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  1. #1
    Super Star Member LD1's Avatar
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    Default hydraulic cylinder (trig-angle 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????
    Attached Thumbnails Attached Thumbnails -dump-bed-jpg  
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  2. #2
    Super Star Member Egon's Avatar
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    Default Re: hydraulic cylinder (trig-angle 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; 09-28-2009 at 08:38 AM.
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  3. #3
    Super Star Member J_J's Avatar
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    Default Re: hydraulic cylinder (trig-angle 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.

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  4. #4
    Super Star Member J_J's Avatar
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    Default Re: hydraulic cylinder (trig-angle calculations)

    Quote Originally Posted by Egon View Post
    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!
    The force of that cylinder is 28.886 lbs, at a straight push, 90 degrees. That force will decrease as you change the angle.
    J.J.

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  5. #5
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    Default Re: hydraulic cylinder (trig-angle calculations)

    Quote Originally Posted by LD1 View Post
    ...
    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.
    ...
    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-

  6. #6
    Super Star Member LD1's Avatar
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    Default Re: hydraulic cylinder (trig-angle calculations)

    Quote Originally Posted by J_J View Post
    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.

    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."
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  7. #7
    Super Star Member LD1's Avatar
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    Default Re: hydraulic cylinder (trig-angle calculations)

    Quote Originally Posted by rus_geek View Post
    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-
    No I don't really understand what you are trying to explain.
    ".........there is only one way to find out."
    "Ok, hold my beer and watch this.........."


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    (4) Sachs-Dolmar 116SI Ported
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  8. #8
    Super Star Member J_J's Avatar
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    Default Re: hydraulic cylinder (trig-angle calculations)

    Quote Originally Posted by LD1 View Post
    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,
    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; 09-28-2009 at 01:44 PM.
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  9. #9
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    Default Re: hydraulic cylinder (trig-angle 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-
    Attached Thumbnails Attached Thumbnails -truck-jpg  

  10. #10
    Super Star Member J_J's Avatar
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    Default Re: hydraulic cylinder (trig-angle 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.

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