Re: The drag force -- a correction squared

[brian whatcott <inet@INTELLISYS.NET>]

At 12:34 10/7/00 -0500, you wrote:
> Am I being particularly dense this Saturday?  My free-body diagram of an
> object in free fall at terminal velocity shows two force vectors; they are
> air drag up and mg down.  The two are equal, so air drag =mg.  Have I
> missed something?  Or is Brian nit-picking on my use of the word
> "general"?
>                                        Regards,
>                                                Jack

Nit picking? Hehe...

Here is Jack's train of argument reduced to a premise/conclusion.

 Is air drag proportional to m.v?

 Drag equals m.g (at terminal velocity)
    THEREFORE
 air drag is proportional to mass.

I cannot recall the name of this logical argument.
I think it is called generalizing from the particular.

Here is the crucial insight.
First, rework the critical mass (ha!) into a cubic sponge,
then reshape it to the popular (mis)conception of a raindrop,
and consider how its terminal velocity varies.

Next review the usual expression for drag -
Drag force equals half.rho. v squared. Cd. A
  You will see mass is noticeably absent.

I suppose I could mention the conjunction
of two curves, one a square law in speed, the other linear,
but surely that much is clear?





brian whatcott <inet@intellisys.net>  Altus OK
                Eureka!