Tires A physics question of leverage

[orezok]

That was the point as to why your Hulk example was not realistic to the real situation [in which a tractor is centered between its support points - wheel/wheel.] It was not centered for Hulk. Ref 2nd quote above.



Again : Extending support points increases bearing load because the extended support points exert more leverage on the bearing pair.

So the tractor bearing locations got wider in addition to gaining weight :confused2:

 

[Egon]

,,,,,,,,,,,When talking to yourself, try not to contradict.

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]

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.

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]

So the tractor bearing locations got wider in addition to gaining weight :confused2:

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.

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.

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]

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.

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.

 

[SPYDERLK]

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.

No. OK.

 

[AxleHub]

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.

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]

You really need to know the spacing between each axle bearing pair, and the placement of wheel center wrt the outer bearing to get a good handle on % load increase. ... The proportion increase in bearing load seems it would be much higher -- ~ 50%

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]

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

 

[jenkinsph]

OK, I talked to the dog and she gave me some good ideas. She's smarter than most.

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.

 

[Egon]

Talking to the Dawg sounded good so I did. She said stop this nonsense and let's go for a beach walk. Smart Dawg!

 

[LD1]

Late to the party....

Cannot for the life of me even figure out the original question. If the wheels have the ability to go to 65" without spacers, and with spacers.....re-configure the hub/rim to also be at 65"......debating the difference???????

Only difference I see is money wasted on spacers.

As to extending the axle.....it most certainly does add load to both the inner and outer bearings. An upward force on the outer, and downward force on the inner.

If the tractor weighs 2000# in the rear, Thats 1000# over each tire. According to newtons 3rd law that also means that the ground is ALSO pushing UP on the tire with 1000# of force. And THAT will no change no matter the spacing.

So...with orezok's broom handle, Grab it with your right hand near the end, and the left hand maybe 15" up the handle. These represent the bearings (right hand being inner and left hand being outer). Now have someone press upward a few inches beyond your left hand. Your left hand has an upward force trying to pull it out of your hand, and your right hand (inner bearing) is being forced downward.

Now have your friend go further out, and apply the SAME upward force......You WILL see alot more force on your hands trying to hold the broom.

 

[orezok]

If the tractor weighs 2000# in the rear, Thats 1000# over each tire. According to newtons 3rd law that also means that the ground is ALSO pushing UP on the tire with 1000# of force. And THAT will no change no matter the spacing.

So...with orezok's broom handle, Grab it with your right hand near the end, and the left hand maybe 15" up the handle. These represent the bearings (right hand being inner and left hand being outer). Now have someone press upward a few inches beyond your left hand. Your left hand has an upward force trying to pull it out of your hand, and your right hand (inner bearing) is being forced downward.

Now have your friend go further out, and apply the SAME upward force......You WILL see alot more force on your hands trying to hold the broom.

Is someone beginning to see the light?

Yes, there is 1000# of upward force exerted by the tire no matter the axle length, but with a longer axle it takes LESS upward force at the tire to raise the tractor at the longer length axle, hence less shock load on the bearing. Now this is not realistic as the tractor is NOT a fixed item but of a fixed movable weight. If you wanted to move a rock with a lever, which would take less force to move it? A short lever or a long one?

 

[LD1]

Is someone beginning to see the light?

Yes, there is 1000# of upward force exerted by the tire no matter the axle length, but with a longer axle it takes LESS upward force at the tire to raise the tractor at the longer length axle, hence less shock load on the bearing. Now this is not realistic as the tractor is NOT a fixed item but of a fixed movable weight. If you wanted to move a rock with a lever, which would take less force to move it? A short lever or a long one?

I think you mis-read my post. As nothing you have said so far I agree with.

You have your thinking all backwards.

IF the tractor rear weighs 2000#, thats 1000# over each rear tire no matter how the wheels are spaced (as long as they are the same).

So if there is 1000# over that wheel, there is NO way to lift that wheel unless 1000# is applied to that wheel in an upward direction. Sure, you could use a lever or whatever, but you are still applying 1000# upward AT THE WHEEL.

The issue isnt what happens beyond the wheel. But between the wheel and the CL of the tractor. The further out the wheel is, the more leverage it has on the bearings.

Remember, no matter the axle/wheel spacing, it takes 1000# to lift that tire regardless. If that wheel is 12" away from the first bearing that 1000# multiplies more than if the wheel is 6" from that first bearing

 

[orezok]

I think you mis-read my post. As nothing you have said so far I agree with.

You have your thinking all backwards.

IF the tractor rear weighs 2000#, thats 1000# over each rear tire no matter how the wheels are spaced (as long as they are the same).

So if there is 1000# over that wheel, there is NO way to lift that wheel unless 1000# is applied to that wheel in an upward direction. Sure, you could use a lever or whatever, but you are still applying 1000# upward AT THE WHEEL.

The issue isnt what happens beyond the wheel. But between the wheel and the CL of the tractor. The further out the wheel is, the more leverage it has on the bearings.

Remember, no matter the axle/wheel spacing, it takes 1000# to lift that tire regardless. If that wheel is 12" away from the first bearing that 1000# multiplies more than if the wheel is 6" from that first bearing

No, I think you do get it. It takes 1000# at the wheel to raise the corresponding 1000# at the bearing. It doesn't matter how long the "lever" is, it only takes 1000# to raise the tractor so no additional load on the bearing. The problem most people are having in their thinking is that the tractor is fixed object and a longer lever exerts more force on the bearing. The bearing moves as soon as the 1000# required to raise it occurs.

 

[LD1]

No, I think you do get it. It takes 1000# at the wheel to raise the corresponding 1000# at the bearing. It doesn't matter how long the "lever" is, it only takes 1000# to raise the tractor so no additional load on the bearing. The problem most people are having in their thinking is that the tractor is fixed object and a longer lever exerts more force on the bearing. The bearing moves as soon as the 1000# required to raise it occurs.

But if the 1000# is further from the bearing, the axle acts as a lever on the bearing. So exert 1000# on the tire connected to a short axle/lever that is close to the bearing means not much load on the bearing. Add spacers and distance, now the axle acts as a longer lever. Still requires 1000# exerted at the tire, but leverage means the bearing sees more force.

 

[LD1]

Are you saying the load at the bearings is the same on the top drawing as they are on the bottom?

 

[orezok]

View attachment 473720

Are you saying the load at the bearings is the same on the top drawing as they are on the bottom?

Yes. When the upward force reaches 1000# at the bearing, the static load amount, the tractor begins to rotate. The force cannot exceed 1000#, it can only match it. Newtons Third law - To every action (force applied) there is an equal but opposite reaction (equal force applied in the opposite direction).

 

[LD1]

Yes. When the upward force reaches 1000# at the bearing, the static load amount, the tractor begins to rotate. The force cannot exceed 1000#, it can only match it. Newtons Third law - To every action (force applied) there is an equal but opposite reaction (equal force applied in the opposite direction).

You need to re-think that then.

1000# pushing up on a tire/axle 10" from outboard bearing and 25" from inboard bearing somehow has the same bearing load as...
1000# pushing up on a tire/axle 20" from outboard bearing and 35" from inboard bearing......

Since when did leverage no longer apply? Leverage makes the load.....in both cases.....exceed 1000# on the outboard bearing. It just exceeds it by a greater amount on the one with the longer lever arm.

 

[orezok]

You need to re-think that then.

1000# pushing up on a tire/axle 10" from outboard bearing and 25" from inboard bearing somehow has the same bearing load as...
1000# pushing up on a tire/axle 20" from outboard bearing and 35" from inboard bearing......

Since when did leverage no longer apply? Leverage makes the load.....in both cases.....exceed 1000# on the outboard bearing. It just exceeds it by a greater amount on the one with the longer lever arm.

Leverage never applied. Forget that premise. That is something that numerous people confused themselves with. We're not moving an immobile rock here. It's strictly an action and reaction situation on the bearing. Do you think Newton is wrong?