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Thread: Help with rollers lifting an object

  1. #1

    Help with rollers lifting an object

    I'm trying to design a hydraulic system that that will lift a tire on a car off the ground with a fixed roller and a sliding (laterally) roller driven by a hydraulic cylinder. In the picture below is a basic layout of what is going on. The black circle is the tire and the two blue circles are roller tubes. Before I determine the pressure of my system, pump speeds, HP, etc I want to find out how much force will be required to lift the tire (one corner of the car) but since the rollers will rotate about its axis, I think I might get a mechanical advantage so my hydraulic system won't have to work as hard.
    Black tire = we'll say it weighs 2000 lbs and can only move up and down (27" diameter)
    Blue circle (1) = 1.5" roller that is fixed to the machine but it can rotate about its axis (looks like it will rotate CCW)
    Blue circle (2) = same 1.5" roller but it slides laterally (left to right) by means of a hydraulic cylinder. It too can rotate about its axis (rotate CW)
    F = The force applied to the sliding roller tube that is pushing in the direction towards the other roller (#1)
    The center of the rollers are about 1.25" off the ground and the tire is on the ground. For now I'm assuming that the tire is not deforming while it is on the ground and also the tire will not deform when the rollers engage the tire.
    jGPtQNp.jpg

    What I am trying to determine is the equation that takes into account the fact that those rollers will be spinning (outward) as roller #2 moves inward and begins to lift the tire.

    I started my design by taking a 1500 PSI system and a 1.5" bore cylinder and I calculated that I could produce about 2650lbs of force which seemed like I was where I wanted to be. But then I was thinking about how those rollers with respect to the tires should be making it easier to lift with each millimeter more a lateral travel. Is there an equation that I can put in the tangent relationship between the rollers and the tire with a given force applied?
    Last edited by Kelly Bramble; 02-14-2014 at 11:17 PM.

  2. #2
    OK, here's what I've come up with. I can treat the interaction with the tire and the roller like a ramp. I can take my known hydraulic pushing force (2651 lbs) and multiply it by the SIN Θ of the initial angle where the tire touches the roller, right?

    2651 * SIN 30 = 1325 lbs or force acting on the tire in the vertical direction at the beginning. As the hydraulic continues to push the roller to the left, the tangential angle gets smaller and force acting on lifting the tire goes down, right? 2651 * SIN 10 = 460 lbs of force lifting the tire. Then if the roller is completely under the tire there is no lifting resultant force (2651 * SIN 0 = 0 lbs).

    So now I'm stuck on the variable of the weight of the car (tire). If I need to lift 2000 lbs, did my equation above say I can only lift 1325 lbs? What about the roller on the other side of the tire, is it helping the same way as the roller that is being moved hydraulically? So do I have 2650 pounds of lifting force at the beginning (the resultant force on both rollers added together)?

  3. #3
    Technical Fellow jboggs's Avatar
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    How are the rollers supported? Are they in contact with the ground? If so, you have a conflict in the moving roller. Its contact with the tire will cause CW rotation. Its contact with the ground will cause CCw rotation. Can't do both.

    You say these rollers will be spinning outward. Is that spinning driven by anything? Or is it just a result of their horizontal motion inward? If the spinning is driven by some external source it might buy you little bit of advantage but not much. If it isn't driven externally it is of no value to you, other than decreasing friction with the tire.

  4. #4
    Quote Originally Posted by jboggs View Post
    How are the rollers supported? Are they in contact with the ground? If so, you have a conflict in the moving roller. Its contact with the tire will cause CW rotation. Its contact with the ground will cause CCw rotation. Can't do both.
    The rollers are supported on the same frame as the hydraulic system. The center of the roller is about 1.25" off the ground.


    Quote Originally Posted by jboggs View Post
    You say these rollers will be spinning outward. Is that spinning driven by anything? Or is it just a result of their horizontal motion inward? If the spinning is driven by some external source it might buy you little bit of advantage but not much. If it isn't driven externally it is of no value to you, other than decreasing friction with the tire.
    The rollers are on a bearing system, they'll spin with the tire as they are pushed together and should help reduce the friction as you mentioned.

  5. #5
    Technical Fellow jboggs's Avatar
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    Drawing1.bmpYou are pulling yourself off track by considering the free-spinning rollers as an additional factor to consider. Actually in an initial analysis you would assume friction to be zero so you can understand the forces. If friction is zero that means your rollers rotate (spin) as they travel along the circumference of the tire. If they were restricted from turning, that would create friction, and that would create a need for additional calculations.

    Simple physics and force vectors will solve your problem. The resultant force at each roller must be perpendicular to its contact point on the tire. That resultant force has a vertical component and a horizontal component. The weight on the tire is W. The vertical component of the resultant force at each roller is 1/2W. The horizontal component is what you are trying to determine (as the force from the hydraulic cylinder pulling the rollers together).

    The vertical component is 1/2W. The horizontal component is determined by the amount of the vertical component and the angle from the horizontal of the resultant force.

  6. #6
    Thanks for the response. So you're thinking that the PSI required by the hydraulic system is 1/2 * W * COSΘ?
    W = 2000 lbs
    Θ = 35
    0.5*2000*COS 35 = 819.15 lbs of horizontal force. Then as roller continues to move towards the middle of the tire the horizontal angle should go down, right? If I plug a lower angle in I get a higher value. I think I have it backwards, it should be SIN Θ, right?

    W = 2000 lbs
    Θ = 35
    0.5*2000*SIN 35 = 573.58 lbs of horizontal force. As the angle gets shallower (roller getting more under the tire) the force required goes down because the roller is not doing as much lifting as it was earlier, right? With SIN 25 = 422.62 lbs, this seems like I'm going in the right direction.

    Now, is the 1/2W because each roller is doing half of the work? Even though roller #1 is fixed in its position (but it can still rotate with respect to the tire's surface) and roller #2 is sliding to the left (and can rotate with respect to the tires surface), that's where I get the 1/2W, right?

    Assuming I have everything right above (and ignoring any friction and stuff like that), I only need a cylinder that provides at least 600 lbs of force, right?

    So I found a chart from a specific company that says their 1.5" bore cylinder at 500 PSI can provide 884 lbs of force. Again ignoring friction, that cylinder would do the job, right? It seems to easy, or is it that easy? I understand that there are factors like friction and final design that play a role but am I headed in the right direction?

  7. #7
    Quote Originally Posted by jboggs View Post
    Drawing1.bmpYou are pulling yourself off track by considering the free-spinning rollers as an additional factor to consider. Actually in an initial analysis you would assume friction to be zero so you can understand the forces. If friction is zero that means your rollers rotate (spin) as they travel along the circumference of the tire. If they were restricted from turning, that would create friction, and that would create a need for additional calculations.

    Simple physics and force vectors will solve your problem. The resultant force at each roller must be perpendicular to its contact point on the tire. That resultant force has a vertical component and a horizontal component. The weight on the tire is W. The vertical component of the resultant force at each roller is 1/2W. The horizontal component is what you are trying to determine (as the force from the hydraulic cylinder pulling the rollers together).

    The vertical component is 1/2W. The horizontal component is determined by the amount of the vertical component and the angle from the horizontal of the resultant force.
    Am I on the right track?
    R = Resultant Force
    F = Lateral Hydraulic Force
    W = Tire Weight

    W=R*COSΘ
    F=R*SINΘ

    Moving variables around and I get F=W*TANΘ, does that seem right?

  8. #8
    Technical Fellow jboggs's Avatar
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    Drawing2.bmp
    OK. Point by point.

    The "PSI required by the hydraulic system" is pressure, not force. First you calculate the
    force required, then you calculate the pressure required to produce that force.

    "Θ = 35" Is that measured from the horizontal or the vertical?

    "0.5*2000*COS 35" This is not a SIN/COS function. SIN and COS are related to the hypoteneuse.
    This function is related to the opposite and adjacent sides only, therefore TAN and COT but I think you caught that later.

    "Then as roller continues to move towards the middle of the tire the horizontal angle should go down, right?" No. The
    angle measured from the horizontal increases to a max of 90 degrees. Yes the horizontal force decreases as the angle increases.

    "is the 1/2W because each roller is doing half of the work?" No. Its because each roller is lifting half the weight.
    Its semantics, but the word "work" in engineering terminology means something slightly different.

    "I only need a cylinder that provides at least 600 lbs of force, right?" That depends on what your starting angle is.
    The horizontal force will be a maximum at your starting position.

    "So I found a chart from a specific company that says their 1.5" bore cylinder at 500 PSI can provide 884 lbs of
    force." All 1.5" bore cylinders produce 884 lbs force at 500 psi - but only when extending (pushing).
    When retracting (pulling), the output force is reduced due to the size of the piston rod. If you intend to pull with
    this cylinder you must consider that factor.

    I can't speak to your trigonometric equations because I don't know how you are measuring your angle.

    Refer to the diagram I'm attaching.

    Also, even though you have to ignore friction in your initial calculations, you cannot ignore it in your final answer. You are not going to lift 2000# without friction being a significant factor. How much of a factor depends on your design. I suggest you contact a good local machine shop and/or hydraulic distributor.

  9. #9
    Quote Originally Posted by jboggs View Post
    "Then as roller continues to move towards the middle of the tire the horizontal angle should go down, right?" No. The
    angle measured from the horizontal increases to a max of 90 degrees. Yes the horizontal force decreases as the angle increases.

    I can't speak to your trigonometric equations because I don't know how you are measuring your angle.
    qgg9zwe.jpg

    Here's a basic picture of the 35 angle I am talking about. The tire is the large diameter circle and the two smaller circles are the rollers. The horizontal solid line is the ground and the horizontal dotted line above it is just showing that the rollers sit 1/2" inch about the ground (or 1.25" form the center of the roller to the ground). Initially the tire is on the ground (yes, I know the tire would slightly deform if on the ground but for this illustration it's fine). I then made a dotted line that is perpendicular to the dotted line that goes between the center of the roller and the center of the tire. This line I created is tangent to the tire and goes to the ground (solid line). The angle I am referring to is that tangential line touching the tire with respect to the ground which you can see is 35.

    Looking at your diagram it is telling me that to get the horizontal force I need the equation (0.5*W)/TANΘ. After reviewing your response I realize the angle I have illustrated will decrease as the roller moves under the tire but when put into your equation the horizontal force increases. That means I have the wrong angle for your equation. If I find the angle you are referring to in your diagram based off how I determined my angle of 35, then your angle is 55. When I put that angle into your equation I get 700 lbs of horizontal force. That is the same answer I get when I put my 35 angle into my equation I listed in a few posts above (0.5*W*TANΘ). This makes sense to me because your equation is the alternate interior angle of my angle and your equation is the inverse of mine.

    OK, that was confusing but we're getting the same answers, but your equation is probably the proper way to do it so that's what I'll use.

    Thanks for the reminder about the frictional forces. I have no intention of leaving them out but I needed to make sure I was starting with the correct equation to determine the horizontal force before I start throwing in variable like friction.

    I know I sound ignorant about this stuff but my problem is that I am horrible at trying to get my point across in these posts. Also, I haven't really done any of this stuff since 2006 when I was in school and it's a pure case of "use it or lose it" for me.
    Last edited by Kelly Bramble; 02-18-2014 at 04:23 PM. Reason: Attach image directly

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