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Thread: combined trailing arm mount/body outrigger

  1. #71
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    Bill, from what I have read, you need to:

    Draw 2 wheels representing front and rear (wheelbase)

    Draw a horizontal line representing the height of COG

    Draw the 2 lines through the rear upper and lower links, where they meet is the IC

    Draw a line from the rear tyre contact patch to meet the IC

    Draw a vertical line through the center of the front wheel.

    the height of the line from tyre contact patch to IC, at the front wheel vertical line is the point that represents the amount of anti-squat. If I am correct, if the tyre CP to IC line was to disect the front wheel vertical line AT the height of COG, this would be 100% anti-squat, above is more below is less....

    I cannot comment on D2 at all.

    I have read and seen pictures of LR with A frame rears hill climbing where they want to "pig root" or have the axle start to drive under. Throttle control can help but not the best answer.

    Im guessing a RRC on 29's and stock springs would climb pretty well, but throw 32's, 34's or bigger on and a spring lift and you have change the geometry alot

  2. #72
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    Quote Originally Posted by Slunnie View Post
    True for more and less than 1G also.
    but gravity isnt applying it horizontally.....

    I must be missing the point

    or confused (very likely)

    but gravity doesnt change (well much unless you start talking about the mass of the surface materials in the earths crust), so if you change every thing else but gravity, my gut tells me the COG must change.....

  3. #73
    slug_burner is offline TopicToaster Gold Subscriber
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    No, CoG does not change. Ignoring the change of the position of the mass of the axles/wheels due to suspension travel.

    The CoG is the point around which all other points moments cancel out. Therefore it does not matter if you place the collection of components on a horizontal or inclined surface the CoG is the same.

    The change in CoG due to the suspension travel would be small and for most applications probably ignored.

    Then again I am not an automotive or mechanical engineer.

  4. #74
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    As I have said, looking at a static vehicle to keep the variables out. This is NO suspenion or acceleration movements.... Just the same vehicle on flat level groun and then on a 45 degree incline.

    Im not saying that it definitely does change....no way, just trying to get my head around it

  5. #75
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    Suspension dynamics are definately not my forte,so can someone correct me and demonstrate where I am going wrong with this admittedly simplified example of what I understand antisquat forces to be?
    Without getting too down and dirty under my vehicle tonight, I am estimating the instant centre of the RRC 3 link is about mid wheelbase and top of chassis rail. This would give an effective equivelant torque tube/radius arm length of about 4 ft .If you sunk the rear tyres in concrete and the rear axles were transmitting 4000ftlbs of torque, the lifting force at the front of the radius arm,mid chassis would be only around 1000lbs.
    An unladen RRC has around a 50/50 front/rear weight distribution, and weighs around 4000lbs,Delete weight of axle assemblies, lets say sprung mass of 2800lbs. The radius arm/torque tube applying a lifting force at mid chassis is attempting to lift the whole 2800lbs with a force of only 1000lbs.
    Seems to me that 1000lbs is insufficient to provide any antisquat at all.
    Bill.

  6. #76
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    Bill, I think I can visualise what you are describing, but I dont think its that simple....because if the upper and lower arms were parrallel to each other, there would be no IC, and in your example create a huge lever....but in reality the AS is lower in this case.

    It must have to do with the IC AND its relation to tyre contact patch, front axle line and COG...so while the 2 links make an imaginary lever, the pivot point is the front axle...so maybe its the distance from the front axle reward to the IC...the longer this distance (which would mean the IC being closer to the rear axle, which would be achieved by greater angle between upper and lower links) the greater the leverage ie higher AS....

    I think thats about as clear as mud

  7. #77
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    2nd go:

    what im seeing in my head is the forces from tyre, pushing the end of the lever which is the IC, but the lever is from the front axle center rear to the IC....

  8. #78
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    Quote Originally Posted by uninformed View Post
    Bill, I think I can visualise what you are describing, but I dont think its that simple....because if the upper and lower arms were parrallel to each other, there would be no IC, and in your example create a huge lever....but in reality the AS is lower in this case.

    It must have to do with the IC AND its relation to tyre contact patch, front axle line and COG...so while the 2 links make an imaginary lever, the pivot point is the front axle...so maybe its the distance from the front axle reward to the IC...the longer this distance (which would mean the IC being closer to the rear axle, which would be achieved by greater angle between upper and lower links) the greater the leverage ie higher AS....

    I think thats about as clear as mud
    If the links are parrallel, then the IC is at infinity and the effective swingarm is one acting from the tyre contact patch forwards parrallel to those links. The point of most interest is where that point passes over your front tyre contact patch.

    The below site is a bit dated and needs an overhaul (parts of it are 12 years old), it's also about mountainbikes where the drive is taken by a chain rather than axle reaction. But the pictures and diagrams may help you.
    Dougal.co.nz
    browse through "suspension", then "pedalling effects"

    I have to say. Suspension dynamics gets VERY complicated very quickly.

  9. #79
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    Thanks dougal,

    Btw you were and others were right about COG....bloody pea brain of mine.

  10. #80
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    Quote Originally Posted by Slunnie View Post
    True for more and less than 1G also.
    Quote Originally Posted by uninformed View Post
    but gravity isnt applying it horizontally.....

    I must be missing the point

    or confused (very likely)

    but gravity doesnt change (well much unless you start talking about the mass of the surface materials in the earths crust), so if you change every thing else but gravity, my gut tells me the COG must change.....
    Gravity is an acceleration and has an approximate value of 9.81 m/s2 on earth's surface. Being a vector quantity, it also has a direction - toward centre of earth for our purposes.

    Weight is a force derived form mass times gravitational acceleration.

    G denotes a magnitude of acceleration equivalent to gravity on earth's surface but often in some other direction, such as how Slunnie used it. It gives a feel of the effect of acceleration in comparison to weight. For example when changing direction (which is acceleration because vectors have direction), pilots of high performance aircraft experience high G forces.

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