loads created by the tillage implement. By the same token, overloading a tractor can lead to serious mechanical failure. This can lead to down time and when a tractor is down, all work associated with that tractor stops.
SELECTION OF TRACTOR SIZE
Using the procedure outlined, let’s determine the tractor size needed for a specific tillage operation.
SITUATION: A small farm with 60 acres of sandy soil planted in row crops. During the spring five weeks of calendar time is available to twice disk the land with a tandem disk harrow prior to planting. The owner has 5 days (Saturday’s) available to prepare the land. Number of productive hours per work day is assumed to 8.
How large should the tractor and harrow be in order to complete the soil preparation during this five week period?
Step 1. Determine the critical high draft tillage operation. In this example disking prior to planting is that operation.
Step 2. Determine available time. There are 35 days of calendar time allotted to this job. During this time span, 5 days (40 hours) are estimated to be available for field work.
Step 3. Determine Field Capacity Needed (Acres per Hour).
Step 4. Determine Width of Implement Needed
In order to disk 3.0 acres per hour at a speed of 5 mph, a disk of the following minimum width is needed:
A 6 foot disk would be bought to provide a margin for field capacity.
Step 5. Tractor Selection
(Source – http://www.caes.uga.edu/departments/bae/extension/pubs/documents/farm%20tractor.pdf)