Lift Capacity @ 24" for SubCompacts

[LD1]

I am going to try again. Please have patience with me and try to follow me through this. I think we are all getting a little tired...

I am going to talk about the top link again because I think that is a sticking point. I would like to use the A, B, B' C, D picture we used before. It would probably be best if whoever reads this draws a little picture on a piece of paper like the picture I posted before. Without the picture no one could follow what I have written. So Please, please, get the picture in front of you when you read this.

Lets say we have a weight box hooked up at AC. Lets say top link AB is horizontal and lets also say lower link CD is horizontal. When everything is horizontal like this the 3ph lift links (which I never showed in my picture) are holding up the weight box and there is no weigh shift onto D. The only thing top link AB is doing is preventing the weight box from pivoting about ball eyes C and dumping out on the ground.

If we wanted, would could have a friend walk to the back of the weight box and push on it to hold it up. We could then even take the top link off. The amount of force our friend has to push is the same as the tension needed in the top link. We always need that amount of horizontal force to keep the box from falling. We could raise or lower our weight box and our buddy would always have to keep pushing with the same amount of horizontal force.

Now lets pretend point B on the tractor, where the top link connects to, is actually on a sliding vertical rail. With this sliding rail we can keep the top link always horizontal. Now that we can have the top link point B stay horizontal, we can hook it up and tell our friend to let go.

We can raise and lower our box, with our sliding link that always stays horizontal. The tension in that top link is always the same. Like you have said, if we raise the box such that C is higher than D, weight gets shifted through the lower link onto D.

Remember we always need that same amount of horizontal force at A to keep out box from falling over, either by the tension in the top link or our friend pushing against the box.

With our box raised (C is above point D) lets say we want to lower top link point B. We get our friend back out to push on the back of the box (with that same horizontal force) and unhook our top link, make it longer so we can hook it up at a point lower, something like our original point B. We put the pin back in and tell our friend to let go. The tension in the top link, now that we have connected it back at point B, has gone up. Even though our box has remained level (the line AC is still vertical).

Lets tell our friend to come around to the front of the box, stand between the box and the tractor, and have him hold onto that top link while we pull the pin out at B. He is going to have to pull on the top link pretty hard. In fact, the lower we tell him to pivot that top link, the closer he gets it to vertical, the harder he has to pull on it.

If he gets that link at a 45% angle to the ground, half of his pulling force is in the horizontal direction (to keep that box from tipping around the ball eye). The other half of his pulling force is in the vertical direction. In effect, he is pulling down on the weight box.

This is how tension in the top link has a contribution to the load at the ball eyes. When the angle of the top link is the same as the angle of the lower link, the weight shift that you have spoken of still happens. However, the friend pulling on the top link is pulling down with the same amount of weight (force) as what is being shifted.

If the top link angle is closer to horizontal than the lower link, your weight shift is greater than the force from the top link. If the top link is closer to vertical than the lower link, your weight shift is less than the force added from the top link.

Excellent way of explaining it. I dont think I have worded it any better.:thumbsup:

 

[arrabil]

Just about everything you said is true and yet none of it has anything to do with the lift capacity. This is Newton's 3rd plain and simple. Force F, the box trying to tip over, equals force -F, the toplink counteracting that. The force vector always cancels itself out regardless of the orientation of the toplink. The reason it does not affect the lift is because the forces you refer to are working around the ball eye rotation while the lift forces are working around the lower link rotation. They are on different "planes" if you will allow that way of saying it. Another way to say it.... the force you are adding by pulling down down on the box is transferred to the toplink point, not the lower link arms.

You should draw it out with all your forces marked as arrows. You will see how the arrows for rotation never intersect or point at the same spots as the arrows for lift.

I am simplifying it some but not for the better of your argument. If the toplink attachment point was in the middle of the box (or below some of it's mass anyway) the toplink point would be holding the X vector of the weight of the CoG of the box that exists above the attachment point. That could make it easier to lift, not harder. Basically in this situation the box has two points to transfer weight to. I mentioned this scenario a while back as the only time the toplink has any effect on lift (but it's not quite the scenario you describe).

This is in essence our disagreement. The toplink cannot add force to the equation. Why? A force has to be generated from somewhere and the toplink is unpowered. Counteracting gravity is not generating force, just Newton's 3rd. Why is this our disagreement? Because if the toplink were powered, this could be parallel linkage. But because it is not powered, it cannot be a parallel linkage. And since it cannot be a parallel linkage, the toplink's only purpose is to orient the load on the ball eyes. Also, when I say powered I do not mean a hydraulic link. It has to be powered from the same point as the other half of the parallel linkage to be part of it.

 

[arrabil]

I need to update that..... You are not adding weight by pulling with the top link. You are preventing rotation. Preventing the rotational forces is absorbed entirely by the toplink attachment point. The weight cannot change just because it's putting more strain on the toplink mount because you chose a less efficient toplink orientation.

(Sorry, I'm doing this on my phone while my kid is at practice.)

 

[LD1]

The force of the toplink does NOT have to be in the same plane to have an effect on the lower arms.

Let me try to explain what tsteahr said in a few different ways. We all knows newtons 3rd. So do we agree that the implement is "pulling" on the toplink putting it in tension??

If we agree with that, according toe newtons 3rd, the toplink is "pulling" back with the exact same amount of force.

Now if the toplink and the lower links are parallel and the same length, through out the entire range of motion of the 3PH they maintain the exact sae angle(remain parallel). Do we agree here???

If both arms are horizontal, there is no vertical component to either. IE, the toplink isnt adding any downforce and there is no weight being transfern to where the lower link connects to the tractor.
Through out the raise of the implement, both the lower and toplinks start to angle downward toward the tractor. Thus creating both a horizontal AND vertical component. The vertical component of the toplink and the lower links are going to be exactally the same since they are on the exact same angle, thus the amount of weight transfered to the lower arms is = to the downward force of the toplink. They cancel each other out.

Now, if the toplink has more downward angle to start with, it will maintain a greater downard angle through out the entire travel. Do we agree here??

If we again break the lower link and toplink into its horizontal and vertical components, the toplink is going to have a greater vertical force. Thus ADDING more than is being transfered onto the lower link @ the tractor side. This reduces capacity. By breaking the vectors into their vertical and horizontal components puts them on the same "plane".

I said I had a few ways to try to explain it, so here is number two

Imagine if the 3PH arms were long enough or had enough range of motion to lift a 4" bushhog into the air WITHOUT the toplink. What would happen?
Answer, the bushhog would swing down and be hanging with its COG below the pins. Do we agree??

Now, suppose you hooked a hydraulic cylinder to the toplink hole of the bushhog on a 45* angle downward toward the tractor and applied force. What woudl happen. The bushhog would pivot back into "normal" position. But that cylinder is applying pressure on a downward angle. Would this ADD to the downward force on the ball ends??? The answer is yes, because the cylinder is not pulling horizontally, but rather at an angle, so horizontally AND downward at the same time. Do you agree here??

 

[tsteahr]

Just about everything you said is true and yet none of it has anything to do with the lift capacity. This is Newton's 3rd plain and simple. Force F, the box trying to tip over, equals force -F, the toplink counteracting that. The force vector always cancels itself out regardless of the orientation of the toplink. The reason it does not affect the lift is because the forces you refer to are working around the ball eye rotation while the lift forces are working around the lower link rotation. They are on different "planes" if you will allow that way of saying it. Another way to say it.... the force you are adding by pulling down down on the box is transferred to the toplink point, not the lower link arms.

You should draw it out with all your forces marked as arrows. You will see how the arrows for rotation never intersect or point at the same spots as the arrows for lift.

It is correct that the force pulling down on the box is transferred to the toplink point. The toplink point on the box. That force HAS to go somewhere. That force added by pulling down is transmitted through the weight box to the eyes on the lower link. The big strong guy is pulling down on the top link. He is pulling down on the top of the weight box using the top link. That force goes straight to the lower link eyes.

I am simplifying it some but not for the better of your argument. If the toplink attachment point was in the middle of the box (or below some of it's mass anyway) the toplink point would be holding the X vector of the weight of the CoG of the box that exists above the attachment point. That could make it easier to lift, not harder. Basically in this situation the box has two points to transfer weight to. I mentioned this scenario a while back as the only time the toplink has any effect on lift (but it's not quite the scenario you describe).

This is in essence our disagreement. The toplink cannot add force to the equation. Why? A force has to be generated from somewhere and the toplink is unpowered. Counteracting gravity is not generating force, just Newton's 3rd. Why is this our disagreement? Because if the toplink were powered, this could be parallel linkage. But because it is not powered, it cannot be a parallel linkage. And since it cannot be a parallel linkage, the toplink's only purpose is to orient the load on the ball eyes. Also, when I say powered I do not mean a hydraulic link. It has to be powered from the same point as the other half of the parallel linkage to be part of it.

This is correct. Your reason is wrong. The top link changes the BALANCE of forces in the linkage system. Every link can and does IMPART a force onto its neighbor. All these forces throughout all the linkages have to balance. When any link is changed, it affects the balance of forces throughout the rest of the links. IE some may see more, some may see less. That downward force of the guy pulling (the link connected to B', its the same thing) is balanced by increased force required by the hydraulic lift links. The link B'D (the tractor) is in tension.

If you raise the point B up high, now we have shifted the force between B and D (the tractor) from tension to compression and we have reduced the force at the lift links. This makes sense because if the toplink attachment point B on the tractor is up high, the weight box is now somewhat "hanging" off the top link but the link BD (the tractor) had gone into compression to hold up the box.

The orientation of all the links relative to each other determine HOW the forces are balanced (distributed) throughout the links.

 

[arrabil]

The only force we're working against is gravity. No matter how the load is oriented only gravity is acting against the object and hence the lift arms. So if the load wants to rotate, is there more force? Or is the object being acted on in new vectors but the total force is the same?

I'm pretty sure the total force is still the same. If thats true, no matter how the counteracting force is distributed/balanced, the lift arms can still see the same maximum amount of force.... the weight of the object.

Yes or no? You guy are saying no, right? Convince me again please cause I got lost with your hydraulic link example, LD1.

 

[tsteahr]

I need to update that..... You are not adding weight by pulling with the top link. You are preventing rotation. Preventing the rotational forces is absorbed entirely by the toplink attachment point. The weight cannot change just because it's putting more strain on the toplink mount because you chose a less efficient toplink orientation.

(Sorry, I'm doing this on my phone while my kid is at practice.)

If a system is static, forces are never absorbed. They are transfered such that all the forces in the system are balanced.

It is correct that the weight can not change, but the force balance changes. There is no such thing as a "less efficient" or "more efficient" toplink orientation. There are only tradeoffs with different orientaions.

 

[tsteahr]

The only force we're working against is gravity. No matter how the load is oriented only gravity is acting against the object and hence the lift arms. So if the load wants to rotate, is there more force? Or is the object being acted on in new vectors but the total force is the same?

I'm pretty sure the total force is still the same. If thats true, no matter how the counteracting force is distributed/balanced, the lift arms can still see the same maximum amount of force.... the weight of the object.

Yes or no? You guy are saying no, right? Convince me again please cause I got lost with your hydraulic link example, LD1.

The total weight (mass of the load times gravity) is still the same. But the forces that result from that weight vary throughout the system, depending on the configuration of that system.

EDIT: When I say forces, you can take that to mean tensions (or compressions) in the components of the system.

 

[tsteahr]

The only force we're working against is gravity. No matter how the load is oriented only gravity is acting against the object and hence the lift arms. So if the load wants to rotate, is there more force? Or is the object being acted on in new vectors but the total force is the same?

I'm pretty sure the total force is still the same. If thats true, no matter how the counteracting force is distributed/balanced, the lift arms can still see the same maximum amount of force.... the weight of the object.

Yes or no? You guy are saying no, right? Convince me again please cause I got lost with your hydraulic link example, LD1.

The amount of force the lift arms see can vary because the amount of force between BD (the rear casting of the tractor) can vary. How much force goes to the lift arms and how much force goes to the rear casting is determined by the configuration of the system. The rear casting can see either tension or compression.

EDIT: When the casting is in compression, the weight is hanging off the casting and the lift arms see less force. When the casting is in tension (when the top link is connected to B') the lift arms see more force.

 

[LD1]

The only force we're working against is gravity. No matter how the load is oriented only gravity is acting against the object and hence the lift arms. So if the load wants to rotate, is there more force? YESOr is the object being acted on in new vectors but the total force is the same?

I'm pretty sure the total force is still the same. If thats true, no matter how the counteracting force is distributed/balanced, the lift arms can still see the same maximum amount of force.... the weight of the object.

Yes or no? You guy are saying no, right? Convince me again please cause I got lost with your hydraulic link example, LD1.

Because the load rotates, this cause the CoG of the load to be raised higher. Again, you cant get something for nothing. In order for the same amount of weight to be lifted higher, MORE force IS required. Because there is only a given amount of force avaliable at the lift arms (this is determined by the cylinder size, PSI, etc.) the load MUST be reduced in order to be lifted. The lower the toplink's position, the higher the mass is trying to be raised, thus the less weight it can lift.

Back to the W=F x D formula. It really is that simple. The 3PH is only capable of a given aount of work before Hydraulic cylinder size/PSI become the limiting factor. If it is only able to lift 1200lbs to a height of 24", it cannot lift that same mass to a height of 30". The mass MUST be reduce for work to remain the same. If the toplink and the lower links are equal length and parallel, ANY point is only trying to be raised 24", thus it can lift 1200lbs and still be within the avaliable work that the 3PH can do. With a lower/shorter toplink, you are asking that mass to be moved a greater distance with only the same avaliable work. Thus the Mass MUST be reuced. It really doesnt matter how much weight gets transfered or where or what is under compression or what is under tension. The only way to move that 1200lb mass a greater distance is to either increase the work avaliale or shed some weight.