orezok
Elite Member
So the tractor bearing locations got wider in addition to gaining weight :confused2:
Tires A physics question of leverage
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So the tractor bearing locations got wider in addition to gaining weight :confused2:
Egon
Epic Contributor
Well that's where the challange portion comes in. Got to thinking about one side of the axle assembly. Namely, inner race, outer race with the weight being applied between the two. Then the axle with a wheel got added in. With that senario I figured it reasonable that the inner race would be a pivot point for the axle. Using this configuration the whole lot is a lever with a pivot point on the inner race. Can't recall the class of lever though.
Tomorrow may see more revaluations from overnight dreams.
orezok
Elite Member
Since the bearings are mounted in a rigid assembly, the pivot point is actually the other tire. Both bearings should receive the same force, one up one down.
SPYDERLK
Super Star Member
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- JD2010, Kubota3450,2550, Mahindra 7520 w FEL w Skid Steer QC w/Tilt Tatch, & BH, BX1500
The tractor still presses down on the wheels with the same force, but if theyre further out that force levers harder on the bearing pairs on each rear axle. A longer lever pushes up/down harder even tho the same force is applied. Think pry bar.
No. Each half axle tries to crush its bearing pair - one upward one downward. The bearing being pushed upward [outer] resists a load the same as the downward force on the inner bearing plus the weight of the load being supported by that axle.
orezok
Elite Member
So what you are saying that if you use a longer pry bar, you have to put more force on it to move something. That is what you are saying.
I give up, I think I'll go talk to the dog. She will understand.
AxleHub
Elite Member
Orezok, your refusal to recognize deflection makes no sense. "Ridgid" is a relative term based on a continous and unchanged position. But adding length between the outer bearing and the rim is an alteration of the position and as a result side pressure on bearings is created as deflection of the axle changes. I've read your responses to several posters and none of those responses of yours seems to indicate you are understanding the dynamics involved.
I'm done trying to communicate. The other posters are clearly recognizing points and issues you are not.
dmac128
New member
Agreed - it's all about geometry. If you look at my calculations, the 4" adjustment increases loads by 23% above 'baseline' and the 4" adjustment + 3" spacer increases loads by 34% above baseline for the assumed geometry. The 9% increase is only for the 4" adj + 3" spacer relative to the 4" adjustment only (which is allowed by the OEM by design).
orezok
Elite Member
OK, I talked to the dog and she gave me some good ideas. She's smarter than most.
We will now do one final demonstration on shock loading of the bearings dependent upon axle length. We are simulating running over a rock with the rear tire. Now this is just a demonstration of concept so no yada yada about well that isn't reality.
Take a broom and hold it horizontally. Hold it at the balance point so that the weight of the head balances the weight of the broomstick. This is so we start out in a no torsional condition. Your hand is now representing the inner and outer bearing in the tractor case, about 4" apart in this demonstration. Now have someone take a 3# hammer and hit the broomstick 4" from your hand which represents the distance from the outer bearing to the tire. OUCH that hurt and maybe even knocked the broom out of your hand.
When you hand stops hurting pick up the broom again and have that same person hit it 2' out from your hand which represents a longer axle in an exaggerated way. Hmmm... that wasn't bad. A little sting but I could handle that all day.
That's it folks. Your final physics lesson.
I'm out of here. Bye
We will now do one final demonstration on shock loading of the bearings dependent upon axle length. We are simulating running over a rock with the rear tire. Now this is just a demonstration of concept so no yada yada about well that isn't reality.
Take a broom and hold it horizontally. Hold it at the balance point so that the weight of the head balances the weight of the broomstick. This is so we start out in a no torsional condition. Your hand is now representing the inner and outer bearing in the tractor case, about 4" apart in this demonstration. Now have someone take a 3# hammer and hit the broomstick 4" from your hand which represents the distance from the outer bearing to the tire. OUCH that hurt and maybe even knocked the broom out of your hand.
When you hand stops hurting pick up the broom again and have that same person hit it 2' out from your hand which represents a longer axle in an exaggerated way. Hmmm... that wasn't bad. A little sting but I could handle that all day.
That's it folks. Your final physics lesson.
I'm out of here. Bye
jenkinsph
Super Star Member
Talk to the dog some more, you may eventually listen to what was said and develop a better understanding of what was said. Explain that a large dog has two bones in its jaws with one sticking out out each side of its mouth. If the bones are longer she has a better chance of prying the jaws open. The large dog may then drop one of the bones and she can take off one of them.
