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Thread: Debate of the Day: Experimental Chain Tensioner

              
   
   
  1. #21
    Senior Member Stevo's Avatar
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    LOL... ya but they're round corners!!
    Current rides: '96 Honda Ohlins VFR, '03 Cannondale C440R, '03 Cannondale Cannibal, '06 Yamaha 450 Wolverine 4x4
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  2. #22
    Senior Member Stevo's Avatar
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    Glad to see you back at it Andrew! Cheers!
    Current rides: '96 Honda Ohlins VFR, '03 Cannondale C440R, '03 Cannondale Cannibal, '06 Yamaha 450 Wolverine 4x4
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  3. #23
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    What I like to do is to disconnect the rear suspension, place the bike on a stand so that the rear wheel is off the ground and free to move and then move the rear wheel and swing arm through its maximum travel to visualize what is happening with the slack at the two extremes of movement and when the axle is in line with the swing arm pivot and the counter shaft sprocket. With that information you can set the chain slack appropriately when the rear wheel is resting on the ground or wherever you typically check and adjust the chain slack.
    Richard - Current bikes: 2018 16.6 kWh Zero S, 2016 BMW R1200RS, 2011 Royal Enfield 500, 2009 BMW F650GS, 2005 Triumph T-100 Bonneville, 2002 Yamaha FZ1 (FZS1000N) and a 1978 Honda Kick 'N Go Senior.

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  5. #24
    Senior Member Stevo's Avatar
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    Correct, for initial set-up without any chain tensioners. I need to find a replacement for the stationary idler when it wears down...
    How about these, Ted Dillard and the ElectroFlyers? https://www.ebay.com/itm/Skateboard-...kAAOSwcZFbLBm3

    Rasta Man!!
    Current rides: '96 Honda Ohlins VFR, '03 Cannondale C440R, '03 Cannondale Cannibal, '06 Yamaha 450 Wolverine 4x4
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  6. #25
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    Quote Originally Posted by Stevo View Post
    Correct, for initial set-up without any chain tensioners. I need to find a replacement for the stationary idler when it wears down...
    How about these, Ted Dillard and the ElectroFlyers? https://www.ebay.com/itm/Skateboard-...kAAOSwcZFbLBm3

    Rasta Man!!
    Very colorful.
    Richard - Current bikes: 2018 16.6 kWh Zero S, 2016 BMW R1200RS, 2011 Royal Enfield 500, 2009 BMW F650GS, 2005 Triumph T-100 Bonneville, 2002 Yamaha FZ1 (FZS1000N) and a 1978 Honda Kick 'N Go Senior.

  7. #26
    Senior Member Ted Dillard's Avatar
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    OMG SO AWESOME
    Power in Flux: The History of Electric Motorcycles
    www.powerinflux.com

  8. #27
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    Quote Originally Posted by Spoonman View Post
    As requested Ted (and in nice Crayola colours too ).

    It all comes down to the ratio of the lengths and the manitude of the Cos scalar. As long as the L1 << L2 then you're never going to be too sensitive to any reasonable deflection.

    Here's my take on this debate:
    Your analysis and formula oversimplifies the relationship between the swing arm(SA) pivot axis and axes of the drive and driven sprockets. Think of there being a virtual SA(VSA) between the drive sprocket axis and the driven sprocket axis, keeping the chain at a fixed, ideal tension. In the set-up you show, the VSA would always have a greater radius of its swing than the actual SA. And, of course, the length of the radius would increase with the distance the drive sprocket axis is away from the SA pivot axis.

    It follows that for any normal given vertical travel away from the inline position, the position of the points of travel of the driven sprocket axis of the VSA would be further away from the actual SA's pivot axis and the points of travel of the axis of its driven sprocket. A longer radius forms a bigger circle. This means that slack in the chain increases, within limits, with the swing of the SA(the actual one) and the distance the drive sprocket axis is from the SA pivot axis. The difference in the slack would not be great. But, the ideal tension of a chain is usually determined by a couple of mm or a small fraction of an inch of adjustment.

    I guess this would be covered by a variation of angle alpha in your formula.
    Last edited by Electro Flyers; 1 Week Ago at 1750.

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  10. #28
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    Quote Originally Posted by Stevo View Post
    That's nice Spoonman!
    Here is my setup (not to scale):
    Attachment 7958

    The stationary "idler" is there to guide the chain between the swingarm pivot and chassis frame crossmember in order to prevent chain wearing on either. There is and needs to be chain slack. As the rear wheel travels up into compression, the chain tightens. That's why when you adjust chain tension for routine maintenance, there must be set slack and your adjustment shouldn't be too tight. Right??!!
    OK - now I see the problem.

    ...and even with the humour intended, the below indeed outlines the missing factor (the undesirable design feature if you like) which I wasn't seeing.

    Quote Originally Posted by Ted Dillard View Post
    LOL

    ...it sure looks like that chain goes around a lot of corners. Like I said, not helpful...
    Except that it is helpful.
    The slack that's appearing in your chain under compression of the suspension is as a result of the need to be able to accommodate the deflection of the chain by that lower idler when the suspension is extended (this is obvious of course, I simply wasn't aware of it/hadn't twigged it). So we can't model your swingarm system in the straight forward sense I've used above, you have a additional component in play. In design, the ideal solution would be to remove that component; but given we're dealing with a hack on an existing design we need to work with it - which explains the need for the tensioner.


    Quote Originally Posted by furyphoto View Post
    In your formula, do you have an ball park idea of what actual ratio is represented by "much less than" ( << ) for an averageish motorcycle, before slack becomes excessive?
    I don't but it will be geometry dependent and you can work it out [strike] quite easily* [/strike] if you know how much net shortening you're willing to accept in the compound length (hypotenuse of the triangle formed by the three points with the swingarm at maximum displacement).

    *edit: not since I revised the formula you can't - I'll try and come up with a restatement of the equation which will relate a single term angular deflection of the swingarm to net length of the system (which will make it a lot easier to use)... expect there's an elliptical equation which will do it.. it'll likely be something second order in any case but leave it with me.
    Last edited by Spoonman; 1 Week Ago at 1304.

  11. #29
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    Quote Originally Posted by Electro Flyers View Post
    Think of there being a virtual SA(VSA) between the drive sprocket axis and the driven sprocket axis, keeping the chain at a fixed, ideal tension. In the set-up you show, the VSA would always have a greater radius of its swing than the actual SA. And, of course, the length of the radius would increase with the distance the drive sprocket axis is away from the SA pivot axis.
    Agreed

    Quote Originally Posted by Electro Flyers View Post
    It follows that for any normal given vertical travel away from the inline position, the position of the points of travel of the driven sprocket axis of the VSA would be further away from the actual SA's pivot axis and the points of travel of the axis of its driven sprocket. A longer radius forms a bigger circle. This means that slack in the chain increases, within limits, with the swing of the SA(the actual one) and the distance the drive sprocket axis is from the SA pivot axis. The difference in the slack would not be great. But, the ideal tension of a chain is usually determined by a couple of mm or a small fraction of an inch of adjustment.

    I guess this would be covered by a variation of angle alpha in your formula.
    You're right!
    I had actually meant to encompass that into the formulaic result but I've not actually done that successfully in that first equation.
    Despite this, for L1 'much less than' L2 (must be some forum code overlap with the double less than notation, can't get to display properly) it would still produce quite an acceptable result but the more accurate approach is as follows:

    L=sqrt[(L1 + L2Cos(alpha))^2 + (L2Sin(alpha)^2]

    Where:
    L -> net length between sprocket centers
    L1 -> distance from drive sprocket center to swingarm pivot center
    L2 -> distance from swingarm pivot center to driven sprocket center
    alpha -> deflection of swingarm from 'inline' with line formed by the pivot and drive sprocket centers

    I'll revise that graphic and reupload.
    Last edited by Spoonman; 1 Week Ago at 0628.

  12. #30
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    Proper formula amended.


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