Abstract
Some famous formulas of engineering and physics carry a large integer, 32, 64 or 128, next to π. Examples are the second moment of area of a round bar πD⁴/64, the flow through a pipe πD⁴Δp/(128ηL), the friction factor of a pipe 64/Re, and the 128π⁵/3 in the cross-section with which a small particle scatters light. This paper rewrites them with radius and wavenumber and counts where the integers come from. No new theorem or law is claimed.
First, the large integers are powers of two that came in by writing with the diameter. Written with the radius, the four formulas of the round bar (the second moments for bending and torsion and the stresses at the rim) and the two formulas of the pipe carry the numbers 1/4, 1/2, 4, 2, 8 and 8. In all six formulas the ratio to the number of the diameter form was two multiplied by itself as many times as the power of the radius. The power of π does not move.
Second, the small numbers that remain come from integrals and move with the dimension of the cross-section. The 1/4 of the second moment of area is the reciprocal of the dimension d of the cross-section plus 2 (a circle has d = 2). The polar second moment is d times that. The 8 of the flow through a pipe is d × (d + 2): it is $\mathbf{3.000000000}$ for the gap between two plates (d = 1), $\mathbf{8.000000000}$ for the pipe (d = 2), and 15 and 24 beyond. This was confirmed by numerical integration.
Third, and this is the core of the paper, the 8 of the pipe is 4 × 2, where the 4 does not depend on the shape of the cross-section and the 2 belongs to the circle. The flow through a duct, multiplied by the viscosity and divided by the pressure gradient, is one quarter of the torsion constant of that cross-section, whatever its shape, because the speed of the flow and the function that gives the stresses in torsion satisfy the same equation. For a circle the torsion constant is πR⁴/2, so 8 appears. The torsion constant equals the polar second moment only for the circle: it was $\mathbf{0.683973703}$ times it for an ellipse (semi-axes 1.7 and 0.9) and $\mathbf{0.843462090}$ times it for a square.
Fourth, the 64 of the friction factor is 8 × 2 × 4 and moves with the shape of the cross-section. The 8 is the denominator of the flow, the 2 is the one half in the dynamic pressure, and the 4 comes from using the diameter twice. It becomes $\mathbf{96.000000000}$ for the gap between two plates and $\mathbf{56.908309}$ for a square duct. In the convention that builds the factor from the wall stress it is 16, and with a Reynolds number built on the radius it is 32.
Fifth, the 128π⁵/3 of Rayleigh scattering is 8π/3 times the fourth power of 2π. The 8π/3 is the transverse share 2/3 of the full solid angle 4π (Paper 1084), and the fourth power of 2π appears because the wavenumber is written with the wavelength. The 8π³/3 of the formula for a gas is this number divided by the square of the 4π of Gaussian units, and 32π³/3 is four times that (an approximation of the refractive index). The attenuation coefficient of air came out as the same $\mathbf{1.039926}$×10⁻⁵ per metre in three ways of writing (with the same input as Paper 1023).
Sixth, the diameter carries 2, and the wavelength and the circumference carry 2π. Writing with the diameter does not change the power of π. Written with the circumference C, the second moment of area becomes C⁴/(64π³), with π moved to the cube in the denominator. The 8 of the spring constant Gd⁴/(8D³n) is 32 ÷ 4, the number left after the π of the cross-section and the π of the length of the wire cancel.
To be honest, the integers 32, 64 and 128 themselves carry no physics. What carries it is the small numbers that remain when the formulas are written with radius and wavenumber. That torsion and flow satisfy the same equation is known and is not a finding of this paper.
Keywords: second moment of area, Hagen-Poiseuille law, friction factor, torsion, Rayleigh scattering, diameter and radius, wavenumber.
Scope (what is not claimed)
No new theorem or law is claimed. The formulas for bending and torsion of a round bar, the formula for the flow through a pipe (Hagen 1839, Poiseuille 1846), the fact that torsion and flow satisfy the same equation (Timoshenko and Goodier 1970), the friction factors of different cross-sections (tabulated by Shah and London 1978) and the formula for the scattering of light by small particles (Rayleigh 1899) are all known. Turbulent flow is not treated. The friction factor is fixed by a formula only in the range where the flow moves in layers. The flow near the entrance of a pipe is not treated either. Cross-sections with holes are not treated. The torsion constant can be written with a single function only for a cross-section without holes. Thin-walled tubes need a different treatment. Particles as large as the wavelength are not treated. Rayleigh’s formula holds when the particle is much smaller than the wavelength. No comparison with tabulated values is made. The 56.908309 of the square duct is the calculation of this paper, and no tabulated value is quoted. No way of writing is said to be the right one. The diameter is easy to measure and the radius gives shorter formulas. It is a choice. The choice of mass plays no part. The formulas treated here contain no mass, so the choice of mass of this corpus (m = 4π) changes nothing. Relation to earlier papers: Paper 942 used the second moment of area πd⁴/64 for the bending stiffness of a string and counted how far the overtones stray. Here that 64 is split into 4 × 16. Paper 558 counted the three powers that live together in the spring constant Gd⁴/(8D³n). Here that 8 is seen to be 32 ÷ 4, with π cancelled. Paper 794 used the formula for the resistance to flow in a thin tube for the limit to the height to which a tree pulls water. Here the 8 of that formula is split into 4 × 2. Paper 1023 showed that Rayleigh’s formula with molecules and the formula written with density fluctuations return the same attenuation coefficient for air. Here the same number is obtained in a third way of writing, and the π of the coefficient is split into three origins. Paper 1084 showed that 4π splits into a longitudinal 4π/3 and a transverse 8π/3. Here the 8π/3 of Rayleigh scattering is used as that transverse part. Paper 905 showed that the 2/3 of Larmor’s formula is the spherical average 1 − 1/d of the angular distribution of radiation. Here the 1/(6π) of the scattering formula written in SI is seen to be that 2/3 divided by 4π. Paper 1086 showed that the π in the formulas of gravity appears when they are written with density and vanishes when they are written with mass. Here it is set out that writing a length with the diameter brings powers of 2 and writing it with the wavelength brings powers of 2π. What is added is the splitting of 32, 64 and 128 into a number from an integral times a number from the way the length is measured, the splitting of the 8 of the pipe into a 4 that does not depend on shape and a 2 of the circle, the count of how the 64 of friction moves to 96 and 56.908309 with the shape of the cross-section, and the splitting of the π in the coefficient of Rayleigh scattering into three origins.
1. The question
Textbooks of machines and of flow keep showing formulas in which a large integer stands next to $\pi$. For a round bar of diameter $D$, the second moment of area $I$, the polar second moment $J$, and the stresses at the rim under a bending moment $M$ and a torque $T$ are
$$I=\frac{\pi D^4}{64},\qquad J=\frac{\pi D^4}{32},\qquad \sigma=\frac{32M}{\pi D^3},\qquad \tau=\frac{16T}{\pi D^3}.$$When a fluid of viscosity $\eta$ flows slowly through a pipe of diameter $D$ and length $L$ under a pressure difference $\Delta p$, the flow $Q$, the mean speed $V$ and the friction factor $f$ are
$$Q=\frac{\pi D^4\,\Delta p}{128\,\eta L},\qquad \Delta p=\frac{32\,\eta L V}{D^2},\qquad f=\frac{64}{Re}$$($Re$ is the Reynolds number). The cross-section with which a small particle of radius $a$ and refractive index $n$ scatters light of wavelength $\lambda$ is
$$\sigma_{\mathrm{sc}}=\frac{128\pi^5}{3}\,\frac{a^6}{\lambda^4}\left(\frac{n^2-1}{n^2+2}\right)^2.$$What do 32, 64 and 128 count? Paper 1086 counted how the $\pi$ in the formulas of gravity appears when they are written with density and vanishes when they are written with mass. Here the same is done for the integers. Lengths are rewritten with the radius instead of the diameter and with the wavenumber instead of the wavelength, to see what remains.
2. Bar and pipe: powers of two from the diameter
Replacing the diameter by the radius $R=D/2$ turns the formulas above into the following.
| Quantity | Radius form | Diameter form | Power of radius | Ratio of numbers |
|---|---|---|---|---|
| Second moment of area | $\pi R^4/4$ | $\pi D^4/64$ | 4 | 16 |
| Polar second moment | $\pi R^4/2$ | $\pi D^4/32$ | 4 | 16 |
| Bending stress | $4M/(\pi R^3)$ | $32M/(\pi D^3)$ | 3 | 8 |
| Torsional stress | $2T/(\pi R^3)$ | $16T/(\pi D^3)$ | 3 | 8 |
| Denominator of the pipe flow | 8 | 128 | 4 | 16 |
| Mean-speed form | 8 | 32 | 2 | 4 |
The right-hand column is the number of the diameter form divided by the number of the radius form. In all six formulas it is two multiplied by itself as many times as the power of the radius. The power of $\pi$ does not move in any row. Since the diameter is twice the radius, this is obvious. But set out like this, 64 reads as 4 × 16, 32 as 2 × 16 or 4 × 8 or 8 × 4, and 128 as 8 × 16. What carries the physics is the 1/4, 1/2, 4, 2 and 8 on the left.
The formulas were also checked numerically. For a circle of radius $1.3$, the second moment of area obtained as a numerical integral over the area was $2.243175694$, in agreement with both $\pi R^4/4$ and $\pi D^4/64$. The polar second moment was $4.486351389$, and the ratio is $0.500000000$. The stress at the rim under a moment of 7 was $4.056748663$ whether computed as $MR/I$ or from either formula, and the torsional stress under a torque of 7 was $2.028374332$.
3. The small numbers come from integrals
Where, then, do 1/4 and 1/2 come from? The cross-section was replaced by a $d$-dimensional ball of radius $R$ ($d=2$ is the circle, $d=1$ a segment of length $2R$). Writing $V_d$ for the volume of the unit ball,
$$I=\int x_1^{\,2}\,dV=\frac{V_d\,R^{d+2}}{d+2},\qquad J=\int r^2\,dV=d\,I.$$This was compared with numerical integrals at radius 1.
| $d$ | $I$ (numerical) | $V_d/(d+2)$ | $J$ (numerical) | $I\div J$ |
|---|---|---|---|---|
| 1 | $0.666666667$ | $0.666666667$ | $0.666666667$ | $1.000000000$ |
| 2 | $0.785398163$ | $0.785398163$ | $1.570796327$ | $0.500000000$ |
| 3 | $0.837758041$ | $0.837758041$ | $2.513274123$ | $0.333333333$ |
| 4 | $0.822467033$ | $0.822467033$ | $3.289868134$ | $0.250000000$ |
| 5 | $0.751969859$ | $0.751969859$ | $3.759849296$ | $0.200000000$ |
The 1/4 in $\pi R^4/4$ of the circle is the value of $1/(d+2)$ at $d=2$. The $d+2$ is the power that results from integrating $r^2$ times the shell of volume $r^{d-1}dr$. The 1/2 of the polar second moment is $d/(d+2)$, and the ratio of the two is $1/d$. Since no direction is preferred, the mean square along one axis is one $d$-th of the sum over all axes. This is the same average that Paper 1084 used when it split $4\pi$ into a longitudinal $4\pi/3$ and a transverse $8\pi/3$ (there $d=3$, giving 1/3 and 2/3).
4. The 8 of the pipe is $d(d+2)$
The speed $u$ of slow flow in a duct satisfies $\eta \nabla^2u=-\Delta p/L$ and vanishes at the wall. In a round pipe $u=(\Delta p/L)(R^2-r^2)/(4\eta)$, and integrating over the cross-section gives
$$Q=\frac{\pi R^4\,\Delta p}{8\,\eta L}.$$With radius $1.3$, pressure gradient $2.0$ and viscosity $0.7$, the numerical integral was $3.204536706$, in agreement with both the radius form and the diameter form. The mean speed was $0.603571429$; the pressure gradient times $D^2$ divided by the viscosity and the mean speed was $32.000000000$, and with $R^2$ it was $8.000000000$. The speed at the centre was $2.000000000$ times the mean.
When the cross-section is replaced by a $d$-dimensional ball, the speed is $u=(\Delta p/L)(R^2-r^2)/(2d \eta)$, and the mean speed is $(\Delta p/L)R^2$ divided by $\eta$ times the number in the table below.
| $d$ | Number in the denominator (numerical) | $d(d+2)$ | Speed at centre ÷ mean speed |
|---|---|---|---|
| 1 | $3.000000000$ | 3 | $1.500000000$ |
| 2 | $8.000000000$ | 8 | $2.000000000$ |
| 3 | $15.000000000$ | 15 | $2.500000000$ |
| 4 | $24.000000000$ | 24 | $3.000000000$ |
The case $d=1$ is flow in the gap between two flat plates ($R$ is half the gap). The 8 of the pipe becomes 3 for the gap between plates. The 8 has nothing to do with $\pi$ or with the diameter; it counts the fact that the cross-section is a round shape in two dimensions.
5. 8 = 4 × 2: torsion and flow satisfy the same equation
The 8 can be split once more. When a bar is twisted, the stresses in the cross-section can be written with a single function $\psi$. With the shear modulus $G$ and the rate of twist $\theta$, $\psi$ satisfies $\nabla^2\psi=-2G\theta$ and vanishes at the edge. The torque is $T=2\int\psi dA$. The flow in a duct, on the other hand, satisfies $\nabla^2u=-(\Delta p/L)/\eta$ and vanishes at the edge. Both are constant multiples of the same equation $\nabla^2\varphi=-1$ (with $\varphi=0$ at the edge). So the constant $J_t$ that fixes the torsional stiffness ($T=G\theta J_t$) and the flow are written with the same integral.
$$J_t=4\int\varphi\,dA,\qquad Q=\frac{\Delta p}{\eta L}\int\varphi\,dA=\frac{\Delta p}{\eta L}\cdot\frac{J_t}{4}$$This 4 does not depend on the shape of the cross-section. For a circle $J_t=\pi R^4/2$, so 4 × 2 = 8 appears in the denominator of $Q$. This correspondence has long been known (Timoshenko and Goodier 1970).
It was checked for three cross-sections. For the ellipse $\varphi$ can be written in closed form. For the square, $\nabla^2\varphi=-1$ was solved by finite differences (with the side divided into 100 and into 200 parts, and the step extrapolated to zero).
| Cross-section | Torsion constant $J_t=4\int\varphi dA$ | Polar second moment $J$ | $J_t\div J$ |
|---|---|---|---|
| Circle (radius $1.3$) | $4.486351389$ | $4.486351389$ | 1 |
| Ellipse (semi-axes $1.7$ and $0.9$) | $3.041042160$ | $4.446139003$ | $0.683973703$ |
| Square (side 1) | $0.140577010$ | $0.166666667$ | $0.843462090$ |
For the circle, the flow multiplied by $\eta$ and divided by the pressure gradient was $1.121587847$, exactly one quarter of the torsion constant. The torsion constant of the ellipse agreed with the formula $\pi a^3b^3/(a^2+b^2)$ and is smaller than the polar second moment $\pi ab(a^2+b^2)/4$. The ratio is $4a^2b^2/(a^2+b^2)^2$. For the square, the finite-difference value $0.140577010$ agreed with the value $0.140577015$ of the series formula found in textbooks, with a difference of $4.46\times10^{-9}$. The integral $\int\varphi dA$ was $0.035132831$ with 100 parts, $0.035141397$ with 200 parts, and $0.035144253$ with the step extrapolated to zero.
The torsion constant equals the polar second moment only for the circle. A cross-section that is not a circle cannot stay flat when twisted, and is that much easier to twist. The “2” in the 8 of the pipe comes from the fact that only for the circle can one write $J_t=J=\pi R^4/2$. In the family of ball-shaped cross-sections of the previous section, the polar second moment divided by “flow × $\eta$ ÷ pressure gradient” is $d^2$ (1, 4, 9, 16), and the 4 is also its value at $d=2$.
6. The 64 of the friction factor
The friction factor $f$ of a duct is the pressure gradient times the width of the duct, divided by the dynamic pressure $\rho V^2/2$. For a cross-section that is not a circle, the hydraulic diameter $D_h$ (four times the area of the cross-section divided by its perimeter) is used as the width. If the Reynolds number is also built on $D_h$, then with the $\varphi$ of the previous section
$$f\cdot Re=\frac{2\,D_h^{\,2}\,A}{\int\varphi\,dA}$$($A$ is the area of the cross-section). Neither the density of the fluid nor its viscosity nor its speed remains in this product.
| Cross-section | Hydraulic diameter | $f\cdot Re$ |
|---|---|---|
| Circle | diameter | $64.000000000$ |
| Gap between two plates | twice the gap | $96.000000000$ |
| Square | side | $56.908309$ |
For the circle, $D_h^{ 2}=4R^2$, $A=\pi R^2$ and $\int\varphi dA=\pi R^4/8$, so $2\times4\times8=64$. The 2 is the one half in the dynamic pressure, the 4 comes from using the diameter twice, and the 8 is the denominator of the flow. For two plates the hydraulic diameter is four times half the gap and the number in the denominator is 3, so $2\times16\times3=96$. For the square it is $56.908309$, not an integer.
Even for the same round pipe, the number changes with the convention. In the convention that builds the factor from the shear stress at the wall (one quarter of the $f$ used here), the product is $16.000000000$. With a Reynolds number built on the radius it is $32.000000000$. The number 64 is the product of three conventions: a round cross-section, the diameter, and the one half in the dynamic pressure.
7. The $128\pi^5/3$ of Rayleigh scattering
The cross-section with which a particle much smaller than the wavelength scatters light is, with the polarizability $\alpha$ of the particle (in Gaussian units, with the dimension of a volume) and the wavenumber $k$,
$$\sigma_{\mathrm{sc}}=\frac{8\pi}{3}\,k^4\alpha^2.$$The $8\pi/3$ is the integral of the angular distribution $\sin^2\theta$ of the scattered light over all directions; the numerical integral was $8.377580410$. It is the transverse share 2/3, with respect to the direction of oscillation, of the full solid angle $4\pi$. It is the same as the transverse $8\pi/3$ counted in Paper 1084, and the 2/3 of Larmor’s formula counted in Paper 905 is this share too.
Writing the wavenumber with the wavelength brings in the fourth power of $k=2\pi/\lambda$.
$$\frac{8\pi}{3}\,(2\pi)^4=\frac{128\pi^5}{3}$$$(2\pi)^4$ is $1558.545456544$. The 128 is 8 × 16, and $\pi^5$ is $\pi\times\pi^4$. Putting in the polarizability of a sphere of radius $a$, $\alpha=a^3(n^2-1)/(n^2+2)$, gives the formula at the beginning. With wavelength $0.55$ and polarizability $0.013$, the ratio of the two forms was $1.000000000000$.
For a gas with number density $N$ of molecules, $n^2-1=4\pi N\alpha$, so the fraction scattered per unit length (the attenuation coefficient) is
$$N\sigma_{\mathrm{sc}}=\frac{8\pi^3}{3}\,\frac{(n^2-1)^2}{N\lambda^4}\;\approx\;\frac{32\pi^3}{3}\,\frac{(n-1)^2}{N\lambda^4}.$$Paper 1023 used the form on the left. The coefficients split as follows.
| Way of writing | Coefficient | Split |
|---|---|---|
| Wavenumber and polarizability | $8\pi/3$ | full solid angle $4\pi$ × transverse share 2/3 |
| Wavelength and polarizability | $128\pi^5/3$ | $(8\pi/3)\times(2\pi)^4$ |
| Wavelength and $n^2-1$ | $8\pi^3/3$ | $(8\pi/3)\times(2\pi)^4\div(4\pi)^2$ |
| Wavelength and $n-1$ | $32\pi^3/3$ | four times the above (square of $n^2-1\approx2(n-1)$) |
| SI polarizability and wavenumber | $1/(6\pi)$ | transverse share 2/3 ÷ $4\pi$ |
$8\pi^3/3$ is $82.683404481$, $32\pi^3/3$ is $330.733617923$, and $1/(6\pi)$ is $0.053051648$. The $\pi$ in the coefficient has three origins. There is the $\pi$ from summing over all directions, the $\pi$ from writing the wavenumber with the wavelength, and the $\pi$ from the $4\pi$ of Gaussian units.
With the same input as Paper 1023 (refractive index $n-1=2.78\times10^{-4}$, number density $2.6868\times10^{25}$ per cubic metre, wavelength 550 nm), the attenuation coefficient of air was computed in three ways of writing. The polarizability of one molecule is $1.646986\times10^{-30}$ cubic metres. Both the form with wavenumber and polarizability and the form with $8\pi^3/3$ gave $1.039926\times10^{-5}$ per metre, the same as the value of Paper 1023. The form with $32\pi^3/3$ gave $1.039637\times10^{-5}$, and the ratio is $1.000278019$. This equals $((n+1)/2)^2$, the amount by which the approximation is off.
8. What the way of measuring a length carries
To sum up so far:
| Rewriting | Number carried | Power of $\pi$ | Example |
|---|---|---|---|
| Radius → diameter | 2 | does not move | $\pi R^4/4\to\pi D^4/64$ |
| Radius → circumference | $2\pi$ | moves | $\pi R^4/4\to C^4/(64\pi^3)$ |
| Wavenumber → wavelength | $2\pi$ | moves | $(8\pi/3)k^4\to128\pi^5/(3\lambda^4)$ |
The diameter is twice the radius, so it carries only powers of 2. The circumference $C=2\pi R$ and the wavelength $\lambda=2\pi/k$ carry powers of $2\pi$. Written with the circumference, the second moment of area is $C^4/(64\pi^3)$; for a circle of radius $1.3$ it is $2.243175694$, the same value as the radius form. The 64 appears here too, but $\pi$ has moved to the cube in the denominator. The area of the cross-section is $C^2/(4\pi)$, which is $5.309291585$.
There is also a case in which the integer and $\pi$ cancel. The spring constant of the coil spring treated in Paper 558 is $k=Gd^4/(8D^3n)$ ($d$ is the diameter of the wire, $D$ the diameter of the coil, $n$ the number of turns). Stretching the coil twists the wire, so this formula is the polar second moment of the wire $\pi d^4/32$ times $G$, divided by the length of the wire $\pi Dn$ and the square of the lever arm $(D/2)^2$. With shear modulus 79, wire diameter $0.9$, coil diameter $7.5$ and 6 turns, both forms gave $0.002559600$. The 8 is 32 ÷ 4, and the $\pi$ of the cross-section and the $\pi$ of the length of the wire have cancelled. The formula for the spring constant has no $\pi$ not because no circle is involved, but because a circle is involved twice and the two cancel.
9. Relation to earlier papers, and what should be said honestly
Paper 1086 counted how the $\pi$ in the formulas of gravity appears with density and vanishes with mass. This paper applied the same view to the integers. Writing with the diameter brings powers of 2, and writing with the radius removes them. Writing with the wavelength brings powers of $2\pi$, and writing with the wavenumber removes them. The numbers that remain however one writes are the content of the formula.
Paper 942 used $\pi d^4/64$, Paper 558 used $Gd^4/(8D^3n)$, and Paper 794 used the formula for the resistance to flow in a thin tube, each as a tool. This paper split the numbers inside those tools. Nothing changes in the results of those papers.
To be honest, the integers 32, 64 and 128 themselves carry no physics. They are the trace left by the practical convention of measuring with the diameter. That torsion and flow satisfy the same equation, the friction factors of different cross-sections, and the formula for Rayleigh scattering are all known. What this paper did was to split the numbers into a number from an integral and a number from the way the length is measured, and to count how the former moves with the dimension and the shape of the cross-section. In the finite-difference calculation for the square, the tolerance for agreement was set before running, but the actual difference was $4.46\times10^{-9}$, more than two orders of magnitude below the tolerance.
Closing
Closing: the 32, 64 and 128 that stand next to π are small numbers from integrals times powers of two from the diameter and the wavelength. The πD⁴/64 of the round bar is πR⁴/4, and the 1/4 is the reciprocal of the dimension of the cross-section plus 2. The 128 of the flow through a pipe is 8 × 16, and the 8 becomes $\mathbf{3.000000000}$ for the gap between plates. The 8 splits further into 4 × 2: the 4 comes from torsion and flow satisfying the same equation and does not depend on the shape of the cross-section, and the 2 comes from the torsion constant of the circle. The 64 of the friction factor becomes $\mathbf{96.000000000}$ for two plates and $\mathbf{56.908309}$ for a square. The 128π⁵/3 of Rayleigh scattering is the transverse share 8π/3 times the fourth power of the 2π carried by the wavelength. To be honest, the large integers are the trace of conventions, and the content of the formulas lies in the small numbers that remain when they are written with radius and wavenumber. The separator is whether the number is moved by how the length is measured or by the shape of the cross-section.
References
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Reference
Yamagishi, Y. (2026). The 32, 64 and 128 That Stand Next to π Are Small Numbers from Integrals Times Powers of Two from Diameter and Wavelength [Paper 1089]. YOI: https://www.yamagishiyuuki.com/yoi/1089
ORCID: 0009-0006-8587-3450
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The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 1089. This paper is a preprint and has not been peer reviewed.