In article ,
alexy wrote:
(Robert Bonomi) wrote:



2) for the 'special case' where the two ends are the same shape, differing
only in size, and are parallel to each other --
This seems to be the case the OP was asking about. At least that is
the prevailing assumption in this thread (which of course does not
make it valid). This specificity is good, since I had been falling
into the trap of assuming this formula for any two end shapes, not
just identical ones leading to a "cone". BTW, is it accurate to say
that this does not have to be a frustrum of a right cone?

the answer to that is 'yes'. as long as you're measuring the 'height'
as the perpendicular distance between the ends.

And can the
shapes be rotated?

Yuppers, but proving it takes calculus. consider an infinite series
of plane sections, parallel to the base. nothing changes if they are
'aligned' relative to each other, or skewed. thus the integral (volume)
will be the same.

(I think the answer to both of those is "yes", but
am I thinking of it correctly?)

It would appear so. grin

You can 'extend' the edges that join the ends, until those edges meet.
(they are guaranteed to all come together at a single point.)

You now have a *cone*. You can calculate the volume between the two
"ends" by calculating the volume of the entire cone above the larger
end, and then subtracting the volume of the cone above the smaller end.

the 'height' of the cone, from the 'big end', can be computed by
using _any_ *linear* dimension of the big end, and the corresponding
dimension from the little end, as follows:

To keep things manageable, lets invent some names:
b is a linear dimension of the big end
l is a linear dimension of the little end
d is the distance between the ends, measured perpendicular to the
planes of the ends.
c is the distance that the 'convergence point' is above the big end
A(b) is the _area_ of the big end
A(l) is the _area_ of the little end
To keep it even more manageable (for me) I used "s" for "small"
instead of "l" for "little", to keep me from assuming that (k-1)*(k+l)
is k^2-1. g

c = b/(b-l) * d

so, the volume of the full cone above the big end is:
A(b)*c
You forgot the 1/3 factor. In a sense, it gets "picked up" in your k
later, but probably better not to mix the two, IMHO.

eek! you're absolutely correct.

and the volume of the full cone above the little end is:
A(l)*(c-d)
thus, obviously, the volume of the 'truncated cone', between the
two ends is:
A(b)*c - A(l)*(c-d)
eliminating the intermediate calculation, by back-substituting for c, gives:
A(b)*(b/(b-l)*d) - A(l)*((b/(b-l)-1)*d)
The area for any given shape is proportional to the square of the
linear dimension of that shape, differences in shape are accounted for
by a difference in the proportionality constant, *only*.

so we get:
k*b^3/(b-l)*d - k*l^2*(b/(b-l)-1)*d
or
k*b*d/(b-l)*(b^2 - (l^2+l-b))
I don't follow this step. And the term on the right does not add up
dimensionally.

lessee, with all the intermediate steps:
k* b^3/(b-l)*d - k*l^2*(b/(b-l)-1*d
k*(b^3/(b-l)*d - l^2*(b/(b-l)-1)*d)
k*(b^3/(b-l)*d - l^2*(b/(b-l)-b/b)*d)
k*b*(b^2(/b-l)*d - l^2(1/(b-l)-1/b)*d)
k*b*d*(b^2(/b-l) - l^2(1/(b-l)-1/b))
k*b*d*(b^2(/b-l) - l^2(1/(b-l)-((b-l)/(b-1))/b)))
k*b*d*(b^2(/b-l) - l^2(1/(b-l)-(b-l)/((b-1))*b)))
k*b*d/(b-l)*(b^2 - l^2(1-(b-l)/b))
k*b*d/(b-l)*(b^2 - l^2(1-(1-(l/b)))
k*b*d/(b-l)*(b^2 - l^2(l/b))
k*b*d/(b-l)*(b^2 - l^3/b)
k*d/(b-l)*(b^3 - l^3)

Yup. I got it wrong first time. I lost a parenthesization, at step 4. *sigh*

Definitions: USENET -- open mouth, insert foot, echo internationally.


I think this step should be k/3*d/(b-s)*(b^3-s^3) (I also reinjected
the 1/3 factor here, so my k is 3 times as large as yours.)

Dividing by (b-s) gives me 1/3 *d *k *(b^2+bs+s^2)
And substituting back to the easily calculated or measured areas using
A(b) = kb^2 and A(s) = ks^2, we get
1/3 *d *(A(b)+sqrt(A(b)*A(s))+A(s))
Which conveniently agrees with the formulas in my CRC Standard
Mathematical Tables book. Glad those formulas haven't changed in the
last 35 years!

That's cheating! grin

Appreciate that you caught my errors, nonetheless..



doing it "step-wise" (computing height of cone, and end areas, separately)
is easier grin



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