Re: OT - Rotations of a low tire?

In article <trfhg1hqf0mdad0hm091h8vr646f93sq9p@xxxxxxx>,
no_spam@xxxxxxxxx says...
> Ned Simmons <news@xxxxxxxxxx> wrote:
>
> >In article <9vbhg192a0q762nt4paihda3g3cfocj0i3@xxxxxxx>,
> >no_spam@xxxxxxxxx says...
> >> Ned Simmons <news@xxxxxxxxxx> wrote:
> >>
> >> >In any case, I just can't accept that the car travels anything other
> >> >than 2*pi*r per rev (as before, r is the distance between the axle and
> >> >the road), regardless of what the tread does.
> >>
> >> As I explained in my previous post (and as others have subsequently
> >> explained), the equation c=2*pi*r applies only if (a) you're dealing
> >> with a circle and (b) r is measured from the center of the circle.
> >> Neither of those are true in this case. Why would you expect to be
> >> able to use that equation when neither of the basic premises are met?
> >
> >You *are* dealing with a circle;
>
> Nonsense.

>From the frame of reference of the axle, r sweeps out a
circle. The shape the tire assumes where it's not
contacting the road has no effect on the velocity of the
vehicle - it's irrelevant.

>
> >r is constant regardless
> >of the shape the tire assumes.
>
> First you say that r (defined by you as the axle-to-ground distance)
> varies with pressure, and now you say it's constant. You can't have it
> both ways.

Huh? r is a function of pressure and I never said
otherwise. I assumed it was a given that the pressure in
the tire would not vary as a function of angular position.

>
> If you define r = c/2*pi, your statement above true. But that's not
> what *you* are calling r.
>
> >Would you argue that you
> >can't use the distance from the driven axle of a tracked
> >vehicle to the road surface to calculate the speed of the
> >vehicle as a function of axle RPM simply because the track
> >isn't circular in shape?
>
> Not really an analogous example because the track is not affixed to
> the drive wheel, but yes, I would argue that you can't compute speed
> based solely on RPM and the distance from the axle to the ground, and
> furthermore, that the distance from axle to ground is irrelevant. You
> *can* compute the vehicle speed as a function of axle RPM *if* you
> know the radius of the drive wheel. Since the wheel is circular, the
> speed is (rotational rate) * radius. But this calculation is valid
> *only* because the drive wheel is circular and *only* if the radius of
> the drive wheel (measured from center to edge) is used, NOT the
> distance from the axle to the ground. It doesn't matter where the axle
> is relative to the ground; for a given wheel size and RPM the vehicle
> speed is the same.
>
> With your definition of radius as the distance from the axle to the
> ground, the tracked vehicle would go faster or slower (at the same
> axle RPM) depending on whether the drive wheel is at the bottom of the
> track, inside the top of the track, or above the track.

In the case of the tracked vehicle r is the distance from
the axle to the surface of the track that contacts the
road. In other words, assuming rigid track and wheel, r is
the pitch radius of the drive wheel plus the distance from
the pitch line of the track to the driving surface.


>
> >> > Think about it from the standpoint of torque.
> >>
> >> Sorry, but an analysis of the physics isn't going to do you much good
> >> unless you first get the geometry right.
> >
> >Then where is the propelling force F acting, if it's not at
> >the interface between the tire and road, i.e.,
> >perpendicular to r and distance r from the axle?
>
> You're asking the wrong question. The problem with your analysis isn't
> where the force is acting, it's in your attempt to use equations
> derived for a circle when you're not dealing with a circle. Once the
> tire is deformed, it is NOT a circle, and you can no longer use
> equations derived for a circle.

I don't know how many more ways to say it. The car isn't
bouncing up and down as the tire rotates - r is constant;
from the axle's frame of reference r sweeps out a circle;
the only propelling force is applied perpendicular to r at
surface of the road; the formulae for circular motion are
applicable.

Plug some numbers into this calculator, it will agree with
my model.

<http://www.club80-
90syncro.co.uk/Syncro_website/TechnicalPages/TRC%
20calculator.htm>

Don, if your following, I'll hopefully get a chance to
respond to you tomorrow night.

Ned Simmons
.

Relevant Pages

  • Re: OT - Rotations of a low tire?
    ... >> with a circle and r is measured from the center of the circle. ... First you say that r (defined by you as the axle-to-ground distance) ... >vehicle as a function of axle RPM simply because the track ... know the radius of the drive wheel. ...
    (rec.crafts.metalworking)
  • Horizon Problem Resolution (2/28/09)
    ... Inflation theory has been offered as a way to overcome this `Horizon Problem'. ... The inner circle of radius $r$ represents space at an earlier time. ... The cosmological redshift, $z$, is ... \section{Velocity Redshift and Distance Redshift Relations} ...
    (sci.physics.particle)
  • Horizon Problem Resolution (2/28/09)
    ... Inflation theory has been offered as a way to overcome this `Horizon Problem'. ... The inner circle of radius $r$ represents space at an earlier time. ... The cosmological redshift, $z$, is ... \section{Velocity Redshift and Distance Redshift Relations} ...
    (sci.physics)
  • Horizon Problem Resolution (2/28/09)
    ... Inflation theory has been offered as a way to overcome this `Horizon Problem'. ... The inner circle of radius $r$ represents space at an earlier time. ... The cosmological redshift, $z$, is ... \section{Velocity Redshift and Distance Redshift Relations} ...
    (sci.astro)
  • Re: De Broglies Mistake
    ... > Copyright By Owner Osher Doctorow Ph.D. ... > or vibration of wave motion. ... > point moving with constant angular velocity w along a circle in the x-y ... > total distance increases linearly proportional to time t, ...
    (sci.physics)