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Productivity and Deck Size ?'s

Discussion in 'Homeowner Assistance Forum' started by dbear, Apr 12, 2006.

  1. dbear

    dbear LawnSite Senior Member
    Posts: 606

    Thinking of replacing my venerable Ransomes 61" three-wheeler currently used to maintain 1.5 acre lot with a ztr. Reasons being (1) age of Ransomes is beginning to be a concern with regard to unit reliability and also parts availability, (2) greater ease at adding collection unit, preferably a clam shell type, for spring/fall cleanups, (3) as I will be adding another 1.5 acre lot to my mowing responsibilities (mom-in-law), easier transport with full-size pickup, and (4) possibly cutting down on mowing time.

    According to my calculations (see Productivity formula below), my mower at max speed (6 mph) and 80% efficiency can cut right around 3 acres/hour, although in practice it usually take me about 1 hour to mow my property (is it possible to only be getting 40% efficiency?). A 48" ZTR at 9 mph and 80% can cut around 3.5 acres/hour.

    Now my questions:
    1. Do these numbers sound right?
    2. What is a realistic cutting ground speed with a commercial ztr (looking at CC Tanks with 12 mph max)?
    3. Price is definitely an issue, hence the 61" to 48" comparison; I would like to get the most bang for my buck, and still get a heavy commercial unit. If going with a 48" ztr, will I miss 12" of mower deck, or should I try, at all cost, to step up to a 54" ztr?

    Productivity = ((Deck Width (in)/12 * Ground Speed (mph)*Ft per Mile) / ft^2 per acre) * Efficiency (%)
    Ft per mile = 5280
    ft^2 per acre = 43559.66

  2. Lumberjack

    Lumberjack LawnSite Member
    Posts: 180

    the production charts assume certain things that may not be true in your yard. Your actual cutting time will depend on the yards layout, how rough it is and how many obstacles you have. as I understand it a z mower of the same size as a tractor will mow about a third faster so your 60 minutes should drop to about 40 minutes with a 61" z. At 48" your cut time will stay about the same. You might want to take a tour of the yard and look for things that slow the tractor down with forced stops/reverses ect. and check for areas that may slow the z down like rough terrain. On wide open cutting areas the charts are telling you that the smaller z mower isnt going to cut any faster.
  3. dbear

    dbear LawnSite Senior Member
    Posts: 606

    Thanks for the reply, Lumberjack...
  4. dfischer

    dfischer LawnSite Member
    from Il
    Posts: 114

    intriguing idea. I do think the formula has several flaws, but may not fully understand what your trying to do. It SEEMS like you're trying to take the efficiency factor and account for starting and stopping, cutting overlap, and the like. But:

    1) You should'nt be applying the efficiency factor to the entire formula as you are. At present it's also reducing your area, and that's screwing up your results.

    2) I don't think applying one factor will give you a good result. An example.

    10x10=100, and an overall 80% efficiency correction would give you (your way) 80.

    But if you applied the 80% to both factors, it would be (10x.8) x (10x.8) = 64. The only point being some care needs to be applied to make sure your factor is being applied in a way that works. If you don't mind, I'm going to play with this for a moment and see if I see an alternate formula that might work better.
  5. dbear

    dbear LawnSite Senior Member
    Posts: 606

    dfischer, go for it! I was attempting to duplicate numbers from the 2005 CC Commercial sales literature, but failed to do so. CC claims a 48" ZTR running at 9 mph with 80% efficiency can do 4.1 acres/hour, not the 3.5 I come up with.

    You got me thinking so I redid my formula to be:
    Productivity = (((Deck Width (in)/12) * Efficiency (%)) * Ground Speed (mph) * Ft per Mile) / ft^2 per acre

    thereby applying the 80% directly to the deck width. The interesting thing is this revision gives the same values as rev 1.

    I see if I can rework it again...
  6. lucforce

    lucforce LawnSite Member
    Posts: 223

    let me check this out
  7. lucforce

    lucforce LawnSite Member
    Posts: 223

    my math gets me to 3.5 acres per hour as well
  8. dbear

    dbear LawnSite Senior Member
    Posts: 606

    I'd really like to know the formula CC used. I reworked it again a couple of other ways - applying efficiency directly to ground speed, and also only directly to the area covered - and still got the same numbers that don't match even their numbers for 100% efficiency.

    I then decided to take efficiency completely out of it. To further simplify, I made my ground speed 1 mph, basicly making it 100% efficiency for one hour at 1 mph. That makes the formula, for all intents and purposes, a simple area calculation with a conversion for units:

    Productivity = ((Deck Size/12) * 5280 ft per mile) / 43,560 ft^2 per acre

    This worked out, for 60" deck, to be 0.6 acres per hour @ 100% efficiency and 1 mph. Using this factor, if you then multiply by your actual mowing speed, let's say the 9 mph CC uses, this gives us the value of 5.5 acres per hour, not the 6.4 acres stated in their literature.

    I've heard of misprints, but their entire chart would be a misprint!
  9. dfischer

    dfischer LawnSite Member
    from Il
    Posts: 114

    Well, I don't think I can defend marketing numbers, so I won't try. A reality based number might be found from:

    (Deck Width-Cutting Overlap)/12 X (Average Cutting Speed X 5280/60 X Minutes@Speed per Hour)

    I'll use my data:

    60' deck, I probably overlap close to 6" per pass, I estimate I'm cutting @ 5
    mph average, and I think I spend maybe 10 minutes in either real slow manuevering (house, tree's etc) or other non-avg. speed cutting things (reversing, turning in a non-z turn (don't want to tear the grass).

    I find it takes me about 20 minutes to cut about 30k sq. feet, but I see the formula shows (I took my sq feet/formula result*60) it should take 18 minutes. Still close enough...

    Don't know if it helps, but I took a shot at it. Good luck..

  10. dfischer

    dfischer LawnSite Member
    from Il
    Posts: 114

    Yeah I can't get their numbers to work either. Whats really funny is if you take the 80% literally and apply it to my formula.

    80 %cut width is 9.6" overlap. 80% of speed is 7.2 mph. 80% of time spend cutting at speed is 48 minutes. That gives 2.23 acres per hour.

    Funny how it adds up ain't it.

    btw, it seems you need to use about 94% of any ONE of ther factors I show to get CC's numbers. Take 94% of 9 mph, as an example and plug it in with 0 overlap and 60 minutes spent at speed (right, that'll happen)

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