|
|
|
|
In article ,
says...
Ned Simmons wrote:
In article ,
says...
Ned Simmons 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
|