Tires A physics question of leverage

   / A physics question of leverage #41  
Agree w Egon and others. Just imagine the axle is ten feet long rather than 6" further out than normal. Just like a breaker bar w a pipe slipped over it. More torque to break free the [bearing] nut.
Jim
 
   / A physics question of leverage #42  
The total load stays the same. The internal bearing forces will change.

Egon . . Your post is correct.

And now for thread participants in general: The initial topic created by the OP was (for sidehill benefit and stability) either moving the rim outward or using spacers. In either scenario we all pretty much agreed they would be similar result.

But the OP also indicated he would use both. For those who say this is was off topic . . they were wrong because the OP generated that issue as the result of the original issue. But total load against the rear axle stays the same . However going from 0 inches of extension on each side . . Then 3 or 4 inches oon each side . . And then 6 or 8 inches of extension on each side alters the load pressure points. Whether flat land or sidehill there is a shift in angular presdure of the same load. Longer distances of gap between rim and bearings shifts the load to the outer edges of the bearings and alters the pressure points on the lugs. But add in sidehill pressure changes even further increases the shift of the load weight . . . Same weight . . . But instead of lugs and bearing having the weight spread evenly across them . . the weight is shift to more weight in specific areas of the lugs and bearings.

Now 4 inches of spacing adjustment of the rims (engineered by manufacturer) on each side (left rim) and 4 inches on right side (right rim) may be calculated by maker to be acceptable (with limited sudehill use and recommended load weight). But 4 added inches of spacers (4" on left and 4" right) is no whete in the engineering calculations too.

Because an axle supports a weight does not mean it does not deflect from perfectly straight . . and sidehills change pressure points and deflection if an axle changes pressure points.

Those pressure point changes impact lugs and bearings.

Had the thread OP only addresses rim widening OR spacers . . this thread would have been completed long ago. But the OP stated whatever the answers were (rims or spacers) he was doing both and others also said they'd do both . . and so BOTH stories are on topic and both are issues to be considered imo.

If someone wants to do both . . that's their decision . . but leverage changes effect parts and longevity of parts . . and claiming there is no harm to doing both imo is not factual . . its an opinion that an engineer could determine factually as correct or incorrect . . but not us.
 
   / A physics question of leverage #43  
I believe that some people ave viewing this situation as a lever. It is not. It should be viewed as a simple beam supported at both ends by tires and with a concentrated load at the center.
 
   / A physics question of leverage #44  
Think you'd have to make that several cantilered beams with fixed ends.
 
   / A physics question of leverage #45  
Think you'd have to make that several cantilered beams with fixed ends.

I don't think so. The rigid case of the transmission makes it one continuous beam. The axles are not connected, but the assembly is a beam.
 
   / A physics question of leverage #46  
Sorry, two cantilevered beams. There is no continuity for one beam. The axle housing has independent forces acting within it.
 
   / A physics question of leverage #47  
I’ll try to explain it in a simpler example.

I want to turn my tractor over on its side so I can easily polish the undercarriage. I bring in my good friend Hulk Hogan to turn it over. Using an example of 1000# centered on the rear axle, Hulk can’t lift enough at the existing 2’ axle stub (500#) to turn it over. Hulk adds a 2’ cheater to the axle making it possible for him to exert 250# of upward force to rotate the tractor (250# x 4’=1000# of rotational torque). Shine away.

Next time I need to do this Hulk isn’t available and I can’t lift 500 or even 250#. So I add a 6’ cheater to the axle and I only have to lift 125#, which in my dreams I can do.

The bottom line is that the length of the axle has no effect on the load on the bearing at the end of the transmission case stub. It was 1000# with the factory axle, 1000# with Hulks 2’ cheater and 1000# with my 6’ cheater.

Now this is not a perfect example as because of the rigid transmission case, the “beam” would rotate on the opposite tire, not the center of the transmission as if it was a lever condition. Tractors do not have independent rear suspension.
 
   / A physics question of leverage #48  
Orezok, I totally disagree.

1. That axle is not a single item . . It is based on a series of bearing and differential components which is why there is such a thing as a diff lock.

2. Engineering is based on directional capacity to determine load testing and limits. The bearings are based on a vertical solidarity in order to support those axle areas that you consider "ridgid" but in reality are merely supported components.

3. Deflection is a part of everything we have contact with. Even gravity or light are subject to deflection and certainly any metal rod is a slave to deflection. Bearings support the axle and every inch the wheels/rims are moved away from those bearing surfaces adds to deflection.

4. I or others could mention bearing drag, or could mention bearing surface design, or the angular stresses that deflection has on vertical bearings. Or the fact that the frame is what supports the transmission/differential/axle assembly.

In effect in your example . . You don't want hulk hogan or a cheater bar to lift 1 side of your tractor . . you want 2 hogan s so each side can be equally lifted upward . . not side to side . . Because the axle or wheel rims or bearings or lugs are NOT designed for side force as your example implies imo.
 
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   / A physics question of leverage #49  
Orezok, I totally disagree.

1. That axle is not a single item . . It is based on a series of bearing and differential components which is why there is such a thing as a diff lock.

2. Engineering is based on directional capacity to determine load testing and limits. The bearings are based on a vertical solidarity in order to support those axle areas that you consider "ridgid" but in reality are merely supported components.

3. Deflection is a part of everything we have contact with. Even gravity or light are subject to deflection and certainly any metal rod is a slave to deflection. Bearings support the axle and every inch the wheels/rims are moved away from those bearing surfaces adds to deflection.

4. I or others could mention bearing drag, or could mention bearing surface design, or the angular stresses that deflection has on vertical bearings. Or the fact that the frame is what supports the transmission/differential/axle assembly.

In effect in your example . . You don't want hulk hogan or a cheater bar to lift 1 side of your tractor . . you want 2 hogan s so each side can be equally lifted upward . . not side to side . . Because the axle or wheel rims or bearings or lugs are NOT designed for side force as your example implies imo.

Given all of that, when you run over a rock with one rear wheel the tractor lifts only one side and the whole ASSEMBLY acts like a single beam.
 
   / A physics question of leverage #50  
I’ll try to explain it in a simpler example.

I want to turn my tractor over on its side so I can easily polish the undercarriage. I bring in my good friend Hulk Hogan to turn it over. Using an example of 1000# centered on the rear axle, Hulk can’t lift enough at the existing 2’ axle stub (500#) to turn it over. Hulk adds a 2’ cheater to the axle making it possible for him to exert 250# of upward force to rotate the tractor (250# x 4’=1000# of rotational torque). Shine away.

Next time I need to do this Hulk isn’t available and I can’t lift 500 or even 250#. So I add a 6’ cheater to the axle and I only have to lift 125#, which in my dreams I can do.

The bottom line is that the length of the axle has no effect on the load on the bearing at the end of the transmission case stub. It was 1000# with the factory axle, 1000# with Hulks 2’ cheater and 1000# with my 6’ cheater.

Now this is not a perfect example as because of the rigid transmission case, the “beam” would rotate on the opposite tire, not the center of the transmission as if it was a lever condition. Tractors do not have independent rear suspension.
The flaw here is that in real life both wheels are extended outward. Not like Hulks cheater on one side only.
 
 
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