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Discussion Starter #1 (Edited)
I apologize if this has been talked about b4, but I can't find specific answers to my question in the threads I was able to find here that involve 3 Pt Hitches, and or quick hitches.

I have decided to go with a set of Pat's easy hitch hooks. The LS J2030H that is now happily on the way in the next couple weeks has a lift capacity of 1433 LBS at the ball ends. Now it's not a question of rather or not I will go with the Pat's, that's been decided. The question is, how much capacity do you lose for every inch further out from the ball ends you go? My assumption is that it's not a heck of a lot. We are talking 3 to 4 Inches that the Pat's add to the length of the arms. Obviously you have to reduce the amount of weight you attempt to lift the further out from the ball ends you get. But I have been told that I can easily pick 500 LBS with a boom pole, and that is considerably further away from the ball ends on the draft arms. Yet in all the threads that I have read, this 3 - 4 inch of extension keeps coming up as a concern/negative. And yes, I have already figured in the extra length needed for the top link and the PTO shafts. (That's the beauty of buying new stuff)

I do not have a boom pole, and the heaviest bit of equipment I will be getting with the tractor is prob the tiller, or maybe the brush hog. So for the sake of argument, lets say the the tiller is 300 lbs. most of that weight is not centered on the ball ends, its effectively some distance behind them, which means, (If I am thinking correctly), that the required force to lift the tiller is something more then 300 lbs. Or should I say, the force at the ball ends is more then 300 lbs. Moving the hitch pins an additional 3 - 4 inches out would have the effect of amplifying this. The question is, what kind of amplification are we talking about? Is it a linear scale, such that each Inch you move out, the weight is increased by a fixed amount, or is is a non linear scale, such that each Inch you move out, the weight is increased by an amount greater then the previous inch increase.

The problem is, I don't know how much force my 300 LB tiller actually requires from my 3 point hitch in stock configuration. Nor do I properly understand how much of an increase in force will be required from the 3 Pt by moving th weight out 1, 2, 3, 4 inches further.

The fact that it keeps coming up as a concern in threads about the Pat's indicates that the increase is substantial. But is this just a matter of over estimation, or is the increase really that substantial? Does anyone know how to calculate this?
 

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Yes, it is pretty much linear. I am not familiar with you tractor model, but if it is like most, it has upper lift arms that actually supply the lifting power to the lower arms through some sort of linkage. If you know what the lift capacity is at the ball ends, then you should be able to use simple physics to determine the maximum lift capacity at any point out beyond the balls.
 

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Ok.. You need to measure two things to find your lift capacity;
* first is the distance from the ball center on the lift arm to the ball center on the other end (mount end) of the SAME lift arm (you're measuring the length of the arm, but only to the ball centers),
* the the second distance is the length from center of the MOUND END ball to the center of the pin of the linkage to the drop arms.

We will call the length of the lift arm A and the distance to the linkage B.

The third distance is A + the distance of the Pat's hitch measured from the large bolt center to center of the hook; this will be C.

This is a two step process, first we need to find the force at the linkage pin. As an example I will say my lift arm length (A) is 20 and the pin distance (B) is 10. I will assume the Pat's system moves the attachment exactly 4 inches.

A quick calculation to give us C is this:

C = A + 4
C = 20 + 4
C = 24

Ok, to find the force at the linkage pin we do thusly;

PinForce = 1433 (your original capacity) * A / B
PinForce = 1433 * 20 / 10
PinForce = 2866 lbs.

To find your new force we now reverse it;
NewForce = PinForce * B / C
NewForce = 2866 * 10 / 24
NewForce = 1194 lbs


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NOTE: Some tractors have 2 or 3 holes for the linkage pin location; moving this pin out one hole (usually an inch or two) can easily gain back what you loose to the Pat's system and perhaps even more! (ie: moving the pin in the example above out 2 inches gives you 1433 lbs force on the Pats!)
 

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Discussion Starter #5
Jim,

Thank you very much for this explanation. I have book marked this for future ref.

I take away two things from this.

1) 4 inches makes a larger difference then I thought, and I don't see how I could possibly lift 500lbs at the end of a 8 foot boom pole as was suggested by the seller.

2) It should be of little consequence to the routine use of the tractor. If i were to put maybe a set of forks on the 3 pt at some point, it might well eat away at its overall ability. But then, removing the Pat's is not gaining me a heck of a lot. I sort of feel like if I can't do what I need to with the Pat's on, then gaining a couple hundred pounds of lift capacity won't make the critical difference. Seems to me as though the tractor will either be able to handle the load, or it won't regardless of rather or not the Pat's are in place.


This has been very helpful and greatly appreciated.
 

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....I don't see how I could possibly lift 500lbs at the end of a 8 foot boom pole as was suggested by the seller....
I'm not sure how much of the lift capacity of the boom pole comes from the 3PH? The ones I've seen have a block and tackle lifting mechanism that provides most of the range of lift as well as the capacity. In all these calculations, I would be more concerned about the amount of front ballast necessary to counteract the leverage from 500 lbs at the end of an 8' boom pole!
 

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The LS J2030H .......... has a lift capacity of 1433 LBS at the ball ends. The question is, how much capacity do you lose for every inch further out from the ball ends you go? My assumption is that it's not a heck of a lot. We are talking 3 to 4 Inches that the Pat's add to the length of the arms. Obviously you have to reduce the amount of weight you attempt to lift the further out from the ball ends you get. But I have been told that I can easily pick 500 LBS with a boom pole, and that is considerably further away from the ball ends on the draft arms. Yet in all the threads that I have read, this 3 - 4 inch of extension keeps coming up as a concern/negative. And yes, I have already figured in the extra length needed for the top link and the PTO shafts. (That's the beauty of buying new stuff)

I do not have a boom pole, and the heaviest bit of equipment I will be getting with the tractor is prob the tiller, or maybe the brush hog. So for the sake of argument, lets say the the tiller is 300 lbs. most of that weight is not centered on the ball ends, its effectively some distance behind them, which means, (If I am thinking correctly), that the required force to lift the tiller is something more then 300 lbs. Or should I say, the force at the ball ends is more then 300 lbs. Moving the hitch pins an additional 3 - 4 inches out would have the effect of amplifying this. The question is, what kind of amplification are we talking about? Is it a linear scale, such that each Inch you move out, the weight is increased by a fixed amount, or is is a non linear scale, such that each Inch you move out, the weight is increased by an amount greater then the previous inch increase.

The problem is, I don't know how much force my 300 LB tiller actually requires from my 3 point hitch in stock configuration. Nor do I properly understand how much of an increase in force will be required from the 3 Pt by moving th weight out 1, 2, 3, 4 inches further.

The fact that it keeps coming up as a concern in threads about the Pat's indicates that the increase is substantial. But is this just a matter of over estimation, or is the increase really that substantial? Does anyone know how to calculate this?
While a boom pole may be quite long, it's primary purpose is to lift from a greater height rather than from further behind the tractor. At 90* to the ground it will have (in theory) 100% of the lift capability available at the hitch balls. As it is tilted back, height is traded for reach, at the expense of lift, until it is horizontal. The cosine (COS on a calculator) of the angle multiplied with the overall length of the boom will give the distance of the lifting point behind the balls. That distance factored into Jim's calculations will give available lift capability by the 3PH at the lifting point on the boom.

At the risk of causing some confusion, here's a slightly different way of looking at what Jim has already laid out. This is a problem involving leverage and ratios.

Given:
A - The center of the pin connecting the lower arm to the tractor, aka, the fulcrum.

B - The center of the pin on the lower arm where lifting force is applied by the rockshaft through the lifting links.

C(x) - The center of the pin on the lower arm to which load is applied, "C" being the center of the balls at the end of the lower arm, and the suffix (x) being the number of inches to be added for load points beyond "C".

Maximum load capability at point C = 1433 lb.

....................................A____________B_______________C(x)
...............................Fulcrum...........Lifting Force...................Load

Lifting Force at B (called F) = [The length in inches of AC / The length in inches of AB] X The load (called L), or,

F = [AC / AB] X L

Once solved, then the equation is reversed for the unknown L(x), (x) being the designator to correspond the Load with C(x).

L(x) = [F X AB] /AC(x)

eg.
F = [AC / AB] X L
F = [30 / 10] X 1433
F = 3 X 1433
F = 4299 pounds of force

To continue, for the extra 3" for the Pat's Hitch;

L(3) = [F X AB] / AC(3)
L(3) = [4299 X 10] / [30 + 3]
L(3) = 42990 / 33
L(3) = 1303 pounds of load 3" beyond the center of the balls.

As with any equation any unknown can be calculated if all other values are known, such as how far beyond the balls can 500 lb be lifted.

AC(+?) = F X AB / L
AC(+?) = 4299 X 10 / 500
AC(+?) = 86", or,
AC(+56) = 30" + 56"

Your actual numbers will be different than these invented numbers.

I hope that I didn't confuse the issue. It's the same equation as Jim's, only with different numbers and an added simple graphic illustration for clarity (I hope :praying: ).
 

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Something else to remember about boom poles... most generally the angle of the pole changes as the 3PH is raised and lowered. You can generally see that while the pole may be level or nearly so with the hitch lowered, it will usually incline to 30-45 degrees with the hitch fully raised (the amount of inclination is dependent on your hitch geometry).

There are two ways to calculate the lift you would have at the boom; the quite complex way and the quick&dirty way. I'll go with the quick and dirty way....

The basis here is that the 3PH does work, think weight lifted over a height. Mathematically we will represent this as Weight Lifted x Height. This number will be same at the 3PH balls and at the end of the boompole, meaning the boompole it simply a giant lever that allows you to trade off capacity for lifting height.

Lets say your Lift arms travel 10" at the hook centers (or ball canters without the Pat's) and the hook of the boompole travels 40", also we will assume your boompole weighs 50 lbs. Using the 3PH capacity we found earlier of 1194 lbs, our calculation is this:

(3PH Capacity - BoompoleWeight) * LiftArmTravel / BoomPoleHookTravel = BoomPoleLiftCapacity

So plugging in our numbers:
(1194 - 50) * 10 / 40 = ?
(1144) * 1/4 = 286 lbs

This is just the lifting force applied by the 3PH, adding a block and tackle will give you more capacity, BUT...
1) You'll overload your 3PH and it will not lift the load
2) You'll likely lift the front of the tractor without a lot of ballast
 

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Discussion Starter #9
WOW,,, Excellent. College math came flooding back to me reading this stuff. And after taking a day or so to let it soak in, I am happy to say I understand it to boot. I don't know how many times people ever asked or wondered about this, but if its not already, this short thread should be a sticky.

I inferred from the first explanation from Jim, that picking 500 LBS off the ground with a 8' boom pole would be highly unlikely. Without actually doing the math out, the numbers clearly would not work out. Jim, in your second plug, If I am actually understanding things properly, your math makes that assumption clear. With or without the Pat's installed, there is no way I could pick up a 55 gallon drum of water off the ground, and into the back of my truck. Which would be about the worst case scenario I could ever envision. Not that this was even really a concern. This stems from a cocamamy Idea I conceived for watering my garden with lake water should the weather turn off as dry as last summer. But I have a 150 gallon sap tank that sits in my trailer that would be just as easy to fill and pump back out on the garden, all without having to lift anything, other then maybe the water pump.

At any rate, point well taken here. The pat's is not reducing the capacity of my 3 point enough to make it a critical issue. Anything that I could not handle with the Pat's on, I likely could not handle with them off any how. I have already forwarded this thread to a few friends. Thanks a lot guys. And excellent mathematical explanations.
 
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