gearing for 44" tires?

[mhughes165]

dear dickhead,

you obliviously dont know what a 200lb tire can do to the pencil axles in a 28 spline 8.8. heck one of the jeep guys i know runs a 31 spline 8.8 on 37inch iroks and has snapped shafts. read this http://www.therangerstation.com/tech_library/axleguide.html
other than the front d60 being slightly under rated its COMPLETELY correct. NOTE they say 35 inch tires max. for the d44, you want to go 11 inches more!
read up and learn before you start asking retarded questions of the good folks around here.

o yea, comparing a mustang to a wheeling rig just dosent work.

86

uhm, dude, were u pissed or somehting when u replied to this, weve already made it clear to him that he needs to run 60's atleast for him to drive it on the road, hes been lookin for them and now knows hell need to gear the hell out of them aswell

 

[Evan]

you obliviously dont know what a 200lb tire can do to the pencil axles in a 28 spline 8.8.

The weight of a tire doesn't have much to do with breaking axle shafts. Strain on the shafts comes when you increase the diameter of the tire. This results in much more leverage the driveline needs to work against in order to rotate the tire.

 

[Toreadorranger]

The weight of a tire doesn't have much to do with breaking axle shafts. Strain on the shafts comes when you increase the diameter of the tire. This results in much more leverage the driveline needs to work against in order to rotate the tire.

And how do you think that leverage works? The more weight there is the more "force" on your lever. Weight has just as much do with it as diameter.

 

[bobbywalter]

weight is a huge factor...especially once spinning...then slipping and biting.

 

[Evan]

weight is a huge factor...especially once spinning...then slipping and biting.

No.

A heavy tire that is spinning quickly and then bites puts no more load on the axle shaft than a light tire would. When the tire "bites" the rotational energy of the tire is absorbed by the tire and the ground, NOT the axle shafts.

EDIT: By the way, I love the quote you left on my Cardomain.

 

[Evan]

And how do you think that leverage works? The more weight there is the more "force" on your lever. Weight has just as much do with it as diameter.

Wrong. For practical purposes the "leverage" against the axle shaft is not determined by the weight of the tire. It's determined by the length of the tire radius. The force on the axle shaft is determined by the torque you're applying to the driveline and the friction coefficient between the tire and the ground. A longer radius multiplies this force.

 

[JohnnyU]

A heavy tire that is spinning quickly and then bites puts no more load on the axle shaft than a light tire would. When the tire "bites" the rotational energy of the tire is absorbed by the tire and the ground, NOT the axle shafts.

Wrong.

It's called inertia. Look up Impulse and Momentum in a Physics, Kinematics, or Machine Element Design text book.

 

[Evan]

Wrong.

It's called inertia. Look up Impulse and Momentum in a Physics, Kinematics, or Machine Element Design text book.

I have. In college I had the highest grade in my class in both Physics I and II. And in this case we're concerned with ANGULAR momentum.

A tire that is spinning and then is suddenly stopped by gaining traction looses it's angular momentum to the ground and vehicle. The axle shaft has NOTHING to do with the process.

I can jack the front of a 2WD truck up and spin the tire rapidly by hand, then drop the jack so the tire hits the ground and is stopped. In this case there is no axle shaft involved at all.

This is why the weight of a tire means squat compared to the radius, when analyzing the force on an axle shaft.

 

[krugford]

How heavy are these tires? 150lbs heavier than the stockers? That's adding close to 15% onto the stock vehicle. Not as much of an increase in forces as the increase in radius will give you, but not negligible either.

If a tire is spinning and catching, it's the inertia of the rest of the driveline that's going to snap the shaft, not the weight of the tire. Spinning the tire by hand does not reflect the same situation as crawling. The difference is how the power is being transmitted through the system. In a running vehicle, the power is being transmitted to the tire through the axle. In your example, the tire is being powered by your hand on the outside of the tire. The transition from a dynamic to static condition on the tire will be accentuated by the larger tire. With a larger tire, the force at the ground required to snap a shaft decreases as the tire increases in diameter.

Evan,
You should remember that having the highest grade in an entry level physics class is not a qualification that relates to this discussion. It does not automatically mean you are right and the fact that you attempted to qualify your statement with an example that is not seeing the same forces as what is being discussed means you have not completely thought through the problem. The tire and wheel will all see some forces from that rapid stop due to their own inertias, small as they may be.

 

[Evan]

How heavy are these tires? 150lbs heavier than the stockers? That's adding close to 15% onto the stock vehicle. Not as much of an increase in forces as the increase in radius, but not negligible either.

If a tire is spinning and catching, it's the inertia of the rest of the driveline that's going to snap the shaft. Spinning the tire by hand does not reflect the same situation as crawling. The difference is how the power is being transmitted through the system. In a running vehicle, the power is being transmitted to the tire through the axle. In your example, the tire is being powered by your hand on the outside of the tire. The transition from a dynamic to static condition on the tire will be accentuated by the larger tire. With a larger tire, the force at the ground required to snap a shaft decreases as the tire increases in diameter.

Right, but the weight of the tire when it catches is irrelevant. That's what my example was trying to show. The only thing that matters is the length of the radius and the amount of skinny pedal.

A heavy tire might be even easier on the axles in this case, because it takes longer to stop and would result in less impulse to the axles, as the time the force is applied is increased.

 

[JohnnyU]

Evan, you're still missing the fundamental concepts at work here.

What is (angular) momentum comprised of?

L = r x p

Where L is the angular momentum, r is the position vector (tire radius in this case) and p is the linear momentum.

p = m*v

Where m is the mass and v is the velocity that the mass is traveling at.

Moreover, Impulse is defined as the integral of force over time:
I = ∫ F dt

It can also be derived:
I = ∫(dp/dt) dt simplifying...
I = ∫ dp simplifying again gives:
I = ∆p You may know this as the "Impulse-Momentum Theory"

I = F∆t = m∆v = ∆p

You can see that even angular momentum is DIRECTLY proportional to the mass. The mass of the tire is VERY relevant to this scenario.


I'm not going to list my qualifications in an effort to make my case, the theory/principles should speak for themselves.

 

[mhughes165]

all i know, and i agree with u johnny, is think of it this way, what hurts more when u drop it on ur foot.....a tire weighing 200pnds, or a tire weighing 100 pnds but the same height.....

 

[Evan]

Evan, you're still missing the fundamental concepts at work here.

What is (angular) momentum comprised of?

L = r x p

Where L is the angular momentum, r is the position vector (tire radius in this case) and p is the linear momentum.

p = m*v

Where m is the mass and v is the velocity that the mass is traveling at.

Moreover, Impulse is defined as the integral of force over time:
I = ∫ F dt

It can also be derived:
I = ∫(dp/dt) dt simplifying...
I = ∫ dp simplifying again gives:
I = ∆p You may know this as the "Impulse-Momentum Theory"

I = F∆t = m∆v = ∆p

You can see that even angular momentum is DIRECTLY proportional to the mass. The mass of the tire is VERY relevant to this scenario.


I'm not going to list my qualifications in an effort to make my case, the theory/principles should speak for themselves.

You just did a lot of typing for nothing. Nobody is saying angular momentum is not directly proportional to mass, or questioning the method of derivation for the Impulse-Momentum Theory. I am saying that when a tire is spinning rapidly and suddenly grabs, the force on the axle shaft does not come from the momentum of the tire, it comes from the energy being applied through the driveline which is multiplied by a factor of the tire radius and the coefficient of friction between the tire and the ground. The energy stored in the rotating tire is transferred to the ground and the frame of the truck.

I listed some qualifications because you implied that I wasn't familiar with high school Physics concepts.

 

[bobbywalter]

evan, you will have to figure out the math for me bud.


because the weight of the tire does effect the axle shaft.


for instance.


truck has 38 inch tires. rarely ever breaks a shaft.




same truck...same tires.


add 15 gallons of water, bust every shaft in minutes. why? i dont know why, seems the water doesnt add resistance right? it just literally goes with the flow in my tiny little mind.


same truck same axles...add 42's....cant keep shafts in the fawker. 38's that are dry...no issues.













my truck....33 in bfg at's....i can pound the fawk out of it with all my ponies...rarely an issue.....but i get stuck alot.


33 in swamper sx's......sheeeeeittttt..snap crackle fawking pop buddy.


i dont know the equation...and i dont want to know the equation...


i just know that i have to drive smarter with heavy ass tires. or have alot of parts.

i know what your saying about the tire stopping not hurting the axle with a free spinning tire....

no shit...how could it.


and not what i am saying either.... the same load of a tire under 300 pounds of instant torque slamming into it....the heavier tire is comming down harder and getting more traction due to its nature and adding more resistance because of increased traction/friction and even more due to the extra weight....the lighter less aggressive tire is easier and less ressitive to turn and will break traction sooner allowing even less weight to be a factor.

i could be wrong.

but a sx and a a/t have completely different affects in my experience. in every way...they fawk up my brakes,spindles bearings and shafts at alarming rates taking it easy where i can let my small block roar and hardly bust a thing.


i dunno...its crazy...maybe its just me.

 

[85_Ranger4x4]

Weight in tires is like a big flywheel, it takes more effort to get spinning and more effort to stop spinning. Catch something good and it will snap everything behind it that isn't up to the task, but it will be harder to stop spinning once it starts, and it will be spinning much faster than a smaller tire (power permitting) and less likely to do anything but keep on spinning. But if you don't have the power to get that flywheel up to speed you are going to have trouble.

Fluid/more weight will help the tire dig down and go places it normally wouldn't without it, and will make life a lot rougher for the axle/everything else involved. I use fluid in my tractor for traction, but it came from the factory with it back in 1953 and was built with it in mind.