Translation:Allgemeine Auflosung der Aufgabe die Theile einer gegebenen Flache auf einer andern gegebnen Flache so abzubilden, dass die Abbildung dem Abgebildeten in den kleinsten Theilen ahnlich wird.

by Carl Friedrich Gauss

1.

The nature of a curved surface is determined by an equation involving the coordinates $x,\ y,\ z$ relative to each of its points. By virtue of this equation, each of these three variables can be considered as a function of the other two. The generality is even greater if we introduce two new variables $t,\ u$ and represent each of the variables $x,y,z$ as a function of $t$ and $u,$ so that, at least in general, specific values of $t$ and $u$ always correspond to a specific point on the surface and vice versa.

2.

Let $X,\ Y,\ Z,\ T,\ U$ be quantities that play, with respect to a second surface, the same role as $x,\ y,\ z,\ t,\ u$ play with respect to the first.

3.

Representing the first surface onto the second means establishing a rule whereby each point on the first surface corresponds to a specific point on the second. This is achieved by stipulating that ${T}$ and ${U}$ are equal to certain functions of the two variables $t$ and $u.$ If one wants the representation to satisfy certain conditions, these functions can no longer be arbitrary. As $X,\ Y,\ Z$ then also become functions of $t$ and $u,$ these functions, in addition to the condition prescribed by the nature of the second surface, must also satisfy those required in the representation.

4.

The problem proposed by the Royal Society of Sciences requires that the representation be similar to the original in its smallest parts. The first step is to express this condition analytically.

By differentiating the functions of $t,\ u$ by which $x,\ y,\ z,\ {X},\ {Y},\ {Z}$ are expressed, we can derive the following equations:

${\begin{alignedat}{3}dx&=adt&&+a'du,\\dy&=bdt&&+b'du,\\dz&=cdt&&+c'du,\\d{X}&={A}dt&&+{A'}du,\\d{Y}&={B}dt&&+{B'}du,\\d{Z}&={C}dt&&+{C'}du,\end{alignedat}}$

The prescribed condition firstly demands that all infinitely small lines emanating from a point on the first surface and lying on that surface be proportional to the corresponding lines on the second surface, and secondly that the former lines include the same angles among themselves as the latter.

Such a linear element on the first surface has the expression

${\sqrt {(a^{2}+b^{2}+c^{2})dt^{2}+(aa'+bb'+cc')dtdu+(a'^{2}+b'^{2}+c'^{2})du^{2}}},$

and the corresponding element on the second surface is

${\sqrt {({A}^{2}+{B}^{2}+{C}^{2})dt^{2}+({A}{A}'+{B}{B}'+{C}{C}')dtdu+({A}'^{2}+{B}'^{2}+{C}'^{2})du^{2}}}.$

If these two elements are to be in a determined ratio, regardless of $dt$ and $du,$ then the three quantities

$a^{2}+b^{2}+c^{2},\qquad aa'+bb'+cc',\qquad a'^{2}+b'^{2}+c'^{2}$

must obviously be proportional to the following three

${A}^{2}+{B}^{2}+{C}^{2},\qquad {A}{A}'+{B}{B}'+{C}{C}',\qquad {A}'^{2}+{B}'^{2}+{C}'^{2}.$

If the ends of a second element on the first surface correspond to the values

$t,u\quad {\text{and}}\quad t+\delta t,u+\delta u,$

then the cosine of the angle that this element makes with the first is

${\frac {(adt+a'du)(a\delta t+a'\delta u)+(bdt+b'du)(b\delta t+b'\delta u)+(cdt+c'du)(c\delta t+c'\delta u)}{\sqrt {\left[(adt+a'du)^{2}+(bdt+b'du)^{2}+(cdt+c'du)^{2}\right]\left[(a\delta t+a'\delta u)^{2}+(b\delta t+b'\delta u)^{2}+(c\delta t+c'\delta u)^{2}\right]}}},$

and for the cosine of the angle between the corresponding elements on the second surface, we obtain a similar expression, where $a,\ b,\ c,\ a',\ b',\ c'$ are replaced by ${A},\ {B},\ {C},\ {A'},\ {B'},\ {C'}.$ The two expressions will be equal to each other when the proportionality in question occurs, and the second condition is therefore already included in the first, which a moment of reflection makes evident.

The analytical expression of the condition of our problem is thus as follows

${\frac {{A}^{2}+{B}^{2}+{C}^{2}}{a^{2}+b^{2}+c^{2}}}\ =\ {\frac {{A}{A}'+{B}{B}'+{C}{C}'}{aa'+bb'+cc'}}\ =\ {\frac {{A}'^{2}+{B}'^{2}+{C}'^{2}}{a'^{2}+b'^{2}+c'^{2}}}$

and this must be a finite function of $t$ and $u$ that we will set $=m^{2}.$ Thus $m$ expresses the ratio by which the linear quantities on the first surface are increased or decreased in their representation on the second (depending on whether $m$ is greater or less than $1$). In general, this ratio will vary with location. In the particular case where $m$ is constant, there will be complete similarity even in finite parts, and when moreover $m=1,$ perfect equality will occur, and each of the surfaces can be developed upon the other.

5.

If, for the sake of brevity, we set

$(a^{2}+b^{2}+c^{2})dt^{2}+2(aa'+bb'+cc')dtdu+(a'^{2}+b'^{2}+c'^{2})du^{2}=\omega ,$

we notice that the differential equation $\omega =0$ will have two integrals. Indeed, if we decompose the trinomial $\omega$ into two factors, linear with respect to $dt$ and $du,$ one or the other must be $=0,$ which will give two different integrals. One of the integrals will correspond to the equation

$0=(a^{2}+b^{2}+c^{2})dt+$

$+\left[aa'+bb'+cc'+i{\sqrt {(a^{2}+b^{2}+c^{2})(a'^{2}+b'^{2}+c'^{2})-(aa'+bb'+cc')^{2}}}\right]du$

(where we have written $i$ for brevity instead of ${\sqrt {-1}},$ since it is easy to see that the irrational part of the expression must be imaginary); the other integral corresponds to a similar equation in which $i$ has been replaced by $-i$.

Therefore, if the integral of the first equation is

$p+iq={\text{Const.,}}$

with $p$ and $q$ denoting real functions of $t$ and $u,$ then the other integral will be

$p-iq={\text{Const.,}}$

from which it follows by the very nature of the question that

$(dp+idq)(dp-idq)$

or equivalently

$dp^{2}+dq^{2}$,

must be a factor of $\omega ,$ and hence

$\omega =n(dp^{2}+dq^{2})$.

where $n$ is a finite function of $t$ and $u.$

Now let $\Omega$ denote the trinomial obtained by replacing $d{X},\ d{Y},\ d{Z}$ with their values at ${T},\ {U},\ d{T},\ d{U}$ in

$d{X}^{2}+d{Y}^{2}+d{Z}^{2}$.

Assume as above that the two integrals of the equation $\Omega =0$ are

${P}+i{Q}=0,$

${P}-i{Q}=0,$

and that

$\Omega ={N}(d{P}^{2}+d{Q}^{2}),$

where ${P},\ {Q},\ {N}$ are real functions of ${T}$ and ${U}.$ These integrations can obviously be carried out (apart from the general difficulties of integration) prior to the resolution of our main problem.

If we substitute functions of $t,\ u$ such that the condition of our main problem is fulfilled, then $\Omega$ transforms into $m^{2}\omega ,$ and we will have

${\frac {(d{P}+id{Q})(d{P}-id{Q})}{dp+idq)(dp-idq)}}={\frac {m^{2}n}{N}}$.

But it is easy to see that the numerator of the first member of this equation can only be divisible by the denominator when

$d{P}+id{Q}$ is divisible by $dp+idq$

and

$d{P}-id{Q}$ is divisible by $dp-idq,$

or when

$d{P}+id{Q}$ is divisible by $dp-idq$

and

$d{P}-id{Q}$ by $dp+idq.$

In the first case, it follows that $d{P}+id{Q}$ will vanish when $dp+idq=0,$ or else ${P}+i{Q}$ will be constant whenever $p+iq,$ is constant; that is, ${P}+i{Q}$ will simply be a function of $p+iq,$ and similarly ${P}-i{Q}$ will be a function of $p-iq.$ In the other case, ${P}+i{Q}$ will be a function of $p-iq,$ and ${P}-i{Q}$ a function of $p+iq.$ It is also easy to see that the converses of these conclusions are valid; that is, when we take for ${P}+i{Q},\ {P}-i{Q}$ (either respectively, or in the reverse order) functions of $p+iq,\ p-iq,$ the exact divisibility of $\Omega$ by $\omega ,$ and thus the required proportionality will occur.

Moreover, it is easy to see that if we set, for example,

${P}+i{Q}=f(p+iq),\qquad {P}-i{Q}=f'(p-iq).$

then the nature of the function $f'$ is determined by that of the function $f.$ Indeed, when among the constant quantities that the latter may contain, none of them are imaginary, the function $f'$ must be completely identical to $f,$ so that real values of $p,\ q$ always correspond to real values of ${P},\ {Q};$ otherwise $f'$ will only differ from $f$ insofar as in the imaginary elements of $f'$ we must replace $i$ everywhere by $-i.$

Consequently, we have

${\begin{alignedat}{3}P&={\tfrac {1}{2}}f(p+iq)&&+{\tfrac {1}{2}}f(p-iq),\\i{Q}&={\tfrac {1}{2}}f(p+iq)&&-{\tfrac {1}{2}}f(p-iq),\end{alignedat}}$

or equivalently, when the function $f$ is chosen completely arbitrarily [with imaginary constant elements chosen at will], ${P}$ will be equal to the real part and $i{Q}$ ($-i{Q}$ in the second solution) will be equal to the imaginary part of $f(p+iq),$ and consequently we can solve for ${T}$ and ${U}$ as functions of $t$ and $u.$

Therefore, the proposed problem has been solved in a complete and general manner.

6.

If $p'+iq'$ represents any determined function of $p+iq$ (with $p',\ q'$ being real functions of $p,\ q$), then it is easy to see that

$p'+iq'={\text{Const.}}\quad$ and $\quad p'-iq'={\text{Const.}}$

also represent integrals of the differential equation $\omega =0,$ since they are completely equivalent to the equations

$p+iq={\text{Const.}}\qquad {\text{and}}\qquad p-iq={\text{Const.}}$

Similarly, the integrals of the differential equation $\Omega =0,$

${P'}+i{Q'}={\text{Const.}}\qquad {\text{and}}\qquad {P'}-i{Q'}={\text{Const.}}$

will be completely equivalent to the equations

${P}+i{Q}={\text{Const.}}\qquad {\text{and}}\qquad {P}-i{Q}={\text{Const.}},$

if ${P'}-i{Q'}$ represents any determined function of ${P}+i{Q}$ (where ${P'},\ {Q'}$ are real functions of ${P},\ {Q}$). Hence it is clear that in the general solution given in the previous article, $p',\ q'$ can replace $p,\ q,$ and ${P'},\ {Q'}$ can replace ${P},\ {Q}.$ Although the solution of the problem does not gain any generality in this way, it is nevertheless sometimes more convenient in applications to use one of these forms instead of the other.

7.

If the functions obtained by differentiating the arbitrary functions $f,f'$ are denoted by $\varphi$ and $\varphi '$ respectively, so that

$df(v)=\varphi (v)dv,\qquad df(v)=\varphi _{1}(v)dv,$

then as a consequence of our general solution, we will have

${\frac {d{P}+id{Q}}{dp+idq}}=\varphi (p+iq),\qquad {\frac {d{P}-id{Q}}{dp-idq}}=\varphi '(p-iq),$

and consequently

${\frac {m^{2}n}{N}}=\varphi (p+iq)\varphi '(p-iq).$

The magnification ratio will therefore be determined by the formula

$m={\sqrt {{\frac {dp^{2}+dq^{2}}{\omega }}{\frac {\Omega }{d{P}^{2}+d{Q}^{2}}}\varphi (p+iq)\varphi _{1}(p-iq)}}$

8.

We will now clarify our general solution with some examples, both to fully illustrate the mode of application and to further highlight the nature of some of the facts presented.

Let us first consider two plane surfaces; we can then set

${\begin{alignedat}{3}&x&&=t,&&\qquad &&y&&=u,&&\qquad &&z&&=0,\\&{X}&&={T},&&\qquad &&{Y}&&={U},&&\qquad &&{Z}&&=0.\end{alignedat}}$

The differential equation

$\omega =dt^{2}+du^{2}=0$

gives here the two integrals

$t+iu=\ {\text{Const.}},\qquad t-iu=\ {\text{Const.}},$

and similarly the two integrals of the equation

$\Omega =d{T}^{2}+d{U}^{2}=0$

are as follows:

${T}+i{U}=\ {\text{Const.}},\qquad {T}-i{U}=\ {\text{Const.}},$

Therefore, the two general solutions of the problem are

(I)	${\begin{alignedat}{2}&{T}+i{U}&&=&&f(t+iu),&&\qquad &&{T}-i{U}&&=&&f(t-iu),\end{alignedat}}$

(II)	${\begin{alignedat}{2}&{T}+i{U}&&=&&f(t-iu),&&\qquad &&{T}-i{U}&&=&&f(t+iu).\end{alignedat}}$

This result can also be expressed as follows: Let the letter $f$ denote any function, then for ${X}$ we should take the real part of $f(x+iy),$ and for ${Y}$ we should take either the real or imaginary part (after removing the factor $i$).

If we use the letters $\varphi ,\ \varphi '$ in the sense explained in Article 7, and if we set

$\varphi (x+y)=\xi +i\eta ,\qquad \varphi '(x-y)=\xi -i\eta ,$

where $\xi$ and $\eta$ are obviously to be real functions of $x$ and $y,$ then for the first solution we will have

${\begin{alignedat}{2}&d{X}+id{Y}&&=&&(\xi +i\eta )(dx+idy),\\&d{X}+id{Y}&&=&&(\xi +i\eta )(dx+idy),\\\end{alignedat}}$

and consequently

${\begin{alignedat}{2}&d{X}&&=&&\xi dx-\eta dy,\\&d{Y}&&=&&\eta dx+\xi dy.\end{alignedat}}$

Now, if we set

${\begin{alignedat}{3}&\xi &&=&&\sigma \cos \gamma ,&&\qquad &&\eta &&=&&\sigma \sin \gamma ,\\&dx&&=&&ds\cos g,&&\qquad &&dy&&=&&ds\sin g,\\&d{X}&&=&&d{S}\cos G,&&\qquad &&dY&&=&&d{S}\sin G,\end{alignedat}}$

so that $ds$ represents a linear element on the first plane, $g$ its inclination to the x-axis, $d{S}$ the corresponding linear element on the second plane, and ${G}$ its inclination to the x-axis, the previous equations yield

${\begin{alignedat}{2}&d{S}\cos {G}&&=&&\sigma ds\cos(g+\gamma ),\\&d{S}\sin {G}&&=&&\sigma ds\sin(g+\gamma ),\end{alignedat}}$

and if $\sigma$ is considered to be positive, which is allowed, then we have

$d{S}=\sigma ds,\qquad {G}=g+\gamma .$

We see, therefore (in accordance with Article 7), that $\sigma$ represents the magnification ratio of the element $ds$ in the representation $dS$, and that it is, as it should be, independent of $g;$ similarly, the fact that the angle $\gamma$ is independent of $g$ shows that all linear elements originating from a point on the first plane will be represented by elements on the second plane, forming the same angles as the former, and, we may add, in the same sense.

If we choose a linear function for $f$, so that $f(v)={A}+{B}v$, where the constant coefficients are of the form

${A}=a+bi,\quad {B}=c+ei,$

then we will have

$f(v)=c+ei$

and consequently,

$\sigma ={\sqrt {c^{2}+e^{2}}},\quad \gamma =\operatorname {arctan} {\frac {e}{c}}.$

The magnification ratio is thus constant at all points, and the representation is everywhere similar to the original.

For any other function $f$ (as is easy to demonstrate), the magnification ratio will not be constant, and consequently, similarity will only occur in the smallest parts.

If the points that must correspond to a given number of points given in the first plane are assigned on the representation, one can easily, using the ordinary method of interpolation, find the simplest algebraic function that fulfills this condition. Indeed, if we denote the values of $x+iy$ for the given points by $a,b,c$ etc., and the corresponding values of ${X}+i{Y}$ by ${A},{B},{C}$ etc., we must set

$f(v)={\frac {(v-b)(v-c)\dots }{(a-b)(a-c)\dots }}{A}+{\frac {(v-a)(v-c)\dots }{(b-a)(b-c)\dots }}{B}+{\frac {(v-a)(v-b)\dots }{(c-a)(c-b)\dots }}{C}+\dots \ {\text{etc.}}$

which is an algebraic function of $v$ whose order is one less than the number of assigned points. In the case of two points, where the function is linear, complete similarity is thus achieved.

This method can be usefully applied in geodesy to improve a map made from moderately accurate measurements, good in small details but slightly distorted on a large scale, when one knows the exact position of a certain number of points. It goes without saying, however, that one can hardly venture far from the regions surrounding these points.

If we treat the second solution in the same way, we find that the only difference lies in the fact that the similarity is inverse; all elements on the representation make angles with each other equal in magnitude to those of the original, but in the opposite sense, so that what was on the right is now located on the left. This distinction is not essential, and it disappears if on one of the planes the side previously considered to be the upper one is now taken as the lower. Moreover, one can always apply this remark, when one of the two surfaces is a plane; thus, in the examples of this nature that follow, we can simply adhere to the first solution.

9.

Let us now consider (as a second example) the representation of the surface of a right cone on the plane. Let us take the equation of the first surface as

$x^{2}+y^{2}-k^{2}z^{2}=0,$

and then set

${\begin{alignedat}{2}x&=kt\cos u,\\y&=kt\sin u,\\z&=t,\end{alignedat}}$

and as before,

${X}={T},\quad {Y}={U},\quad {Z}=0.$

Then the differential equation

$\omega =(k^{2}+1)dt^{2}+k^{2}t^{2}du^{2}=0$

gives the two integrals

$\log t\pm i{\sqrt {\frac {k^{2}}{k^{2}+1}}}u={\text{Const.}}$

Thus, we have the solution

${\begin{alignedat}{2}{X}+i{Y}&=f\left(\log t+i{\sqrt {\frac {k^{2}}{k^{2}+1}}}u\right),\\{X}-i{Y}&=f\left(\log t-i{\sqrt {\frac {k^{2}}{k^{2}+1}}}u\right);\end{alignedat}}$

or in other words, if $f$ denotes an arbitrary function, we take for ${X}$ the real part of

$f\left(\log t+i{\sqrt {\frac {k^{2}}{k^{2}+1}}}u\right)$

and for ${Y}$ the imaginary part, after removing the factor $i$.

For example, if we take $f$ to be an exponential function,

$f(v)=he^{v}$

where $h$ is a constant, and where $e$ denotes the base of the hyperbolic logarithm, then we have the simplest representation

${X}=ht\cos \left({\sqrt {\frac {k^{2}}{k^{2}+1}}}u\right),\quad {Y}=ht\sin \left({\sqrt {\frac {k^{2}}{k^{2}+1}}}u\right).$

Applying the formulas of Article 7, we have here

${\begin{alignedat}{2}n&=(k^{2}+1)t^{2},\\{N}&=1.\end{alignedat}}$

And, since $\varphi (v)=\varphi '(v)=he^{v},$ we have

$\varphi \left(\log t+i{\sqrt {\frac {k^{2}}{k^{2}+1}}}u\right)\varphi '\left(\log t-i{\sqrt {\frac {k^{2}}{k^{2}+1}}}u\right)=h^{2}t^{2},$

and consequently, the magnification ratio will be

$m={\frac {h}{\sqrt {k^{2}+1}}}$

and thus constant. If we further set

$h={\sqrt {k^{2}+1}},$

then the representation becomes a complete development.

10.

Thirdly, let us represent the surface of the sphere of radius $a$ on a plane. Here we set

${\begin{alignedat}{2}x&=a\cos t\sin u,\\y&=a\sin t\sin u,\\z&=a\cos u,\end{alignedat}}$

from which we obtain

$\omega =a^{2}\sin ^{2}udt^{2}+a^{2}du^{2}.$

The differential equation $\omega =0$ then gives

$dt\mp i{\frac {du}{\sin u}}=0,$

and integrating this gives

$t\pm i\log \ \operatorname {cotang} {\tfrac {1}{2}}u={\text{Const.}}.$

Therefore, if we continue to let $f$ denote an arbitrary function, we should take for ${X}$ the real part and for $i{Y}$ the imaginary part of

$f\left(t\pm i\log \ \operatorname {cotang} {\tfrac {1}{2}}u\right).$

We intend to treat some special cases of this general solution.

If we choose for $f$ a linear function, setting $f(v)=kv,$ we have

${X}=dt,\quad {Y}=k\log \ \operatorname {cotang} {\tfrac {1}{2}}u.$

Applied to the Earth, where $t$ denotes the geographical longitude and $90^{\circ }-u$ the latitude, this representation is clearly nothing other than the Mercator projection. For the scale factor, the formulas from Article 7 give

$m={\frac {k}{a\sin u}}.$

If we choose for $f$ an imaginary exponential function, starting with the simplest one $f(v)=ke^{iv}$, we have

${\begin{alignedat}{2}f\left(t+i\log \operatorname {cotang} {\tfrac {1}{2}}u\right)&=ke^{-\log \operatorname {tang} {\tfrac {1}{2}}u+it},\\&=k\operatorname {tang} {\tfrac {1}{2}}u\ (\cos t+i\sin t),\end{alignedat}}$

and

${X}=k\operatorname {tang} {\tfrac {1}{2}}u\cos t,\quad {Y}=k\operatorname {tang} {\tfrac {1}{2}}u\sin t;$

which is easily recognized as the stereographic projection.

If we generally set $fv=ke^{i\lambda v},$ we have:

${X}=k{\operatorname {tang} {\tfrac {1}{2}}u}^{\lambda }\cos \lambda t,\quad {Y}=k{\operatorname {tang} {\tfrac {1}{2}}u}^{\lambda }\sin \lambda t.$

For the scale factor, we obtain

$n=a^{2}\sin ^{2}u,\quad {N}=1,\quad \varphi (v)=i\lambda ke^{i\lambda v},$

hence

$m={\frac {\lambda k{\operatorname {tang} {\tfrac {1}{2}}u}^{\lambda }}{a\sin u}}.$

We see that here the representations of all points for which $u$ is constant occur along a circle, and the representations of all points for which $t$ is constant along a straight line, moreover, the circles corresponding to all different values of $u$ are concentric. This provides a very useful projection for maps when it is only necessary to represent a part of the surface of the sphere, and then the best thing to do is to choose $\lambda$ so that the scale factor is the same for the extreme values of $u,$ so that it takes its smallest value towards the middle. If the extreme values of $u$ are $u^{0}$ and $u',$ then we should set

$\lambda ={\frac {\log \sin u'-\log \sin u^{0}}{\log \operatorname {tang} {\tfrac {1}{2}}u'-\log \operatorname {tang} {\tfrac {1}{2}}u^{0}}}.$

The sheets of celestial maps Nos. 19 to 26 by Prof. Harding are constructed using this projection.

11.

The general solution of the example treated in the previous article can also be presented in another form, which we think should be added here due to its elegance.

According to the considerations presented in Article 6, since

$\operatorname {tang} {\tfrac {1}{2}}u(\cos t+i\sin t)$

is a function of

$t+i\log \operatorname {cotang} {\tfrac {1}{2}}u,$

and

$\operatorname {tang} {\tfrac {1}{2}}u(\cos t+i\sin t)={\frac {\sin u\cos t+i\sin u\sin t}{1+\cos u}}={\frac {x+iy}{a+z}},$

the general solution can also be given by

${X}+i{Y}=f\left({\frac {x+iy}{a+z}}\right),\quad {X}-i{Y}=f'\left({\frac {x-iy}{a+z}}\right)\ ;$

meaning that ${X}$ should be set equal to the real part, and $i{Y}$ should be set equal to the imaginary part of $f\left({\frac {x+iy}{a+z}}\right),$ where $f$ denotes an arbitrary function. Instead of $f\left({\frac {x+iy}{a+z}}\right)$, it is easy to see that one can also take an arbitrary function of ${\frac {x+iz}{a+x}}$ or ${\frac {z+ix}{a+y}}.$

12.

Fourthly, let us consider the representation of the surface of an ellipsoid of revolution on the plane. Let $a$ and $b$ be the two principal semi-axes of the ellipsoid, so that we can set

${\begin{alignedat}{2}&x&&=a\cos t\sin u,\\&y&&=a\sin t\sin u,\\&z&&=b\cos u,\end{alignedat}}$

Then we have

$\omega =a^{2}\sin ^{2}udt^{2}+(a^{2}\cos ^{2}u+b^{2}\sin ^{2}u)du^{2},$

and the differential equation $\omega =0$, if we set ${\sqrt {1-{\tfrac {b^{2}}{a^{2}}}}}=\varepsilon$ for short (with the revolution semi-axis $b<a$), is

$0=dt\mp idu{\sqrt {\operatorname {cotang} ^{2}u+1-\varepsilon ^{2}}}.$

If we then set

${\sqrt {1-\varepsilon ^{2}}}\operatorname {tang} u\ =\ \operatorname {tang} w$

where, when applied to the terrestrial spheroid, $90^{\circ }-w$ represents the geographical latitude and $t$ the longitude, this equation transforms into

$0=dt\mp idw{\frac {1-\varepsilon ^{2}}{(1-\varepsilon ^{2}\cos ^{2}w)\sin w}}$

the integration of which gives

${\text{Const.}}=t\pm i\log \left[\operatorname {cotang} {\frac {1}{2}}w\left({\frac {1-\varepsilon \cos w}{1+\varepsilon \cos w}}\right)^{{\tfrac {1}{2}}\varepsilon }\right].$

Therefore, with $f$ denoting an arbitrary function, we must take for ${X}$ the real part and for $i{Y}$ the imaginary part of

$f\left(t+i\log \left[\operatorname {cotang} {\tfrac {1}{2}}w\left({\frac {1-\varepsilon \cos w}{1+\varepsilon \cos w}}\right)^{{\tfrac {1}{2}}\varepsilon }\right]\right).$

If we choose a linear function for $f$, i.e., $fv=kv,$ we will have

${X}=kt,\quad {Y}=k\log \operatorname {cotang} {\frac {1}{2}}w-{\tfrac {1}{2}}k\varepsilon \log \left({\frac {1-\varepsilon \cos w}{1+\varepsilon \cos w}}\right),$

which provides a projection analogous to that of Mercator.

If, on the other hand, we choose an imaginary exponential function for $f$, i.e., $f(v)=ke^{i\lambda v},$ we will have

${\begin{alignedat}{2}&{X}&&=k{\operatorname {tang} {\tfrac {1}{2}}w}^{\lambda }\left({\frac {1-\varepsilon \cos w}{1+\varepsilon \cos w}}\right)^{{\frac {1}{2}}\varepsilon \lambda }\cos \lambda t,\\&{Y}&&=k{\operatorname {tang} ^{\lambda }{\tfrac {1}{2}}w}^{\lambda }\left({\frac {1-\varepsilon \cos w}{1+\varepsilon \cos w}}\right)^{{\frac {1}{2}}\varepsilon \lambda }\sin \lambda t,\end{alignedat}}$

which, for $\lambda =1$, provides a projection analogous to the stereographic polar projection, which in general is very advantageous for representing a part of the earth's surface in cases where flattening needs to be taken into account.

As for what remains to be said about the other case where $b>a,$ it would indeed be easy to deduce it from the preceding discussion where, while retaining the same notation, epsilon is imaginary, but where $\left({\frac {1+\varepsilon \cos w}{1-\varepsilon \cos w}}\right)^{{\frac {1}{2}}\varepsilon }$ will still be real. However, to be complete, let us indicate the formulas specific to this case, and to begin, let us set ${\sqrt {{\tfrac {b^{2}}{a^{2}}}-1}}=\eta .$ We must then determine $w$ using the equation

${\sqrt {1+\eta ^{2}}}\operatorname {tang} u=\operatorname {tang} w,$

and the differential equation

$0=dt\mp idw{\frac {1+\eta ^{2}}{(1+\eta ^{2}\cos ^{2}w)\sin w}}$

yields the integral

${\text{Const.}}=t\pm \left[\log \operatorname {cotang} {\tfrac {1}{2}}w+\eta \ \operatorname {arc\ tang} (\eta \cos w)\right],$

so we must take for ${X}$ the real part and for $i{Y}$ the imaginary part of

$f\left(t+i\left[\log \operatorname {cotang} {\tfrac {1}{2}}w+\eta \ \operatorname {arc\ tang} (\eta \cos w)\right]\right).$

We can then immediately derive the analogues of the two particular cases considered earlier. For the first one, we must set

${X}=kt,\quad {Y}=k\log \operatorname {cotang} {\frac {1}{w}}+\eta k\operatorname {arctang} (\eta \cos w),$

and for the second one

${\begin{alignedat}{2}&{X}&&=k{\operatorname {tang} {\tfrac {1}{2}}w}^{\lambda }e^{-\eta \lambda \operatorname {arctang} (\eta \cos w)}\cos \lambda t,\\&{Y}&&=k{\operatorname {tang} {\tfrac {1}{2}}w}^{\lambda }e^{-\eta \lambda \operatorname {arctang} (\eta \cos w)}\sin \lambda t,\end{alignedat}}$

13.

As a final example, we will discuss the general representation of the surface of an ellipsoid of revolution on that of the sphere.

We will retain the notations for the ellipsoid from the previous article, and denote the semi-diameter of the sphere by ${A}$, and we will set

${\begin{alignedat}{2}&{X}&&={A}\cos {T}\sin {U},\\&{Y}&&={A}\sin {T}\sin {U},\\&{Z}&&=\cos {U}.\end{alignedat}}$

Applying the general solution from Article 5, with $f$ denoting an arbitrary function, we find that we must set ${T}$ equal to the real part and $i\log \operatorname {cotang} {\tfrac {1}{2}}{U}$ equal to the imaginary part of

$f\left(t+i\log \left[\operatorname {cotang} {\tfrac {1}{2}}w\left({\frac {1-\varepsilon \cos w}{1+\varepsilon \cos w}}\right)^{{\frac {1}{2}}\varepsilon }\right]\right).$^[1]

The simplest solution is obtained by setting $f(v)=u,$ hence

${T}=t,\quad \operatorname {tang} {\tfrac {1}{2}}{U}=\operatorname {tang} {\tfrac {1}{2}}w\left({\frac {1+\varepsilon \cos w}{1-\varepsilon \cos w}}\right)^{{\frac {1}{2}}\varepsilon }.$

This provides an extremely useful transformation in higher geodesy, of which we can only indicate a few details here in passing. In particular, if we consider points on the ellipsoid's surface and on the sphere's surface as corresponding when they have the same longitude and their respective latitudes, $90^{\circ }-{U},\ 90^{\circ }-w,$, are related by the equation above, then a system of relatively small triangles (which will always be the ones that can be used for actual measurements), formed on the surface of the ellipsoid by geodesic lines, will correspond to a system of triangles on the surface of the sphere, whose angles are exactly equal to the corresponding angles on the ellipsoid, and whose sides differ so little from arcs of great circles that, in most cases where the strictest accuracy is not required, they can be regarded as such; however, when the greatest accuracy is necessary, the deviation from the arc of a great circle can be easily evaluated with all the necessary precision using simple formulas. The entire system can be calculated using angles, as if it were on the sphere itself (after appropriately transferring one side of one of the triangles to the sphere, and making the necessary adjustments), then determining the values of ${T}$ and ${U}$ for all points in the system, and from these latter values one can go back to the corresponding values of $w$ (the simplest way to do this will be to use an auxiliary table, which is easy to construct).

As a network of triangles only extends over a very small part of the Earth's surface, the desired goal will be achieved even more completely if we further generalize the general solution by taking not $f(v)=v$ but $f(v)=v+{\text{Const.}}$ Obviously, there would be absolutely nothing gained if we attributed a real value to this constant, since then ${T}$ and $t$ would simply differ by that constant, and consequently only the origins of longitudes would change. But it is quite different when the constant is attributed an imaginary value. If we set it $=i\log k,$ then we will have

${T}=t,\qquad \operatorname {tang} {\tfrac {1}{2}}{U}=k\operatorname {tang} {\tfrac {1}{2}}w\left({\frac {1+\varepsilon \cos w}{1-\varepsilon \cos w}}\right)^{{\frac {1}{2}}\varepsilon }.$

In order to recognize the most suitable value of $k$ here, we must first determine the magnification ratio.

In this case, with the notation of Articles 5 and 6, we have

${\begin{alignedat}{2}n&=a^{2}\sin ^{2}u,\\{N}&={A}^{2}\sin ^{2}{U},\\\varphi v&=1.\end{alignedat}}$

Moreover,

${\begin{alignedat}{2}&m&&={\frac {{A}\sin {U}}{a\sin u}}={\frac {{A}\sin {U}}{a\sin w}}{\sqrt {1-\varepsilon ^{2}\cos ^{2}w}}\\&&&={\frac {A}{a}}{\frac {k(1-\epsilon ^{2}\cos ^{2}w)^{{\frac {1}{2}}+{\frac {1}{2}}\varepsilon }}{\cos ^{2}{\frac {1}{2}}w(1-\varepsilon \cos w)^{2}+k^{2}\sin ^{2}{\frac {1}{2}}w(1+\varepsilon \cos w)^{2}}},\end{alignedat}}$

which consequently depends only on the latitude. The smallest possible deviation from complete similarity is obtained if we determine $k$ so that $m$ takes equal values for the extreme latitudes, which will clearly cause $m$ to take a value very close to its maximum or minimum for the average latitude. If we denote the extreme values of $w$ by $w^{0}$ and $w',$ we obtain in this way

$k={\sqrt {\frac {{\frac {\cos ^{2}{\frac {1}{2}}w^{0}(1-\varepsilon \cos w^{0})^{\varepsilon }}{(1-\varepsilon ^{2}\cos ^{2}w^{0})^{{\frac {1}{2}}+{\frac {1}{2}}\varepsilon }}}-{\frac {\cos ^{2}{\frac {1}{2}}w'(1-\varepsilon \cos w')^{\varepsilon }}{(1-\varepsilon ^{2}\cos ^{2}w')^{{\frac {1}{2}}+{\frac {1}{2}}\varepsilon }}}}{{\frac {\sin ^{2}{\frac {1}{2}}w'(1+\varepsilon \cos w')^{\varepsilon }}{(1-\varepsilon ^{2}\cos ^{2}w')^{{\frac {1}{2}}+{\frac {1}{2}}\varepsilon }}}-{\frac {\sin ^{2}{\frac {1}{2}}w^{0}(1-\varepsilon \cos w^{0})^{\varepsilon }}{(1-\varepsilon ^{2}\cos ^{2}w^{0})^{{\frac {1}{2}}+{\frac {1}{2}}\varepsilon }}}}}}$

To determine at which latitude $m$ reaches its maximum or minimum value, we have

${\begin{alignedat}{2}{\frac {dm}{m}}&=\operatorname {cot} {U}d{U}-\operatorname {cot} wdw+{\frac {\varepsilon ^{2}\cos w\sin wdw}{1-\varepsilon ^{2}\cos ^{2}w}},\\{\frac {d{U}}{\sin {U}}}&={\frac {dw}{\sin w}}-{\frac {\varepsilon ^{2}\sin wdw}{(1-\varepsilon ^{2})\cos ^{2}w}}={\frac {(1-\varepsilon ^{2})dw}{(1-\varepsilon ^{2}\cos ^{2}w)\sin w}},\end{alignedat}}$

from which

${\frac {dm}{m}}={\frac {1-\varepsilon ^{2}dw}{(1-\varepsilon ^{2}\cos ^{2}w)\sin w}}(\cos {U}-\cos w).$

From this, it is clear that $m$ reaches its maximum or minimum value when ${U}=w;$ if we denote the value of $w$ at this point by ${W},$ we have

$k=\left({\frac {1-\varepsilon \cos {W}}{1+\varepsilon \cos {W}}}\right)^{{\frac {1}{2}}\varepsilon }\qquad {\text{or}}\qquad \cos {W}={\frac {1-k^{\frac {2}{\epsilon }}}{\varepsilon (1+k)^{\frac {2}{\varepsilon }}}}$

from which $W$ can be derived, once $k$ has been calculated using the previous formula. However, in practice, it matters little whether there is a perfectly rigorous equality of the values of $m$ at the extreme latitudes, and one can be content with approximately taking the average latitude for $90^{\circ }-{W}$, and deriving $k$ from there. The general relation between ${U}$ and $w$ is then given by the formula

$\operatorname {tan} {\tfrac {1}{2}}{U}=\operatorname {tan} {\tfrac {1}{2}}w\left\{{\frac {(1-\varepsilon \cos {W})(1+\varepsilon \cos w)}{(1+\varepsilon \cos {W})(1-\varepsilon \cos w)}}\right\}^{{\frac {1}{2}}\varepsilon }.$

For effective numerical computations, it is more advantageous to use series, which can take several forms, but we will not delve into their development here.

Moreover, as one can easily see, for $w<{W},$ we have $\ {U}>w,$ so $\cos {U}-\cos w$ and therefore ${\frac {dm}{dw}}$ is negative; for $w>{W},$ we have ${U}<w,$ and thus ${\frac {dm}{dw}}$ is positive. Thus it is clear that for $w={U}={W},$ the value of $m$ will always be a minimum, which moreover is

$={\frac {A}{a}}{\sqrt {1-\varepsilon ^{2}\cos ^{2}{W}}}.$

Therefore, if we take the radius of the sphere to be

$A={\frac {a}{\sqrt {1-\varepsilon ^{2}\cos ^{2}{W}}}},$

then the smallest parts of the representation of the ellipsoid for latitude $90^{\circ }-{W}$ are not only similar but also equal to the original; but for other latitudes, it is larger.

It is advantageous to expand the logarithm of $m$ into a series of powers of $\cos {U}-\cos {W},$ whose first terms, which suffice in practice, are as follows:

${\begin{alignedat}{3}&\log \operatorname {hyp.} m=&&&&\log \left({\frac {A}{a}}{\sqrt {1-\varepsilon ^{2}\cos ^{2}{W}}}\right)\\&&&+&&{\frac {\varepsilon ^{2}}{2(1-\varepsilon ^{2})}}(\cos {U}-\cos {W})^{2}\\&&&-&&{\frac {2\varepsilon ^{4}\cos {W}}{3(1-\varepsilon ^{2})^{2}}}(\cos {U}-\cos {W})^{3}+\dots \end{alignedat}}$

Therefore, if we map e.g. the Danish monarchy onto the surface of the sphere between the boundaries defined by latitudes $53^{\circ }$ and $58^{\circ },$ and if we set ${W}=34^{\circ }30',$ then, with a flattening of ${\tfrac {1}{303}},$ the representation at the boundaries, linearly evaluated, will only be enlarged by ${\tfrac {1}{530000}}.$

We must be content here to have given only a brief indication of one method of representing figures in higher geodesy, reserving a properly detailed exposition for another occasion.

14.

We still need to consider here in a slightly more detailed manner a circumstance that arises in our general solution. We have shown in Article 5 that there are always two solutions; either ${P}+i{Q}$ must be equal to a function of $p+iq$ and ${P}-i{Q}$ equal to a function of $p-iq,$ or ${P}+i{Q}$ must be equal to a function of $p-iq$ and ${P}-i{Q}$ to a function of $p+iq.$ We now want to further prove that in one of the solutions the parts of the representation have a position similar to that of the original, and that conversely in the other solution they are placed inversely. We also want to indicate a criterion by which this can be recognized a priori.

But before that, let us note that there can only be direct or inverse similarity if we distinguish two sides for each of the two surfaces, one of which will be regarded as the upper side and the other as the lower side. Since this distinction is somewhat arbitrary, the two solutions, in fact, are not essentially distinct, and inverse similarity becomes direct similarity, as soon as for one of the surfaces, the side previously considered as the upper is now regarded as the lower. Therefore, this distinction could not arise in our solution, since the surfaces are simply determined by the coordinates of their points. If one wishes to introduce this distinction, one must first determine the nature of the surface by another method that entails this distinction. To this end, we assume that the nature of the first surface is determined by an equation $\psi =0,$ where $\psi$ is a given uniform function of $x,\ y,\ z.$ Thus, at all points on the surface, the value of $\psi$ vanishes, and at all points in space that do not belong to the surface, it does not vanish. Therefore, when crossing the surface, the value of $\psi$ changes, in general at least, from positive to negative, and vice versa during the reverse passage; that is to say, on one side of the surface, the value of $\psi$ will be positive, and on the other, negative; we will consider the former side as the upper side, and the latter as the lower side. We will make the same conventions for the second surface, with its nature determined by an equation $\Psi =0,$ where $\Psi$ denotes a given uniform function of the coordinates ${X},\ {Y},\ {Z}.$ Differentiation then yields

${\begin{alignedat}{3}&d\psi &&=&&edx&&+gdy&&+hdz\\&d\Psi &&=&&{E}d{X}&&+{G}d{Y}&&+{H}d{Z},\end{alignedat}}$

where $e,\ g,\ h$ are functions of $x,\ y,\ z$ and ${E},\ {G},\ {H}$ of ${X},\ {Y},\ {Z}.$

Since the considerations we must employ for our envisaged purpose, although not inherently difficult, are of a somewhat unusual nature, we will strive to provide them with the greatest clarity. Between the two representations that correspond on the surfaces $\psi =0$ and $\Psi =0$, we consider six intermediate representations on the plane, so that we have to consider eight representations, namely:

${\begin{array}{llc}&&\scriptscriptstyle {\begin{array}{l}{\text{where points are regarded}}\\{\text{as corresponding if their}}\\{\text{respective coordinates are}}\end{array}}\\1^{st}&{\text{The original on the surface given by the equation }}\psi =0&x,\ y,\ z\\2^{nd}&{\text{The representation on the plane}}&x,\ y,\ 0\\3^{rd}&\qquad {\text{''}}&t,\ u,\ 0\\4^{th}&\qquad {\text{''}}&p,\ q,\ 0\\5^{th}&\qquad {\text{''}}&P,\ Q,\ 0\\6^{th}&\qquad {\text{''}}&T,\ U,\ 0\\7^{th}&\qquad {\text{''}}&X,\ Y,\ 0\\8^{th}&{\text{The representation on the surface given by }}\Psi =0&X,\ Y,\ Z\end{array}}$

We will now compare these various representations based only on the sense of their infinitely small linear elements, leaving the magnification ratio aside entirely; thus two representations will be considered to have the same sense when, given two linear elements issuing from a point, the one on the right in one representation corresponds to the one on the right in the other: otherwise, they will be said to have the opposite sense. For planes 2 through 7, the side where the positive value of the third coordinate is found will always be regarded as the upper side; for the first and last surfaces, on the other hand, the distinction between the upper and lower sides depends on where the values of $\psi$ and $\Psi$ are positive and negative, as agreed upon earlier.

Here it is clear that at each point on the first surface, where for constant $x$ and $y$, one moves to the upper side by a positive increase in $z,$ the representation on 2 will have the same sense as that on 1. This will obviously happen whenever $h$ is positive, and for negative $h$, the opposite will occur, and the representations will then have the opposite sense.

Similarly, the representations on 7 and 8 will have the same or opposite sense depending on whether ${H}$ is positive or negative.

To compare representations 2 and 3, consider on the first one an infinitely small line of length $ds$ drawn from the point with coordinates $x,\ y$ to another point with coordinates $x+dx,\ y+dy,$ and let $l$ be the angle that this line makes with the x-axis, measured in the direction from the x-axis to the y-axis, so that

$dx=ds\cos l,\ dy=ds\sin l.$

In representation 3, let $d\sigma$ be the length of the line corresponding to $ds,$ and let $\lambda$ be the angle it makes with the x-axis, counted as before, such that $dt=d\sigma \cos \lambda ,\ du=d\sigma \sin \lambda .$

Consequently, with the notation from Article 4, we have

${\begin{alignedat}{3}&ds\cos l&&=\ d\sigma (a\cos \lambda +a'\sin \lambda ),\\&ds\sin l&&=\ d\sigma (b\cos \lambda +b'\sin \lambda ),\end{alignedat}}$

and thus

$\operatorname {tang} l={\frac {b\cos \lambda +b'\sin \lambda }{a\cos \lambda +a'\sin \lambda }}.$

If we now consider $x$ and $y$ as constants and $l,\lambda$ as variables, differentiation gives us

${\frac {dl}{d\lambda }}={\frac {ab'-ba'}{(a\cos \lambda +a'\sin \lambda )^{2}+(b\cos \lambda +b'\sin \lambda )^{2}}}=(ab'-ba'){\frac {d\sigma ^{2}}{ds^{2}}}.$

Thus we see, that depending on whether $ab'-ba'$ is positive or negative, $l$ and $\lambda$ always increase in the same direction or vary inversely, and thus, representations 2 and 3 have the same sense in the former case and the opposite sense in the latter.

Combining this result with the one previously found, it is recognized that representations 1 and 3 have the same or opposite sense depending on whether ${\frac {ab'-ba'}{h}}$ is positive or negative.

Since on the surface with the equation $\psi =0$ we have

$edx+gdy+hdz=0,$

we will also have

$(ea+gb+hc)dt+(ea'+gb'+hc')du=0,$

no matter the ratio of $dt$ and $du,$ thus it is obviously identical,

$ea+gb+hc=0\qquad ea'+gb'+hc'=0.$

From this it follows that $e,\ g,\ h$ are proportional to the quantities $bc'-cb',\ ca'-ac',\ ab'-ba'$ respectively, and therefore

${\frac {bc'-cb'}{e}}\quad =\quad {\frac {ca'-ac'}{g}}\quad =\quad {\frac {ab'-ba'}{h}}.$

Therefore, as a criterion for whether the representations 1 and 3 have the same or opposite since, any of these three expressions, or alternatively, multiplying them by the essentially positive quantity $e^{2}+g^{2}+h^{2},$ the symmetric expression thus obtained

$ebc'+gca'+hab'-ecb'-gac'-hba'.$

Similarly, the representations 6 and 8 will have the same or opposite sense depending on whether

${\frac {BC'-CB'}{E}},\quad {\frac {CA'-AC'}{G}},\quad {\frac {AB'-BA'}{H}},$

or if preferred, the symmetric expression

${EBC'}+{GCA'}+{HAB'}-{ECB'}-{GAC'}-{HBA'}.$

has a positive or negative value.

The comparison of the senses of representations 3 and 4 relies on principles entirely similar to that of 2 and 3, and depends on the positive or negative sign of the quantity

${\frac {\partial p}{\partial t}}{\frac {\partial q}{\partial u}}-{\frac {\partial p}{\partial u}}{\frac {\partial q}{\partial t}}\ ;$

likewise, the positive or negative sign of

${\frac {\partial {P}}{\partial {T}}}{\frac {\partial {Q}}{\partial {U}}}-{\frac {\partial {P}}{\partial {U}}}{\frac {\partial {Q}}{\partial {T}}}$

will determine the relationship between the senses of representations 5 and 6.

Finally, concerning the comparison of representations 4 and 5, we can refer to the analysis of Article 5, which clearly shows that they have the same or opposite sense depending on whether the first or the second solution is chosen, that is to say, depending on whether we set

${P}+i{Q}=f(p+iq)\quad$ and $\quad {P}-i{Q}=f'(p-iq)$

or

${P}+i{Q}=f(p-iq)\quad$ and $\quad {P}-i{Q}=f'(p+iq).$

From all this, we conclude that when the representation on the surface with equation $\Psi =0$ must not only be similar in its smallest parts to the original with equation $\psi =0,$ but must also have the same sense, one must consider the number of negative quantities among the following four:

$\qquad {\frac {ab'-ba'}{h}},\quad {\frac {\partial p}{\partial t}}{\frac {\partial q}{\partial u}}-{\frac {\partial p}{\partial u}}{\frac {\partial q}{\partial t}},\quad {\frac {\partial {P}}{\partial {T}}}{\frac {\partial {Q}}{\partial {U}}}-{\frac {\partial {P}}{\partial {U}}}{\frac {\partial {Q}}{\partial {T}}},\quad {\frac {AB'-BA'}{H}}.$

If this number is zero or even, the first solution should be chosen; if among the above quantities one or three are negative, the second solution should be chosen. For the opposite choice, inverse similarity will always be found.

Moreover, it can be further demonstrated, denoting the four previous quantities respectively by $r,\ s,\ S,\ R,$ that we always have

${\frac {r{\sqrt {e^{2}+g^{2}+h^{2}}}}{s}}=\pm n,\qquad {\frac {R{\sqrt {{E}^{2}+{G}^{2}+{H}^{2}}}}{S}}=\pm {N},$

where $n$ and ${N}$ have the same meaning as in Article 5. We omit the easily found demonstration of this theorem; indeed, this theorem is not necessary for our intended purpose.

1. Here we will partly neglect the second solution from Article 5, which differs only in that $+{T}$ is replaced with $-{T},$ and would correspond to an inverse representation, and also the case of an elongated ellipsoid, which obviously reduces to that of the flattened ellipsoid according to what was done in the previous article.