Re: Dimensionality Checking (Ada 20XX)

[Stephen Leake <stephen.a.leake.1@gsfc.nasa.gov>]

"Mark Lundquist" <mlundquist2@attbi.com> writes:

> "Stephen Leake" <stephen.a.leake.1@gsfc.nasa.gov> wrote in message
> news:upu5ln22l.fsf@gsfc.nasa.gov...
> >
> > It is simply wrong to try to define units for trig and exponential
> > functions. Remember the Taylor expansion for Sin:
> >
> > Sin (x) = x - 1/6 x**3 ...
> >
> > If x has dimensions of meters (shudder :), then what are the
> > dimensions of Sin (x)?
> 
> Hokey smokes, it's the ratio of two sides of a triangle, so its
> dimensions are Distance/Distance! Forget the Taylor series!

True. But Sin also appears in places that have nothing to do with
triangles, and other functions in Ada.Numerics are also unrelated to
geometry. So the point about Taylor expansions is more general, and I
think makes it clearer about why it is simply wrong to try to define
Sin (meters).

> > This is why angles must be dimensionless, as
> > radians are.
> 
> Angles are not dimensionless.  They are the ratio of distance around the
> circumference to the radius.  That's why for radian, there's 2 * Pi of 'em
> in a circle.  The SI defines the units of radians as (m * m**-1) ('m' stands
> for 'meters').  I don't get why people keep saying radians are
> dimensionless.

Well, in my book m * m ** -1 == 1, so that's dimensionless. I think
this SI definition is silly, but clearly we have to live with it. We
are arguing over a minor point in the meaning of "dimensionless".

The point is, to be meaningful for Sin, the argument x must have no
dimensions. Another way to say this is that the dimensionality must
resolve to 1 (as in m * m ** -1 == 1). 

Can we agree that any argument passed to Sin (or other transcendental
functions) must have dimensionality that resolves to 1, in the above
sense? Then I promise to not say "radians are dimensionless", instead
I'll say "the dimensions of radians resolve to 1" :).

-- 
-- Stephe