2020: Two Warranty Case Studies & Good Video

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Jakenbake

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If we're going to compare in a vacuum, I will tackle the static case. :) It makes absolutely no difference whatsoever.

If you want a comparison under actual driving conditions, you need to specify those conditions and which part of the steering, suspension, or drivetrain interests you.
Once again I am probably being dense, but I view this really as a deviation to the OP's original post. I know for sure I am not advocating deceiving an OEM on warranty work (I don't even have a warranty). I tell the drive through person that I ordered a medium instead of a large, rather than say "ahhh it is their fault on that one"

I viewed this as @MrT was asking for a comparison of a hypothetical two tires, not mounted on a vehicle and possibly even using just the equations that @KAH24 listed. Honestly, for real production tires I expect the radius term, for example in rotational inertia, will carry more "weight" than the actual weight term (mass) as the radius term is squared. Assuming that is the case when others who will want to actually argue for purposes of deceiving OEM's will know why @KAH24 is correct. It is important to know "why" as much as know "what".

This is just my curiosity, and I while I am not dying to sit back down and crack open the dynamics text book again, no promises but I will try and look into it just for our collective knowledge. But not before I get in the pool and enjoy the evening! :p



EDIT: this should probably be a champagne room topic and I bet if it went to engineering-ey then no one would tune in lol!
 

MrT

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If we're going to compare in a vacuum, I will tackle the static case. :) It makes absolutely no difference whatsoever.

If you want a comparison under actual driving conditions, you need to specify those conditions and which part of the steering, suspension, or drivetrain interests you.
I will specify, the axle shaft, lets go with 2" diameter, oem wheels, one has a 35" tire weighing 89lbs and the other a 37" tire weighing 69lbs. Total weight to make things easier 35 with wheel 120lb, 37 with wheel 90lb. What force is put on said axle to get each one rolling all other aspects being equal.

People seam to be getting upset and reading between the lines going as far as accusing someone of wanting to defraud the dealership.

Keep in mind KAH said he knew and there was math involved that he used to base his decision for warranty denial. I happen to be researching the affect of weight on rotational inertia and unsprung weights affect on a vehicle in general. I thought he could be a resource for information and there may be a good conversation but instead ... well apply the statements he made to me to yourself and see what you think. I will hold off posting what I think because I am still interested in an actuall response to my question
 

MrT

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It's hard to tell what applies to your question because you never actually stated it as a question. Care to try again with a little specificity?
Sure, what is the difference in force needed to rotate a 120lb 35" tall tire and a 90lb 37" tall tire. Use the force applied to or from the axle at 2" diameter if that helps.
 

MrT

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Using math can you show the overall affect of each example? I haven't found much on the leverage aspect vs weight yet(looking)

OEM 100lb
Stock diameter +10lb at the tire. 110lb
37" 90lb

My question is as follows:

The lighter 37" tire vs heavier oem height aftermarket tire. At some point the weight difference will negate the leverage. My question is do you know where that occurs?
And here is where I asked earlier in the conversation. As you can clearly see I am currently running 40's I jumped my truck over a church bus full of nuns, bent the frame, returned it to stock and I am currently at the dealership denied warranty so instead I am trading it in for 75k. It's crazy, 60k miles, beat and broken but bring on the TRX! As long as people are just coming up with off the wall stuff I guess I will play too.

Edited to add: I also blame Ford for this and stuff that happened to me before I even owned a Ford. A class action lawsuit is on the horizon (me, my wife and dog)
 
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MrT

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Once again I am probably being dense, but I view this really as a deviation to the OP's original post. I know for sure I am not advocating deceiving an OEM on warranty work (I don't even have a warranty). I tell the drive through person that I ordered a medium instead of a large, rather than say "ahhh it is their fault on that one"

I viewed this as @MrT was asking for a comparison of a hypothetical two tires, not mounted on a vehicle and possibly even using just the equations that @KAH24 listed. Honestly, for real production tires I expect the radius term, for example in rotational inertia, will carry more "weight" than the actual weight term (mass) as the radius term is squared. Assuming that is the case when others who will want to actually argue for purposes of deceiving OEM's will know why @KAH24 is correct. It is important to know "why" as much as know "what".

This is just my curiosity, and I while I am not dying to sit back down and crack open the dynamics text book again, no promises but I will try and look into it just for our collective knowledge. But not before I get in the pool and enjoy the evening! :p



EDIT: this should probably be a champagne room topic and I bet if it went to engineering-ey then no one would tune in lol!
You're not being dense, welcome to clown world. If you are able to look into it that would be awesome. I think the answer would be pretty interesting.
 

GordoJay

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Sure, what is the difference in force needed to rotate a 120lb 35" tall tire and a 90lb 37" tall tire. Use the force applied to or from the axle at 2" diameter if that helps.
Whatever force is needed to overcome the friction in the system. If you want to accelerate the tire, that's different. If you want to accelerate a vehicle while it's sitting on those tires, that's really different. Things get complicated quickly. My point is that you haven't specified the problem tightly enough, and when you do it will be obvious that the answer isn't easy to calculate.
 

GordoJay

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And here is where I asked earlier in the conversation. As you can clearly see I am currently running 40's I jumped my truck over a church bus full of nuns, bent the frame, returned it to stock and I am currently at the dealership denied warranty so instead I am trading it in for 75k. It's crazy, 60k miles, beat and broken but bring on the TRX! As long as people are just coming up with off the wall stuff I guess I will play too.
@Oldfart jumped a bus full of nuns too ...
 

DFS

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Engineering and isolating these stressors isn’t as simple as plugging numbers into some magical calculator for you. This is a system of components that will all respond differently to the larger diameter or heavier tire (though the mass is closer to the center of rotation). I don’t want to speak for KAH as I’m not a West Point educated engineer, though I did go to a respectable school for mechanical engineering. And what you are asking is not a simple task without requisite information, this isn’t something that is easily reduced to the lowest common denominator ( the tire in this scenario). Seeing as OEM automotive engineering is KAH’s wheelhouse maybe he can get you what you’d like. But from my perspective there’s zero chance I’m attempting to get you meaningful answers when I’m missing a plethora of information (I’m a piping guy). I don’t think there’s any ill intentions from any party here, I just don’t think a meaningful answer is simple here.
 
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