Mowing efficency with a rotary mower

[tmc_31]

Hi all,

I was finishing up mowing 8.5 acres this afternoon that I had started yesterday. I got to wondering how efficient I was being in getting the job done. I've been mowing small acreages now for a couple of years so this is not my first rodeo. Some time ago I put together a spreadsheet to help me price my mowing jobs. One of the factors in the calculation is mowing efficiency. If one were to mow a single strip the width of the mower with no stopping or time outs, one could say he was mowing at 100% efficiency.

Everybody who mows knows that mowing at 100% efficiency would be a very rare thing. There are many factors that affect how efficiently we mow such as overlap, speed, obstacles in the field, mowing pattern and the number of times during the mowing cycle you have to stop and take a leak etc. (human factors).

When I built my spreadsheet I used 90% as an efficiency factor. And I have based my pricing accordingly. I haven't really wondered to much about how efficient I was mowing until this afternoon. Based on my spreadsheet, I should have been able to mow that field in 2.7 hours at 90%. It took me 3.4 hours. This would be an efficiency of more like 70%.

The question is, what is a reasonable efficiency to shoot for? And what are the things one can do to improve efficiency?

Merry Christmas,

Tim

 

[smstonypoint]

Tim,

I stumbled upon this info from Exmark: Exmark Lawn Equipment. They say 80% is reasonable for landscapers, who are presumably using ZTRs but who likely have to deal with more obstacles than is typical for someone mowing a field.

Iowa State provides estimates of average field efficiencies for selected machines here: http://www.extension.iastate.edu/Publications/PM696.pdf. They don't list rotary mowers, but the estimates for mower conditioners range from 80 to 83%. The highest efficiencies are only 85% (for offset disks and seedbed conditioners).

BTW, there was a thread a few months ago dealing with the use of a Zamboni pattern for increasing mowing efficiency. I will try to find it if you are interested.

Steve

 

[RaydaKub]

I would think you'd have to take into account how thick the grass is. In a drought year, you could drive full speed and knock down the weeds. In a reasonably wet year, you might have to slow way down. Yet you're actually mowing a lot more by volume, though not be area.

 

[tmc_31]

Hey guys,

Thanks for the links Steve, I have a ZTR that I use in our yard. Because of it's extreme maneuverability, I would guess that the efficiency would be very high compared to a shredder on the back of a tractor. Most of the shredding jobs that I do have varying degrees of obstacles that slow down the mowing, unlike a cultivated field might. Still, the comparison to say a haybine (mower conditioner) is interesting. I think most agricultural shredding/mowing that is done is in pasture rather than cultivated field. I know my work has been.

Raydacub, you are right. The thickness of what you are mowing will have great effect on how much ground you can cover in a given time. In my calculation, speed is a separate factor from efficiency. I have found that I will vary my speed to maintain the quality of cut to accommodate heavier growth. If the ground is rough I may have to slow down even further just to stay on the tractor. In our area most of the pasture is mesquite, prickly pear and native grasses. It is not usually very thick. Most of the time I can get 4.5-5 mph and do a reasonably good job of cutting. I have mowed a few cultivated fields at as much as 6 mph. Much more that an the quality of the cut begins to suffer.

Tim

 

[keegs]

Interesting post...thanks for sharing. I haven't done much hay mowing yet...but plan to start this summer on a few fields we have in Maine.

I seems the metric you're most interested in is time. Perhaps you've got or plan to have someone do the mowing...?? I think I'd add some others....e.g. fuel ....how much does it cost to mow 40 acres with a 55 hp gas versus a 45 hp diesel tractor...in mid-July versus late-Sept?

As well, how does the business end of the operation i.e. the kind of mower, impact efficiency? e.g. How efficient is a 7' sickle bar mower compared to a 7' brush hog .... a 7' disk mower.... ? understanding of course that each have different mowing characteristics.

What's the the average service time on this make/ model mowing equipment component compared to some other make/model and how do their costs compare.

How does time of day, temp, weather conditions.... factor into efficiency.

 

[riptides]

I calculate my efficency on the tractor as time spend away from other chores. if you know what I mean.

 

[smstonypoint]

As well, how does the business end of the operation i.e. the kind of mower, impact efficiency? e.g. How efficient is a 7' sickle bar mower compared to a 7' brush hog .... a 7' disk mower.... ? understanding of course that each have different mowing characteristics.


How does time of day, temp, weather conditions.... factor into efficiency.

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

 

[gwdixon]

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. :laughing: :laughing:

 

[Chilly807]

Part way through reading the first post, I saw 90% and immediately thought the number was likely closer to 70% for me. Funny that number came up at the end. It varies a lot depending on the terrain too, for example if you have to mow up a hill or down because of the slope.

Generally speaking, the longer the run the more efficient it'll be, less turning time in relation to actual mowing. With the gear I have, which is 30 HP and a 5 foot cutter, I average about an acre per hour or a little better, up to about an acre and a half depending on the field shape and condition. I find mowing faster than about 4 MPH doesn't do a good job, and will pound the crap out of you anyway on most fields. That drops to 3 MPH or less on rough fields. The tractor burns roughly a gallon of diesel per hour in average conditions.

I think there are too many variables to have any kind of meaningful formula other than field size and condition. If I were pricing jobs, I'd base it on worst-case scenarios for the estimates, and tell the customer it MAY be less at the end of the day depending on how it goes. As an example, a 10 acre field will take 10 hours at whatever hourly rate, which is the most you'll pay. I don't do this for a living either, so if I scared some away by overpricing the job it wouldn't bother me in the least.

Sean

 

[LD1]

It all depends on how you define efficiency. And for whatever reason, it always works out FAR less than I like, so I dont even bother trying to figure it anymore. All I know is I can cut at a rate of between 1-1/2 and 2 acres per hour, depending on how thick it is, and I price according to that after seeing the job. If I quote over the phone, I assume worst case.

What I mean by efficiency being way worse than one things....take Sean's example above. He has the same tractor as me, and says he can do 1 acre per hour or a little better at ~4MPH. Which is 4th gear on our tractors.

4 miles in an hour x 5280 ft x 5' cutter = roughly 2.4 acres. and he can "actually" do 1.5 (best case) per hour. Thats only about a 63% efficiency.

One of MY examples. My largest job was a 13 acre wide open field. I did manage to do it in 6.5-7 hours and maintain a 2 acre per hour rate.

But, I was using my 6' cutter, and in 5th gear most of the time which is 5MPH.

5mph x 5280 x 6' = Tells me I "should" have been able to do 3.6 per hour @ 100% 2/3.6 puts me at a 56% eff.

All in all, I say 60% is probabally pretty average. But again, I really dont figure eff. All I know is under reasonable conditions, how many acres per hour I can do, and price accordingly. 80% is VERY unlikely to ever happen with a bushhog.

 

[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. :laughing: :laughing:

Ha, truer words were never spoken:drink:. 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 FC[SUB]12[/SUB] and FC[SUB]7[/SUB] 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. FC[SUB]12[/SUB] should be higher than FC[SUB]7[/SUB].

Let AVC[SUB]12[/SUB] and AVC[SUB]7[/SUB] 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. AVC[SUB]12[/SUB] should be lower than AVC[SUB]7[/SUB].

Calculate Q* = (FC[SUB]12[/SUB] - FC[SUB]7[/SUB])/(AVC[SUB]7[/SUB]-AVC[SUB]12[/SUB]).

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..:laughing:

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