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35's to 37's...too much for my axles?

Discussion in 'The Garage' started by gruveb, Jul 15, 2003.

  1. gruveb

    gruveb Registered Member

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    35\'s to 37\'s...too much for my axles?

    I have a 12 bolt rear and a 10 bolt front, 4.11 gears, detroit locker in the rear diff.

    Right now i have 35x15.5x15 Super Swamper SX's. I'm thinking of going with the SSR's and am trying to decide between 35 14.5's or 37 12.5's. How much difference in centrifugal force would there be?
     
  2. Bubba Ray Boudreaux

    Bubba Ray Boudreaux 1 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    [ QUOTE ]
    How much difference in centrifugal force would there be?

    [/ QUOTE ]

    I've got a dollar says that no one answers that as posted.......
     
  3. zakk

    zakk 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    couldn't one compare the weight of the tire to compare (roughly) the centrifical force exerted onto the axle?

    warm up that $1 bubba! i'll send you my paypal /forums/images/graemlins/grin.gif
     
  4. Bubba Ray Boudreaux

    Bubba Ray Boudreaux 1 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    I don't have a clue on what centrifugal force is. Isn't that the force that is exerted on something, say the human body that is in one of those super duper merry go-rounds at NASA that put something like beaucoup G Forces on the astronauts?

    If that's the case, then wouldn't there be some involved calculation using the weight of the tire (for example, a 33 BFG MT at app. 45 pounds) multiplying the revolutions the tire makes say at 60 mph?

    Looks like we may break this whole 1/2 ton versus 1 ton debate down to levels never seen before (thank you Tony Bruno for that line).

    Then we get there, wouldn't we have to include the properties of the material used for the axle shaft, and then we would have to worry about if the tire/wheel assembly is balanced out perfectly, cause the slight unbalance of that assembly would exert a slightly more force onto the bearing assembly which the axleshaft rides on, thereby bringing out any structual weakness in the bearings...............

    /forums/images/graemlins/thinking.gif /forums/images/graemlins/thinking.gif /forums/images/graemlins/thinking.gif /forums/images/graemlins/thinking.gif

    But wait, could we actually make an equal comparsion since semi floaters and full floaters are of a different design, therefore the force is exerted on something that is uncomparable to an axle bearing?
     
  5. zakk

    zakk 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    my head exploded at sentance #2 /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif
     
  6. Bubba Ray Boudreaux

    Bubba Ray Boudreaux 1 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    [ QUOTE ]
    my head exploded at sentance #2 /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif

    [/ QUOTE ]
    /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif

    I'm still trying to figure out what the hell I said. I bet Greg could sort it all out for us.

    But I think if one reads between all the BS I typed, I think the answer was "STFU and get a 14FF," said by anonymous NCBD member to myself numerous occassions. /forums/images/graemlins/rotfl.gif
     
  7. RustBuket

    RustBuket 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    I think the centrifugal force aplied to an object is increased in an exponential way as the rpm's are increased @ a linear rate. I think that the distance away from the the center point also has something to do with C-force. /forums/images/graemlins/rolleyes.gif
     
  8. gruveb

    gruveb Registered Member

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    Re: 35\'s to 37\'s...too much for my axles?

    /forums/images/graemlins/rotfl.gif

    Well, let's dumb it up a bit:

    Forgetting about the mass circling about an inch further from the center..........

    Now I'm comparing TSL SX's and SSR's....

    I have a 35x15.5" tire (13.1" ..I think tread width).
    The 37x12.5's have around a 9" treat width.

    So.......It seems to me that the drop in over 3 inches in actual width (sizes are decieving, I looked 'em up) would more than compensate for the added 2" in height.

    And then that brings the question, which would be more beneficial, 3" of width or 1" height (remembering that the actual tire height is going to be divided by 2)?
     
  9. RustBuket

    RustBuket 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    If you add width, you are adding more of the heaviest part (the tread istself), right? I figure an in (in height) of sidewall would be alot lighter than an inch (in width) of tread. But something we may be forgetting, as tires get taller the patterns stay the same but each block (for lack of a better word) gets bigger. So, its hard to say.
     
  10. gruveb

    gruveb Registered Member

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    Re: 35\'s to 37\'s...too much for my axles?

    Well, also what happens is the further away from the center the weight gets, the more force there is.

    If you're on a merry-go-round and sit in the middle, the force pulling you outward is minimal. The further out you go, the more centrifugal force there is, yet your actual weight hasn't changed.

    That's why I'm thinking that a 37X12.5 tire would exert more centrifugal force than a 35X15.5" tire would.
     
  11. Greg72

    Greg72 "Might As Well..." Staff Member Super Moderator

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    Re: 35\'s to 37\'s...too much for my axles?

    [ QUOTE ]
    [ QUOTE ]
    my head exploded at sentance #2 /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif

    [/ QUOTE ]
    /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif /forums/images/graemlins/rotfl.gif

    I'm still trying to figure out what the hell I said. I bet Greg could sort it all out for us.

    But I think if one reads between all the BS I typed, I think the answer was "STFU and get a 14FF," said by anonymous NCBD member to myself numerous occassions. /forums/images/graemlins/rotfl.gif

    [/ QUOTE ]


    BUWAHAHAHAHAAAAAA


    Well Bubba, I don't know if I can win that $1 prize (maybe I could but I don't think the time it would take to do the calcs is financially justified!)

    ...but I think your "rant" was closer than you give yourself credit for.

    My Take:

    Who cares about centrifugal forces really? The forces that are pressing against the outer tire carcass as it spins have very little effect on anything axlewise.....in fact I would argue that in terms of "pure centrifugal forces" the only people who should care about those are the tire manufacturer themselves. They need to insure that "at speed" the tire can withstand all those G's as it spins.

    The underlying question (and where I think Bubba made some insightful comments) is in the area of twisting forces. We ARE talking about strength at LOW rpms here....the ability of an axleshaft to withstand the twisting forces applied against it as torque runs it's course.....from the engine, through gear reductions in the tranny (doubler) and differentials.

    In the case of larger tires, do you want to know what I think causes breakage? It's not the extra MASS of the tire.....it's TRACTION!!! (ooooh Greg's getting controversial! /forums/images/graemlins/grin.gif)

    Let's think about it, that axleshaft has a limit to how much torsional force it can absorb (as it gets twisted up with LOTS of torque) before it finally breaks. At slow speed, the only thing that is increasing torque on that axleshaft is extra engine RPM (more torque) coming down the driveline.....if the tires slip, that torque is allowed to "bleed off" through the spinning tire(s) and it does not reach that breaking point. HOWEVER, if the tire is larger....and hence has more TRACTION....there is no bleeding off of torque, and the axleshaft must absorb those forces (and break if they go too high)!!!

    Any takers on this answer? /forums/images/graemlins/thumb.gif
    ......it sounded good when I typed it! /forums/images/graemlins/rotfl.gif
     
  12. Pookster

    Pookster 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    Centrafugal? or Centripetal? (sp on both)

    THere is a difference, One is the tendency for particle matter to want to go to the outside circumfrence of the tire, the other is the tendency of the tire to want to continue to rotate on its axles. THink of it like whipping a weight tied to the end of a string, vs a spinning top.

    Yes, larger tires do put more stress on the axles. It provides a greater leaverage force on the axles themselves.

    Fc = (rmv)2/(mr3)= L2/(mr3)

    If you acutally want to calculate it (which I dont) click here . it will explain all the vector forces involved. /forums/images/graemlins/thumb.gif
     
  13. gravdigr

    gravdigr 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    centripital force only plays a factor when slowing the vehicle. You are fighting the inertia the wheel and tire have built up.
     
  14. Bubba Ray Boudreaux

    Bubba Ray Boudreaux 1 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    /forums/images/graemlins/thumb.gif /forums/images/graemlins/thumb.gif /forums/images/graemlins/thumb.gif /forums/images/graemlins/thumb.gif
     
  15. gravdigr

    gravdigr 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    good reply, but I think you need to take one step beyond
    [ QUOTE ]
    if the tire is larger....and hence has more TRACTION

    [/ QUOTE ]

    not only is traction an issue here, the distance from the center of the axle to the edge of the tire plays a huge role. Given the same traction for a 35" tire and a 37" tire, at ALL times more torsional force is applied to the axle with the bigger tire (accellerating, braking, cruising, climbing). It seems to me the main factors affecting your axle with bigger tires are:

    Centripital force - felt when braking (with motor, since braking with the actual brakes the axle does not absorb the centripital force). The axle is taking the force of the big tires bleeding off the inertia created from rotating at speed.

    Torsion - Felt simply because there is more distance from the axle to the outside of the tire when running bigger tires. Think of this distance as a lever, and when you get bigger tires you increase the length of the lever without moving the fulcrum (in essence there is no fulcrum). You will move more distance with less initial movement, but it takes more power to do so...hope that made sense.

    E-mail my dollar to gravdigr@adelphia.net /forums/images/graemlins/deal.gif
     
  16. Greg72

    Greg72 "Might As Well..." Staff Member Super Moderator

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    Re: 35\'s to 37\'s...too much for my axles?

    I can understand that concept of spinning more "mass" with the bigger tires (like having larger, heavier flywheels!).....I just don't think it's a major factor in why axleshafts break.

    HYPOTHETICAL:

    What if you had two identical trucks, one with 35's and one with 37's......put them both on a frozen lake that is covered with oil and other slippery stuff....

    When you mash the gas pedal, my sense is that BOTH trucks will spin the tires instantly. Sure, the 37" tires will create a higher "instant" load against the axles than the 35's would, but it's not going to snap axles.... (insert complex calculations about acceleration of tires vs. tensile and yield strengths of axleshafts of varying cross-sectional diameters here)/forums/images/graemlins/grin.gif

    Add a little traction though, and things WILL go "boom".

    Has anyone heard the story about the guy who ran an open D44 up front in his K5 for years with NO breakage? Then one day, he installed a locker......suddenly, he's breaking axleshafts and u-joints non-stop. Why do you suppose that is....? I'm going to perservere and say that it's because of his increased TRACTION, and the inability of the axle to "bleed off" torque through a spinning wheel......


    Fun stuff. No idea if anyone (including myself) is actually learning anything from it. /forums/images/graemlins/eek.gif /forums/images/graemlins/laugh.gif /forums/images/graemlins/cool.gif
     
  17. gruveb

    gruveb Registered Member

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    Re: 35\'s to 37\'s...too much for my axles?

    It makes more sense to think that what causes breakage is the amount of torque required to turn a heavier tire with lots of traction than the breakage is caused by the weight being amplified by centrifugal force.

    Simply, the axles are the weakest point and can only handle X amount of torque before they snap. Sometimes the amount of torque they can handle is LESS than what it takes to turn a larger tire.

    So, that would lead me to believe that running numberically higher gears would compensate. With a numerically lower gear, it takes more throttle to move the tires from idle and it sounds like at the extremely low end is where the breaks occur. Larger tires essentially change the gear ratio making it numerically lower.

    Any thoughts?
     
  18. Pookster

    Pookster 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    that oil on ice works very well in theory, but in practice, I think the following was left out:

    You are still dealing with rotational mass. It takes more kenetic energy to spin a 37 than a 35. Its just that on oil( and ice) the amount of friction that needs to overcome traction is virtually the same (for arguments sake).

    Yes, if you add a locker, you increase loads on the axles. Remember the old subaru commercial? From the wheels that slip to the wheels that grip? Well, as we all know, with an open diff, its from the wheels that grip to the wheels that slip. The one with the least traction receives maximum kenetic energy delivery.

    However, take for example a rocky rooty uphill climb. A slipping tire suddenly grips, and *wham* axle or u-joint or sometihng goes. Two forces trying to counteract each other.

    I think you should be alright, your tire weights are not significantly different since your dropping 2 " of width.
     
  19. gravdigr

    gravdigr 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    hehe, OK...say I have a magical 35" tire that has the SAME traction as a 37" (we are getting really hypothetical here aren't we?). We are at a drag strip. We both mash the gas, even with the same traction, the axle with the 37" tire will take more abuse. So while the traction of a 37" tire over a 35" tire may play a small role, the main reason it is harder on the rig is the extra distance from the center of the axle (or fulcrum of the lever) and the edge of the tire (where the lever moves the load). This part is basic physics, I made a diagram but it looks crappy. I can post it if needed.
     
  20. eldon519

    eldon519 1/2 ton status

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    Re: 35\'s to 37\'s...too much for my axles?

    Friction(traction in the oil and ice example) is the product of the friction coefficient(mu) multiplied by the normal force(on a level surface, normal force equals weight). The friction coefficient between tire rubber and oil covered ice is going to be almost exactly the same between a 35" tire and a 37" tire. The normal force will be almost the same (the 37" tire will have slightly more since it is slightly heavier). So pretty much, two tires of the same compound and similar design will have about the same traction on oil covered ice regardless of their size. The reason the 37" tire will experience a higher instant load is because it is heavier(has more mass) and mass is a measure of an objects inertia. So the reason it loads more is cause it weighs more, not cause it has more traction. This example wouldn't hold true on very many other surfaces though because then the larger tire is able to dig into the earth more. I guess what I'm getting at is that friction is not affected by surface area.

    You are right partially about traction being the reason a larger tire will break an axle shaft. Like you mentioned with the locker concept, an open differential can only apply as much torque as the loosest tire can handle. An open diff isn't likely to break parts cause in many high-stress situations, it can't sustain hardly any torque. With a locker though, the axles actually bear the strain of the torque. To further explain this though, the torque sustained by the axle shaft would be higher on the bigger tire because it torque is the product of force x radius arm. With a bigger tire, the radius arm(half of diameter) is larger as well as the force of friction(traction) so consequently, the larger tire can exert more torque on the axle shaft leading to failure.
     

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