Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-L] rms / conic / arithmetic / geometric averages

Thanks again to David Bowman for his (5/29/21) clear, detailed, and helpful comments on my 5/29/21 post regarding the flared horn paraboloid and spherical segment equivalent cylindrical radii.

1) Flared Horn

David is quite right. I neglected the minus sign option when taking the square root and ended up mistakenly revolving the outer side of the parabola about the y-axis (which produces a bulbous horn) instead of the inner side (which produces a flared horn). Again, a plot of my expression for x as a function of y would have shown this error (I did plot y vs x but not my chosen x vs y) I agree with his results for the equivalent cylindrical radius (R) for the flared horn formed by the inner side of my chosen parabola. (See David's response below for my notation and more expressions)

R = sqrt((R_1^2 + 2*R_1*R_2 + 3*R_2^2)/6)

2) Spherical Segment with Two Bases

I'm glad that we have reached consensus that my results of 5/24/21 are correct despite the ambiguities of the English language.

Finally, I have not been able to find a mathematical expression for the shape of the flared bell of a trumpet. I have found internet articles for bell sizes, bore sizes, and how trumpets are made but no mathematical expression for the shape of a trumpet bell. If any of you know of such an expression, I would be most appreciative if you would send it to me.

Don Polvani

-----Original Message-----
From: Phys-l <> On Behalf Of David Bowman
Sent: Saturday, May 29, 2021 10:23 PM
Subject: Re: [Phys-L] rms / conic / arithmetic / geometric averages

Regarding Don P's 'average' radius calculation for his new updated paraboloid
of revolution:

1) My results for a "proper" flared (parabolic) horn (similar to a
trumpet bell) are as follows:

Parabola : y = a*(x - c)^2
a = h/(R_1 - R_2)^2
c = R_1
V = pi*h*((17*R_1^2 - 14*R_1*R_2 + 3*R_2^2)/6)
R = sqr((17*R_1^2 - 14*R_1*R_2 + 3*R_2^2)/6)

Where: h = vertical height between bottom and top of the paraboloid
of revolution
R_1 = radius (larger) of bottom base
R_2 = radius (smaller) of top base
V= volume of paraboloid of revolution
R = radius of equivalent cylinder with same
height and same volume

As a high school trumpet player (long, long ago!), I am curious if
anyone on the list knows the exact mathematical shape of a trumpet

I suspect that Don's (volume & 'average' radius) calculation above is not for the
paraboloid he may have intended. But is rather for a paraboloid formed by
revolving the 'other half' of the parabola he describes above around the y axis.
In order for his new parabola to actually describe a flared bell horn when
revolved he needs to revolve the 'inside' half of the above parabola whose
inverse form is: x = R_1 - (R_1 - R2)*√[(y/h)]. *That* flared paraboloid has an
average radius:
R_avg = √[(R_1^2 + 2*R_1*R_2 + 3*R_2^2)/6].

But the paraboloid Don describes above having the formula:
R_avg = √[(17*R_1^2 - 14*R_1*R_2 + 3*R_2^2)/6] is for revolving the other
'outer' half of the parabola whose inverse form is: x = R_1 + (R_1 - R2)*√[(y/h)].

Both halves, i.e. x = R_1 ± (R_1 - R2)*√[(y/h)], when solved for y = y(x), give
Don's above full parabola function, i.e. y = h*((R_1 - x)/(R_1 - R2))^2. But the
revolved half having Don's above average radius formula does not make for a
flared bell paraboloid. Rather it is, again, for a blunted/bulbous one. For both
half versions the parabola's vertex is at the outer edge of bottom base (i.e. the
one with radius R_1) at the coordinates (x_v, y_v) = (R_1, 0). But the half of the
parabola for which Don's above volume & average radius formulas apply
actually has a top base with a radius of 2*R1 - R2 rather than his desired radius
of R_2. This makes the top even wider than the bottom (because R_1 > R_2),
and again that wrong/outer half paraboloid has a blunted/bulbous shape rather
than a flared one. The top base formed from the 'inner' half of the parabola
does, indeed, have a upper end radius of R_2 and has an overall flared shape,
but its average radius is R_avg = √[(R_1^2 + 2*R_1*R_2 + 3*R_2^2)/6].

I would point out that even both of Don's updated paraboloids (those formed
from each of the 2 halves of the parabola) again have the parabola vertex at one
of the bases where, this time, the slope of the figure is now zero (rather than
infinite as in his earlier version), and again both ends of the horn are thus treated
asymmetrically, and this asymmetry shows up in the average formulas. In his
latest paraboloids the parabola has essentially been turned sideways compared
to his earlier paraboloid and compared to my paraboloidal horn.

When, in an earlier post, I mentioned my power law solid of revolution being
flared for p > 1 what I had in mind is for the independent variable to be along the
symmetry axis direction, and for the power law to open up away from that axis
as a power law as one moves along that axis from the neck toward the flared

(distance from axis) = (constant)*(distance along axis from sharp tip)^P

where p > 1 for a flared end (& p = 2 for my paraboloidal horn). The
vertex/singular point of the power law is *on* the symmetry axis corresponding to
the sharp point zero radius tip end. The actual sharp tip singular point is on a
portion of the symmetry axis that is the cut off so the narrow end has a nonzero
finite radius at the 'neck' end of the horn. This makes the whole horn be made of
regular nonsingular points of the power law/parabola where where the slope is
finite and nonzero everywhere on the horn, and the corresponding formulas are
symmetric in both ends (and thus don't care which end is the wide/flared end
and which one is the narrow/neck end).

Don continues:

2) My results for the zone and segment of two bases of a sphere, as
far as I can see, do not assume that one base is in one hemisphere and
the second base is in the other hemisphere.
David, if I interpret his comment correctly, seems to feel that the
two bases must be in different hemispheres to result in the R shown.
My derivation does not assume this ("opposite
hemisphere") restriction. Numerical evaluations of my result also
show that the two bases may be in the same or different hemispheres
and still yield the correct equivalent R. Also, when the two bases
are in the same hemisphere then:

min(R_1,R_2) <= R <= max(R_1,R_2)

But, I agree that this inequality need not be satisfied when the two
bases are in different hemispheres. However, in that case, we still
have (at least, in the numerical case that I tried):

R < R_sphere
Perhaps, I misunderstood David's comment about the opposite hemisphere

I agree that Don's formula is correct for the situation for when both ends are in
the same hemisphere and for the situation for when the ends are in opposite
hemispheres. I didn't mean to require that the later situation must always be the
case. I just thought that was the situation Don had in mind because he wrote

This case explicitly involves the height h in the expression for R.
It does check in the hemispherical case
(R_1 = R_sph, R_2 = 0, h = R_sph ) and full sphere case
(R_2 = R_1 = 0, h = 2*R_sph).

Because both of Don's given examples of a hemisphere and the whole sphere
can be understood as limiting cases of having both ends in opposite
hemispheres (even the hemisphere case can be understood this way because
the equator can be thought of as belonging to both hemispheres) I just thought
Don had that ends-in-opposite- hemispheres situation in mind. Apparently I
misread Don's intent. Since Don's formula is also correct when both ends are in
the same hemisphere, and since in that situation min(R_1, R_2) ≤ R_avg ≤
max(R_1, R_2) does indeed hold, the formula *does* have this central tendency
property for *that* situation. But in the other ends-in-opposite-hemispheres
situation that property can break down, and does so maximally in the whole
sphere case (R_1 = R_2 = 0, h = 2*R_sphere).

All my results simply find the radius of a cylinder of the same height
and same volume as the object in question. This "equivalent"
cylindrical radius may, or may not, agree with someone's definition of
the "average" radius, but it will always produce the same volume for
the same height.


Don Polvani

David Bowman
Forum for Physics Educators