Re: A weighty subject

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

Robert Carlson wrote:

> [John Mallinckrodt]  writes:
>
> >  I see lots of potential here for confusion down the line.  I happen to
> >  like the identification of weight with the magnitude of m*a_freefall
> >  because that *is* what a scale reads (i.e., they are using my definition),
>
> Scales will read whatever you calibrate them to read.

I simply don't know what to make of this response in the context of our
discussion.  It doesn't seem intended to promote clarity.

> >  We know and that is certainly your right.  We would have it be otherwise
> >  since there is already a perfectly good name for the force of gravity
> >  (i.e., "the force of gravity") and a very meaningful alternate use for the
> >  name "weight."
>
> I prefer weight, it is much shorter than "the force of gravity."

Robert. You can believe me when I say, I know that.

> > > Would those advocating not using weight also advocate not using
> > > acceleration, but instead always say the time rate of change in
> > > velocity?
> >
> >  Why would we do that?  I see no advantage to be gained.
>
> The advantage is that it is shorter and means the same thing.  Why do
> mathematicians define m = (delta y)/(delta x)?  I conclude it is shorter, and
> easier to say slope, than delta y divided by delta x.

Let me get this straight.  In response to your question (perhaps you would
like to read it again), I express puzzlement as to why you would suggest
such a thing.  Your response leaves me even more puzzled and I can only
read it as saying that *you* think "it" (i.e., "time rate of change of
velocity") is a shorter phrase than "acceleration."  I suspect that you
simply didn't read carefully.

> >  Everyone that I
> >  know agrees that acceleration IS the time rate of change of velocity.
> >  I've never heard anyone suggest a good alternate use for the word
> >  acceleration.
>
> Everyone would also agree that weight is the force of gravity if they would
> read their physics texts.

Correct me if I am wrong, but I sense that you are saying you really
aren't interested in my *reasons* for promoting a different definition.

> It is not my definition.  I do not presume to be on the level to make
> fundamental physics definitions.  I will leave that to those with much more
> knowledge than myself.  Perhaps this is you John, convince me.

I have tried my best to explain rationally the reasons that I prefer a
different definition.  You seem to want to have it that I am simply
"wrong."  I see no point in beating my head against the wall.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm