Re: A newcomer here

On Tue, 13 Sep 2005 10:20:56 GMT, Jonathan L Cunningham
<spam@xxxxxxxxxxxxxxxxx> wrote in
<news:4326a326.2248153@xxxxxxxxxxxxx> in
rec.arts.sf.composition:

> On Mon, 12 Sep 2005 23:03:07 -0400, "Brian M. Scott"
> <b.scott@xxxxxxxxxxx> wrote:

[...]

>>Alternatively, let the inner radius be r, the outer radius
>>kr. The volume is proportional to (k^3 - 1)r^3, the force
>>to (k^3 - 1)r = (1 + k + k^2)(k - 1)r = (1 + k + k^2)t,
>>since t = (k - 1)r. Thus, the force is proportional to
>>1 + k + k^2. But if t is fixed, k = 1 + t/r is decreasing
>>as r increases, so the force is decreasing.

> Force goes as the inverse square of the *outer* radius -

Ouch. Yes. Modify the alternative version by taking r to
be the outer radius and kr the inner radius. Then you get
volume ~ (1 - k^3)r^3, and you can legitimately divide by
r^2 to get force ~ (1 + k + k^2)(1 - k)r = (1 + k + k^2)t.
But now k = 1 - t/r is increasing with r, so force is as
well.

[...]

Brian
.

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