3 Point hitch upper link

[Chewwy]

True, it does compute the change in the tension at an angle, but it misses the basic question. What is the force to begin with?

That was not the question Runner asked in the post I replied to, (i.e., is 1000 lbs of tension on the top link support correct?).

 

[MoArk Willy]

It's not too hard to get an estimate if you make a few simplifying assumptions:

If the weight of the implement is 1,000 lb. the tension in the top link is 2,000 lb., assuming:

1. The center of gravity of the implement is 4 ft. behind the 3-point pins. Probably reasonable for a box blade or flail mower. If it is longer, the tension in the top link is greater.

2. The height of the riser for the top link is 2 ft. That may be a bit high. If so the top link tension would be grater.

3. The top link and lower links are parallel. They probably are not, which would increase the tension in the top link.

4. The implement is raised until the lower and upper links are horizontal. When they are not the tension in the top link increases.

Thank you for that drawing and explanation.
Now let me see if I understand how you got your numbers.
The load on the top link is relative to the height between the lower arms and the top link, assuming it was parallel.
The distance between the pivot point on the lower arms and the pivot point on the top link are how you calculated the effort needed to lift any given weight.
The hypotenuse is irrelevant in the drawing and the calculation, except to show a connection.
So the only thing that would lessen the pull on the top link would be a greater distance between the bottom pivot point and the upper pivot point (lower arms and top link)
Is that on the right track?

 

[ning]

I have nothing to add other than a mental illustration.

Note that the bottom links lift the close end of the implement, which as noted in the barbell case (or a very well balanced ballast box) could be the entire thing.
Without a top link, in the case of a rotary cutter, the bottom links are lifting about 1/2 the weight of the cutter, with the other half being on the wheel (more or less, depending on distances from the link pins to the cutter center of mass, etc); the weight is bridged from the 3ph links to the wheel. Exactly how much the bottom links lift depends on the geometry and mass of the attachment.

Now attach the top link with slack, and there's no change in the weight distribution.
As you put tension in the top link (retract it), the geometry may change slightly so that the weight on the bottom links may change a bit, but the weight that was on the wheel is now moved to the top link - which has the tension force on it which is composed of both a horizontal force and a vertical force, so the top link is now carrying weight.

To visualize that, imagine:

• you're standing with your back to a wall
• a 10' 2x6 is on the ground, butt to the wall (like you) and it's pointed away from the wall
• rope from your hand to the end of the board (this is the top link)
• when rope is slack, there's no tension and no weight affecting your hand
• if you put tension on the rope and lift the end of the board off of the ground, you're pulling in two directions - horizontally towards the wall as well as upwards; the upwards pull is the weight of half of the board
• notice how if you pull at ground level - only horizontally - the end of the board doesn't rise (duh), because you're not pulling up; you have to pull up enough to raise the board. if you pull straight back from hip level, the rope is transferring that force in both a horizontal and vertical component.

 

[Runner]

Terry, thanks so much for the diagram and analysis. Because of my background, I tend to think of things in terms of being part of a building structure. This is obviously WAY more complicated than I assumed. I was basically looking at it as if the mower was a canopy hanging on the side of a wall.

But if I understand what you are saying, this is closer to a cantilevered truss design with 1000 lbs hanging on the end of the truss. The reason for the 2000 lbs is because of the moment arm induced by the separation of the top and bottom "truss chords" (which in this case are actually the draft arms and the top link).

Is that sort of it?

 

[TerryR]

Thank you for that drawing and explanation.
Now let me see if I understand how you got your numbers.
The load on the top link is relative to the height between the lower arms and the top link, assuming it was parallel.
The distance between the pivot point on the lower arms and the pivot point on the top link are how you calculated the effort needed to lift any given weight.
The hypotenuse is irrelevant in the drawing and the calculation, except to show a connection.
So the only thing that would lessen the pull on the top link would be a greater distance between the bottom pivot point and the upper pivot point (lower arms and top link)
Is that on the right track?

You're welcome.

Your are right about the distance between the lower pivot point and the end of the top link being part of the calculation, but that's relative to the distance between the lower pivot point and the center of gravity. In my illustration I made the distance from the lower pivot point to the top link half the distance from it to the load. Thus the tension in the top link is twice the load.

Yes, the tension in the top link would be reduced if the connection to the top link is raised. It would also be reduced if the load is moved closer to the lower pivot point.

 

[TerryR]

But if I understand what you are saying, this is closer to a cantilevered truss design with 1000 lbs hanging on the end of the truss. The reason for the 2000 lbs is because of the moment arm induced by the separation of the top and bottom "truss chords" (which in this case are actually the draft arms and the top link).

Is that sort of it?

Yes, the key is the distance between the top and bottom links, but only relative to the distance between the lower attachment point (outer end of the lower arms) and the center of gravity.

As others have noted, if that distance is zero, for example with a balanced counterweight, there is zero tension in the upper link no matter what the distance between the upper and lower arms.

 

[Smokeydog]

Add a bump and watch those numbers jump.

 

[MoArk Willy]

You're welcome.

Your are right about the distance between the lower pivot point and the end of the top link being part of the calculation, but that's relative to the distance between the lower pivot point and the center of gravity. In my illustration I made the distance from the lower pivot point to the top link half the distance from it to the load. Thus the tension in the top link is twice the load.

Yes, the tension in the top link would be reduced if the connection to the top link is raised. It would also be reduced if the load is moved closer to the lower pivot point.

Let me ask you this...or better yet question the math, and stick with a 1000# load.
You can attach the lower arms of a 3 point and lift a ballast box...tip it so to speak.
But in order to lift it, you need to attach the top link to keep the load from falling over.
Let's put the math aside for a bit.
Since the lower arms do all of the lifting of the 1000# weight.....why would the top link have to support any of the weight in addition to what is needed to balance it?
Obviously if the lower arms connected in the middle of the weight and not at one end....then the top link would be doing nothing but a balancing act to keep the weight from falling over.
Now back to the math.
Although it makes sense...it doesn't seem logical.....if the lower arms are lifting the 1000#....the top link is balancing that 1000#.....not lifting it.
That's where I start questioning the numbers.
I'd like to be able to prove this one way or another....but I'm not about to spring (pun intended) for a scale to replace the top link when lifting a load of a specific weight.
But here's an experiment for you to try.
Pick up a can of anything, beer, pop or whatever...or a carton of milk.
Put a finger down on the table and set the edge of the container on your finger.
Now take your other hand and put a finger on the top of the container and pull it (for push it) until the weight is resting only on your finger on the table.
If the math was correct....wouldn't it take the same amount of force as the weight of the container to keep it in balance on your finger?
Yet it doesn't..........See my dilemma?

 

[CalG]

The first rule of load analysis is

"YOU MUST TRUST THE MATH"

Force vectors have no will of their own.

 

[Williy]

That's like X = Y2 -t + g =ab its all GREEK to me;

willy

 

[SPYDERLK]

Let me ask you this...or better yet question the math, and stick with a 1000# load.
You can attach the lower arms of a 3 point and lift a ballast box...tip it so to speak.
But in order to lift it, you need to attach the top link to keep the load from falling over.
Let's put the math aside for a bit.
Since the lower arms do all of the lifting of the 1000# weight.....why would the top link have to support any of the weight in addition to what is needed to balance it?
Obviously if the lower arms connected in the middle of the weight and not at one end....then the top link would be doing nothing but a balancing act to keep the weight from falling over.
Now back to the math.
Although it makes sense...it doesn't seem logical.....if the lower arms are lifting the 1000#....the top link is balancing that 1000#.....not lifting it.
That's where I start questioning the numbers.
I'd like to be able to prove this one way or another....but I'm not about to spring (pun intended) for a scale to replace the top link when lifting a load of a specific weight.
But here's an experiment for you to try.
Pick up a can of anything, beer, pop or whatever...or a carton of milk.
Put a finger down on the table and set the edge of the container on your finger.
Now take your other hand and put a finger on the top of the container and pull it (for push it) until the weight is resting only on your finger on the table.
If the math was correct....wouldn't it take the same amount of force as the weight of the container to keep it in balance on your finger?
Yet it doesn't..........See my dilemma?

Your table finger is the load/lift point. The COM of the container/load extends out beyond this lift point. That extension is a lever that must be overcome by pulling on the top to lift COM. Your top finger works on a different lever (container height). So the pull, tension, there will be different than the container weight.

 

[Runner]

Since the lower arms do all of the lifting of the 1000# weight.....why would the top link have to support any of the weight in addition to what is needed to balance it?

The thing is, this is a complicated system and the draft arms ONLY do ALL of the lifting of the 1000# IF the top link is connected. If the top link weren't connected the draft arms would only be lifting HALF the weight of the box (just like the weight of the can sitting on your finger).

Then, when you DO attach the top link, it is not only lifting the other half of the weight, but that amount of weight at some distance from the lifting point, which is what multiplies the apparent weight, like the difference between tightening a bolt with a short wrench or using a 10 foot cheater bar.

What blows my mind is the idea that the tension on the top link comes out to TWICE the actual weight of the box, but that's why Engineers get the big money.

 

[TerryR]

You can attach the lower arms of a 3 point and lift a ballast box...tip it so to speak.
But in order to lift it, you need to attach the top link to keep the load from falling over.
Let's put the math aside for a bit.
Since the lower arms do all of the lifting of the 1000# weight.....why would the top link have to support any of the weight in addition to what is needed to balance it?
Obviously if the lower arms connected in the middle of the weight and not at one end....then the top link would be doing nothing but a balancing act to keep the weight from falling over.

The top link never supports any supports any of the weight of the implement. It can't because it's hinged at the front. Try unhooking the top link from the implement and pushing down on it. It just swings away.

The lower arms always support all of the load. All the top link does is resist the tendency of the implement to rotate around the pivot point, the pins at the rear of the lower links. But only if the implement does tend to rotate.

Lets look at a ballast box (one with the pins above its center):

The load doesn't want to tip, so there is no load at all in the top link. In fact, you don't even need to attach it.

Now, let's imagine an implement whose center of gravity is two feet back from the pins:

Now the implement wants to tip backwards. The load exerts a downward load of 1000 lb., with a two-foot lever arm behind the pins. The top links is connected two feet above the pins, so it has a two-foot lever arm too. Thus it has to exert a 1000 pound force to counter then tendency to rotate.

Now consider an implement whose center of gravity is four feet behind the pins:

The load is still 1000 lb. but now it has a four-foot lever arm to rotate the implement. The top link still has only a two-foot lever arm, so it has to work twice as hard to keep the implement from rotating, that is it has to apply a 2000 lb. force.

 

[SPYDERLK]

The top link never supports any supports any of the weight of the implement. It can't because it's hinged at the front. Try unhooking the top link from the implement and pushing down on it. It just swings away.

The lower arms always support all of the load. All the top link does is resist the tendency of the implement to rotate around the pivot point, the pins at the rear of the lower links. But only if the implement does tend to rotate.

Lets look at a ballast box (one with the pins above its center):
View attachment 1317206

The load doesn't want to tip, so there is no load at all in the top link. In fact, you don't even need to attach it.

Now, let's imagine an implement whose center of gravity is two feet back from the pins:
View attachment 1317203

Now the implement wants to tip backwards. The load exerts a downward load of 1000 lb., with a two-foot lever arm behind the pins. The top links is connected two feet above the pins, so it has a two-foot lever arm too. Thus it has to exert a 1000 pound force to counter then tendency to rotate.

Now consider an implement whose center of gravity is four feet behind the pins:
View attachment 1317202

The load is still 1000 lb. but now it has a four-foot lever arm to rotate the implement. The top link still has only a two-foot lever arm, so it has to work twice as hard to keep the implement from rotating, that is it has to apply a 2000 lb. force.

I though it might be interesting to link a grope we did on this a dozen or so yrs ago. Different perspectives may help things click with everybody.
Conversion Factor for 3PH load capacity

 

[CalG]

IT might be noted, that when resolving vector forces (loads), that if the values for the directional force arrows do not sum to ZERO, the objects are in motion! ;-)

 

[MoArk Willy]

It just gets more fascinating.
The ballast box example is perfect....assuming such a ballast box exists where the load is balanced.
But the center of gravity explanation........that loses me.
If the implement is attached at one end....the pivot point....does that put the center of gravity at the far end of the implement?
The 3 point attachments of some implements is not at the bottom of the implement....in fact it rarely is.
A box blade will have the lower arm attachment point at the top of the box blade frame, not at the bottom.
And the top link attachment point is about 32" from the ground.....I can easily roll the box blade forward while it is sitting on the ground....it's about 350#, by pulling on the upright top link support.
That will affect the balance / center of gravity. I think the formula is solid....but implements are varied in their lower arm attachment point....which changes the balance point.
I think I'm about to get a digital scale and experiment. The drawings as presented are 100% accurate I believe.
But in all reality they aren't a realistic example of how the implements are attached.....with the exception of a mower where all the weight is behind the attachment point and the attachment point is very low on the implement. So I think the answer to the original question is "it depends".....rather than a fixed equation.
Now part of the reason I started this thread was my wonderment of how the top link can be relatively slight in construction compared to the robust arms and lower end of the 3 PT mechanism.

 

[CalG]

It just gets more fascinating.
The ballast box example is perfect....assuming such a ballast box exists where the load is balanced.
But the center of gravity explanation........that loses me.
If the implement is attached at one end....the pivot point....does that put the center of gravity at the far end of the implement?
The 3 point attachments of some implements is not at the bottom of the implement....in fact it rarely is.
A box blade will have the lower arm attachment point at the top of the box blade frame, not at the bottom.
And the top link attachment point is about 32" from the ground.....I can easily roll the box blade forward while it is sitting on the ground....it's about 350#, by pulling on the upright top link support.
That will affect the balance / center of gravity. I think the formula is solid....but implements are varied in their lower arm attachment point....which changes the balance point.
I think I'm about to get a digital scale and experiment. The drawings as presented are 100% accurate I believe.
But in all reality they aren't a realistic example of how the implements are attached.....with the exception of a mower where all the weight is behind the attachment point and the attachment point is very low on the implement. So I think the answer to the original question is "it depends".....rather than a fixed equation.
Now part of the reason I started this thread was my wonderment of how the top link can be relatively slight in construction compared to the robust arms and lower end of the 3 PT mechanism.

Some setups replace the upper link with a chain. (brush hogs!)

Tension only works one way, ;-)

The upper link can be less robust than the lift arms because that link is only supposed to see tension, along with "light" compression loads. Buckling is the result of too much compression. Most any shape has greater tension capabilities than buckling values. Depending on constraint. Pure compression can blow the roof off "what will it hold" ideas.
(I had to straighten the top link on a recently purchased tractor, I had to straighten the sway links too ;-)
The lift arms see BENDING loads based on the fixing pins, the lift link locations, and the load. Many tractors have multiple pin positions for the lift links on the lift arms to achieve various performance objectives. (That is a whole another set of pictures, arrows and lines l-)

 

[SPYDERLK]

Now part of the reason I started this thread was my wonderment of how the top link can be relatively slight in construction compared to the robust arms and lower end of the 3 PT mechanism.

Some setups replace the upper link with a chain. (brush hogs!)

Tension only works one way, ;-)

The upper link can be less robust than the lift arms because that link is only supposed to see tension, along with "light" compression loads. Buckling is the result of too much compression.

 

[ning]

I drew these up earlier, then deleted, but after yet more "the math doesn't matter" kind of posts I figure I'll post it anyways.

In these illustrations, pretend the implement is a barbell, 100# each end. Really a rotary mower isn't much different from this example but making things as simple as possible is good.

First pic: Ends are on the ground, no tractor involved.
(Not pictured: ground pushing up 100# on the weights, this is called "the normal force"):

Next pic:
Tractor lifts one end with no top link:

Funky cylinder thing is the lift arm.
The 100# of the close end is on lift links. The other end's 100# is still on the ground.

Last pic: just like second pic, but top link lifts the other end:

Other funky cylinder thing at the angle is the top link... part of it. The rest of it extends to the far end weight, whether via piston or chain or what.
There’s no more weight on the lift links - still 100.
Top link has all 100 of the other end.

Final pic: top link pulled strongly enough to raise the end to be level with the top link itself (I got tired of drawing 100's and arrows sorry)

In this case, the bottom link is now carrying a chunk of the weight of the far end as well as its end.
The tension in the link is shown by the diagonal line going from top right to bottom left; the components are the horizontal compression and the vertical weight pulling down
Top link is carrying some; the higher the angle that the implement has with the bottom link, the more goes onto the bottom link (as soon as the end of the implement goes above the bottom arms, the bottom arms start getting some of the end load).

Consider if the top link lifts the barbell completely vertical (assuming it would fit, ok) - then the top link would end up with zero tension and the bottom arms would be carrying the weight of both ends of the barbell (200#).
That's the situation with a perfectly balanced ballast box.

 

[404nouser]

Is the distance between upper and lower pins on the tractor the same as the implement.