3 Point hitch upper link

[SPYDERLK]

It's not too hard to get an estimate if you make a few simplifying assumptions:
View attachment 1288750

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.

Also note that this shows a true 4 bar system. The 1000# lift/holding force would be the only unchanging value as the load was moved in or out.

 

[Runner]

Depends on the geometry of the load.
Lifting the top link is in tension. Just like in rigging the tension in lifting rigging depends on the angle. Can exceed load weight. Plowing or grading in compression. Dynamic loading, in motion or influenced by outside source, can exceed static load many many times.

If there were load gauges on our 3pt or FEL components most folks would be shocked by forces they are subjected to.

This seems to be the most accurate answer to the question as stated as far as I can see. Just for fun (but not really fun for me) I assumed the attachment to be a finish mower and the angle of the top link to the rear support points to be 30 degrees and 500 lbs supported by the front (lower pins) and 500 lbs supported by the top link (rear pins).

Checked an old structural calculation book and it gave me the formula sin30 = 500 (weight supported by top link)/A(tension in top link). I came up with 1000 lbs of tension on the top link support. Not sure if I believe that answer because it doesn't seem possible, but that's what I came up with.

Hopefully one of the actual engineers on this site will tell me if this is anywhere near correct....

 

[TerryR]

This seems to be the most accurate answer to the question as stated as far as I can see. Just for fun (but not really fun for me) I assumed the attachment to be a finish mower and the angle of the top link to the rear support points to be 30 degrees and 500 lbs supported by the front (lower pins) and 500 lbs supported by the top link (rear pins).

How far back is the center of gravity of the mower from the 3-point pins? How high above the center of the lower link links is the attachment point for the upper link?

Without those there is no way to calculate the load on the upper link. Applying the formula for the angle of the top link is meaningless unit you know the horizontal load on the top link. See the diagram in my post above (by an actual mechanical engineer, me).

 

[Chewwy]

This seems to be the most accurate answer to the question as stated as far as I can see. Just for fun (but not really fun for me) I assumed the attachment to be a finish mower and the angle of the top link to the rear support points to be 30 degrees and 500 lbs supported by the front (lower pins) and 500 lbs supported by the top link (rear pins).

Checked an old structural calculation book and it gave me the formula sin30 = 500 (weight supported by top link)/A(tension in top link). I came up with 1000 lbs of tension on the top link support. Not sure if I believe that answer because it doesn't seem possible, but that's what I came up with.

Hopefully one of the actual engineers on this site will tell me if this is anywhere near correct....

You are correct.

Fx=F∗cos(a)
Fy=F∗sin(a)

Where F is the magnitude of the force

a is the angle (degrees)
Fx is the horizontal force
Fy is the vertical force

The force in tension is the square root ot the sum of the squares of the horizontal and vertical forces.

This is just the static load (i.e., no movement).

 

[TerryR]

You are correct.

Fx=F∗cos(a)
Fy=F∗sin(a)

Where F is the magnitude of the force

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?

 

[Runner]

How far back is the center of gravity of the mower from the 3-point pins? How high above the center of the lower link links is the attachment point for the upper link?

Without those there is no way to calculate the load on the upper link. Applying the formula for the angle of the top link is meaningless unit you know the horizontal load on the top link. See the diagram in my post above (by an actual mechanical engineer, me).

Thank you. Believe it or not, it took me 1.5 hours to come up with my answer, and during that time, you posted your answer, so I didn't see it until after I was done

2000lbs is even more unbelievable, but I would definitely take the word of someone who knows what they are talking about over me.

One question though, assuming the center of gravity is dead in the center of the unit, why wouldn't you divide the load between the front and rear support points, i.e., 500 on the lower pins and 500 on the top link (rear support connections)?

 

[SPYDERLK]

It's not too hard to get an estimate if you make a few simplifying assumptions:
View attachment 1288750

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. Believe it or not, it took me 1.5 hours to come up with my answer, and during that time, you posted your answer, so I didn't see it until after I was done

2000lbs is even more unbelievable, but I would definitely take the word of someone who knows what they are talking about over me.

One question though, assuming the center of gravity is dead in the center of the unit, why wouldn't you divide the load between the front and rear support points, i.e., 500 on the lower pins and 500 on the top link (rear support connections)?

Looking at TerryR diagram you can see that the top link supports no load - only tension. When above or below the position shown the vertical line at front of implement comes into play causing forces to interact on all links because of the cantilevered weight of the implement COM. The sum of these effects at each and every position preserve the constant 1000# lift force applied at the lift balls.

 

[Runner]

Looking at TerryR diagram you can see that the top link supports no load - only tension. When above or below the position shown the vertical line at front of implement comes into play causing forces to interact on all links because of the cantilevered weight of the implement COM. The sum of these effects at each and every position preserve the constant 1000# lift force applied at the lift balls.

Would a quick hitch make any difference in the load distribution?

 

[TerryR]

One question though, assuming the center of gravity is dead in the center of the unit, why wouldn't you divide the load between the front and rear support points, i.e., 500 on the lower pins and 500 on the top link (rear support connections)?

No, the whole 1000# load will be split between the outboard ends of the two lower links. That's what holds up the implement.

There is a 1000# load in the opposite direction at the front end of the lower links. That's what holds them up. Then the same load reversed again is carried by the wheels, which is what holds up the front of the lower links. The load at the wheels is increased because the tractor frame is rigid, and the load at the front of the lower arms tries to lift the front of the tractor -- I guessed that might be 200 lb. in this diagram:

But all of that is irrelevant to the load on the upper link.

No vertical load carried by the upper link. Remember, it is connected by pins on both ends, so it is free to rotate. If a vertical load were applied to the outer end it would just rotate downward until the load hits the ground. It's the lower arms that carry the load.

The upper link keeps the implement from rotating around the pins in the lower link, which is what it wants to do (just leave the upper link off and raise the lower arms and see what happens). So the upper link just carries a horizontal load resisting the tendency of the implement to rotate until it's rear end is on the ground. It's the multiplier (lever) effect of the difference between the distance between the pins and the center of gravity and the height of the upper link connection that causes the tension in the upper link to get so high.

 

[TerryR]

Would a quick hitch make any difference in the load distribution?

It would shift the implement back a few inches, thus moving the center of gravity back a bit and increasing the tension in the upper link proportionally. If it placed the outer end of the top link higher or lower than it would be with a direct connection to the implement, that would also lower or raise the tension in the link a bit.