Flail Mower Let's talk flail mowers

[jimmysisson]

So, I guess the upshot is I should fashion a rear debris shield for my Mott SHD88 flail? Right now it's open, just as I bought it a couple years ago well used. It does produce (and spread all over) lots of fine clippings, and cuts grass very well as is. The rear is open about 8-10" so I guess it doesn't develop much of a gradient, if its supposed to.
It would be great not to emerge from mowing with inches of clippings on the mower, drawbar, PTO shield, etc.



This is one section not cleaned off yet, after one hour cutting 16" grasses.
Jim

 

[jimmysisson]

Just checked, and it's 13" open. Guess I'll close itin and see how that works.




Jim. Hmm, hope to get rid of 2nd pic

 

[hunt4570]

Just checked, and it's 13" open. Guess I'll close itin and see how that works.


View attachment 433807

Jim. Hmm, hope to get rid of 2nd pic

So you have tires back there instead of a roller? I like that!

 

[whirly]

Get a piece of old conveyor belt and bolt it on. They also sell a "rubber" flap for it. That is what mine came with but I still get chaff on the top because my roller is set high.

 

[IslandTractor]

The only mention I saw of suction was when they were talking about flails with scoop knives, but I didn't try to go over every line. Then again, we know that promotional materials are always based upon scientific data right? :laughing:

Well, just don't tell them that LouieJunior can't move feathers with his flail...

Here you go GMan, an honest to goodness experimental data containing article that looks at power needed to generate rotation and VACUUM in a commercial flail mower. You might have fun working out the calculations to double check theirs.

Note that while this article (PDF link below) is in reference to a commercial flail, the drawings indicate they are using the typical Y style blades not the scoop blades. And, all kidding aside, virtually ANY blade and mount will generate rotation of air when spinning at 60-90 mph.


"a model for determining the power in kW at the PTO:"

(SORRY TBN cut and paste didn't include the elegant equation but you can see it in the article here: http://www.scielo.cl/pdf/ciagr/v36n1/art05.pdf)

where, n is the total number of cutting blades, p the air density (p = 1,225 kgｷm-3), Fa the surface facing the air during the cutting blades' move-ment (m2), Cx the non-dimensional coefficient of air resistance that includes the effects of the cutting blade shape (Cx = 0,1.. .1), R the radius of the extreme cutting blades (m), r the ratio of the cutting apparatus of the shredder (m), ωthe angular speed of the cutting apparatus (rad-s-1), Y the mass of remnants in humid base (kgｷm-2), B the width of machine work (m), V the shredder's work speed (mｷs-1), M? the moment of the friction force in the supports (Nm), ns the number of cutting blades cutting the stemｷssimultaneously, ε the work consumed for cutting per area unit (Jｷm-2), bc the projection in the horizontal plane of the cut width of a cutting blade (m), k the relationship between the actual cutting area and the working area, Rh the distance from the rotor's cutting zone (m), and η PTO PTO the efficiency of the transmission from the tractor's PTO to the cutting mechanism of the shredder.

The equation that determines the power needed to overcome the opposing forces during the vacuum work (air resistance and friction in the supports of the cut mechanism) and the power needed to make the technological process (cutting and spreading of the shredded material) can be stated as:

(SEE PDF for equation)

where a, c, b and d are coefficients related to the power losses by air resistance, spread of shredded material, and the moment of the friction force in the supports and cut, respectively.

Have fun!:thumbsup:

 

[GManBart]

Here you go GMan, an honest to goodness experimental data containing article that looks at power needed to generate rotation and VACUUM in a commercial flail mower. You might have fun working out the calculations to double check theirs.



GMan, found an article for you with real live experimental data to determine power needed to generate both rotation and vacuum with a flail mower. Get out your calculator to check their math.
Note that while this article (PDF link below) is in reference to a commercial flail, the drawings indicate they are using the typical Y style blades not the scoop blades. And, all kidding aside, virtually ANY blade and mount will generate rotation of air when spinning at 60-90 mph.


a model for determining the power in kW at the PTO:

(SORRY TBN cut and paste didn't include the elegant equation but you can see it in the article here: http://www.scielo.cl/pdf/ciagr/v36n1/art05.pdf)

where, n is the total number of cutting blades, p the air density (p = 1,225 kgｷm-3), Fa the surface facing the air during the cutting blades' move-ment (m2), Cx the non-dimensional coefficient of air resistance that includes the effects of the cutting blade shape (Cx = 0,1.. .1), R the radius of the extreme cutting blades (m), r the ratio of the cutting apparatus of the shredder (m), ωthe angular speed of the cutting apparatus (rad-s-1), Y the mass of remnants in humid base (kgｷm-2), B the width of machine work (m), V the shredder's work speed (mｷs-1), M? the moment of the friction force in the supports (Nm), ns the number of cutting blades cutting the stemｷssimultaneously, ε the work consumed for cutting per area unit (Jｷm-2), bc the projection in the horizontal plane of the cut width of a cutting blade (m), k the relationship between the actual cutting area and the working area, Rh the distance from the rotor's cutting zone (m), and η PTO PTO the efficiency of the transmission from the tractor's PTO to the cutting mechanism of the shredder.

The equation that determines the power needed to overcome the opposing forces during the vacuum work (air resistance and friction in the supports of the cut mechanism) and the power needed to make the technological process (cutting and spreading of the shredded material) can be stated as:

(SEE PDF for equation)

where a, c, b and d are coefficients related to the power losses by air resistance, spread of shredded material, and the moment of the friction force in the supports and cut, respectively.

Have fun!:thumbsup:

You do realize that they never mention anything about "vacuum" right? They compare friction losses to two different kinds of blades...that's the "air resistance" part. They're actually quite clear when they compare the mechanical resistance caused by throwing the cut material over the rotor....their term, not mine. Notice they describe it as a mechanical throwing, not a pressure-related movement. They also point out how that's all controlled by the rotation speed, the blade size/type, and the ground speed or feed rate of the cut material. No mention of pressure gradients at all....funny.

A blade the size and shape of a typical Y flail blade will generate minimal air movement, which is why they won't generate enough air movement to lift a feather....pretty low standard really. Will it move some air? Sure, but it's not much.

The next time you're driving on the highway, lower the window, and put your hand out into the breeze at whatever legal speed is there....in the 60-90mph range like you mentioned above. Keep your hand on edge and see how much resistance you feel. Then imagine the resistance being something like 1/10th of that (hands and arms aren't sharp/smooth, and are far larger in total cross-section).

 

[biodegraded]

Does someone here know what model Vrisimo flail this is that showed up on my local CL? It's listed as an "8 foot" cutter, though that may be its overall width. It looks to be a precursor to the current Super Series. The most distinctive feature is what appears to be a completely rounded belt housing top with vents. Would parts be available for it? I'm trying to decide if it's worth the 3 hour round trip to scope it out. Thanks.

 

[IslandTractor]

You do realize that they never mention anything about "vacuum" right? They compare friction losses to two different kinds of blades...that's the "air resistance" part. They're actually quite clear when they compare the mechanical resistance caused by throwing the cut material over the rotor....their term, not mine. Notice they describe it as a mechanical throwing, not a pressure-related movement. They also point out how that's all controlled by the rotation speed, the blade size/type, and the ground speed or feed rate of the cut material. No mention of pressure gradients at all....funny.

A blade the size and shape of a typical Y flail blade will generate minimal air movement, which is why they won't generate enough air movement to lift a feather....pretty low standard really. Will it move some air? Sure, but it's not much.

The next time you're driving on the highway, lower the window, and put your hand out into the breeze at whatever legal speed is there....in the 60-90mph range like you mentioned above. Keep your hand on edge and see how much resistance you feel. Then imagine the resistance being something like 1/10th of that (hands and arms aren't sharp/smooth, and are far larger in total cross-section).

1) It was mostly a joke, I just loved the equation. This article doesn't directly address our question.

2) Next time you are driving down the highway at 60-90mph, hold out 50-100 1/8"x3" blades plus the surface area of mounting points and clevises etc. I suspect it would generate quite a bit of resistance (and low pressure area behind).

3) In the article discussion they state:
"The results obtained in this investigation, the
analysis of the linear and cubic components for
the demand of the vacuum power and the linear
and quadratic components for the power of the
technological process, as well as the influence
of the angular speed of the cutting mechanism
with respect to those components (Figures 3, 4
and 5) have not been reported in the literature
so far."

I'm neither a physicist nor mechanical engineer but my suspicion is that other than the power needed to actually cut the vegetal matter, a good portion (well at least some) of the PTO power needed to run a flail at operating speed (while not actually cutting for example) is the power needed to move the air (which generates vacuum). The rest would be mechanical friction losses.

Maybe we'll have to call an engineer at Vrismo or JD to see what they have to say.....

 

[hunt4570]

Ok, Island Tractor, Leonz, and Gmanbart... I've been following this conversation about gradient pressures ect since it started. Now I'm just trying to make sense of all this. Nobody is mowing turkey feathers anyway right?
So I'm thinking... what you guys are trying to determine here is " do our flail mowers suck" is this correct?:laughing:

 

[IslandTractor]

Ok, Island Tractor, Leonz, and Gmanbart... I've been following this conversation about gradient pressures ect since it started. Now I'm just trying to make sense of all this. Nobody is mowing turkey feathers anyway right? So I'm thinking... what you guys are trying to determine here is " do our flail mowers suck" is this correct?:laughing:

You got it! I need a tee shirt that says "My flail sucks"