Appendix/Ramblings/T6CoordinatesPt3

Concentric Ellipsoidal (T6) Coordinates (Part 3)

Best Thus Far

Part A

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$3$	---	---	---	---	---	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

 

$A\equiv {\ell }_{3D}^{-2}$ $=$ $(x^2+q^4{y}^2+p^4{z}^2),$       and,       $B\equiv {{𝒟}}^2$ $=$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)$ $\Rightarrow \frac{\partial A}{\partial x}$ $=$ $2x,$ $\frac{\partial A}{\partial y}$ $=$ $2q^{4}y,$ $\frac{\partial A}{\partial z}$ $=$ $2p^{4}z;$ $\frac{\partial B}{\partial x}$ $=$ $2x(4q^4{y}^2+p^4{z}^2),$ $\frac{\partial B}{\partial y}$ $=$ $2q^{4}y(p^4{z}^2+4x^2),$ $\frac{\partial B}{\partial z}$ $=$ $2p^{4}z(q^4{y}^2+x^2).$

Try …

${\lambda }_3$	$\equiv$	$(\frac{B}{A}{)}^{m/2}=(D{\ell }_{3D}{)}^m$
$\Rightarrow \left[\frac{AB}{m{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x_i}$	$\equiv$	$\frac{1}{2}\{A\cdot \frac{\partial B}{\partial x_i}-B\cdot \frac{\partial A}{\partial x_i}\}$
$\Rightarrow \left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln x_i}$	$\equiv$	$\frac{x_i}{2}\{A\cdot \frac{\partial B}{\partial x_i}-B\cdot \frac{\partial A}{\partial x_i}\}.$

In this case we find,

$\frac{x}{2}\{{\}}_x$	$=$	$x^2(2q^4{y}^2+p^4{z}^2{)}^2,$
$\frac{y}{2}\{{\}}_y$	$=$	$q^4{y}^2(2x^2+p^4{z}^2{)}^2,$
$\frac{z}{2}\{{\}}_z$	$=$	$p^4{z}^2(x^2-q^4{y}^2{)}^2.$

The scale factor is, then,

$h_3^{-2}$	$=$	$\sum _{i=1}^3(\frac{\partial {\lambda }_3}{\partial x_i}{)}^2$
	$=$	$\sum _{i=1}^3\left\{\right[\frac{m{\lambda }_3}{2AB}\left]\right[A\cdot \frac{\partial B}{\partial x_i}-B\cdot \frac{\partial A}{\partial x_i}]{\}}^2$
	$=$	$\left[\frac{m{\lambda }_3}{AB}{]}^2\right\{[x(2q^4{y}^2+p^4{z}^2{)}^2{]}^2+[q^{4}y(2x^2+p^4{z}^2{)}^2{]}^2+[p^{4}z(x^2-q^4{y}^2{)}^2{]}^2\}$
$\Rightarrow h_3$	$=$	$\left[\frac{AB}{m{\lambda }_3}\right]\left\{\right[x(2q^4{y}^2+p^4{z}^2{)}^2{]}^2+[q^{4}y(2x^2+p^4{z}^2{)}^2{]}^2+[p^{4}z(x^2-q^4{y}^2{)}^2{]}^2{\}}^{-1/2}.$

Part B (25 February 2021)

Now, from above, we know that,

$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2=AB$

$=$

$[x(2q^4{y}^2+p^4{z}^2){]}^2+[q^{2}y(p^4{z}^2+2x^2){]}^2+[p^{2}z(x^2-q^4{y}^2){]}^2.$

Example: $(q^2,p^2)=(2,5)$     and     $(x,y,z)=(0.7,\sqrt{0.23},0.1)\Rightarrow {\lambda }_1=1.0$
$[x(2q^4{y}^2+p^4{z}^2){]}^2$	$[q^{2}y(p^4{z}^2+2x^2){]}^2$	$[p^{2}z(x^2-q^4{y}^2){]}^2$	$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$
2.14037	1.39187	0.04623	3.57847

As an aside, note that,

$AB$	$=$	$\left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln x}+\left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln y}+\left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln z}$
$\Rightarrow m$	$=$	$\frac{\partial \ln {\lambda }_3}{\partial \ln x}+\frac{\partial \ln {\lambda }_3}{\partial \ln y}+\frac{\partial \ln {\lambda }_3}{\partial \ln z}.$

We realize that this ratio of lengths may also be written in the form,

$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$

$=$

$6x^2{q}^4{y}^2{p}^4{z}^2+x^4(4q^4{y}^2+p^4{z}^2)+q^8{y}^4(4x^2+p^4{z}^2)+p^8{z}^4(x^2+q^4{y}^2).$

Same Example, but Different Expression: $(q^2,p^2)=(2,5)$     and     $(x,y,z)=(0.7,\sqrt{0.23},0.1)\Rightarrow {\lambda }_1=1.0$
$6x^2{q}^4{y}^2{p}^4{z}^2$	$x^4(4q^4{y}^2+p^4{z}^2)$	$q^8{y}^4(4x^2+p^4{z}^2)$	$p^8{z}^4(x^2+q^4{y}^2)$	$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$
0.67620	0.94359	1.87054	0.08813	3.57847

Let's try …

$\frac{\partial {\lambda }_5}{\partial x}$	$=$	$-x(2q^4{y}^2+p^4{z}^2),$
$\frac{\partial {\lambda }_5}{\partial y}$	$=$	$q^{2}y(p^4{z}^2+2x^2),$
$\frac{\partial {\lambda }_5}{\partial z}$	$=$	$p^{2}z(x^2-q^4{y}^2).$

This means that the relevant scale factor is,

$h_5^{-2}$	$=$	$[-x(2q^4{y}^2+p^4{z}^2){]}^2+[q^{2}y(p^4{z}^2+2x^2){]}^2+[p^{2}z(x^2-q^4{y}^2){]}^2=(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$
$\Rightarrow h_5$	$=$	$\left(\frac{{\ell }_{3D}}{{𝒟}}\right),$

and the three associated direction cosines are,

${\gamma }_{51}=h_5\left(\frac{\partial {\lambda }_5}{\partial x}\right)$	$=$	$-x(2q^4{y}^2+p^4{z}^2)\left(\frac{{\ell }_{3D}}{{𝒟}}\right),$
${\gamma }_{52}=h_5\left(\frac{\partial {\lambda }_5}{\partial y}\right)$	$=$	$q^{2}y(p^4{z}^2+2x^2)\left(\frac{{\ell }_{3D}}{{𝒟}}\right),$
${\gamma }_{53}=h_5\left(\frac{\partial {\lambda }_5}{\partial z}\right)$	$=$	$p^{2}z(x^2-q^4{y}^2)\left(\frac{{\ell }_{3D}}{{𝒟}}\right).$

These direction cosines exactly match what is required in order to ensure that the coordinate, ${\lambda }_5$, is everywhere orthogonal to both ${\lambda }_1$ and ${\lambda }_4$. GREAT! The resulting summary table is, therefore:

Direction Cosine Components for T10 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$4$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$5$	---	$\frac{{\ell }_{3D}}{{𝒟}}$	$-x(2q^4{y}^2+p^4{z}^2)$	$q^{2}y(p^4{z}^2+2x^2)$	$p^{2}z(x^2-q^4{y}^2)$	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

Try …

${\lambda }_5$	$=$	$x^{-2q^4}\cdot y^{2q^2}+y^{q^2{p}^4}\cdot z^{-q^4{p}^2}+x^{-p^4}\cdot z^{p^2}$
	$=$	$\frac{y^{2q^2}}{x^{2q^4}}+\frac{y^{q^2{p}^4}}{z^{q^4{p}^2}}+\frac{z^{p^4}}{x^{p^2}}$
	$=$	$\frac{1}{x^{2q^4+p^2}{z}^{q^4{p}^2}}\{[x^{p^2}]y^{2q^2}[z^{q^4{p}^2}]+[x^{2q^4+p^2}]y^{q^2{p}^4}+[x^{2q^4}]z^{p^4+q^4{p}^2}\}$
	$=$	$\frac{1}{x^{2q^4+p^2}{z}^{q^4{p}^2}}\left\{{\mathfrak{F}}_{\mathfrak{5}}\right\}.$

This gives,

$\frac{\partial {\lambda }_5}{\partial x}$	$=$	$-\frac{2q^4}{x}\left(\frac{y^{2q^2}}{x^{2q^4}}\right)-\frac{p^4}{x}\left(\frac{z^{p^2}}{x^{p^2}}\right)$
	$=$	$-\frac{1}{x^{2q^4+p^2+1}}[2q^4{y}^{2q^2}{x}^{p^2}+p^4{x}^{2q^4}{z}^{p^2}].$

Or, given that,

$x^{2q^4+p^2+1}$

$=$

$\frac{x}{z^{q^4{p}^2}}\left\{\frac{{\mathfrak{F}}_{\mathfrak{5}}}{{\lambda }_5}\right\},$

we can also write,

$\frac{\partial {\lambda }_5}{\partial x}$

$=$

$-\frac{z^{q^4{p}^2}}{x}\left\{\frac{{\lambda }_5}{{\mathfrak{F}}_{\mathfrak{5}}}\right\}[2q^4{y}^{2q^2}{x}^{p^2}+p^4{x}^{2q^4}{z}^{p^2}]$

Similarly,

$\frac{\partial {\lambda }_5}{\partial y}$	$=$	$\frac{2q^2}{y}\left(\frac{y^{2q^2}}{x^{2q^4}}\right)+\frac{q^2{p}^4}{y}\left(\frac{y^{q^2{p}^4}}{z^{q^4{p}^2}}\right)$
	$=$	$\frac{1}{x^{2q^4}{z}^{q^4{p}^2}}\left[\frac{2q^2}{y}\right(y^{2q^2}{z}^{q^4{p}^2})+\frac{q^2{p}^4}{y}(y^{q^2{p}^4}{x}^{2q^4}\left)\right]$
	$=$	$\frac{x^{p^2}}{y}\left\{\frac{{\lambda }_5}{{\mathfrak{F}}_{\mathfrak{5}}}\right\}\left[2q^2\right(y^{2q^2}{z}^{q^4{p}^2})+q^2{p}^4(y^{q^2{p}^4}{x}^{2q^4}\left)\right];$
$\frac{\partial {\lambda }_5}{\partial z}$	$=$	$-\frac{q^4{p}^2}{z}\left(\frac{y^{q^2{p}^4}}{z^{q^4{p}^2}}\right)+\frac{p^4}{z}\left(\frac{z^{p^4}}{x^{p^2}}\right)$
	$=$	$\frac{1}{x^{p^2}{z}^{q^4{p}^2}}\left[\frac{p^4}{z}\right(z^{p^4+q^4{p}^2})-\frac{q^4{p}^2}{z}(x^{p^2}{y}^{q^2{p}^4}\left)\right]$
	$=$	$\frac{x^{2q^4}}{z}\left\{\frac{{\lambda }_5}{{\mathfrak{F}}_{\mathfrak{5}}}\right\}\left[p^4\right(z^{p^4+q^4{p}^2})-q^4{p}^2(x^{p^2}{y}^{q^2{p}^4}\left)\right]$

Understanding the Volume Element

Let's see if the expression for the volume element makes sense; that is, does

$(h_1{h}_4{h}_5)d{\lambda }_{1}d{\lambda }_{4}d{\lambda }_5$

$=$

$dxdydz?$

First, let's make sure that we understand how to relate the components of the Cartesian line element with the components of our T10 coordinates.

Line Element

MF53 claim that the following relation gives the various expressions for the scale factors; we will go ahead and incorporate the expectation that, since our coordinate system is orthogonal, the off-diagonal elements are zero.

$d{s}^2=d{x}^2+d{y}^2+d{z}^2$

$=$

$\sum _{i=1,4,5}{h}_i^{2}d{\lambda }_i^2.$

Let's see. The first term on the RHS is,

$h_1^{2}d{\lambda }_1^2$	$=$	$h_1^2\left[\right(\frac{\partial {\lambda }_1}{\partial x})dx+(\frac{\partial {\lambda }_1}{\partial y})dy+(\frac{\partial {\lambda }_1}{\partial z})dz{]}^2$
	$=$	$h_1^2\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^{2}d{x}^2+(\frac{\partial {\lambda }_1}{\partial y}{)}^{2}d{y}^2+(\frac{\partial {\lambda }_1}{\partial z}{)}^{2}d{z}^2$
		$+\cancel{2\left(\frac{\partial {\lambda }_1}{\partial x}\right)\left(\frac{\partial {\lambda }_1}{\partial y}\right)}dxdy+\cancel{2\left(\frac{\partial {\lambda }_1}{\partial x}\right)\left(\frac{\partial {\lambda }_1}{\partial z}\right)}dxdz+\cancel{2\left(\frac{\partial {\lambda }_1}{\partial x}\right)\left(\frac{\partial {\lambda }_1}{\partial y}\right)}dydz];$

the other two terms assume easily deduced, similar forms. When put together and after regrouping terms, we can write,

$\sum _{i=1,4,5}{h}_i^{2}d{\lambda }_i^2$	$=$	$\left[h_1^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2+h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2]d{x}^2$
		$+\left[h_1^2\right(\frac{\partial {\lambda }_1}{\partial y}{)}^2+h_4^2(\frac{\partial {\lambda }_4}{\partial y}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial y}{)}^2]d{y}^2$
		$+\left[h_1^2\right(\frac{\partial {\lambda }_1}{\partial z}{)}^2+h_4^2(\frac{\partial {\lambda }_4}{\partial z}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial z}{)}^2]d{z}^2.$

Given that this summation should also equal the square of the Cartesian line element, $(d{x}^2+d{y}^2+d{z}^2)$, we conclude that the three terms enclosed inside each of the pair of brackets must sum to unity. Specifically, from the coefficient of $d{x}^2$, we can write,

$h_1^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2$

$=$

$1-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2.$

Using this relation to replace $h_1^2$ in each of the other two bracketed expressions, we find for the coefficients of $d{y}^2$ and $d{z}^2$, respectively,

$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right](\frac{\partial {\lambda }_1}{\partial y}{)}^2$	$=$	$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial y}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial y}{)}^2\right](\frac{\partial {\lambda }_1}{\partial x}{)}^2;$
$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right](\frac{\partial {\lambda }_1}{\partial z}{)}^2$	$=$	$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial z}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial z}{)}^2\right](\frac{\partial {\lambda }_1}{\partial x}{)}^2.$

We can use the first of these two expressions to solve for $h_4^2$ in terms of $h_5^2$, namely,

$(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial y}{)}^2$	$=$	$(\frac{\partial {\lambda }_1}{\partial x}{)}^2-h_4^2(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2$
$\Rightarrow h_4^2\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]$	$=$	$(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial y}{)}^2+h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial y}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2$

Analogously, the second of these two expressions gives,

$h_4^2\left[\right(\frac{\partial {\lambda }_4}{\partial z}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]$

$=$

$(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial z}{)}^2+h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial z}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial z}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2$

Eliminating $h_4$ between the two gives the desired overall expression for $h_5$, namely,

$0$	$=$	$\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial z}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial z}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial z}{)}^2\left(\frac{\partial {\lambda }_1}{\partial x}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]$
		$-\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial y}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial y}{)}^2\left(\frac{\partial {\lambda }_1}{\partial x}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial z}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]$
	$=$	$h_5^2\left\{\right[\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial y}{)}^2-\left(\frac{\partial {\lambda }_5}{\partial y}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2\left]\right[\left(\frac{\partial {\lambda }_4}{\partial z}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-\left(\frac{\partial {\lambda }_4}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial z}{)}^2]$
		$-\left[\right(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial z}{)}^2-(\frac{\partial {\lambda }_5}{\partial z}{)}^2\left(\frac{\partial {\lambda }_1}{\partial x}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]\}$
		$-\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]$
		$+\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial z}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]$
	$=$	$\frac{h_5^2}{h_1^4{h}_4^2{h}_5^2}\left\{\right[{\gamma }_{51}^2{\gamma }_{12}^2-{\gamma }_{52}^2{\gamma }_{11}^2\left]\right[{\gamma }_{43}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{13}^2]-[{\gamma }_{51}^2{\gamma }_{13}^2-{\gamma }_{53}^2{\gamma }_{11}^2\left]\right[{\gamma }_{42}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{12}^2\left]\right\}$
		$+\frac{1}{h_4^2{h}_1^2}\left\{\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{43}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{13}^2]-\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right[{\gamma }_{43}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{13}^2]-(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{42}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{12}^2]+\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right[{\gamma }_{42}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{12}^2\left]\right\}$
	$=$	$\frac{h_5^2}{h_1^4{h}_4^2{h}_5^2}\left\{\right[{\gamma }_{51}^2{\gamma }_{12}^2-{\gamma }_{52}^2{\gamma }_{11}^2\left]\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}\left]\right[{\gamma }_{43}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{13}]-[{\gamma }_{51}^2{\gamma }_{13}^2-{\gamma }_{53}^2{\gamma }_{11}^2\left]\right[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]\right[{\gamma }_{42}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{12}\left]\right\}$
		$+\frac{1}{h_4^2{h}_1^2}\left\{\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}][{\gamma }_{43}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{13}]-\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}\left]\right[{\gamma }_{43}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{13}]-(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}][{\gamma }_{42}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{12}]+\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]\right[{\gamma }_{42}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{12}\left]\right\}$
	$=$	$\frac{1}{h_1^4{h}_4^2}\{-[{\gamma }_{51}{\gamma }_{12}+{\gamma }_{52}{\gamma }_{11}\left]{\gamma }_{43}\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}]{\gamma }_{52}+[{\gamma }_{51}{\gamma }_{13}+{\gamma }_{53}{\gamma }_{11}\left]{\gamma }_{42}\right[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]{\gamma }_{53}\right\}$
		$+\frac{1}{h_1^4{h}_4^2}\left\{{\gamma }_{12}^2\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}]{\gamma }_{52}-{\gamma }_{11}^2[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}]{\gamma }_{52}-{\gamma }_{11}^2[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}]{\gamma }_{53}+{\gamma }_{13}^2[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]{\gamma }_{53}\right\}$
	$=$	$\frac{1}{h_1^4{h}_4^2}\left\{\right[(-{\gamma }_{51}{\gamma }_{12}-{\gamma }_{52}{\gamma }_{11}){\gamma }_{43}+{\gamma }_{12}^2-{\gamma }_{11}^2]({\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}){\gamma }_{52}+[({\gamma }_{51}{\gamma }_{13}+{\gamma }_{53}{\gamma }_{11}){\gamma }_{42}-{\gamma }_{11}^2+{\gamma }_{13}^2]({\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}){\gamma }_{53}\}$

… Not sure this is headed anywhere useful!

Volume Element

$(h_1{h}_4{h}_5)d{\lambda }_{1}d{\lambda }_{4}d{\lambda }_5$	$=$	$(h_1{h}_4{h}_5)\left[\right(\frac{\partial {\lambda }_1}{\partial x})dx+(\frac{\partial {\lambda }_1}{\partial y})dy+(\frac{\partial {\lambda }_1}{\partial z}\left)dz\right]\left[\right(\frac{\partial {\lambda }_4}{\partial x})dx+(\frac{\partial {\lambda }_4}{\partial y})dy+(\frac{\partial {\lambda }_4}{\partial z}\left)dz\right]\left[\right(\frac{\partial {\lambda }_5}{\partial x})dx+(\frac{\partial {\lambda }_5}{\partial y})dy+(\frac{\partial {\lambda }_5}{\partial z}\left)dz\right]$
	$=$	$\left[\right({\gamma }_{11})dx+({\gamma }_{12})dy+({\gamma }_{13}\left)dz\right]\left[\right({\gamma }_{41})dx+({\gamma }_{42})dy+({\gamma }_{43}\left)dz\right]\left[\right({\gamma }_{51})dx+({\gamma }_{52})dy+({\gamma }_{53}\left)dz\right]$

COLLADA

Here we try to use the 3D-visualization capabilities of COLLADA to test whether or not the three coordinates associated with the T6 Coordinate system are indeed orthogonal to one another. We begin by making a copy of the Inertial17.dae text file, which we obtain from an accompanying discussion. When viewed with the Mac's Preview application, this group of COLLADA-based instructions displays a purple ellipsoid with axis ratios, (b/a, c/a) = (0.41, 0.385). This means that we are dealing with an ellipsoid for which,

$q\equiv \frac{a}{b}$

$=$

$2.44$

and,

$p\equiv \frac{a}{c}$

$=$

$2.60.$

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2};$
${\lambda }_2$	$=$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}};$
${\ell }_{3D}$	$\equiv$	$[x^2+q^4{y}^2+p^4{z}^2{]}^{-1/2};$
${𝒟}$	$\equiv$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

First Trial

First Trial (specified variable values have bgcolor="pink")
x	y	z	${\lambda }_1$	${\ell }_{3D}$	${𝒟}$
0.5	0.35493	0.00000	1	0.46052	2.11310

Unit Vectors

${\hat{e}}_1$	$=$	$\hat{ı}(x_0{\ell }_{3D})+\hat{ȷ}(q^2{y}_0{\ell }_{3D})+\hat{k}(p^2{z}_0{\ell }_{3D})$
	$=$	$\hat{ı}(0.23026)+\hat{ȷ}(0.97313);$
${\hat{e}}_2$	$=$	$\frac{x_0{q}^2{y}_0{p}^2{z}_0}{{𝒟}}\left\{\hat{ı}\right(\frac{1}{x_0})+\hat{ȷ}(\frac{1}{q^2{y}_0})+\hat{k}(-\frac{2}{p^2{z}_0}\left)\right\}$
	$=$	$-\hat{k}\left(\frac{2x_0{q}^2{y}_0}{{𝒟}}\right)$
	$=$	$-\hat{k}\left(1\right);$
${\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\{-\hat{ı}[x_0(2q^4{y}_0^2+p^4{z}_0^2)]+\hat{ȷ}[q^2{y}_0(p^4{z}_0^2+2x_0^2)]+\hat{k}[p^2{z}_0(x_0^2-q^4{y}_0^2)\left]\right\}$
	$=$	$\frac{2q^2{x}_0{y}_0{\ell }_{3D}}{{𝒟}}\{-\hat{ı}(q^2{y}_0)+\hat{ȷ}(x_0)\}=\left(1\right){\ell }_{3D}\{-\hat{ı}(q^2{y}_0)+\hat{ȷ}(x_0)\}$
	$=$	$-\hat{ı}(0.97313)+\hat{ȷ}(0.23026).$

Tangent Plane

From our above derivation, the plane that is tangent to the ellipsoid's surface at $(x_0,y_0,z_0)$ is given by the expression,

$x{x}_0+q^{2}y{y}_0+p^{2}z{z}_0$

$=$

$({\lambda }_1^2{)}_0.$

For this First Trial, we have (for all values of $z$, given that $z_0=0$) …

$(0.5)x+(2.11310)y$	$=$	$1$
$\Rightarrow y$	$=$	$\frac{(1-0.5x)}{2.11310}.$

So let's plot a segment of the tangent plane whose four corners are given by the coordinates,

Corner	x	y	z
A	x_0 - 0.25 = +0.25	0.41408	-0.25
B	x_0 + 0.25 = +0.75	0.29577	-0.25
C	x_0 - 0.25 = +0.25	0.41408	+0.25
D	x_0 + 0.25 = +0.75	0.29577	+0.25

Now, in order to give some thickness to this tangent-plane, let's adjust the four corner locations by a distance of $\pm 0.1$ in the ${\hat{e}}_1$ direction.

Eight Corners of Tangent Plane

Corner 1: Shift surface-point location $(x_0,y_0,z_0)$ by $(+\mathrm{\Delta }{e}_1)$ in the ${\hat{e}}_1$ direction, by $(+\mathrm{\Delta }{e}_2)$ in the ${\hat{e}}_2$ direction, and by by $(+\mathrm{\Delta }{e}_3)$ in the ${\hat{e}}_3$ direction. This gives …

$x_1$

$=$

$x_0+(\mathrm{\Delta }{e}_1)0.23026-(\mathrm{\Delta }{e}_2)0.97313$

Second Trial

Second Trial … $(q=2.44,p=2.60)$ [specified variable values have bgcolor="pink"]
x_0	y_0	z_0	${\lambda }_1$	${\ell }_{3D}$	${𝒟}$
0.5	0.35493	0.00000	1	0.46052	2.11310

Generic Unit Vector Expressions

Let's adopt the notation,

${\hat{e}}_i$

$=$

$\hat{ı}[e_{ix}]+\hat{ȷ}[e_{iy}]+\hat{k}[e_{iz}]$

for,

$i=1,3.$

Then, for the T6 Coordinate system, we have,

$e_{1x}$	$\equiv$	$x_0{\ell }_{3D};$		$e_{1y}$	$\equiv$	$q^2{y}_0{\ell }_{3D};$		$e_{1z}$	$\equiv$	$p^2{z}_0{\ell }_{3D};$
$e_{2x}$	$\equiv$	$\frac{q^2{y}_0{p}^2{z}_0}{{𝒟}};$		$e_{2y}$	$\equiv$	$\frac{x_0{p}^2{z}_0}{{𝒟}};$		$e_{2z}$	$\equiv$	$-\frac{2x_0{q}^2{y}_0}{{𝒟}};$
$e_{3x}$	$\equiv$	$-x_0(2q^4{y}_0^2+p^4{z}_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3y}$	$\equiv$	$q^2{y}_0(p^4{z}_0^2+2x_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3z}$	$\equiv$	$p^2{z}_0(x_0^2-q^4{y}_0^2)\frac{{\ell }_{3D}}{{𝒟}}.$

Second Trial
	x	y	z
$e_1$	0.23026	0.97313	0.0
$e_2$	0.0	0.0	-1.0
$e_3$	- 0.97313	0.23026	0.0

What are the coordinates of the eight corners of a thin tangent-plane? Let's say that we want the plane to extend …

• From $(-{\mathrm{\Delta }}_1)$ to $(+{\mathrm{\Delta }}_1)$ in the ${\hat{e}}_1$ direction … here we set ${\mathrm{\Delta }}_1=0.05$;
• From $(-{\mathrm{\Delta }}_2)$ to $(+{\mathrm{\Delta }}_2)$ in the ${\hat{e}}_2$ direction … here we set ${\mathrm{\Delta }}_2=0.25$;
• From $(-{\mathrm{\Delta }}_3)$ to $(+{\mathrm{\Delta }}_3)$ in the ${\hat{e}}_3$ direction … here we set ${\mathrm{\Delta }}_3=0.25$.

${\mathrm{\Delta }}_x$	$=$	${\mathrm{\Delta }}_1{e}_{1x}+{\mathrm{\Delta }}_2{e}_{2x}+{\mathrm{\Delta }}_3{e}_{3x}=-0.23177;$
${\mathrm{\Delta }}_y$	$=$	${\mathrm{\Delta }}_1{e}_{1y}+{\mathrm{\Delta }}_2{e}_{2y}+{\mathrm{\Delta }}_3{e}_{3y}=+0.10622;$
${\mathrm{\Delta }}_z$	$=$	${\mathrm{\Delta }}_1{e}_{1z}+{\mathrm{\Delta }}_2{e}_{2z}+{\mathrm{\Delta }}_3{e}_{3z}=-0.25000.$

Tangent Plane Schematic

vertex x y z   x y z 0† $x_0-|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.26823 0.24871 -0.25 1 $x_0-|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.26823 0.46115 -0.25 2 $x_0-|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.26823 0.24871 0.25 3 $x_0-|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.26823 0.46115 0.25 4 $x_0+|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.73177 0.24871 -0.25 5 $x_0+|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.73177 0.46115 -0.25 6 $x_0+|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.73177 0.24871 0.25 7 $x_0+|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.73177 0.46115 0.25

†In the figure on the left, vertex 0 is hidden from view behind the (yellow) solid rectangle.

Third Trial

GoodPlane01

Third Trial … $(q=2.44,p=2.60)$ [specified variable values have bgcolor="pink"]
x_0	y_0	z_0	${\lambda }_1$	${\ell }_{3D}$	${𝒟}$
0.8	0.24600	0.00000	1	0.59959	2.34146

Again, for the T6 Coordinate system, we have,

$e_{1x}$	$\equiv$	$x_0{\ell }_{3D};$		$e_{1y}$	$\equiv$	$q^2{y}_0{\ell }_{3D};$		$e_{1z}$	$\equiv$	$p^2{z}_0{\ell }_{3D};$
$e_{2x}$	$\equiv$	$\frac{q^2{y}_0{p}^2{z}_0}{{𝒟}};$		$e_{2y}$	$\equiv$	$\frac{x_0{p}^2{z}_0}{{𝒟}};$		$e_{2z}$	$\equiv$	$-\frac{2x_0{q}^2{y}_0}{{𝒟}};$
$e_{3x}$	$\equiv$	$-x_0(2q^4{y}_0^2+p^4{z}_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3y}$	$\equiv$	$q^2{y}_0(p^4{z}_0^2+2x_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3z}$	$\equiv$	$p^2{z}_0(x_0^2-q^4{y}_0^2)\frac{{\ell }_{3D}}{{𝒟}}.$

Third Trial
	x	y	z	${\mathrm{\Delta }}_{\mathrm{T}\mathrm{P}}$
$e_1$	0.47967	0.87745	0.0	0.02
$e_2$	0.0	0.0	-1.0	0.25
$e_3$	- 0.87753	0.47952	0.0	0.25

In constructing the Tangent-Plane (TP) for a 3D COLLADA display, we first move from the point that is on the surface of the ellipsoid, ${\overrightarrow{x}}_0=(x_0,y_0,z_0)=(0.8,0.246,0.0)$, to

vertex "m"	${\overrightarrow{P}}_m$	Components
		$x_m=\hat{ı}\cdot {\overrightarrow{P}}_m$	$y_m=\hat{ȷ}\cdot {\overrightarrow{P}}_m$	$z_m=\hat{k}\cdot {\overrightarrow{P}}_m$
0	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1-{\mathrm{\Delta }}_2{\hat{e}}_2-{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}-{\mathrm{\Delta }}_2{e}_{2x}-{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}-{\mathrm{\Delta }}_2{e}_{2y}-{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}-{\mathrm{\Delta }}_2{e}_{2z}-{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) - 0.25 (-0.87752) = 1.00979	0.24590 - 0.02 (0.87753) - 0.25 (0.47952) = 0.10847	- 0.25 (-1.0) = + 0.25
1	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1+{\mathrm{\Delta }}_2{\hat{e}}_2-{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}+{\mathrm{\Delta }}_2{e}_{2x}-{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}+{\mathrm{\Delta }}_2{e}_{2y}-{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}+{\mathrm{\Delta }}_2{e}_{2z}-{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) - 0.25 (-0.87752) = 1.00979	0.24590 - 0.02 (0.87753) - 0.25 (0.47952) = 0.10847	+ 0.25 (-1.0) = - 0.25
2	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1-{\mathrm{\Delta }}_2{\hat{e}}_2+{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}-{\mathrm{\Delta }}_2{e}_{2x}+{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}-{\mathrm{\Delta }}_2{e}_{2y}+{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}-{\mathrm{\Delta }}_2{e}_{2z}+{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) + 0.25 (-0.87752) = 0.56307	0.24590 - 0.02 (0.87753) + 0.25 (0.47952) = 0.34823	- 0.25 (-1.0) = + 0.25
3	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1+{\mathrm{\Delta }}_2{\hat{e}}_2+{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}+{\mathrm{\Delta }}_2{e}_{2x}+{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}+{\mathrm{\Delta }}_2{e}_{2y}+{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}+{\mathrm{\Delta }}_2{e}_{2z}+{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) + 0.25 (-0.87752) = 0.57103	0.24590 - 0.02 (0.87753) + 0.25 (0.47952) = 0.34823	+ 0.25 (-1.0) = - 0.25
4		0.8 + 0.02 (0.47952) - 0.25 (-0.87752) = 1.0290	0.24590 + 0.02 (0.87753) - 0.25 (0.47952) = 0.1436	- 0.25 (-1.0) = + 0.25
5		0.8 + 0.02 (0.47952) - 0.25 (-0.87752) = 1.0290	0.24590 + 0.02 (0.87753) - 0.25 (0.47952) = 0.1436	+ 0.25 (-1.0) = - 0.25
6		0.8 + 0.02 (0.47952) + 0.25 (-0.87752) = 0.59021	0.24590 + 0.02 (0.87753) + 0.25 (0.47952) = 0.38333	- 0.25 (-1.0) = + 0.25
7		0.8 + 0.02 (0.47952) + 0.25 (-0.87752) = 0.59021	0.24590 + 0.02 (0.87753) + 0.25 (0.47952) = 0.38333	+ 0.25 (-1.0) = - 0.25

Tangent Plane Schematic	Vertex Locations via Excel
$x_0=0.8,z_0=0.0,y_0=0.246,{\lambda }_1=1.0$
Tangent Plane Schematic
${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.25,{\mathrm{\Delta }}_3=0.25$

GoodPlane02

$x_0=0.075,z_0=0.0,y_0=0.4089,{\lambda }_1=1.0$

Tangent Plane Schematic

${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.25,{\mathrm{\Delta }}_3=0.25$

GoodPlane03

$x_0=0.25,z_0=0.20,y_0=0.33501,{\lambda }_1=1.0$
Tangent Plane Schematic	Tangent Plane Schematic
${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.25,{\mathrm{\Delta }}_3=0.25$	${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.10,{\mathrm{\Delta }}_3=0.25$
CAPTION:   The image on the right differs from the image on the left in only one way — ${\mathrm{\Delta }}_2$ = 0.1 instead of 0.25. It illustrates more clearly that the ${\hat{e}}_3$ (longest) coordinate axis is not parallel to the z-axis when $z_0\ne 0.$

GoodPlane04

$x_0=0.25,z_0=1/3,y_0=0.1777,{\lambda }_1=1.0$

Tangent Plane Schematic

${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.10,{\mathrm{\Delta }}_3=0.25$

Further Exploration

Let's set: $x_0=0.25,y_0=0.33501,z_0=0.2\Rightarrow {\lambda }_1=1.00000,{\lambda }_2=0.33521.$

$q\equiv \frac{a}{b}$

$=$

$2.43972$

and,

$p\equiv \frac{a}{c}$

$=$

$2.5974.$

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2};$
${\lambda }_2$	$=$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}};$
${\ell }_{3D}$	$\equiv$	$[x^2+q^4{y}^2+p^4{z}^2{]}^{-1/2};$
${𝒟}$	$\equiv$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

Next, let's examine the curve that results from varying $z$ while ${\lambda }_1$ and ${\lambda }_2$ are held fixed. From the expression for ${\lambda }_2$ we appreciate that,

$x$

$=$

$\frac{{\lambda }_2{z}^{2/p^2}}{y^{1/q^2}};$

and from the expression for ${\lambda }_1$ we have,

$x^2$

$=$

${\lambda }_1^2-q^2{y}^2-p^2{z}^2.$

Hence, the relationship between $y$ and $z$ is,

$[\frac{{\lambda }_2{z}^{2/p^2}}{y^{1/q^2}}{]}^2$	$=$	${\lambda }_1^2-q^2{y}^2-p^2{z}^2$
$\Rightarrow {\lambda }_2^2{z}^{4/p^2}$	$=$	$y^{2/q^2}[{\lambda }_1^2-q^2{y}^2-p^2{z}^2].$

Alternatively, $y$ $=$ $[\frac{{\lambda }_2{z}^{2/p^2}}{x}{]}^{q^2}.$ Hence, the relationship between $x$ and $z$ is, $x^2$ $=$ ${\lambda }_1^2-p^2{z}^2-q^2[\frac{{\lambda }_2{z}^{2/p^2}}{x}{]}^{2q^2}$ $\Rightarrow x^{2(q^2+1)}$ $=$ $x^{2q^2}[{\lambda }_1^2-p^2{z}^2]-q^2[{\lambda }_2{z}^{2/p^2}{]}^{2q^2}$ $\Rightarrow x^2$ $=$ $\left\{x^{2q^2}\right[{\lambda }_1^2-p^2{z}^2]-q^2[{\lambda }_2{z}^{2/p^2}{]}^{2q^2}{\}}^{1/(q^2+1)}$

Here are some example values …

${\lambda }_1=1$    and,     ${\lambda }_2=0.33521$
$z$	1st Solution			2nd Solution		lambda_3 coordinate
	$y_1$	$x_1$		$y_2$	$x_2$
0.01	0.407825695	0.0995168		-	-
0.03	0.40481851	0.138		-	-
0.04	0.40309223	0.1503934		-	-
0.08	0.393779065	0.1854283		-	-
0.12	0.37990705	0.2103761		-	-
0.16	0.36067787	0.23111		1.04123×10-4	0.9095546
0.2	0.33500747	0.2500033		2.23778×10-4	0.85448
0.22	0.31923525	0.2592611		3.36653 ×10-4	0.82065
0.24	0.30106924	0.2686685		5.2327 ×10-4	0.78192
0.26	0.2799962	0.2784963		8.53243 ×10-4	0.73752
0.28	0.25521147	0.2891526		1.491545 ×10-3	0.68634
0.3	0.22530908	0.3013752		2.89262 ×10-3	0.62671
0.32	0.1873233	0.3168808		6.6223 ×10-3	0.55579
0.34	0.13149897	0.3423994		2.09221 ×10-2	0.46637
0.343	0.1191543	0.3490285		0.026458	0.4496
0.344	0.1145	0.3517		0.02880	0.4435
0.345	0.1093972	0.354688		0.03155965	0.4371186
0.3485	0.0847372	0.3713588		0.0480478	0.4085204

See Also

• Direction Cosines