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Thread: swaybars and articulation

  1. #21
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    It would have to be, not so uninformed, basic physics - for every action there is an opposite and equal reaction. If this super spring is pushing the axle down it must be pushing the chassis up with equal force.

  2. #22
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    Serg, as you are aware, KC is up himself at times, and that was one of those golden times.

    I go along with him, with his dislike for cones and dislocation suspension. He would have a point if the X-spring was only used for rear suspension together with poorly balanced front articulation.

    There are significant technical difficulties that prevent the use of ideal springs for suspension travel much greater than stock in a Land Rover. The X-spring goes a good way towards addressing some of those. It would be better if the main spring was designed for the purpose, but that is going to increase the cost.

    Coil overs are ideal, and so widely used in the USA, which is possibly why KC is so dismissive of other approaches. I bought four Swayaway racerunners for my bushie many years ago, but couldn't get them engineered here

    Seeing the X-spring when it first came out has given me another idea, which so far looks very promising. Although the design is advanced, I have a lot of other stuff to finish before I start to make it happen.

    I have no idea what the post following yours is about

  3. #23
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    Does the X-spring and more "spring rate" to the springs already in the vehicle?

    I just watched a vid, by the looks of it when the vehicle is at rest the x-spring is being compressed inside its little cup. And to me it is not idependant of the normal spring????

  4. #24
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    Quote Originally Posted by uninformed View Post
    Does the X-spring and more "spring rate" to the springs already in the vehicle?

    I just watched a vid, by the looks of it when the vehicle is at rest the x-spring is being compressed inside its little cup. And to me it is not idependant of the normal spring????
    The X-spring makes the springs dual-rate.

    When the main spring is seated the X-spring isn't doing anything and your spring rate is the same as it always was. But as it articulates and moves off the seat (which will happen earlier than it used to) the second spring comes into play and spring rate is reduced.

    It's a very clever solution to those needing more droop travel and longer springs than you can normally fit between the axle and spring mount.

  5. #25
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    Quote Originally Posted by uninformed View Post
    Does the X-spring and more "spring rate" to the springs already in the vehicle?

    I just watched a vid, by the looks of it when the vehicle is at rest the x-spring is being compressed inside its little cup. And to me it is not idependant of the normal spring????
    The rate changes depending upon the axle travel and spring extension.

    If the spring rates are:
    Ko for the outer spring
    Ki for the inner spring
    Although the inner spring is inside the outer, the top hat effectively couples the springs in series, not parallel.

    Once the outer spring looses direct contact with the axle during rebound/droop the combined spring rate becomes less than either of the individual rates and:
    Combined spring rate = (Ko x Ki) / (Ko + Ki)
    While the axle is in direct contact with the outer spring the spring rate will be the rate of the outer spring, i.e. spring rate = Ko

    Edit: I wrote that while having breakfast and Dougals post came up while I was taking my time typing, etc.

  6. #26
    chook73 Guest
    Quote Originally Posted by Bush65 View Post
    The rate changes depending upon the axle travel and spring extension.

    If the spring rates are:
    Ko for the outer spring
    Ki for the inner spring
    Although the inner spring is inside the outer, the top hat effectively couples the springs in series, not parallel.

    Once the outer spring looses direct contact with the axle during rebound/droop the combined spring rate becomes less than either of the individual rates and:
    Combined spring rate = (Ko x Ki) / (Ko + Ki)
    While the axle is in direct contact with the outer spring the spring rate will be the rate of the outer spring, i.e. spring rate = Ko

    Edit: I wrote that while having breakfast and Dougals post came up while I was taking my time typing, etc.
    How would/could you calculate the point on a side slope where the x-spring could possibly start to push the vehicle over?

  7. #27
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    Being a visual more than mathematical person, i'd use something along the lines of a force diagram where the compression on the main spring is equal or less than the compression rate of the X Spring, you'd need to know your CofG too, I suspect with all your portal problems you'll have a fair idea where that is by now...

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