Lift Capacity @ 24" for SubCompacts

[arrabil]

Didn't find nothing, huh?

How about a non-force transmitting parallel link? Could you find one of those?

 

[LD1]

In trying to keep this civil, as it is a learning experience for everyone and I enjoy these discussions, Arrabil, would you be willing to work through this post with me and answer the questions I ask. This will tell us where out differences lie and may help give eachother a better understanding where the other one is comming from.

First, Do you agree with the basic physics formula of Work=Force x Distance???

Now, in the case of a 3PH, would you agree that the distance is the distance that the implement is lifting, Force is the resistance, or weight of the implement opposing the 3PH, and work is the max amount that the 3PH can generate. Do you agree so far???

Now, to keep #'s simple, suppose we have a 3PH with a ball end rating of 1000 lbs, and a maximum lift travel of 20". With the above formula, would you agree that the 3PH is capable of 20,000 units of work??

If the top link were the same length and exactally parallel to the 3PH, Any point behing the ball ends is going to raise the exact same amount, the 20". So you can pick ANY point behind the ball eyes (which will rais the exact same amount as the ball eyes) and because we know that the 3PH is capable of 20,000 units of work, ANY point will have an avaliable force (lift capacity) of 1000lbs. Again, provided sufficent front ballast and strength of components, do we agree so far??

Now lets drop an shorted the toplink, which will cause the implement to rotate. Now when we raise out implement, The ball ends still only travel that same 20" as before, but a point 24" behind the ball ends raises 30" due to the rotating effect. Since we are still talking about the same hitch with 20,000 units of work avaliable we need to calculate the maximum amount of force it can generate @ 24" back. So...20,000/30=667lbs, of force avaliable.

But if we chose a higher toplink position which only cause the point @ 24" to be raised 25" off the ground instead of the 30" as before, with the same W=F x D formula, our lift force would then be 20000/25, or 800lbs.

What part of this do you not agree with?? A 3PH can only generate so much work. And in order to lift something higher, it has to give up some capacity.

 

[arrabil]

What part of this do you not agree with??

You are not considering the weight transfer onto the lower pins. I don't know why I keep saying it and you guys keep ignoring it. You cannot say you lifted 1000# at 24" because in reality some of the weight is being held by the tractor and it is not being lifted at all.

This is not physics class where they ignored half the problem to teach us the main concept. You must account for weight transfer at 30". You need to figure out the angle, split the forces into X and Y, and you need to subtract X from the 3ph calculation.

Again, I'll use the example of lifting an object with your arms and using your stomach to help hold it. That weight that your stomach supports is not weight that your arms have to lift. All of us can try this at home or work and we know its true. It is not insignificant and cannot be ignored like you are doing.

This will tell us where out differences lie and may help give eachother a better understanding where the other one is comming from.

You are proving that there is a mechanical advantage in the toplink because you can lift more at 24" than your calculations show. This is an end-around explanation. You did not show a mechanical advantage, you just need an answer for why the numbers are off and saying its a parallel linkage makes it sound real good. But your numbers are off because you didn't factor in weight transfer.

What am I saying that is so complicated to understand? You have literally ignored this every time I've brought it up. Why do you think this should be ignored then?

 

[tsteahr]

You are not considering the weight transfer onto the lower pins. I don't know why I keep saying it and you guys keep ignoring it. You cannot say you lifted 1000# at 24" because in reality some of the weight is being held by the tractor and it is not being lifted at all.

This is not physics class where they ignored half the problem to teach us the main concept. You must account for weight transfer at 30". You need to figure out the angle, split the forces into X and Y, and you need to subtract X from the 3ph calculation.

Again, I'll use the example of lifting an object with your arms and using your stomach to help hold it. That weight that your stomach supports is not weight that your arms have to lift. All of us can try this at home or work and we know its true. It is not insignificant and cannot be ignored like you are doing.


You are proving that there is a mechanical advantage in the toplink because you can lift more at 24" than your calculations show. This is an end-around explanation. You did not show a mechanical advantage, you just need an answer for why the numbers are off and saying its a parallel linkage makes it sound real good. But your numbers are off because you didn't factor in weight transfer.

What am I saying that is so complicated to understand? You have literally ignored this every time I've brought it up. Why do you think this should be ignored then?

We discussed weight shift in posts 35 and 42. Weight shift at the lower links is one part to be considered. The other part is the downward force component imparted by the top link tension. You are only looking at half of the system. You need to consider the top as well as the bottom links.

 

[arrabil]

The other part is the downward force component imparted by the top link tension.

 

[LD1]

You are not considering the weight transfer onto the lower pins. I don't know why I keep saying it and you guys keep ignoring it. You cannot say you lifted 1000# at 24" because in reality some of the weight is being held by the tractor and it is not being lifted at all.

This is not physics class where they ignored half the problem to teach us the main concept. You must account for weight transfer at 30". You need to figure out the angle, split the forces into X and Y, and you need to subtract X from the 3ph calculation.

Again, I'll use the example of lifting an object with your arms and using your stomach to help hold it. That weight that your stomach supports is not weight that your arms have to lift. All of us can try this at home or work and we know its true. It is not insignificant and cannot be ignored like you are doing.


You are proving that there is a mechanical advantage in the toplink because you can lift more at 24" than your calculations show. This is an end-around explanation. You did not show a mechanical advantage, you just need an answer for why the numbers are off and saying its a parallel linkage makes it sound real good. But your numbers are off because you didn't factor in weight transfer.

What am I saying that is so complicated to understand? You have literally ignored this every time I've brought it up. Why do you think this should be ignored then?

I understand what you are saying, BUT,
For starters, when the load/implement is on the ground, and the lower arms are at a downward angle to near level, there is NO weight transfered at all. And yet the 3PH will still lift the loads in the way I calculated.

What you are purposing is that through the range of motion of the 3PH, the higher it is lifted, the less force on the lift arm is required. Is that what you are saying??

I actually have two thoughts on this on and I honestly don't know which is correct at the moment without some number crunching that I don't have time for right this moment. So' Ill throw them both out there and maybe Spyderlk or tsteahr can pick up and run with it.

First thought is because as the lift arm raises the lower links, the angle between the lift arm and the lower link is reducing. Thus, with the same amount of hydraulic force being applied to the lift arm and lower link, the less "felt" force at the ball ends negating any weight transfer at all. It is the same concept of mounting a hydraulic cylinder at an angle, like a loader of dump bed. The vertical componet of force is decreased as the angle gets smaller.

So, using my above examples, If the lower arm were 24" long and the lift arm hooked in the middle @ 12", to have a ball end rating of 1000lbs, a force of 2000lbs would need to be exerted on the lift arm. Assuming a 90* angle between the lift link and the lower arm. If at max lift the angle is reduced to 60* this would mean less lift force At the ball ends. Sin60 x force. Or only about 86% of what we started with. Which would negate any load carried by the lower link.

My second thought on this is that the lower link ISNT carrying any load at all.

If you raised you 3PH all the way and the lower arms were on a pretty steep angle, you are saying that the lower mount on the tractor is carring a good bit of the load. But what would happen if you unhooked the arm at that point. I bet it would try to raise up in the air unstead of falling to the ground like if it were holding a bunch of weight. Same can be said for a FEL. If you raised it all the way in the air, you are saying that it would take less force at the cylinder, because the rear loader pin is carrying the weight. Yet if you were to remove that upper rear pin, it would fly up in the air and smack you in the face.

Again, I havent given these "thoughts" much thought. I am just throwing them out there off the top of my head for someone else to run with. I havent had time to do any serious number crunching so please forgive me if I got something wrong here.

 

[arrabil]

You are saying how it takes more effort to lift the higher you raise. I'm saying the weight transfer makes the load smaller. They shoud very well cancel each other out. Or come close to it. That would also give you more capacity than the simple calculations result in.

My second thought on this is that the lower link ISNT carrying any load at all.

My initial instinct is that this is preposterous. So I'll let you explain it again. How can the load be carried by anything else? The lower arms hold ALL the weight. The rockshaft arms lift the lower links so they deal with that weight. If you don't hook up the toplink and let the object come to rest in a natural position, then lift it, you are subject to the same forces, weight transfers, lift heights, etc. So the arms carry all the weight.

But what would happen if you unhooked the arm at that point. I bet it would try to raise up in the air unstead of falling to the ground like if it were holding a bunch of weight.

First the object would smack into the back of the tractor and do damage to you and it. Second, how can the lift arms raise any higher? They have a set travel and you just said we were at the top of it.

Same can be said for a FEL.

I don't believe so. I haven't given the FEL any thought but using the same scenario the weight transfer occurs on the lift cylinders. So the weight is transferred to the pins over the hydraulic fluid. So the cylinders still have to deal with the weight no matter what. There maybe other weight transfer but I don't think it is of the same magnitude as on the 3ph.

 

[LD1]

My initial instinct is that this is preposterous. So I'll let you explain it again. How can the load be carried by anything else? The lower arms hold ALL the weight. The rockshaft arms lift the lower links so they deal with that weight. If you don't hook up the toplink and let the object come to rest in a natural position, then lift it, you are subject to the same forces, weight transfers, lift heights, etc. So the arms carry all the weight.


First the object would smack into the back of the tractor and do damage to you and it. Second, how can the lift arms raise any higher? They have a set travel and you just said we were at the top of it.


I don't believe so. I haven't given the FEL any thought but using the same scenario the weight transfer occurs on the lift cylinders. So the weight is transferred to the pins over the hydraulic fluid. So the cylinders still have to deal with the weight no matter what. There maybe other weight transfer but I don't think it is of the same magnitude as on the 3ph.

Yes the lower arm is carrying the load. But the lower arm is mounted on a pivot at the tractor. THAT is the point I am saying ISNT carrying any load. The LIFT arm is what I was saying is carrying the load in that example. If you raised the 3PH all the way, and removed the pin @ the tractor, It would raise, and the implement would drop, pivoting about the point where the lift arm connects. That is how I compared it to the loader. You have an arm. Connected to a pivot on one end, a load on the other, and you are applying a force to lift it somewhere between the two. The pivot isn't carrying any vertical load, because if you unhooked the pivot, the lift point becomes the "new" pivot, and the old pivot will raise as the load drops.

Again, I havent put much thought into this, I am just kinda thinking out loud here so I may very well be way off base.

 

[arrabil]

Since you're in the mood for thinking about my ideas all of a sudden, I'd like to add a couple more....

1) The rockshaft is capable of far more work than the 0" lift capacity indicates. Otherwise it would be a struggle to lift the max weight. I have seen a BX lift its max load and I wouldn't call it a struggle.

That means the capacity at 24" might not conform to any formula we can come up with. The capacity may be determined by something else.

2) Like what? You guys have said a number of times if the rockshaft can raise more, all we need is more ballast on the front to lift it. Yes and no. What if the axle bearings can't handle the ballast and the load? Or the transmission tunnel casting? Or the R1 or R3 tires? Or...

3) The manufacturer is not considering a 1000# static load. They are also taking into account that it swings, bounces, and hits rocks in the ground. That could result in twice the force or more being generated. What if your ballast plus the load plus the bouncing is enough to snap the tractor in half? We've seen it done from 3pt backhoes. So maybe they limit the 24" load so that you don't ballast against more and then drive through the woods?

But the lower arm is mounted on a pivot at the tractor. THAT is the point I am saying ISNT carrying any load.

No, it has to carry it. Because the lower arm is connected to the lift arms via a bar that pivots at both ends. That bar would have to be rigidly connected for the lowest mount not to carry any weight (in which case it also would be completely unnecessary). AND most importantly, because that lower point is the fulcrum against which the lift arms lever.

 

[tsteahr]

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.