Orange peel question

[Starman]

Can anyone help with this please?

I want to make a hollow wooden ball from segments cut from flexible board. The board will only curve in one direction – ie. You can’t make a dish with it.

What shape should the pieces be?

I’m looking for a formula with variables: (R) for the radius of the sphere, (N) for the number of segments, (W) the width of a segment, (L) for the distance from one end of the segment. So that I can plug in N and R and get the width at each value of L.

If I then mark a straight line on the flat board and mark off all the half widths at intervals I can cut N of these shapes, bend them and fasten the edges to make a sphere(oid). I want the joints to lie on the surface of the sphere.

I think the center line of the curved segment is elliptical (but I'm not sure) so I possibly need to calculate the length of curve of an ellipse from one of it's vertices.

[Guv]

Sorry, Starman, in theory you cannot do this. In practice, you might get a good approximation (see below).

A sphere has nonzero Gaussian (Total) curvature everywhere. This means that no finite piece of it can be flattened out to make a plane, and it cannot be made of of finite pieces of a plane without stretching or tearing the pieces.

Cylinders, cones, and some other surfaces have zero total curvature everywhere, and are called developable surfaces. Entire cones and cylinders (as well as pieces of them) can be created from a single piece of a flexible plane without any stretching or tearing of the plane.

Cylinders and cones can be flattened out to make a piece of a plane without any stretching or tearing. Some developable surfaces are more complicated than cylinders and cones. Perhaps they cannot be made from a single piece of a plane. It should be possible to make them from multiple pieces of a plane.

In practice, you might get a good approximation with pentagons and hexagons. Soccer balls are made this way. If you find a soccer ball and examine it, you will see the pattern. Some balls use two colors: One for the hexagons and one for the pentagons. A multicolored ball makes it easy to see the pattern. It is difficult to explain the shape verbally.

The basic shape is based on a regular dodecahedron, which is made up of 12 pentagons. I forget how many hexagons are used. The idea is that the hexagons and pentagons have sides which are the same length. Each pentagon seems to be surrounded by 5 hexagons. If you look at a soccer ball or a polyhedron made this way and focus on the pentagons, you can visualize a dodecahedron which has been exploded slightly in order to surround each face with 5 hexagons. It requires a lot less than less than 60 (= 12 * 5) hexagons, because each touches more than one pentagon.

I think there is more than one polyhedron made of pentagons and hexagons. I believe that 12 pentagons are always used. The soccer ball uses the minimum number of hexagons.

[Guv]

I did a little research. You need 20 hexagons and 12 pentagons to make the sphere-like polygon described in my previous post. With 32 faces, it would be a not bad approximation to a sphere.

If I wanted to make a hollow sphere out of wood, I would use a lathe to make two hemispherical shells and glue them together.

Another approach would be to make 20-60 pieces corresponding to the areas of the sphere between lines of constant longitude. Remember that you can only approximate a sphere by bending pieces of a plane. Each piece would be identical. Each piece would look like two smoothly curved lines that intersected. You could draw an approximation to the shape of each piece as follows.

• Set a compass to make a circle with a 5 inch radius.
• Draw a straight line 8 inches long.
• Put the compass point at one end of the line and draw an arc.
• Put the compass at the other end of the line and draw an arc.

If the arcs are long enough to intersect, the shape is similar to the shape of the pieces you would have to make. It is not the right shape, but will give you an idea of the correct shape.

Suppose you decided to make a sphere from 30 pieces similar to the curved piece described above. The concept is that the widest part of each piece corresponds to a piece of the equator of the sphere, while the points of each piece correspond to the north and south poles. After making the 30 pieces, you would bend them to form the sphere. If I were doing this, I would make some extra pieces and experiment with filing slanted edges to make the pieces fit after they were bent.

A more accurate size and shape for the pieces is as follows.

• From pointed tip to pointed tip, the pieces are Radius*Pi long (Half the circumference of the sphere). Draw a center line from tip to tip.
• At the widest point, the pieces are 2*Radius*Pi / 30 wide (1/30 of the circumference).
• The width decreases according to the cosine function.
• 2*Radius*Pi*Cosine(Distance / Radius) / 30 is the width formulae, where Distance is measured along the center line from the widest (or middle) part of the piece.

Pi is 3.14159 26535 89793 if you want to be picky about accuracy. If you want to use more or less pieces, substitute another number for 30 in the above formulae.

Note that when the pieces are bent to form a sphere, the bottom side of each piece must form a smaller sphere.

The radius of the inner sphere is the radius of the outer sphere minus the thickness of the wood. If you call this smaller radius Rho (Greek letter R), the above formulae can be used to determine the shape of the other side of each piece. Just substitute Rho for Radius in the above formulae. After making the pieces, you could put guide marks on the bottom side of each piece to aid in the filing or routing of slanted edges.

If I were attempting such a project, I would try some mass production method. Perhaps make metal templates for both sides, and stack 5-10 pieces to be cut at the same time.

Frankly, I would not have the patience to do all of the above accurately enough to make a satisfactory sphere, and I am pretty obsessive-compulsive. I have done many projects and am very handy with both wood working and metal working tools. There was a time when I had a complete set of power tools (Radial arm saw, table saw, several sizes of jig saws and band saws, two lathes, drill presses, planer, router, et cetera) and all sorts of hand tools. When I had all that equipment, I would not have tried this project. I might have tried making a hollow sphere in two pieces using a lathe.

Tell me about your experiences if you try this project.

[Starman]

Hi Guv,
Thanks for the reply.

It’s actually the ‘areas of the sphere between lines of constant longitude’ approach that I am after.
(The pentagons and hexagons idea is great, that may be the next project.)

It all stems from a question someone asked me about 20 years ago who wanted to make a domed ceiling in their bathroom. I have used reams of paper covered in wire frame looking balls and pieces that look like segments of oranges with latitudes, longitudes, triangles emanating from the center of the sphere and loads of trig equations.

Having spent so much time trying I thought it would be fitting to actually construct something roundish just to feel that I have the answer.

What I intend is that if for instance I cut 6 pieces ‘the overlap between two circles shape’ (I’m sure there is a name for this shape – is it a lunette ?) then the ‘ball’ would be hexagonal if viewed from the north pole. Viewed from the side, level with the equator and a flat towards you, you would see a circle and turning the ball 30 degrees (on the pole to pole axis) I think it becomes elliptical (with the pointy ends up and down )– ie. It’s an oval shape but I don’t know if it’s a mathematical ellipse - the tips may come to a point.

I have been ignoring the thickness of the wood due to the original question involving a bathroom and the wood being only about 4mm thick – a file and a bit of filler wont hurt.

It would be ideal to be able to calculate the radius of the two overlapping circles from the radius of the sphere and the number of sections.

Is there any chance now?

[Guv]

The shape you want cannot be made from two intersecting arcs of a circle. I described that shape to provide an approximate shape that can be visualized easily.

The correct shape is described by the formulae in my previous post. You must face some trig calculations which can easily be done using VB or a good hand calculator.

If I were doing this, I would one piece and use it as a template when making the remaining pieces.. There is a lot of calculating and measuring involved to make such a piece. If possible, use a thin sheet of metal for the template.

My intuition (often wrong) says that 6 pieces are not enough to make an accurate sphere. 30 seems about right to me, and that is a lot of work.

[Starman]

That’s not quite the right shape Guv,

The hexagon I mentioned is intended to be inscribed within a circle the radius of the sphere (at the equator).

Don’t worry about the actual number of components and the practical side of things (yet – your idea for calculating for the thickness of the wood and the edge angles is brilliant and I will probably use it).

If the Sphere is radius R, and N is the number of pieces then:

For the flat shapes I have the curved length of each side = Pi*R , when they are bent into shape these curved edges form the longitudes.

At the widest point they are R*sin (Pi/N) – chords across the equator.

At any point the width of the pieces are R * sin(Phi)*sin(Pi/N) where Phi is the angle from the center of the sphere to the point on the curved edge -latitudes.

But what I can’t find is the length of the center line (point to point) of the pieces or the relationship between the distance down the center line and Phi.

[Starman]

Sort of off topic Guv, but I have often wondered why people have two lathes - is it one for inboard and one for outboard or are they just completely different sizes?

[Guv]

First, I had two lathes because the one that could handle bigger items was awkward to use when working on small items. The larger one could be used like a screw machine as well as a lathe.

A screw machine is designed to allow working on a piece which is held only at one end. With a screw machine you can make a hollow hemisphere. When you are finished, you have a chunk of the wood (or metal) at the north pole which must be ground off. It was held by a chuck while you were working on the hemisphere. You can think of a screw machine as a "Chinese drill." Instead of the drill cutting a hole in something, you use tools to change the shape of the drill while it is rotating.

I think lunette is the correct name for the pieces we are talking about. The length of the center line of the lunette from tip to tip is half the circumference of the sphere (R*pi). In theory, this is also the length of the circular arcs forming the sides of the lunette. In a plane, this is impossible. On a sphere, it is normal. On a sphere, the lunette is formed by two longitudes from pole to pole. The center line of the lunette is also a pole to pole semi-circular line of longitude. If you sliced a lunette out of a paper sphere and tried to flatten it out on a plane, you would have to do some stretching or ripping. For a thinner lunette, less stretching/tearing is required.

As I said in a previous post, make a plane lunette with the length of the center line R*Pi. When the lunette is finished, measure the length of one of the circular arcs forming a side of the lunette. The difference between that length and R*Pi will indicate how well (or badly) the lunettes will approximate a sphere when bent. Use one of the following for the width.

• 2*Pi*R*Cosine(Distance / Radius) / N, if you measure distance from the center or widest part of the lunette.
• 2*R*Pi*Sine(Distance / Radius ) / N, if you measure distance from the tip of the lunette.

I think these width formulae are excellent approximations, better for larger N. Note that Distance / Radius is an angle in radians, which is what VB trig functions use as arguments. Note that both formulae give zero for the width at the tip, and 2*R*Pi / N for the widest part of the lunette.

[Starman]

Guv,

Thanks again for your time, and thank you for the description of a screw machine. I have never heard of one of these – it sounds like the cutter is rotating while the work is held fixed – is this right?

I have no doubt that when actually constructing a ball from flat sections, your formulae will be perfectly adequate, but I really would like to know the exact maths for the lunette - the flat area between two longitudes not the curve.

I have been using 6 sections as an example to deliberately accentuate the difference between the sphere and the ball-like shape so it is easier to visualise the center line of the lunette as being inside the sphere – it would be easier to explain if I could write posh like you.

I'm wondering again if the curved edges of the flat lunette do form arcs of circles – but I can’t see how to mathematically unravel them from the sphere ie. I know how long they are (Pi*R) – but when the lunette is flat I don’t know the chord length (tip-to-tip) although I have the height at centre – if I knew the chord length I could work out if these are circular arcs.

[Guv]

Starman: It has been a long time since I worked with power tools. It was a hobby, not my profession. My terminology might be incorrect. There is a device called an automatic screw machine, which is a high production metal working tool. Screw machine might imply automatic screw machine. There is also a device called a turret lathe, which might be the term for a "non-automatic" screw machine.

The basic idea with these devices is that the material being worked on rotates as for a lathe, but is supported at only one end, unlike a lathe which supports it at both ends. With support at only one end, you cannot make items like candle sticks and table legs. The advantage is that tools can be applied from more directions than with a regular lathe, allowing you to make all sorts of complex shapes, which would be impossible to make on a lathe. The tools for these devices are never hand held. There is a tool holder (called a turret) which moves on a track toward or away from the piece being worked on. Aside from motion on the track, the turret can be moved/rotated in various ways. The turret is designed to hold 5-10 tools, only one of which is active at a time. One axis of rotation is used to change the active tool.

I have seen automatic screw machines in action, but never worked with one. They used to be mechanically programmed to do dozens of automatically sequenced operations. I imagine that modern ones have programmable computer chips or a complete PC to control them.

On the subject of the lunettes, I am certain that the flat version is not made from circular arcs. A lunette sliced from a hollow spherical shell is a generatrix of circular arcs, but that is a 3D object. Relative to the center of the sphere, its edges are circular arcs, but I would not expect the flat lunette to have circular arcs as sides. For a 2D object, it is not at all clear what shape is the best approximation. It is my guess that the previous posted formulae describe a good approximation if the sphere is to be made up of 30 or more lunettes.

You seem to be planning to make a 3D object using six lunettes, which I did not realize until your most recent post. As you suggested, this object would have hexagonal cross sections for all planes parallel to the equator, and the center line of each lunette would be inside a circumscribing sphere. Since this object is a poor approximation to a sphere, I do not have confidence in my previously posted formulae.

Assuming that the lunettes are bent without stretching/tearing, the center lines of the lunettes would be on the surface of an inscribed sphere, while the sides of the lunettes would be on the surface of a circumscribing sphere. Using this information, we can come up with some numerical formulae. To help understand the formulae, imagine a regular polygon with N sides inscribed in the equatorial circle of a sphere, and a circle inscribed in the polygon. Some formulae are as follows (Trig functions assumed to use arguments in radians).

• Rho = Radius*cos(Pi / N), radius of inscribed circle/sphere.
• Chord = 2*Radius*sin(Pi / N), Length of a side of the polygon.
• Ark = 2*Pi*Radius / N , Length of equatorial arc from vertex to next vertex.
• SmallChord = 2*Radius*sin(Pi / N)*cos(D / R)
• SmallArk = 2*Pi*Radius*cos(D / N) / N
• Centerline = Pi*Radius*cos(Pi / N ), length of centerline of lunette. (Pi*Rho)
• LunetteSide = Pi*Radius, length of curved line forming side of lunette.

SmallArk and SmallChord relate to a polygon parallel to the equator and distance D from the equator along a line of longitude. They could be used to calculate the width of the lunette distance D from the widest part. Chord and SmallChord are probably better approximations than Ark and SmallArk. For large values of N, it does not matter. In a previous post, I gave Ark and SmallArk formulae for use in computing Lunette width.

If a perfect sphere could be made from lunettes, the following would be true.

• Rho = Radius
• Ark = Chord
• CenterLine = LunetteSide

By doing some calculations, we can estimate how good a sphere can be made for various values of N, by comparing values that would be equal for a perfect approximation. Following are some values I calculated for a sphere of radius 10. For your 6-Lunette object, the shape is not a good approximation to a sphere.

N = 6
Rho = 8.660254
Chord = 10.000
Ark = 10.4719755
Centerline = 27.206990
LunetteSide = 31.415927

N = 30
Rho = 9.948219
Chord = 2.09057
Ark = 2.094395
CenterLine = 31.243827
LunetteSide = 31.415927

N = 60
Rho = 9.986295
Chord = 1.046719
Ark = 1.047198
CenterLine = 31.372872
LunetteSide = 31.415927

I hope there are no typo's or miscalculations in the above. I tried to be careful, but the best laid plans ....

[Starman]

Hi Guv,

Sorry it's taking so long to get back on this but work seems to be getting in the way.

I'm afraid I'm not convinced that the center lines of the lunettes lie on the surface of the inscribed sphere. While they will be in contact at the equator, surely once you move towards the poles the center line must move away from the inner sphere's surface or the distance between the poles would be 2*Rho and not 2*R.

As another weak attempt at explaining the shape I am chasing, imagine a 'spherical' lampshade made up of six longitudenal circles of wire.
Wrap the lampshade with paper ribbon or whatever it is that they make lampshades with. Now cover one strip (lunette) with printers ink and roll the lampshade from South to North pole to make a print on a flat surface.
It is this print that I am trying to create.

If I made six of these prints, cut them out, rolled but not curled them, and fastened them together at the edges I will have a copy of the lampshade.
(This is not what I intend to do - just trying to explain the shape)

They say a picture is worth a thousand words - I reckon we should have a picture each very soon.

[Guv]

Starman: My analysis was in error. It is so obvious to me now that you pointed out the flaw in my logic. Hind sight is so perfect. I believe that I understand what you are trying to construct. After thinking about your last post, it seems that there are three possibilities for the final object.

• The lunette center lines are longitudes of an inscribed sphere, while the sides are longitudes of some larger non-spherical surface of revolution.
• The lunette sides are longitudes of a circumscribing sphere, while the center lines are longitudes of some smaller non-spherical surface of revolution.
• Neither are longitudes of a sphere.

I believe that the third alternative is always possible, but that it can be avoided.

If the lunettes are properly constructed, it is only a matter of bending them to meet at the proper polar points. If you do not bend them enough, you get prolate spheroids. If you bend them too much you get oblate spheroids. If you bend them the correct amount, either the lunette center lines or the sides are longitudes of a sphere, but not both.

I believe that my formulae might result in good approximations, but are incorrect. Better formulae could be developed by thinking in terms of circumscribing polygons instead of inscribed ones.

A formula for lunette widths must be based on lengths measured along either the center line or a side of the lunette. Note that neither the centerline nor the sides are circles when the lunette is flat. When bent, one or the other is expected to be a circle. When flat the lunette center line is known to be a straight line, but the shape of a side is far from obvious.

I am convinced that the correct approach is to force the lunette center lines to be longitudes of a sphere. It seems easier to work with the center line. The length of the center line is easy to control, and it known to be a straight line. Measuring along the center line from either the tip or the middle is not difficult. This approach does not require knowing the length of the sides nor the nature of their curvature.

If you try to design lunettes whose sides are longitudes of a sphere, you can calculate the length of the sides, but determining the shape becomes a problem. If I were to use this approach, I would try to determine the length of the center line and use it as the basis of construction. Once all the analysis is done, this is equivalent to the making the center lines longitudes, but the analysis seems to require more effort.

I will think about a new set of formulae, which are likely to be better than those previously posted. At present I am very busy with installing device drivers and application on my new system. I will probably find time to work on this and post in the next 2-3 days.

[Guv]

Starman: After a little thought, it looks as though the sides of polygons circumscribing circles of latitude could be used.

See previous posts. It seems right to make the centerline of the lunette the longitude of a sphere. The length of the centerline would be half the circumference of a sphere. This will cause the sides of the lunettes to be longitudes of a larger surface of revolution which is not a sphere. The lunettes sides will not be circular arcs.

Lunette width could be the the length of a side of a circumscribing polygon, instead of the length of an inscribed polygon. The formulae can be obtained by substituting the tan trig function for the sin function in the chord formulae previously posted.

I am very busy installing software on my new system, so I cannot elaborate on the above for the time being.

I did some calculations using my MathCad software. For 20-60 lunettes, the inscribed polygon sides, circular arcs, and circumscribing polygon sides vary in the 3rd or 4th significant digit. For 6 lunettes, they vary in the 1st or 2nd digit.

It seems that perhaps, it might be right to use the cicular arc formula for the lunette width function. For a large number of lunettes, it does not matter. For 6 lunettes, my intuition does not suggest which formulae to use, but I am fairly sure that the length of inscribed polygon sides is not correct.

[Starman]

Guv,

I know it probably makes things more difficult but the shape I am after is your second option:

> The lunette sides are longitudes of a circumscribing sphere, while the center lines are longitudes of some smaller non-spherical surface of revolution.


It is this non-spherical surface of revolution that I think is elliptical in cross section.
My best attempt to visualise it is to take the proposed 3d ball object and turn it 30 degrees (for N=6) on it’s polar axis so you have a line of longitude directly towards you.
If you ignore perspective/parallax then I believe the other two longitudes now line up with the center lines of two opposite lunettes. As the longitudes are circular and turned through an angle they will now appear elliptical.

It is about here that my maths fails me so I am appealing to someone with greater ability.

[Guv]

Starman: If you want the lunette sides to be circular longitudes of a sphere, you can calculate their length (assuming you know the radius of the sphere). Determining their shape looks very difficult to me. No matter how you approach this problem, for a flattened lunette the sides are not arcs of circles.

The following is an embellishment on descriptions in previous posts, and corrects some erroneous formulae and/or statements in previous posts.

First imagine an N-sided polygon inscribed in the equator of a sphere, and an N-sided polygon circumscribing the equator. The equator is a special case of a circle of constant latitude. Now consider other circles of constant latitude with inscribed and circumscribing N-sided polygons.

Imagine the two 3D surfaces generated by the N-sided polygons if the latitude is varied continuously from the North pole to the South pole. Either of these surfaces is a reasonable approximation to the sphere for large values of N. For N = 6, they are not good approximations, but might be viewed as interesting and/or esthetically pleasing. Both surfaces have N faces shaped like lunettes.

I am almost certain that the following statements are true (or at least excellent approximations to truthful statements).

• The lunettes described above for either surface can be flattened out without stretching or tearing, allowing the design and construction of flat lunettes which can be bent to approximate a sphere.
• The center lines of lunettes for the circumscribing surface are circles of longitude for the inscribed sphere. When the lunettes are flattened, the center lines are straight lines. The sides are curves of longitude for a larger non-spherical surface (An oblate spheroid, I think).
• The center lines of lunettes for the inscribed surface are curves of latitude for a prolate spheroid, while the sides are circles of longitude for the circumscribing sphere. When flattened, the center lines are straight lines, and the sides are arcs of some curve which is not a circle.

Following are various formulae.

• InnerSide = 2 * Radius * sin(Pi / N) * cos(Latitude)
• CircularArc = 2 * Radius * Pi * cos(latitude) / N
• OuterSide = 2 * Radius * tan(Pi / N) * cos(Latitude)

Formulae for flattened lunettes.

• Pi * Radius is length of the side of an inner lunette.
• I do not know a formula for length of centerline for inner lunette.
• Pi*Radius is length of centerline of outer lunette.
• I do not know formula for length of side of outer lunette.
• Latitude = Distance / Radius: For inner lunette, measure Distance along a side. For outer lunette, measure distance along the centerline.

Above latitudes are in radians, which is the correct unit for Visual Basic trigonometric function. The above assumes Latitude is zero at the equator.

The above formulae for Sides (inner & outer) can be used to construct flat lunettes. It looks much easier to construct lunettes for the outer surface. For the outer surface lunettes, the length of the centerline can easily be calculated. For outer surface lunettes, it is also a conceptually easy task to determine lunette width at various places, allowing the marking of points to be connected by an approximating curve.

I know you have some reason for wanting the sides of the lunettes to be circles of longitude for the constructed surface, but that requires lunettes designed using the inner surface formulae, which seems very difficult compared to a design based on the outer surface formulae.

[Starman]

Guv,

I have paraphrased some of your post to statements that I believe are true:

1) The lunettes can be flattened out without stretching or tearing, allowing the design and construction of flat lunettes which can be bent to approximate a sphere.

2) The center lines of lunettes for the inscribed surface are curves of latitude(?) for a prolate spheroid, while the sides are circles of longitude for the circumscribing sphere. When flattened, the center lines are straight lines, and the sides are arcs of some curve which is not a circle.

3) I do want the sides of the lunettes to be circles of longitude for the constructed surface and this requires lunettes designed using the inner surface formulae.

4) This seems very difficult.

Number 1 is easy – this is the desired result.

Number 2 – Have you a definite reason for stating that the sides are not arcs of circles? It’s not that I strongly believe that they are, just that it may open up some possibilities that I have not explored yet. I wondered if they might be, similar to flattening out the curved surface of a right circular cone and finding the base now forms part of a circle.

Number 3 – About twenty years ago this was the shape and the construction (with varying N) that I imagined and the formula has been eluding me ever since. It has become one of my ‘Unanswered Questions’ and I’m afraid that an approximation will not ease my concience.

Number 4 – I’m glad in a way that it is difficult, I was worried that I may have missed something basic during my search.

It does seem rather silly that the study of maths (math’s ?) should be able to land a satellite on a distant planet, deal with multi-dimensional space and numbers that trail off into infinity but can’t easily describe a simple finite sphere with a few bits shaved off.

Do you think that it is possible to find an answer ?

[Starman]

Ho hum!

The shapes may look identical, but I would know there was a difference.

I shall continue my reading about elliptic integrals of several kinds. The pages I have on these seem to either stop just before they give me my answer, or start at a point just beyond my comprehension.

I have very little spare money and I’m not sure that I dare approach ‘er indoors on the other matter although she may be tempted by a trip to the States – are you into being nagged and just generally blamed for things?

I shall begin cold-calling map makers immediately – my previous idea was to ask the people who make those segmented chocolate oranges.

Thank you most sincerely for your time and patience and for looking at this problem, If I get any further, or stuck further on I will let you know.

Thanks Guv.

Starman.

[Guv]

Starman: Keep me up to date on how your lunette project is going.

Also, about your desire for lunette sides becoming circles of longitude: Do you have an explanation I would understand?

[Guv]

Starman: Have you been doing anthing with your lunette project?

I recently examined a globe and discovered that they are no longer constructed by pasting paper lunettes on a sphere.

It seems to me that they use something like Ink Jet printer technology to print directly on a sphere. This is just a guess. Do you have any idea how they do it these days?

[Starman]

Hi Guv,

I’m afraid I havn’t been looking at my formulae lately, except for drawing a pic in povray.
Pov has a clever way of wrapping a bitmap around a sphere and I suppose that it is a variation of these mathematics that is used for constructing globes.
Other than work, which seems very demanding recently, I have actually been trying to write a vb program to cross your desert with the minimum number of vehicles – you can almost cross the dessert on foot now without touching the sand, by walking on the vehicles I have left laying around.
Part-way across the desert is a pyramid with a cube in the middle and I have spent a little time drawing pictures of this.
Still struggling – I will be back if I find anything.