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09282009, 07:31 AM #1
hydraulic cylinder (trigangle calculations)
I am planning on turning a flat bed into a dump bed with a normal cylinder lift (not a scissor) and was just pondering the formula for the calculation of cylinder force mounted on an angle and hoped you guys might be able to clear some things up.
According to the formula you take the sin of the angle x the cylinder force and you get the vertical lift force of of the cylinder. What I don't inderstand is why this formula is the correct one when to me the numbers just dont add up.
Example: I plan on using a 3.5" x 30" cylinder with a 3000psi pump which = 28,800lbs force. the cylinder is 34" retracted length and I plan on the base being 16" below the rod pivot.
Sin 16/34= ~28* according to the formula that would = 13520lbs lift.
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.
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
16/46 x 28800= 10017lbs vertical lift and
30/46 x 28800= 18782lbs horizontal force
Which totals 28800lbs
I have attached a rough diagram
What am I missing????".........there is only one way to find out."
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09282009, 08:18 AM #2
Re: hydraulic cylinder (trigangle calculations)
The chances are very good that the applied force changes in direct correlation to the hydraulic angle changes.
So for design purposes you would need the complete lift design angles.
You have determined the angle of the force which is 28 degrees and 28,800
so
sin 28 = vertical force/28,800
or
16/34 = 28,800/vertical if you use a ratio method.
Don't guarantee my numbers as I'd always made the top 99% of the class possible!Last edited by Egon; 09282009 at 08:38 AM.
Egon50 years behind the timesLivin in aWorn out skin bag filled with rattlin bones

09282009, 08:39 AM #3
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Re: hydraulic cylinder (trigangle calculations)
That cylinder at 28,863 lbs, pushing at a 28 degree starting angle, will only push with a force of 13,551 lbs and the push force improves as the cylinder raises the trailer bed to a higher angle. When, and if the bed of the trailer is at 90 degrees, then it could push at the cylinder max force of 28,863. Rules are rules. You could push a greater force if the cylinder was used in a lever principle, pushing on a lever that was pushing on something else. What's that saying , get me a lever big enough, and I can move the world.
If you want the cylinder to start out at the max lifting capacity of the cylinder, you will have to mount the cylinder at a 90 degree angle to the bed, and the push force will diminish as the bed is raised.
The length of the cylinder will not have any bearing on the force of the cylinder. The length will determine the distance the bed front will move up.J.J.
When I works, I works hard. When I sits and thinks, I goes to sleep.
Git er done.

09282009, 08:48 AM #4

09282009, 11:15 AM #5
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Re: hydraulic cylinder (trigangle calculations)
Leverage. The total of the forces exerted is greater than the force applied. Think about trying to pull your truck with a rope. If you're bigger than I am, you can probably throw the rope over your shoulder and move the truck a little ways. But tie the other end of the rope to a fixed point (tree) and pull sideways in the middle of the rope. The force you apply directly to the rope is much smaller than the force applied to the truck, and the truck moves more easily.
Make sense?
rus

09282009, 12:18 PM #6
Re: hydraulic cylinder (trigangle calculations)
I understand the formula and the principal you are describing. What I dont understand is how the total output force can be greater than the input, and why using force vectors isn't a more accurate way of measuring lift??
I know that the higher the bed raises the more force the cyl has, but I am only worried about it's starting capacity, If I have enough force to start the dump, the rest of the cycle is easy,".........there is only one way to find out."
"Ok, hold my beer and watch this.........."
MX5100 DIY Remotes
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SachsDolmar 120SI Ported
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09282009, 12:20 PM #7
Re: hydraulic cylinder (trigangle calculations)
".........there is only one way to find out."
"Ok, hold my beer and watch this.........."
MX5100 DIY Remotes
MX5100 Diverter install
Ford 5500 Backhoe
Kubota MX5100HST w/LA844
2005 Dodge 3500 4x4 Diesel
8N Rebuilt and restored
Bushhog 306
3 Homemade wood hauling trailers
Dolmar 6400 84cc ported
SachsDolmar 120SI Ported
(4) SachsDolmar 116SI Ported
Dolmar PS540
SachsDolmar 115i
SachsDolmar 117
SachsDolmar 112

09282009, 01:21 PM #8
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Re: hydraulic cylinder (trigangle calculations)
If you know the load you have to push, which is the trailer bed plus the contents, then if it is less than the 13 ,551 lbs push force on the cylinder, at the angle you mounted it, then it will raise it up and dump.
If the bed weighs 300 lbs, and a load of sand weighs in at 3000 lbs, then you only have to raise 3,300 lbs, and with that cylinder you have, it should be no problem, as you start with a potential to push 13,551, and as the angle increases, more force is available but not necessary.Last edited by J_J; 09282009 at 01:44 PM.
J.J.
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Git er done.

09282009, 01:36 PM #9
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Re: hydraulic cylinder (trigangle calculations)
I'll flesh out my truck example a little (illustrated, even) and see if it helps you understand how the sum of the forces can be greater than the input.
Applying the 100lb load as shown, with a 10 degree angle in the rope will give you 567lb of force at the truck. The distance moved at the load, versus at the truck is the difference. Force*distance remains constant.
Another example would be a pulley attached to a movable load. If you apply 100lb to the rope and move it a foot, you will exert a 200lb force on the load, but only move it six inches. This is harder to calculate for your cylinder example, because the angles/forces are constantly changing, but you get the idea.
rus

09282009, 01:59 PM #10
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Re: hydraulic cylinder (trigangle calculations)
Where did that 39,000 come from?
If LD1 is only using the 3.5 in cylinder he said he wants to use, then the maximum push force is 28,886 lbs. That's it period. That cylinder is really overkill on that trailer bed, unless it is built to support 12,000 lbs. If it is built for that weight, then he is good to go with that cyl, and that angle. Anything other than the cyl itself is some kind of a torque multiplier, and that is the only way to come up with those high figures.
I can see where the 25,485 came from. That is the angle as the bed lifts up and as it gets to the 62 degrees, between the bed and the cylinder angle, the push force is 25,485, and gets stronger as the angle increases to 90 degrees, which is the max push angle and push potential. .J.J.
When I works, I works hard. When I sits and thinks, I goes to sleep.
Git er done.