## The Optimum Machine Size

Optimizing the machine size is one technique for reducing production costs. When the machine is too small, the operating costs may be higher, and its reduced capacity will require a greater number of hours to complete an operation. It is also possible to exceed the design forces when the machine size is small for the available power. When the machine is too large, power is the limiting factor and the machine may not produce the desired results, or the life of the tractor may be shortened, because modern tractors are designed to operate at lighter drafts and faster speeds.

The optimum machine size can be selected from two different points of view: it can be based on the amount of time available to complete an operation, or it can be based on the amount of power available from the tractor. The process used depends on the situation.

When the limiting factor is time, the recommendation is to determine the available time (timeliness) for completing the operation and then determine the size of machine required. Once the machine size is determined, the required tractor size can be determined. A high priority on timeliness may result in a calculation that would require a larger tractor to power the machine. A full discussion of timeliness is beyond the scope of this text, but we will illustrate this concept for determining the size of machine and tractors.

When the limiting factor is power, the recommended procedure is to determine the maximum size of machine that can be powered by the tractor. When power is the limiting factor, the calculations may result in a time requirement for completion of the activity that is not realistic. The first example problem will match tractor and machine size base on timeliness.

Problem: A general recommendation is that the primary tillage for planting corn should not take longer than 1 week. If you anticipate planting 500 acres of corn and only work 12.0 hr per day and 6 days per week, what size of moldboard plow do you need if you can plow at typical speed and efficiency?

Solution: The first step is to determine the capacity of the plow, CE (ac/hr) required to plow the ground in the available time. This can be accomplished by using units cancellation:

To plow 500 acres in 1 week, the plow must have a width and be operated at a speed that will produce a capacity of 6.9 ac/hr. The second step is to determine the width of the plow. This is accomplished by remembering that the capacity of field equipment can be determined by using the effective capacity equation, Chapter 9.

8.25

Rearranging this equation to solve for width (w) and looking up the typical speed and efficiency in Appendix IV results in the width of the plow:

Ce x 8.25 6.90 hT x 56.925 w = —-=-hr-=-= 15.8125 or 16 ft

To be able to plow the field in 1 week, assuming that the plow will be operated at typical speed and efficiency, you will require a plow with an effective width of 16 ft. Plows are not usually sized by feet of width; instead they are sized by the effective width of each plow bottom and the number of bottoms. Sixteen feet is the same as a 12-16 plow (12 bottoms with 16 inches per bottom).

The next step is to determine the size of the tractor needed to pull the plow. This is accomplished by determining the drawbar horsepower required. The first step is to replace the term force in the drawbar power equation with the term draft (DF) or draft force.

PDbHP =

The term draft is used to describe the amount of force required to pull a machine. Appendix V gives the American Society of Agricultural and Biological Engineers (ASABE) standard values for the draft of agricultural machines. Notice that more than one type of unit is used for draft. That is, the draft of plows is listed as pounds of force per square inch (cross sectional area) plus a speed factor, spring tooth harrows are listed as pounds per foot of width, and some rotary power machines are given in horsepower. It is important to pay close attention to the units being used so that the equation can be modified. The draft is calculated by the following equation:

DP = Fi x [Ci + (C2 x V) + (C3 x V2)] x w x Td where DF = Draft in lb or N; Fi = Soil texture factor. For fine soils i = 1, for mediums soils i = 2, and for coarse textured soils i = 3; C1, C2, C3 = machine specific parameters from Appendix V; W = machine width, ft or m; Td = Tillage depth, in or cm.

Problem: What size tractor (Dbhp) will be required to pull the 12-16 (16.0 ft) plow that was calculated in the previous problem?

Solution: The first step is to find the draft of the plow in Appendix V. Notice that the draft for a plow varies with soil type. We will assume that the plow will be used in a clay soil, which is a fine textured soil. From Appendix V, for customary units we find the following parameters:

The tillage depth, Td = 6 in, and the plow width, w = 16 ft. Selecting a depth of 6.00 inches and solving for draft:

2 16 in 1ft

= 1.0[113 + (0 x 4.5) + (2.3 x 4.52)]12 bottoms x -x -x 6.0 in bottom 12 in

Solve for drawbar power in hp

PDbHP =

With typical speed and efficiency, and operating 6 inches deep in clay soil, the 12-16 plow will require a tractor with 180 drawbar horsepower.

One result of using this method to determine the optimum size of machines is now apparent. The calculated tractor size may be larger than the size of tractor that you have. What are your alternatives if you do not want to purchase a larger tractor? Plowing is usually the highest draft tillage operation on a farm so a tractor could be leased for the time required to do the plowing. The plowing could be hired out. Another option, which might allow you to complete the plowing with the tractor available, is to adjust one or more of the factors in the drawbar horsepower equation. A common practice is to extend the amount of time available to complete the tillage operation. It also is possible to reduce the draft of the plow by reducing the operating depth or to reduce the speed. In some situations one of these two choices is the best option, but for this problem the available time will be adjusted.

Problem: Assume that the largest tractor you have available to pull the plow in the previous example is 100 drawbar horsepower. If this tractor is used, how many 12 hr days will it take to plow the 500 acres?

Solution: The first step is to determine the size of plow the 100 Dbhp tractor can pull. This is accomplished by rearranging the drawbar horsepower equation to solve for width [because Area (A) = Width(w) x Depth (d)]:

Dbhp x 375 100.0 Dbhp x 375

Under the conditions set up in this problem, a 100.0 Dbhp tractor is capable of pulling a plow that is 99.2 inches wide. A 99.2-inch wide plow would be 6.2 or 6, 16 inch bottoms. Therefore, a 6-16 plow (8.00 ft) plow is used. The next step is to determine the capacity of a plow 8.00 ft wide.

Determining the number of days using units cancellation: 1 day 1 hr days =-- x -x 500 ac = 11.9047... or 11.9 days y 12 hr 3.5 ac y

This means using the 100 Dbhp instead of the 193 Dbhp, will take 6 extra days (12 days - 6 days = 6 days) to plow the field.

Another situation that often arises in agriculture is determining the size of machine that a tractor can power.

Problem: The opportunity to buy a 185 PTO horsepower tractor is too good to pass up. What size (width) of offset disk harrow (Wdh) and field cultivator (Wfc) will you need for your new tractor? The disc harrow is operated 6 inch deep. Assume a clay soil and the lowest typical value for speed and efficiency. Note: remember to convert from PTO power to drawbar power using the 86% rule.

Solution: To solve this problem PTO power must be changed to drawbar power.

Draft can be calculated by rearranging the drawbar power equation.

375 ^DbHP X 375

159 hp x 375

The following parameters for the ASABE draft equation were obtained from Appendix V:

F1 = 1.0 (fine textured, clay, soil) C = 62 C2 = 5.4 C3 = 0.

Rearrange the ASABE equation for draft and solve for the machine width (W).

The 185 PTO horsepower tractor will be able to pull an offset disk harrow up to 35 ft wide in a clay soil when operating at the low end of the typical speed and efficiency ranges.

The draft of a secondary tillage field cultivator is calculated as pounds of force per tool (shovel, sweep etc.) plus a speed factor. Therefore, the number of shanks determines the size of the cultivator. For a cultivator, the drawbar horsepower equation is converted to:

The first step in determining the size of cultivator than can be pulled by the 159 hp tractor is determining the number of shanks for the cultivator. Assume the tillage depth Td is 3 inches. From Appendix V, the machine parameters for a secondary tillage field cultivator are:

Solving the ASABE draft equation for w (in this case number of tools or shovels):

= 1.0 x (19 + (1.8 x 6) + (0 x 62)) x 3 _ 17,000 = 89.4

= 190.156 or 191 tools or shanks

The 185 PTO horsepower tractor will be able to pull a field cultivator with 191 shanks. The width of the cultivator depends on the number of rows of shanks and the spacing between each shank. For example, one possible arrangement would be a cultivator with three rows of staggered shanks, the second and third row staggered equal distance between the front row shanks, spaced 21 inches apart. This arrangement would have 191 shanks (64,63, and 64), and be 112 ft wide (64 shanks or tools multiplied by 21 inches and divided by 12 in/ft). Note: the tape measure width from shank to shank would be (63 x 21 inches) divided by 12 in/ft or 110.25 ft. The effective width includes one half of the spacing on each end of the row. 110.25 + 10.5 + 10.5 = 112 ft.

Appendix V presents the draft of equipment in the form found in the ASABE Standards. Other sources may provide draft figures for machines in pounds per foot. If these values are used, the drawbar horsepower equation does not need to be modified each time, but the calculations will not be as accurate as this example problem.