after expansion and substitutions: H=R-h and r²=2hR-h².
We now have the total volume for the bi-domed tunnel as
a function of
the radius R=H+h with H the only known variable as H=3 inches.
The trick to the problem is that the
cylindrical part of this tunnel
with radius r is fully inscribed in the circle of radius R.
One can now vary radius
y=r=R.sinθ from 0 to R as a function of the
angle θ with small θ implying small cyclindrical radii and so
more
narrow tunnels.
Corollarily the maximum tunnel would be the sphere itself for r=R and
θ=90°.
For
θ=45°, we have the most symmetric case, where y=x=H=R√2/2=3 and so R=6/√2=3√2 for our given tunnel
size with h=R-H=3(√2-1)=(1-√2/2)R ~ 1.243 inches.
VTotal=(271/1000).4πR³/3; to the encompassing cases, where the
tunnel 'swallows' the sphere.
[The
bi-domed tunnel volume here so represents 27.1% as the sphere's
missing volume and fixing H=3; R=10/3; h=1/3 for:
VSphere-VTotal=(4π/3){R³-3hR²+3Rh²-h³}=(4π/3){729/27}=36π
cubic-inches and the same result as before, now as a remainder of so 72.9%].
A near 'full size' tunnel say
for data set: {R=10h/9; h=9R/10; x=R-h=R/10;
y=R√(99/100); θ=arctan(y/x)=84.261°} gives a calculated Volume
VTotal=(999/1000).4πR³/3 [or a 99.9% volume differential with remainder 0.1% as again the 36π
cubic-inches].
So as perhaps expected, the volume differential again is 36π cubic-inches;
now from a almost
maximal data for x=H=3 in R=30; h=27 and
VSphere-VTotal=(4π/3){R³-3hR²+3Rh²-h³}=(4π/3){27,000-72,900+65,610-19,683}=36π
cubic-inches.
All tunnels allow calculation of x=R-h as the quasi-H of the given
generalisation of the data. As
the size of the sphere increases in R, so
does the size of the tunnel in r. The volume differential remains
however
constant in 36π cube-units. The tunnel varies from narrow to
wide across the angular displacement defined by the Pythagorean
theorem and the described geometries; say the familiar sin²θ+cos²θ=1.
So setting x=H defines a specific family
data set and allows calculation
of R and h as functions of H.
x=H=R-h=cosθ.R and R²=(h+3)²=x²+y²=H²+r²=h²+6h+9.
We
can so substitute R=(h+3) and describe an IMAGINARY SPHERE, which
still 'carries' the bi-domes attached to the tunnel.
Doing this will
crystallize the reason for the constancy of the volume differential of
36π cube-units from above.
VSphere=4πR³/3;
VCylinder=2πHr²; R²=r²+9 with R and r in inches.
R=h+3 and so the difference in volumes is a function of h, the height of
the spherical dome cross-sectional radius
r (of cyclinder in inches).
VSphere-VCylinder=(4π/3){h³+9h²+27h+27}-6π{h²+6h}=
........................4π/3{h³+9h²/2+27}.
(*)
One then maximises/minimises this change of the cylinder inscribed in
the sphere in relating R and r.
We
generalise the 'total height' of the cylinder as 2H=constant.
Then the maximum/minimum condition will be the derivative
of
VSphere-VCylinder as differential set to 0.
4πR²-d(2πH[R²-H²])/dR=0=4πR²-4πHR for H=R.
Or using more formal calculus (chain rule) for cyclindrial volume V=2Hπr²:
dV/dR=dV/dr.dr/dR=4Hπr.dr/dR for a fixed tunnel length of 2H=6 and the relation: R²=r²+9 for 2R.dR=2r.dr for
dr/dR=R/r.
This gives a specific differential equation (DE) as:
R²=rH.dr/dR and which in our limiting case gives: R²=rHR/r=HR for R=H,
The limiting case is a minimum for R=3 with r=0 as the second derivative is positive:
d²(VSphere-VCylinder)/dR²=8πR+4πRH=4πR(2+H)=20πR > 0.
More generally, our DE represents a family of solutions relating R and r by:
R²=rH dr/dR or R².dR=rH.dr for ∫R².dR=∫(rH.dr for 2R³ - 3r²H = Constant
and for which: Constant=2R³-9R²+81 by R²=r²+9.
In the limits: r=0 for R=H=3 & r=H=3 for R=3√2: Constant=54 & 27(4√2-3) respectively.
In the problem, H=3 inches and the problem is solved, as the R=H or diameter 2R=2H necessarily relates the cylindrical
dimensions as maximised relative to the minimised spherical dimensions.
Thence r=3 inches and R=√(r²+9)=√18=3√2 inches
and h=(R-3)=3(√2-1) and about 1.24264..inches.
This recalls our
'most symmetric' case of the 45° from before.
Substituting into equation * then gives the left
over volume as
(4π/3){54√2-81/2}~150.24 cubic inches.
This left over volume here represents two times
the spherical cap of
base-radius r and height h plus the volume encompassed by the sphere
from the outside and the curved
surface of the cylinder from the inside
as our 'imaginary sphere'.
So we subtract the cap volumes from 150.24 cubic
inches for the correct
answer of 113.1 cubic inches.
2VSphericalDome=2π{2R³/3-R²H+H³/3}=2π{36√2-54+9)=37.144 cubic-inches and 150.24-37.14~113.1.
The
problem as described is one of inscribing the symmetry of the square
into the symmetry of the circle, taking account of
the six faces of a
cube; four of which are curved and two of which become truncated.
An interesting and
pertinent extension of the above engages the surface
area of the tunneled cylindrical prism. We have used the derivative
of
the volume to minimise/maximise the volumes and to derive their h-dome
function and so a brief analysis of the surface
area might further serve
for illustration.
One obtains the surface area of a sphere ASphere from
first principle as
summation of perimeters and must take care as to the curvilinear nature
of the line element dl=R.dθ.
For perimeter L and polar coordinates
x=Rcosθ; y=Rsinθ and with dy=Rcosθdθ:
L=∫dl={L}={2πR-0}=2πR
for a full 360 degree transversion of the
line integral.
We again choose the x-axis as the central axis and the
sphere's diameter
about which we define the perimeter summation for the interval [H,R]
defining the spherical dome for
the tunnel and mirrored in the interval
[-H,-R] at the other end of the sphere.
ADome=2π∫y
dl=2π∫Rsinθ.R.dθ=2πR²∫sinθ.dθ=2πR²{-cosθ} for the
intervals [l1=0
,l2=2πR/4] and [θ1=0,θ2=π/2].
So ASemiSphere=-2πR²{0-1}=2πR²
as the surface area of a hemisphere.
Subsequently, ADome=-2πR²{x/R-1}=2πR²{1-H/R}=2πR{R-H}=2πRh.
The
cylinder has curved surface area: ACylinder=2πr.2H=4πr(R-h) for a
total surface area of:
ATotal=4π{h(R-r)+rR}=4π{hR+(R-h)√[2Rh-h²]}.
As
can be easily ascertained by inspection; the limiting scenario for
the surface area to reduce to the familiar spherical
4πR² is again
the boundary condition R=h or H=0 for which R=r and for which the
tunneling cylindrical prism is
the sphere itself.
Tony B.
