Conversion Factor for 3PH load capacity

[LD1]

Boy Im glad youre here. Now I can take a rest!

mwb, there are other ways to to get a handle on how much force can be applied where. Use an energy calculation combined with empirical observation. As you know E=FxD. In lifting a load the tractor eyes apply a force for a distance. This Energy is stored in the load. Thus if the load rises more than the eyes it has to weigh proportionately less than the upward force provided by the eyes. [Since there will be rotation in this case it will be necessary to measure the rise of the estimated center of mass.] Alternately, if the load rises the same as the eyes [as it does w a perfect // linkage] the eyes provided a force of exactly the loads weight.
larry

I'm glad there are at least two of us that understand how things like this work.

I think the forklift example I gave in an earlier post is a perfect comparison for this. A force is applied in an upward motion and a load at some distance away, it doesn't matter what the distance is, it takes the same upward motion to lift, however, the further the load is away, the greater the stress on the components, and more likely to tip over, but same force. A perfect //ogram is the same in that it applies an upward force to a load that will remain leved (ie, move the same distance as the ball ends). It's just the method for applying the force is different, but the same force is required.

 

[SPYDERLK]

THAT is IMPOSSIBLE!

Not be be argumentative or anything...

Cheers,
Mike

the earth was flat

 

[LD1]

The earth was also the center of the universe too.

 

[LD1]

To sum up everything about hydraulics, weather it be loaders, forklifts, 3ph's, dumptrucks, or whatever else you can think of it is pretty simple

The amount that can be lifted and the distance it is lifted is DIRECTALLY proprotional to the force applied and the travel distance of the object of force. (minus any small frictional losses of pivots and bushing and whayever else)

Yes it really is that simple

 

[SPYDERLK]

I'm glad there are at least two of us that understand how things like this work.

I think the forklift example I gave in an earlier post is a perfect comparison for this.

Yes. I was tempted but consciously never used it because I thot there would be a focus on extraneous differences by those acustomed to only simple action reaction mechanisms. Told me so.
larry

 

[LD1]

Maybe we scared off mwb. He hasn't posted in awhile.

I'm kiinda disappointed as I answered his questions and he hasnt answered mine

 

[gpflepsen]

Jeeze, I thought I had a handle on it...

Another example of a // lift mechanism is the car lift that is found on the rear of race car haulers.

 

[SPYDERLK]

Hi,

Thanks everyone. I didn't know this was such in interesting subject.

I was not able to follow this thread completly because it had grow so large while I was buying and driving my new tractor. My comparisions are done.

If a reasonable conversion factor exists it may be useful to others.

At 24" setback, 80+/-5% of the lift available at the eyes seems about right based on the statstics iv seen and heard.
larry

 

[mwb]

Maybe we scared off mwb. He hasn't posted in awhile.

I'm kiinda disappointed as I answered his questions and he hasnt answered mine

Gotta eat and sleep sometime...

I don't know why you two are so aggressive about this. You would rather accuse me of thinking that the world is flat than explain...

Anyway, while I was run off I did some sketches of the problem - and I stand corrected - a perfect //gram does not change the lift force as the load is extended past the ball ends. BUT this is not a practical answer. I have attached four sketches; note I have not included all of the forces (i.e. in the tractor frame) and I did not calculate a Mohrç—´ circle for each component

#1 is a "at ball ends" load
#2 is what SPYDER and LD1 are trying to explain (note that there is no change in the lift force but the top link and drag link are put in tension, compression).
#3 and #4 are what I think is a more practical application.




I don't know what manufactures use to calculate extended loads (this is what the OP was asking for) and your theory does not come close to answering the question. Maybe the extended load is limited by the tension in the top link? This would not help someone trying to calculate a lift capacity of a boom pole though.

My intention was to give the OP an answer that made sense and that could be used in a practical application.

Do you have any constructive comments on this?

Peace,
Mike

 

[LD1]

Well I'm glad you came around and now their are 3 of us that understand how a perfect //ogram work. In previous post too I also mentioned that it would increase the tension and compression of the top and lower arms. I also mentioned that it would be harder on all the pins as well. So now we agree on all that.

As I have mentioned in several posts before, to calculate the load at a given distance away, there is no set number or percentage. It varies with every tractor, and it can even vary on the same tractor by mounting the top link into a different hole. The further you deviate from a //ogram, the more capacity you loose. The best way is to measure the travel distances of the pins, and a know point however far back you want, and then you have your percentage.

On my tractor I have 3 mounting locations for the toplink. The top hole is the closest to parallel. Hooking their and extending the toplink about the same length as the lower arms, I may only loose 5% capacity @ 24".
But on the same implement, If I shorten all the way and mount toplink to lower hole, this causes the implement to raise much faster in proportion to the ball ends, thus reducing my capacity @ 24" by more than the first setup.