hydraulic cylinder (trig-angle calculations)

[rus_geek]

J_J -- 39000 is the approximate sum of force magnitudes. If the force of the cylinder is absolutely the limit of the available forces, then looking at my truck example, if you put in a cylinder to supply the 100lb load, are you saying there's no way it can apply a force to the truck that is greater than 100lb?

 

[IH Steve]

The force at the truck and at the tree would be the same. The force at the point of deflection would be less. My 2 cents..

 

[CurlyDave]

But the sin of 62* (the other angle to equate to rearward push) x 28800 = 25428lbs. How can a cylinder with a force of 28800lbs create a total of about 39000lbs force??? What am I missing.

You are adding the forces as scalar quantities, when they are actually vectors. The correct vector addition in this case is A**2 + B**2 = C**2. (A squared plus B squared = C squared)

Wouldn't it be better to use force vectors. where 16" is the vertical leg of the triangle, 34 is the hypotenuse, which would make the horizontal leg = 30

30+16=46

The horizontal leg of that triangle is 30 inches (just as your diagram shows), not 46 inches. If you do the force vector calculation, it gives exactly the same result as the trig calculation.

You need to make the horizontal structure beefy enough to withstand the 25428 lbs load it will see, plus a safety factor.

 

[LD1]

I understand that the 13xxx squared + 25xxx squared = 28xxx squared. If you use the sin formula or this one.

What is puzzling me is that a cylinder that has a max force of 28000lbs, can create a vertical force of 13000lbs and a horizontal force (trying to push the bed straight back against the pivots) of 25000lbs. I dont understand how the two resulting forces can add up to more than the supplied force.

The 30" side of the triangle is the horizontal force, and 16" is the vertical load. I used 16/46 as the ratio of vertical force to to the total force.

I know that the sin formula works and that I will actually have 13000lbs lift because it has probabally been tested in many situations, but I was wondering mathematically just how exactally it works out. Call me an idiot.

 

[LD1]

Also the reason fo needing 13xxxlbs vertical force is because I want to be able to dump 8000lbs. Yes I can build everything to handle it no problem. But the cylinder pivot will be 36" forward from the rear pivot on an 8.5' bed. to find the lift @ the midpoint of 51" is lift x 36/51.

13,500 x .705 = 9517lbs @ 3000psi

If I were to drop one size to a 3" cyl I could only dump 7000lbs

The larger cylinder also gives me the opportunity to drop to a 2500psi system and still be able to dump right at 8000lbs.

So it really isnt overkill. This is just a field truck that doesnt see the road and has hauled over 7000lbs many times in the past.

 

[CurlyDave]

What is puzzling me is that a cylinder that has a max force of 28000lbs, can create a vertical force of 13000lbs and a horizontal force (trying to push the bed straight back against the pivots) of 25000lbs. I dont understand how the two resulting forces can add up to more than the supplied force.

All I can say is that is just the way vector math works. I wish I had a more understandable explanation, but I don't know any other way to express it.

BTW, I think you are doing pretty well to start out this way, analyze the problem and work it out ahead of time instead of just getting out the welder and putting it together by gosh and by golly.

I know a guy who built a retaining wall that collapsed twice before he finally did it sort of right. I didn't know him for the first time, but I did when he built it the second time. I calculated the way to build it for him and he didn't want to believe me. When it fell down the second time, he was more willing to listen.

His son built a water feature with about a 8-9' waterfall in his front yard, which promptly collapsed. The second time around on that one, he made his son listen to how to build it.

A little engineering up front goes a long way.

 

[J_J]

J_J -- 39000 is the approximate sum of force magnitudes. If the force of the cylinder is absolutely the limit of the available forces, then looking at my truck example, if you put in a cylinder to supply the 100lb load, are you saying there's no way it can apply a force to the truck that is greater than 100lb?

Pushing the load straight on with that cyl, is 28,886 available force. You can use the cyl to gain additional forces, but not pushing the load straight on.
He is starting the push at a low angle and can push only a computed maximum 13,xxx lbs. If he uses that cyl to push on something else, a lever of some kind, yes he can get more force, but the extra is because of the lever principle.

It's sort of like you lifting a certain weight. With another person helping, then you two can lift more. The other guy being the other force required, to lift past your limit.

 

[Egon]

I dont understand how the two resulting forces can add up to more than the supplied force.

They can't.



Right-Angled Triangle Calculator

Try this site. just plug in the numbers!

 

[J_J]

Just to add some more fuel to the fire, think about how the cylinders are mounted on your loader. They are mounted at an angle. Why you say. Because you have to have enough force to get started to lift the lift arms, with a load. They have computed the size of the cylinders to do just that. If you look up the actual rating of those cylinders, you will notice that they will lift straight on about 5000 lbs plus, and where the lifting point on the loader arms, will determine the load limit.

If you could place that cylinder down under a full bucket, and lift straight up, you will notice that you can now lift 5000 lbs, and that is just one culinder. Facts are facts.

Ask yourself this, how can one man move a railroad car by himself. The answer is a device that weighs about 40 lbs and uses the lever principle.

 

[SPYDERLK]

I understand that the 13xxx squared + 25xxx squared = 28xxx squared. If you use the sin formula or this one.

What is puzzling me is that a cylinder that has a max force of 28000lbs, can create a vertical force of 13000lbs and a horizontal force (trying to push the bed straight back against the pivots) of 25000lbs. I dont understand how the two resulting forces can add up to more than the supplied force.

The 30" side of the triangle is the horizontal force, and 16" is the vertical load. I used 16/46 as the ratio of vertical force to to the total force.

I know that the sin formula works and that I will actually have 13000lbs lift because it has probabally been tested in many situations, but I was wondering mathematically just how exactally it works out. Call me an idiot.

LD1, Think about what happens when you have two forces acting together, but not from the same direction. The resultant is not a direct sum of the 2 but a vector sum. Your resultant [unless the forces oppose] is more than either of the 2 but not as much as them added together... and is in a different direction than they. Your cyl acts along a continuously varying vector that is the hypotenuse. It is the force. We are going backward here and are resolving that force into 2 constituent parts at right anges to each other. Same exact idea looking at it from a different perspective.
larry