Flail Mower Confusion
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CurtisC
Bronze Member
Sorry 100 td but all of the implement dead weight plus live/operational vertical loads are supported by the 2 lower or lifting arms. The top link controls tilt of the implement and sees only horizontal dead loads due to the implement cg offset (almost always aft of the lift arm connection) and any additional live rotational loads encountered during use.
100 td
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Half isn't exact, it's general jargon, meaning weight is transferred.
If the weight was "directly" centered above the lower pivot point, it would take all the load. But normally this is not the case. Replace the top link with your hands and see how much load is transferred to the tractor via the top link .
The tractor carries the entire weight but it is carried on the lift arms(transferred back to the tractor) and on the top link ( transferred back to the tractor)
I'm not going to do all the drawings and angles and calculations of weight/force but this may help out.
Think of a box, 1 yard x 1 yard x 2ft high. The box weighs 500lbs (but can weigh anything).
Person 1 lifts one end up 2", leaving the other end on the ground (pivot point), Person 1 is lifting 250lbs. Person 2 now lifts the other end 2" off the ground at the same time, Person 1 and Person 2 are now both lifting 250lbs each.
The 3pt arms connect to the bottom end of the box. You don't have a top link connected
Person 1 raises the lift arms up 2", the tractor has 250lbs transferred onto it. Person 2 now lifts the other end 2", the tractor still has 250lbs and Person 2 has 250lbs. Person 1 attaches the top link to the top of the box.
Tractor now has 500lbs load on it. But it doesn't have 500lbs on the pivot point, as a percentage is transferred to the tractor via the top link. I would need to find my old trigonometry books and draw it up exactly to spec to calculate it.
If the weight was "directly" centered above the lower pivot point, it would take all the load. But normally this is not the case. Replace the top link with your hands and see how much load is transferred to the tractor via the top link .
The tractor carries the entire weight but it is carried on the lift arms(transferred back to the tractor) and on the top link ( transferred back to the tractor)
I'm not going to do all the drawings and angles and calculations of weight/force but this may help out.
Think of a box, 1 yard x 1 yard x 2ft high. The box weighs 500lbs (but can weigh anything).
Person 1 lifts one end up 2", leaving the other end on the ground (pivot point), Person 1 is lifting 250lbs. Person 2 now lifts the other end 2" off the ground at the same time, Person 1 and Person 2 are now both lifting 250lbs each.
The 3pt arms connect to the bottom end of the box. You don't have a top link connected
Person 1 raises the lift arms up 2", the tractor has 250lbs transferred onto it. Person 2 now lifts the other end 2", the tractor still has 250lbs and Person 2 has 250lbs. Person 1 attaches the top link to the top of the box.
Tractor now has 500lbs load on it. But it doesn't have 500lbs on the pivot point, as a percentage is transferred to the tractor via the top link. I would need to find my old trigonometry books and draw it up exactly to spec to calculate it.
CurtisC
Bronze Member
The problem with your example is that person 2 doesn't "lift" the other end. Person 2, or the top link, can only pull (or push) the top of your box horizontally and is unable to provide any vertical ("lift") force. Person 1 has to not only lift the box vertically at the 2 lower connection points but also counteract any horizontal force person 2 exerts on the single point above. Those three connection points are the only external forces applied to your box when its off the ground. To make the math simple lets modify your box to a 500 lbs. box that's 2 ft. X 2ft. X 2ft with the CG dead center. To hold the box off the ground, the lower arms together (person 1) must exert a total of 500 lbs of upward vertical force since the top link can provide none. To keep the top of the box from simply rotating away from the tractor, the top link (person 2) has to pull the box toward the tractor with 250 lbs. of force. To keep the box stationary the lower arms (person 1) subsequently have to push in the opposite direction with a combined force of 250 lbs. in addition doing all the lifting. With all due respect, I'd suggest a book on basic static mechanics not trigonometry.
bkrgi
Bronze Member
The B26TLB handles the Woodmaxx FM62H just fine. In heavy grasses just plan to go slower but she will get the job done
more video
B26TLB Flail Mowing Part 2 - YouTube
more video
B26TLB Flail Mowing Part 2 - YouTube
JWR
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Curtis is correct. Open and shut. Do not get drawn into any kind of complex description or people carrying boxes examples. All that is erroneous.
In case there is an engineer around, this is a grossly simple Statics problem. With the implement attached and lifted just enough to clear the ground, break the lift arm force into vertical and horizontal components. The vertical ( counting both lift arms) is the weight of he implement. The horizontal component is opposite polarity and equal to that of the top link.
Now take the force on the (free to rotate about the attachment pin...) top link. The vertical component is zero. Has to be zero because the rotation about the pin is unhindered. Horiz component is opposite polarity and equal to that shared by the lower lift arms.
Bottom line: Top link provides NO lift. Bottom lift arms provide 50% of the lift force for the implement each. End of story.
p.s.: Ok, sure there is a TINY vertical component at the toplink to the extent that it is not level with the ground. Completely negligible in context of this discussion.
