Mowing efficency with a rotary mower

[tmc_31]

This thread reflects what happens when you sit on a tractor all day with a repetitive job to do. The mind starts to wander and thoughts tend to "How can this task be done easier/quicker/better/more efficiently".

Probably many inventions and implement modifications have come from that situation.
But heck, if you are having fun doing the job maybe you don't want it to be over in a hurry.

Ha, truer words were never spoken. It always amazes me the cool stuff I can think up while riding the tractor.

It seems that there is a lot of factors that affect efficiency. Quite a few that I had not thought of (hence why I started this thread). Ultimately, it comes down to putting the most money into my pocket on any given job.

Tim

[tmc_31]

As I am reading through you guys replies, I am pondering what I could have done to make this mowing job more efficient. For much of the job, I was using the 270deg turn method on the corners. I am fairly certain this not the most efficient method. Maybe a modified zamboni pattern would have been better. I spent some extra time dodging small mesquites that I will grub out later rather than mow down. Human factors played a role (not sure what I could have done differently there except not drinking as much coffee before mowing. Getting old is tough on the body). I have a 7' wide mower and I try to overlap about 1 rear tire width (1.5').

I am pretty sure that a wider mower would be more efficient. I have been toying with buying a 12' flex wing. According to my calculations, the 12' mower is more cost effective ( not including the initial expense) on mowing jobs of about 20 acres or more (compared to my 7' mower). It's that $10,000 hit on the front end that keeps me from going for it. Plus most of my mowing jobs to date have been 10 acres or less.

[smstonypoint]

I have seen this mowing time calculator at TractorData referenced in several threads: TractorData.com - Mowing with tractors.

Out of curiosity, I computed the field efficiency used in the calculator -- 90.94%. Seems a trifle optimistic.

Steve

[tmc_31]

Steve, I think that mowing calculator is what prompted me to develop my own mowing calculator. As I said I started out with a 90% efficiency factor. You are entirely correct, 90% or better is way optimistic. I am going to pay more attention to how I approach a mowing job in the future and try to be as efficient as possible. I will also track the efficiency of each job (it's pretty easy to do). It will be interesting to see how much improvement I can make.

regards,

Tim

[hawkeye08]

sorry if I missed it, but I always go much slower on the first mow than later after I know where things are..

[smstonypoint]

I am pretty sure that a wider mower would be more efficient. I have been toying with buying a 12' flex wing. According to my calculations, the 12' mower is more cost effective ( not including the initial expense) on mowing jobs of about 20 acres or more (compared to my 7' mower). It's that $10,000 hit on the front end that keeps me from going for it. Plus most of my mowing jobs to date have been 10 acres or less.

Tim,

Here's a simple formula that would provide an estimate of the acreage (Q*) you have to mow each year to have total costs (variable and fixed) be equal for the 7' and 12' cutters.

Let FC₁₂ and FC₇ represent the annual fixed costs ($/year) of the 12' and 7' option, respectively. The fixed costs include depreciation, interest on your investment, insurance, property taxes, etc. FC₁₂ should be higher than FC₇.

Let AVC₁₂ and AVC₇ represent the average variable costs ($/acre) of operating the 12' and 7' mowers, respectively. These variable costs include fuel, allowance for repairs and maintenance, and your labor rate. AVC₁₂ should be lower than AVC₇.

Calculate Q* = (FC₁₂ - FC₇)/(AVC₇-AVC₁₂).

If you expect to mow more than Q*, it pays to go with the 12' mower. If you expect to mow less that Q*, it pays to stay with the 7' mower.

Steve

[creekbend]

See http://www.extension.iastate.edu/Publications/PM696.pdf for a discussion of theoretical field capacity, measured in acres/hour, and computed as implement width (ft) * ground speed (MPH)/8.25.

Effective field capacity is computed as the theoretical field capacity * field efficiency (%).

The field efficiency is less than 100% because of the time spent turning, overlapping, etc.

I doubt whether the field efficiencies would differ much between the various implements, but the ground speeds may well differ.

The time of day, etc. may affect ground speed, but I speculate they would not have much effect on field efficiency.

Steve

Thanks for providing the link.

[tmc_31]

An interesting calculation Steve. Thanks for sharing. It clearly shows that I should be using a 12' mower. Now if only I can convince SWMBO.

Tim

[schmism]

An interesting calculation Steve. Thanks for sharing. It clearly shows that I should be using a 12' mower. Now if only I can convince SWMBO.

Tim

I was going to point out that there is a size of mower to size of area being mowed that is important to consider. EG a very large mower on a very small property is fairly inefficient with all the time that is taken up going back and picking up small areas that were missed while at the same time re-mowing areas you already covered. conversely a very small mower on a very large property you have inefficiency associated with overlap etc that builds up very quickly.

I wouldnt know how to do the calculation but someone out there im shure has. and it sounds like the above equation is close to what im talking about.

[Chilly807]

An interesting calculation Steve. Thanks for sharing. It clearly shows that I should be using a 12' mower. Now if only I can convince SWMBO.

Tim

I prefer to call 'em SWRUC.. She Who Remains Un-Convinced..

I actually don't think mine is like that, she thinks we should buy a backhoe attachment sooner rather than later. Go figure. I'm leaning toward an actual backhoe, but take what you can get, right?

Sean