RE: supplementary S.I. units

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

On Fri, 25 Sep 1998, Donald E. Simanek wrote:

A lot of introductory stuff with which I completely concur and then
challenged my statement,

> > All equations in physics are tyrannical with respect to
> > dimensional consistency.  With only a few exceptions, however,
> > they are completely neutral about units.  ANY units carrying
> > appropriate dimensions are perfectly acceptable.  For instance, in
> > the equation
>
> Neutral about units? You mean all that effort physicsts have put in to
> devise *coherent* unit *systems* has been in vain? 

Not at all.  I simply meant what I went on to illustrate in the example
from constant acceleration kinematics. 

> Consider this equation
> from a highway engineering book:
>
>               h
>      d = 67.39  - 0.33
>
> Here d represents the distance in feet at which a road sign is generally
> legible to an automobile driver, h is the height of the lettering in
> inches.

Side point:  Is this *really* used?  Nobody really thinks that 3 inch
lettering is visible (let alone legible) at 58 miles, do they?  And I can
certainly read quarter inch lettering from farther away than two and a
half feet. 

> The equation could be made to work for any choice of units by rewriting it
>
>           ch
>      d = K   - d
>
> where K, c, and d are constants. Each choice of units would require a
> different set of values for the constants. Do we want to deal with this
> sort of thing?!

Whether or not we want to deal with it is not relevant to the point. 
Furthermore, I could quibble and point out that this is not an equation
from physics.  But that is also beside the point; I *still* disagree. I
prefer to distinguish between saying that two quantities have different
units and saying that they have different values.  The value of the speed
of light in a vacuum is the same regardless of the units used to express
it.  Despite the fact that 1 is not 12, 1 foot *is* 12 inches, etc. 

Similarly, in the equation you cite I can specify that K *is* 
67.39 ft^(inch/h), c *is* 1 inch^(-1), and the d on the right side of the
equation *is* 0.33 ft. In the SI system these *same* values could be
*expressed* as K = 67.39 x (0.3048 m)^(0.0254 m/h), c = 39.37 m^(-1), and
the d on the right side of the equation = 0.101 m.  Admittedly all of this
is awkward (even "ugly" if you like, although some perverse side of me
kind of enjoys these things) precisely because the value of K in this not
very "physical" equation actually depends on the value of h--i.e., K isn't
really a constant. 

BTW, what you'd really want to do, of course, is write the equation as
something like

                ch
  distance = a(K   - d)

in which case all of the "constants" really are *constants* and
necessarily take on the values a = 1 ft = 0.101 m, K = 67.39, c = 1
in^(-1) = 39.37 m^(-1), and d = 0.33.  

Given the above, I'll go out on a limb and speculate that the relationship
can't possibly reflect reality outside of a *very* restricted range (which
is likely describable by a far simpler relationship) precisely because of
the fact that two of the constants take on surprisingly simple values
*only* in the British system of units.

> To avoid such ugliness, physicists devised coherent unit systems, such as
> cgs, MKS, English, etc. If you keep everything within one such system, you
> don't have to deal with variable constants in equations. 

Because of the problems noted above, I try strictly to avoid "variable
constants." ;-) 

> F = ma works in
> any coherent system, so long as you express each quantity only in the
> designated units appropriate for that system. 

I would say F = ma works in any system of units--coherent or not--so long
as you don't make the mistake of *omitting* the units.  Starting with a
coherent set of units eliminates the need for unit conversions down the
line, but the equation "works" in any event.  This is the message I try to
stress to my students when they repeatedly approach me in lab with
problems that stem precisely from this mistake. 

> ...
> I'd be interested to see John's method for dealing with the steradian, the
> standard (natural) measure of solid angle.

I confess that I haven't I haven't given it much thought, but the issue
wouldn't be what to do about the steradian; it would be how do we
consistently treat solid angle as a dimensioned quantity.  I really don't
want to push this one, but I suspect that it could go much the same way as
I have done for plane angle and would involve the concept of "tangential
area per unit solid angle" rather than "radius squared."

> I notice John's discussion of radian introduced the unit "cycle". What is
> its precise definition, please? Put the definition in words, or an
> equation. Not so simple, is it? John seems to be using "cycle" as a
> "unit". Is it also dimensionless? What is gained by throwing out one
> dimensionless unit and introducing another, even more problematic? 

Of course in my "preferred" system all angular quantities have both
dimensions (of angle) and units.  But I guess I must be missing your point
about defining the "cycle."  What's wrong with "that angle which brings us
back to the starting point"?  Or, if you don't like that, why can't I just
define it as 2 pi radians?
 
> When I taught the introductory course, I gave students a checklist of info
> about dimensions and units. All of it is obvious, but surprisingly, many
> students don't notice these things.
> ...
> 2. Terms which are added or subtracted must have the same dimensions and
> the same units.

I think I know what you mean by this, but I hesitate to say that one
cannot add 1 inch to 1 foot.
 
> 3. Quantities on either side of the equal sign must have the same
> dimensions and the same units.

Does this mean that it is not proper to write "1 inch = 2.54 centimeters"
or "1 m = 100 cm"?
 
And finally, lest it be forgotten in all of the above, I'll reiterate that
I do not advocate straying from accepted pedagogy regarding the radian (or
steradian.)

John
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