On Tue, 25 Jun 2019 14:31:49 +1000, Xeno wrote:

> I thought that one would stump you. ;-)

It did.

> The clue is in the wording - *with respect to the force vectors*.

It's hard to figure out where the abnormal tire wear is coming from when
the suspension geometry of Gillespie indicate high positive camber on the
inside wheel in a tight turn while the force vectors indicate (high?)
negative camber on that same wheel!

> That diagram you linked to above, I'll use that in an attempt to explain
> what is meant.

(For reference): <https://i.postimg.cc/YqHVb5gY/mount33.jpg>

What throws me off is that the left wheel is positive camber while the
right wheel is negative camber, where, on the vehicle, in a turn, I'm not
sure _that_ is what happens since they both started out with similar static
camber, and when the steering wheel was turned to lock, they each took on a
"more positive" camber, didn't they?

It's really just "one" wheel, in three different situations, right?
a. positive camber on turns (with a small force vector to the right)
b. zero camber on turns (with a larger force vector to the right)
c. negative camber on turns (with an even larger force vector)

> For a start, dismiss the middle wheel, ignore it.

Heh heh. I never _understood_ what it was doing there anyway!

It seems like this is the _outside_ wheel, if we assume the car body is to
the right of each of the three wheels, and that the center of the turning
circle is also to the right.

So that diagram seems to be indicating the _outside_ wheel in a turn, does
it not?

> Now imagine the left hand wheel is on the *outside* of the turn. You can
> see the effect of the lateral forces acting on the tread. The arrow
> indicates the *force vector* direction.

Yes. The tread in the lefthand wheel is firmly on the pavement while the
weight of the vehicle is forcing the tire to the left of the tread, making
the tread 'squirm' due to the lateral forces acting toward the center of
the turn radius (I think).

> Lets now deal with the other wheel. Imagine it is the wheel on the other
> side of the car and the axle is coming out of the left side rather than
> the right.

Like this, right?
<https://i.postimg.cc/vZs6Vm3B/mount35.jpg>

> It has positive camber with respect to its position on the
> car but, naturally, it is leaning the other way.

In the updated diagram above, BOTH wheels now have positive camber if we
look just at their suspension geometries. Is that what you mean?

> The lateral forces,
> however, are acting in the same direction as the outside wheel

This is a good point that, if we assume _both_ wheels "start" out with
positive camber, then the lateral forces are pushing both footprints to the
right in that modified diagram, toward the center of the radius force
vector.

> but the effect on the inner wheel is different. *To these forces*, the inside
> wheel looks to have negative camber because they aren't referencing the
> vehicle as being left or right. The forces see two wheels trying to turn
> in the same direction but are leaning in opposite directions.

My problem, I think, is that I _thought_ the geometric camber of the
outside wheel, in a turn, is nearly neutral, while it's only the inside
wheel which has high geometric positive camber.

However ...

With regard to the force camber (so to speak), I easily comprehend that
it's obvious that, if we assume the two wheels in the modified diagram are
_both_ positive camber (as in the modified diagram), then for sure, it's
easy to see that the radius force vector for the left (outside) wheel is to
the right, away from the direction of lean, and yet, in the case of that
right (inside) wheel, the radius force vector is toward the direction of
lean.

That is, if we assume both wheels start with positive camber...
o The left wheel force vector is _away_ from the direction of camber lean
o While the right wheel force vector is _toward_ the direction of lean

But what confuses me is Gillespie says, I thought, that the geometric
camber _changes_ on a steering-wheel-lock turn such that the left (outside)
wheel _changes_ geometric camber (on that one outside wheel) to be almost
neutral?????

That is, my confusion arises from the fact I thought the situation was
o Both wheels can start out in a straightaway with positive camber
o But when you turn hard right, the steering geometries were such that
o The left (outside) wheel takes on an almost neutral camber,
o While the right (inside) wheel takes on a decidedly positive camber

So my confusion stems from the question of isn't it unrealistic to assume
both wheels have positive camber in the middle of the tight turn?

> This picture will assist you in imagining the scenario.
> http://www.conceptcarz.com/images/Bu...-10-MH-010.jpg
Ah, that's where my confusion lies.
o In a straightaway, both front wheels clearly have a positive camber.

But didn't Gillespie say the camber for each wheel in a turn is DIFFERENT?
o Isn't the inside wheel taking on a very positive geometric camber
o While the outside wheel takes on what seems to be an almost zero camber?

ooooooooooooh.... I think I get it now.

If I put my hands with the pinky on the table out in front of me, and the
thumbs up toward the ceiling in a vertical and then lean both hands outward
to simulate high positive camber on both, and then, I imagine a turn to the
right, BOTH hands tilt to the right.

However, what was a positive camber in the left hand, starts to look more
and more like less positive camber and then neutral camber and then maybe
even slightly negative camber in the left hand, since the force is to the
right.

Meanwhile, what was a positive camber in the right hand, which only gets
more and more and more positive as I lean both hands proportionally to the
right, such that the right hand (inside wheel) decidedly looks vastly more
positive in the right hand, since the force is to the right.

> One wheel, the outer, is showing a positive camber to the lateral force.
> The inner, on the other hand, is showing a negative camber lean with
> respect to the lateral forces. It will look like this;
> https://www.gmforum.com/attachments/...ivecamber1.jpg

Yes! That's what seems to be happening on the INSIDE tire of the turn,
right?

> The above pic shows what the inner wheel likely looks like on the road
> in a sharp turn.

This part I completely understand now, and agree with you, as the wear
pattern is extremely consistent with exactly that situation!
<https://i.postimg.cc/T1HkcsX5/mount31.jpg>

I'm working on responding to the rest of the post, but I have to go to a
meeting a few hours away so it won't be until tonight.