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How would you run a lateral acceleration test in a vehicle on twisty roads at no more than 40mph?



 
 
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  #11  
Old June 25th 19, 09:28 PM posted to comp.mobile.android,alt.comp.freeware,rec.autos.tech
Arlen G. Holder
external usenet poster
 
Posts: 51
Default How would you run a lateral acceleration test in a vehicle on twisty roads at no more than 40mph?

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.
Ads
  #12  
Old June 27th 19, 02:20 PM posted to comp.mobile.android,alt.comp.freeware,rec.autos.tech
Xeno
external usenet poster
 
Posts: 363
Default How would you run a lateral acceleration test in a vehicle ontwisty roads at no more than 40mph?

On 26/6/19 6:28 am, Arlen G. Holder wrote:
> 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?


On the picture, yes. I was wanting you to think outside the square.

> 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!


The article is intended to represent the different possible positions of
a wheel on *one side* of a car. The middle wheel is representative of a
wheel with 0 degrees of camber. It is flat on the road and has more grip
than a wheel with positive camber *on the outside of a turn*.

That is why I added that pic of the old race car - you can see the
positive camber on *both wheels* and their relationship to each other.
What I wanted you to do was look at the force vector arrows that, in a
turn, run in one direction. If you superimposed those arrows under the
race car in the appropriate direction, you get closer to my point.

>
> 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?


Yes.
>
>> 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).


Yes.
>
>> 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>


Perfect.
>
>> 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?


Yes.
>
>> 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.


Yes.
>
>> 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.


Yes, it is limited in the amount of camber roll compared to the inside
wheel. It would look like the wheel you erased, the centre one. It would
still be less effective than the one *presenting itself* to the force
vector as a negative camber. That said, there is a limit to the benefit
of negative camber.


Whilst on the topic of camber, what you need to get squared away in your
mind is the *effect* of camber, as a separate entity, before you look at
its place in a car.
This article is a good start and introduces an important term:

Camber Thrust; https://en.wikipedia.org/wiki/Camber_thrust

Wikipedia describes it well. Moving right along, in a vehicle, the
camber thrust of one side is balanced out by the camber thrust of the
other, through the *steering linkage*, the net effect being the thrust
forces of each side are balanced. You might ask, "Why do we have
*static* camber at all?". The effect of each wheel trying to deviate
from the centre line of the vehicle in opposite directions puts the
steering linkage under tension. That helps reduce the effect of any free
play that could potentially create a *shimmy* effect. That's one effect
of camber, be it positive or negative, that has an influence on straight
running.
SAI. caster, mechanical trail & pneumatic trail also have influences on
the straight running of a vehicle - primarily keeping the vehicle
travelling straight or returning the vehicle to straight running after a
turn. When you want to turn, however, you end up *fighting* these forces.

let's delve a little deeper. The camber thrust I mentioned above leads
to a phenomenon know as *camber steer*. When a vehicle's wheels are
inclined with respect to the vertical, the rolling radius is shorter on
one side of the tread than the other. This link shows it best;

https://ars.els-cdn.com/content/imag...0750651318.gif

The tyre forms the frustrum of a cone and tries to rotate around its
apex. This causes the wheel to deviate from a straight path to produce
the effect known as *camber steer*. Positive camber will, therefore,
make the wheels turn away from each other (toe out). Negative camber
however will make the wheels turn towards each other (toe in). This is
one of the reasons why the wheel track has to be set to match the design
of the suspension to counteract the inherent tendency of the wheels to
either move towards or away from each other. One advantage, as I stated
before, is that it keeps the steering under tension. It is also the
primary reason why camber angles must be matched side to side.

Now, consider this, when you want to turn a corner you are fighting
against caster, SAI, mechanical trail and pneumatic trail, all of which
are trying to return the steering to the straight ahead position. On a
nose heavy car, that can be quite a challenging task when you realise
that, through the effects of SAI you are actually lifting the front of
the car.

Now, knowing the effects of camber steer in the straight ahead position,
what happens when we *turn* the steering? The inside wheel *increases*
positive camber. That means the steering now has an imbalance in camber
steer. The camber thrust at the inside wheel has now increased and is
*assisting* in turning the car against the forces that are trying to
straighten it. What about the outside wheel? Well, it has gone from
having a matched opposite lean to leaning the same way as the inside
wheel though, as you have noted, not as much. Regardless, it is still
assisting through its increased camber steer, in turning the vehicle in
the same direction as the inner wheel's camber steer is going. This is
one of the reasons I showed you this pic;

https://cdn.forconstructionpros.com/...e2c6fd4f. jpg

This is what your car is trying to do, face the force vector resisting
vehicle inertia with a negative camber but, at the same time, use camber
steer to assist in the turn. On some cars, camber steer can be utilised
effectively in aiding steering ease but, on others, and the Toyota Yaris
is one, you can feel the self centering effects of SAI, caster, etc. up
until about 2/3rds steering lock. After that point the increasing camber
steer effects take over and dominate to the point where, when you want
to straighten up, you have to physically turn the steering back to the
centre until you reach the point where the self centering forces again
dominate. Even powering forward (it is FWD) doesn't provide sufficient
self centering forces to counter the camber steer. It is a weird
sensation and it feels as if the steering has hit some *over-centre*
point. It tells you how powerful camber steer can be but it also
indicates that the manufacturer of the Yaris, and similar cars with the
issue, hit a compromise point and you can see, or feel, that it was a
middle of the road option. It also tells you that, on the more heavily
loaded *outside* wheel, too much negative camber can create unforeseen
issues. The manufacturer, in this case, has decided that high lock
situations will only occur in very low speed events. Therefore the
camber steer issue at high lock wasn't an issue and the focus was placed
on low steer angle, high speed situations.

>
> 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.


Yes.
>
> 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


Yes.
>
> 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?????


Yes. Suspension design however plays a significant role here. Some
designs can travel from positive to negative and end up back at positive
by the time full lock is achieved. Some strut types, IIRC, have this issue.
>
> 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


Yes, depending on suspension design.
>
> 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?


In some case, that is the assumption, in others, not. I saw a car today
that had a decided negative cant on the outside front wheel. Not as much
as the inside runner, of course, but enough to be visually noticeable.
>
>> 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.


Yes.
>
> 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.


Wonderful.
>
> 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?


Spot on. Looks quite horrible, doesn't it? The wider the tyre, the worse
it gets. What's more, pumping the tyre up more stiffens the sidewalls of
the tyre and likely will aggravate the issue. Low pressure = more
sidewall flex, high pressure = less.
>
>> 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.
>

Take your time, lots of info to digest and more in this post.

--

Xeno


Nothing astonishes Noddy so much as common sense and plain dealing.
(with apologies to Ralph Waldo Emerson)
 




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