Zak I believe that the equation is much more complicated than that and it
depends on whether you are doing static of dynamic calculations. The dynamic
(moving) calculations would be the worse case, and should be done in a worse
case scenario, ie; 55MPH on a sharp turn on a rough road each wheel moving
thru its maximum and minimum vertical travel, using the maximum and minimum
caster, camber, and toe in, maximum knuckle inside diameter, and minimum hub
outside diameter, same minimum outside and inside diameter of each bearing
and race. Add the minimum axel nut torqe to the mix. Than add the mamimum
tire size and width, adn the rebound ratio of the shocks.
Even more complicated than that, but you get the idea. Looks like you will
be solving an equation with with 10-12 equations and god only knows how many
unknowns. I am to rusty and would have to get out my old statics and
dynamics books to do it.
.
1. Caster and camber of each wheel.
2. toe in of the wheel
3. Upper and lower ball joint alignment
4.
>John,
>
>I'm sorry, but that doesn't do it for me.
>
>What is that 2680 from the chassis applied to, the bearing spacer? It has
>to be applied to one or both of the bearings. If it were applied equally
>to both, then the loadings on the bearings would really be
>-11190+1340=-9850 lbs (down) and +11190+1340=12530 (up). These would
>transmit forces of 9850 lbs (up) and 12530 (down) to the hub. Which would
>be the 2680 different I said were needed to cancel the up force on the hub
>from the wheel.
>
>We are only looking at the loads applied to the hub. These loads include
>the force from the inner bearing, the force from the outer bearing, and the
>force from the wheel. There are no other forces on the hub. Anything
>applied to the bearings (ie your load from the chassis) is already included
>in the forces that the bearings apply to the hub and cannot be included
>again. I am going off of basic statics laws. I do have a BS in mechanical
>engineering, so I'm not making this stuff up.
>
>Zak
>
>
>
>
>>I like your diagram better... so here goes, but it is just what I was
>>trying to say in my post. I'm just adding to your diagram.
>>
>>
>>>Obviously, my diagram came out like crap when all the tabs got dropped
>>
>> from coach 2680 | 11190 on (outer) bearing
>> exerted on B.center| |
>> | |
>> ______v_____V______________
>> ^ o ^
>> | 4.014 from centerline| tire to centerline bearing set
>> | |
>>11190 on (inner) bearing 2680 from wheel
>>
>>
>>--
>>Regards,
>>John 74 Glacier near Washington, DC.
>>
>
>
Tom & Marg Warner
Vernon Center NY
1976 palmbeach
depends on whether you are doing static of dynamic calculations. The dynamic
(moving) calculations would be the worse case, and should be done in a worse
case scenario, ie; 55MPH on a sharp turn on a rough road each wheel moving
thru its maximum and minimum vertical travel, using the maximum and minimum
caster, camber, and toe in, maximum knuckle inside diameter, and minimum hub
outside diameter, same minimum outside and inside diameter of each bearing
and race. Add the minimum axel nut torqe to the mix. Than add the mamimum
tire size and width, adn the rebound ratio of the shocks.
Even more complicated than that, but you get the idea. Looks like you will
be solving an equation with with 10-12 equations and god only knows how many
unknowns. I am to rusty and would have to get out my old statics and
dynamics books to do it.
.
1. Caster and camber of each wheel.
2. toe in of the wheel
3. Upper and lower ball joint alignment
4.
>John,
>
>I'm sorry, but that doesn't do it for me.
>
>What is that 2680 from the chassis applied to, the bearing spacer? It has
>to be applied to one or both of the bearings. If it were applied equally
>to both, then the loadings on the bearings would really be
>-11190+1340=-9850 lbs (down) and +11190+1340=12530 (up). These would
>transmit forces of 9850 lbs (up) and 12530 (down) to the hub. Which would
>be the 2680 different I said were needed to cancel the up force on the hub
>from the wheel.
>
>We are only looking at the loads applied to the hub. These loads include
>the force from the inner bearing, the force from the outer bearing, and the
>force from the wheel. There are no other forces on the hub. Anything
>applied to the bearings (ie your load from the chassis) is already included
>in the forces that the bearings apply to the hub and cannot be included
>again. I am going off of basic statics laws. I do have a BS in mechanical
>engineering, so I'm not making this stuff up.
>
>Zak
>
>
>
>
>>I like your diagram better... so here goes, but it is just what I was
>>trying to say in my post. I'm just adding to your diagram.
>>
>>
>>>Obviously, my diagram came out like crap when all the tabs got dropped
>>
>> from coach 2680 | 11190 on (outer) bearing
>> exerted on B.center| |
>> | |
>> ______v_____V______________
>> ^ o ^
>> | 4.014 from centerline| tire to centerline bearing set
>> | |
>>11190 on (inner) bearing 2680 from wheel
>>
>>
>>--
>>Regards,
>>John 74 Glacier near Washington, DC.
>>
>
>
Tom & Marg Warner
Vernon Center NY
1976 palmbeach
