Subject:
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Re: Motor Sensor (fwd)
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Newsgroups:
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lugnet.robotics
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Date:
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Wed, 9 Jun 1999 22:43:57 GMT
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Original-From:
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Jim Choate <ravage@EINSTEIN.=stopspam=ssz.com>
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Viewed:
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1347 times
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----- Forwarded message from Shawn Menninga -----
Date: Wed, 09 Jun 1999 18:23:38 -0400
From: Shawn Menninga <smq@dwarfrune.com>
Subject: Re: Motor Sensor (fwd)
Exerpted from the appendix of my college physics book: :-)
3) Angular forms
Position (theta) -- angular units (radians) from a fixed angular position
Velocity (omega) -- time rate of change of angular position (d theta/dt)
Acceleration (nu) -- time rate of change of angular velocity (d omega/dt)
So with a simple spinning wheel, you've got angular velocity and vector
acceleration. Case closed. :-)
----- End of forwarded message from Shawn Menninga -----
No, the case isn't closed. You really should read more Louis Agassiz (i.e.
study nature not books).
The problem with yours and Kekoa's interpretation is that the angular
velocity isn't constant, it's dependent upon the radius of the moment arm
it's measured at *and* the angular acceleration. It's a derived function,
not a fundamental one.
This is *not* directly convertable to angular acceleration which is
indipendent of the radius and the frame of reference.
But more importantly on a spinning shaft the angular velocity isn't
constant which is more than a tad problematic.
Bottem line, angular velocity is worthless except in special cases. If
you're talking about how much current is going to flow at a given rpm, etc.
then the *only* value you need is angular accelleration from which one can
derive the angular velocity.
The fact that angular velocity is a derivative function of angular
accelleration makes it less than useful.
Hopefuly a simple example *will* put this particular piece of confusion to
rest...
Let's image your going in a straight line. Point your arm in the direction
of your velocity vector. It's constant.
Now stand in one spot and imagine your arm representing the velocity vector
for the shaft (i.e. the approximage dia. of your body in this case). It
should be tangential to your imaginary radius. Now begin to turn.
The question to you is, now where is your velocity vector pointing and how
big is it? Answer, depends on where one is measuring that velocity and where
they are standing when they do it. Imagine the angular velocity equation
if one is measuring from the end-on of the shaft at the O.D. of the shaft,
where one is on the shaft, and where one is measuring the angular velocity
parallel to the axis of the shaft. All three of them will give you different
numbers for the exact same situation. The *only* constant independant of
frame of reference in angular accelleration.
Because the velocity vector moves continously it's not a fundamental factor
but rather a function of your constant rotation which is a fixed number of
angles/period-of-time and not a particular distance covered in a period of
time.
Another way to look at it:
If I tell you the outer surface of the motor shaft is turning at 5m/s how
many rotations/second are occuring?
There is a reason motors are given in rpms and not m/s (with all due respect
to your physics books and the instructor who didn't make this clear the
first time).
____________________________________________________________________
Never underestimate the stupidity of some of the people we
have to deal with.
William A. Reinsch, BoXA to Congress
The Armadillo Group ,::////;::-. James Choate
Austin, Tx /:'///// ``::>/|/ ravage@ssz.com
www.ssz.com .', |||| `/( e\ 512-451-7087
-====~~mm-'`-```-mm --'-
--------------------------------------------------------------------
--
Did you check the web site first?: http://www.crynwr.com/lego-robotics
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