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