On the Not Spherical


The 19 Century Standpoint

Carl Gauss, living more than 200 years after John Napier thought Napier’s discovery was actually not so regular, but, in fact, rather eccentric!

In the beginning of Gauss' fragments, he is looking at the general trigonometric relationships in the construction. He gives the singular example of a "20" case (20 being the sum of the product of the squares of all of the tangents of the 5 sides of the pentagramma). It is rather irregular looking, on the sphere and on the plane, and it remains undetermined, by this author, if this is but an example (see an image of the projective cone from Gauss’ example Pentagramma Mirificum), although some believe that there is potentially a relationship to the platonic solids ( in addition to the fact that any Pentagramma Mirificum construction can be defined as a compound of 5 octahedra—see work by Ben.)

The hypotenuses of the triangles which bound the pentagon on the sphere in Gauss’ example are:

Hyp. 1- 71 deg. 33’ 56”
Hyp. 2- 39 deg. 13’ 54”
Hyp. 3- 54 deg. 44’ 7”
Hyp. 4- 65 deg. 54’ 19”
Hyp. 5- 30 deg.



Above is skeleton of the cone representing one possible projection from Gauss' "20" case. We can clearly see the teal line which delineates the unit sphere.


Gauss himself does not seem to give you the regular construction, but is given to us in an appendix to the published fragments written by a one R. Fricke, who compiled what is presented to us in Gauss’ Werke (he is rumored to have had access to more than what is published in the Werke on the Pentagramma Mirificum). Fricke says the side of the regular construction is the arcsine of the square route of the golden section (check)- about 52 degrees. The sine of the other side (about 38 degrees) is in the golden section proportion to the sine of 90 (Like this : Side of Cube: Side of Dodecahedron :: Sine of 90: Sine of 38 degrees. The sine of 38 squared will give you the sine of 52.

After much toil, this author determined that this construction is the only case of a pentagramma mirificum where the vertices lie on one plane and hence define a circular base perpendicular to the axis or axial triangle of the cone. The vertices lie on a circle of latitude (we'll call it that), of about 48 degrees.




Although it is yet unclear if Gauss himself makes this explicit, the curious reader can access Gauss’ fragments and reference the section where he establishes “limit values for t”—where we see something resembles the square route of 5; maybe the golden section. (However, it appears that even Gauss’ own example does not lie within his delineated “limit”!) What is the limit? We know that it does not define a range where multiple regular constructions can be found (since there is but one), nor does it define Pentagramma constructions for which all vertices would lie on a plane (the same thing, actually).

If we apply what we know from Kaestner’s lessons on spherical trigonometry, we see why this can't happen (that is, no other construction will have a circular base)




If the five points on the sphere do not define a circle then the equivalent of the elliptical cut of that cone (which in the pentagramma is the cut in which the projection is inscribed) is not necessarily an ellipse, is it?

The operation to determine from the PM vertices what the base of the cone would be is not easy to figure out, but we can flank it from underneath!






The Great Apollonius

From here we can delve into an investigation of conic sections. Do projections from all other constructions have the same property as the regular pentagramma projections? Do the others define an “elliptical cone” which in its oblique form will still inscribe the projection in an ellipse? Gauss himself was at one point seeking the simplest and “most elegant” construction of an ellipse from five points, which he brings up in two of the letters, roughly translated by this author. He uses the utterly confusing and in Gauss' words “magical” construction of Newton from the Principia as an example of what he doesn't mean.

Here is a very rough translation of a letter to one of Gauss’ collaborators, H.C. Schumacher, from 1843. Heinrich Christian Schumaker was an astronomer and a friend of Gauss; the two wrote letters to one another over a period of several decades. This one was written within weeks of the last PM fragments, which are dated (the first fragments are not). The letter in original German is easily accessible.

Dear Schumacher,

I am thankful to you my dear friend for your several letters and interesting communications. You must not, however, count on me to be precise in responding . Since 4-6 weeks I am (beginning with accidental circumstances) drawn into some mathematical speculations where I am again drawn through new prospects and turned in other directions, and have achieved much, but have also been much mistaken. In some such turmoil is one incapable (or at least I am), being drawn into some other subject, which also explains my late responses. These speculations considered ( for the most part familiar things since they pertain to methods worked on by me) last of all concerning the conic sections. To me it came back to my memory when I cam across a half century ago my first reading of Newton's Principia, most of which I found unsatisfactory, namely his treatment of theorems pertaining to the conic sections. But I read always with the feeling that through the learning I did not become the master of it, particularly the straight line which determines a conic section. Newton solves the problem:

(Gauss includes what he considers a more clear rendering of Newton, whose construction and markings he regards as very unhelpful-- since that was hard to translate, I'll include it straight from the horse's mouth)

From Newton's Principia-

Proposition XXII, Problem XIV

to describe a trajectory that shall pass through five given points. Let the five given points be A, B, C, P, D. From any one of these as A, to any other two as B, C which may be called the poles, draw the riestingly enough this construction embodies and is based on one of ght lines AB, AC and parallel to those the lines TPS, PRQ through the fourth point P. Then from the two poles B, C, draw through the fifth point d two indefinite lines BDT, CRT meeting with the last drawn lines TPS, PRQ the former with the former, the latter with the latter in T and R. Then drawing the right line tr parallel to TR cutting off from the right lines PT, PR any fragments Pt, Fr proportional to PT, PR and if through their extremities t, r and the poles BG the right lines Bt, Cr are drawn meeting in d, that point will be placed in the trajectory required. For that point D is placed in a conic section passing through the four points A, B, C, P, and the lines Rr, Tt vanishing , the point d comes to coincide with the point D. Wherefore the conic section passes through the five points A, B, C, P, D.

The same otherwise:

Of the given points join any 3 as A, B, C and abut two of them B, C as poles making the angle ABC, ACB of a given magnitude to revolve apply (?) the legs BA, CA first to the point D, then ?? to the point p and mark the points MN in which the other legs BL, CL intersect each other in both. This letter is amusing, but also confusing, because it seems to me that Apollonius had already figured out how to, in my opinion, “elegantly” construct an ellipse from five points. cases. Draw the indefinite right line MN, and those movable angles revolve about their poles B... in such a manner that the intersection which is now supposed to be m, of the legs BL, CL or BM, CM may always fall in that indefinite right line M... and the intersection which is now supposed to be d, of the legs BA, CA o BD, CD will describe the trajectory required ?AD. For, the point d will be placed in a conic section passing trough the points B, C and when the point comes to coincide with the points L, M, N the point d will be the tangent required. Hence also maybe be found the centers, diameters, and latera recta of the trajectories.


This construction is now admittedly extremely delicate! And as far as concerns this magic straight line, there are many questions to be answered, especially what relations this straight line has to the elements of the conic section. .. If one can determine with ease the position of the straight line and the 3 points ABC... Various things of this sort can I now know in a better way, but I do not know whether I myself can carry out the whole thing-- some other business has not allowed me to continue to work on this problem. Early on have you, if I'm not mistaken, buried yourself with the conic sections, certainly more than I. Already in the Monatliche Correspondence you cite 1810 Nov. s 508 you reference the elegance of the ellipse, and various theorems in this book. (fn. Puissant). Do you know this delicate construction which defined the midpoint of a conic section, without superfluous things? A very difficult question which I have not yet answered; this development I have not yet made--

Yours Truly,


(What are the “methods developed by myself”, and what does he mean in saying he is now in a position to know these things in a better way?) You can see from the cryptic Newton (which \this author was never able to construct per Newton's instructions/hand wavings) what Gauss means by “superfluous”. However, Gauss’ comments at the end may seem perplexing to all those students of the ellipse, as he seeks an elegant construction of the ellipse, and a proof to determine its center. Why would the brilliant Gauss, at a relatively old age, still seek answers to such problems? We can go back to the Greek geometers, mostly Apollonius and Pappus ( the reader can also reference some of the work done on Conics featured on the LaRouche Youth Movement’s Harmonies of the World website) to see what these ancients may have thought about such problems, and if their treatment could be deemed elegant.

Later on, Blaise Pascal (1663-1682) would examine the problem of the six point conic, but if you have no method by which to construct a sixth point, this does not seem to be of much use! The Pascal proof says that if six points lie on a conic, you will be able to achieve a certain geometrical construction, the “Hexagramma Mysticum”. Gauss also cites the the work by Colin MacLaurin (1698-1746) and Braikenridge(1700-1762), which seems similar to Pascal's theorem, although the advance may yet be unrevealed to this author’s eyes. (Work on these problems may also aid the reader in approaching Gauss’ reference to Jacobi later in the fragments, when he uses the Poncelet Porism to examine series of ellipses).

So as not to go off on a tangent, ( a la Poncelet) we will return a bit more directly to the problem. In Kepler's Harmonies of the World, he reviews in full the harmonic divisions of the circle. Something which was examined by the Greeks was an internal type of harmony present in the conic sections, between intersecting chords.



For example, if two chords intersect inside of a circle, the product of the two parts of one chords will equal the product of the two parts of another. In the conic sections, since they are in a projective relationship with one another, points of intersection will carry over, although the distances will be altered. Interestingly we still have this type of proportion in the ellipse, between a pair of chords which are parallel and intersected by a third.



The condition about parallel lines seems to be there so that we can actually construct the sixth point. The test for the ellipse would be, if we start from any
one of the five points, A, B, C, D, or E, will our point F be constructed in same way? Otherwise, we get something like the parabola:

For the parabola, we have: ( note that the data is rough, only included for the sake of an example)
AO*OB= .69, CO*OD= .80, CO '*O'D=.84, E0'*O”F= ..97



The ratio between these is about .8625, however for another set entirely the proportion would be different. If our first five points lie on an ellipse, locating other points on the curve would reveal a curve which is closed, unlike what we get with the parabola. A more thorough investigation of the conic sections would be necessary to say more about these relationships.

This author tried to complete the procedure ( which can be found in Apollonius’ Conics) to find the axes and center, and at the point where Pappus uses an equation of the second degree, ran into the problem of needing to define PQ' in terms of PQ.



Solve: PQ squared + x PQ – QV-Q'W=0

What is seemingly problematic is the issue of defining PQ' in some ratio to PQ, when the potential ratios seem to be infinite.

For all those elliptical experts try this yourself!!


The Elusive Elliptical Cone

In one way, this can enable us to tackle the problem of “those other cones” from the back door, by examining them in their oblique form. In another way, we may be left laughing at ourselves! Remember that only in the case of the regular or “Golden Pentagramma”, as some call it, is the “cone” a cone with a circular base (just to clarify, the cone has its apex at the center of the sphere, and its oblique base circumscribing the projection. The base which is perpendicular to the axial triangle, is in the case of the regular pentagramma mirificum, a circle). Perhaps the best way to investigate the other cones was to look at their oblique cuts, i.e., projections, since investigating their bases proved troublesome on the sphere.

Since any of the diagonals which are projected from the PM are 90 degree arcs, and, with the erection of the altitude lines intersecting at the “eye point” (point of tangency of the sphere) divided in two, any two portions any projected altitude line will have the same product. Why?

Tangent of alpha x Cotangent of Alpha (Tangent of Beta, its complement) = 1



This won’t greatly facilitate the Pappus construction, although it’s knowledge puts us in a position to determine many of the necessary elements.

Only chords which are divided at the Gauss eye point have two parts with a product equal to one. Although any of the projected vertices from the PM were 90 degrees apart, unless in the plane they are divided at the eye point, that line will not be divided into two parts, representing complementary tangents. This is only the case at the eye point. Such is the funny nature of tangents. The tangent of ninety degrees is infinite, but the tangent of, for example, 40 degrees and 50 degrees is graspable. So the construction of other points on the curve (O'F) will not be so simple as constructing the “complementary chord” to EO'.



One altitude line will at one end give us a vertex which lies on our ellipse, and at the other end, a point which is collinear with two other vertices. This property will exist for any projection from any PM. However, the point here is not so much how many points can we construct on the curve, in fact we only need six to figure out the other important elements, center and axes.

So, after this, although some insight has hopefully been gained, or questions posed, we return to our problem of the elliptical cone. In fact, since the orientation of the five projected pentagramma vertices will define a closed curve, they can define an ellipse, although to get the elements we will need at least another point.

The other cones are cones that have elliptical bases perpendicular to the axial triangle, and oblique cuts of them are still elliptical, although this could admittedly be demonstrated in a more direct fashion, for example, determining the base from the orientation of the points on the sphere itself, and from there determine what an oblique cut would be. A future challenge will be to construct the elliptical base of the cone once its oblique cut were defined.

We know from Kepler's lucid presentation in the Optics that the conics are in a projective relationship to one another, with a discontinuity between the ellipse and parabola. In projective geometry, things can be proved or examined from that standpoint, although it may seem to lack a quality of rigor we associate with other types of problems. I.e. if the base of the cone were defined as an oval, could it project as an ellipse? If you are holding an oval, be it a cut in something or an oval you've cut out, can you tilt it and get what looks like a 2-d ellipse?



It is rather amazing that there is an unlimited potential to generate different irregular pentagons which have their altitude lines all intersecting in one point. Think of the intersection of the altitude lines in a planar pentagon as being a derivative of the fact that they can intersect almost anywhere inside of the Spherical Pentagramma. Try to construct many irregular planar pentagons whose altitude lines intersect, it's damn hard! But with the PM, you could cheat and do it much easier. You could even say that there is an infinite potential to project such pentagons from the PM, and you would be right, without constructing anything! (Question- is this made even more clear with the use of complex numbers?)


Gauss, Pascal and Mobius

Gauss continues to look at permutations of the PM, referencing some work of Jacobi, which developed upon the idea of Poncelet's Porism, which is looking at circles, but because of their projective relationship, certain properties, like points of intersection will be invariant, so the ellipse can also be examined. It may be fun to get a little sidetracked by this and decided to figure out Pascal's theorem somewhere along with way. Given 6 points, why are the intersection points of intersecting chords collinear? (Why is it so mystified by Wikipedia et al?)



If we remember than Pascal is a student of Desargues, one of the father's of projective geometry, then how he would have gone about proving this theorem becomes clear. It involves knowing almost nothing about angles in the construction, but maintaining harmonic divisions. Even one of the editors of the Conics says that much of what we have from Desargues is in Apollonius. Gauss' student Mobius generalized Pascal's theorem, saying that if any two of the three points are collinear the third must be, because he understood that what determined this “mystical hexagram” wasn't mystical, but Desargues anharmonic ratios. Try to figure it out using angles etc., may put you in a bind!



Gauss also applies the barycentric coordinates of Mobius to the projection, and subsequently discusses it in a letter (not yet fully translated). Gauss discusses the Barycentric Calculus of Mobius, which gives one a very interesting insight into Gauss' thoughts at this late date . Mobius develops a very interesting physically determined coordinate system, and his Barycentric Calculus seems to be a very interesting thing to translate and work through. Mobius, an astronomer and student of Gauss( head of the Astronomy Dept. at the Univ. of Leipzig) solves the simple problem of identifying the location of one point with respect to a number of others. Rather than slapping a Cartesian grid on the plane, he says let the point P, which is sought, be the centroid or barycenter of the four other fundamental points or “net”. If ,for example, P is the center of gravity (Archimedes has some papers on this for polygons) of the net of points, the coordinates of P are(0,0,0,0). The weights at the “net of points” would differentiate as as P moved away from being the center of gravity.



Gauss, in the 1840's thought his student had really come up with something important. These were amongst his thoughts toward the end of his life, and 11 years before Riemann's Habilitation Dissertation. Here is a rough translation of part of a letter to Schumaker, which follows the previous one cited. Gauss is talking about the Barycentric Calculus.

Overall, in relation to any such new calculus, one can achieve through it nothing that wasn't able to be achieved without it. But if the essential nature of it is necessary, the use of it can only be seen in the same way that an inspiration of genius, which can be forced by no one can solve particular tasks which can't otherwise be solved but mechanically, and without the help of such an inspiration of genius, one is left powerless. ...

Here we have a perplexing and at the same time revealing insight into Gauss, give that it is only a part of one letter. The utility of a new calculus is not that it allows problems to be solved mechanically in a way which would otherwise be impossible, but it is of a conceptual utility like that of a creative insight.


Why the Eccentric Anomaly?



Gauss, in one of the last fragments, inscribes the projection in a circle, a la Kepler and determines the eccentric anomaly of different positions. Doesn't he already know how to find an orbit? From 3 observations? Probably he is not trying to solve some practical problem, and the investigation may be more in the vein of what Gauss is investigating with the Arithmetic Geometric mean, and elliptical series.

Although having such questions unanswered leaves my mind rather unsettled, we cannot ignore overall what Gauss did with Napier's construction as a creative act. Although the fragments seem to be dealing with seemingly different aspects, (at one point, looking through Gauss' letters and seeing one's which referred to surveying, you may ask if that wasn't related in some way) there is a unity in the overall thrust, which departs from a domain of Euclidean geometry to one which takes this self-polar characteristic of the sphere and seems to exhaust its implications, and we are left to think about much more than a set of rules.

If you have grasped some elements of this investigation, you too may be left with thoughts, seemingly unrelated, as some others have. Maybe you are left thinking about musical composition and a theme which “defines the universe” of the piece. With the first spherical triangle many possible relations are already well-defined.

If you were a believer in empty space or if you thought Gauss’ work was limited to a more sophisticated domain than that of sphaerics, hopefully you have been provoked to think otherwise!

As Lyndon LaRouche once said at a cadre school, there is no one “physical reality” because it exists in your mind and your idea of it changes! There is much more to figure out about the Pentagramma Mirificum, and even thoughts about how Gauss may have continued or concluded his investigation of the fragments if he had lived longer and kept working on it.

Given Lyndon LaRouche’s recent insistence on the superiority of spherical functions, Gauss’ treatment of the Pentagramma Mirificum appears to be a very apt example of that which seems to deviate so much at times from the purely spherical, that even old Napier may have been shocked!