As with my second Instructable, this third in the series derives from a reference made in my first effort, “(Un)Folding the Mysteries of A4 Paper - There’s Some Pretty Far Out Geometry In There.”* In its introduction, I provide links to websites which detail some of the mathematics behind the paper model we learn to create from a single sheet of A4 paper - a Square Dipyramid. I then proceed with the Instructable, without any further explicit mathematical analysis. Though by no means a mathematician myself (more of a tinkerer), in this Instructable, we will revisit some of that underlying math as best as I can express it, and extrapolate from those findings a strategy for discovering how to fold and form more dipyramids, this time with bases of any regular polygonal shape – pentagon, hexagon, octagon, etc., collectively referred to as ‘Regular Dipyramids’.
One of the webpages I referenced, http://mathworld.wolfram.com/SquarePyramid.html, draws attention to a mathematical formula describing a special case Square Pyramid, where (h) height, is equal to (a) length of side divided by 2, or h=a/2. I proceeded to highlight that formula, recognizing that it describes the particular model that we were learning to make, or I should say, one half of it, as we are always creating the doubled form, dipyramid. As a dipyramid, that formula reduces to the simple expression: h=a, for the height of a dipyramid is twice the height of its single form. In this Instructable we will delve deeper into the significance of that formula, and see how it relates to A series paper. Then, as noted above, we will proceed by looking at all regular pyramids, http://mathworld.wolfram.com/SquarePyramid.html, , or in our case, regular dipyramids, that conform to the same formula: h=a. Finally, we will discover how all such regular dipyramids can (theoretically) be modeled from a rectangular sheet, with a fold pattern similar to the square based example of our initial studies.
Step 1: Materials and Tools
Unlike my two previous Instructables, this one does require some simple mathematical calculations, as well as some measuring, marking, and trimming of our A4 sheets. So in addition to the usual tape, straight edge and scoring tool, we will need a scale for measuring, pencil for marking, knife and straight edge for cutting, and a calculator. We will be taking advantage of some online math shortcuts as well, in particular, a Triangle Calculator, as my trigonometry skills are not what they used to be. (Any mathematicians out there, please feel free to jump in).
I will be going back to the use of 100 gsm kraft paper for this exercise, but again any A4 paper will work.
Step 2: Labeling Parts and Defining Terms
For the purpose of this Instructable, I will use terms and symbols found on “mathworld” webpages, such as:
From these sources, we get the following, as well as the diagram shown:
n=number of sides of base
r=inradius or apothem*
α=central angle (not shown on diagram)
Step 3: Unfolding the Mystery - Part 1
An examination of our standard fold pattern of the A4 paper shows how we are always creating similar rectangles with sides approximating the proportion: 1:√2, and right triangles with sides of proportions: 1:√2:√3. These proportions, by design, are inherent and unique to A series paper sizes. The link below describes the rationale for the paper being designed with these proportions and what makes them unique:
As A4 paper is always measured at 210mm x 297mm, it is easy to see that the fold pattern defines our smallest right triangle with legs equal to one quarter of those lengths, or 52.50 x 74.25. And the hypotenuse can be calculated using the Pythagorean Theorem, giving us a length of 90.94mm. Checking our stated proportions we can confirm: 52.50 x √2 = 74.25 and 52.50 x √3 = 90.93 by my calculator. Close enough.
Step 4: Unfolding the Mystery – Part 2
Taking one of our models, I have marked each of those small right triangles with a red dot on the top, and yellow on the bottom. When we unfold the model, a tally shows that sixteen of these triangular units combine to form its total surface area, two per each of the eight faces, half on the top pyramid, and half on the bottom. Those paired right triangles form the isosceles triangle of each face. Each has a base equal to 2 times the short leg, or 105mm, defining one of the four sides of the pyramids’ base (a). The lateral sides (e) are equal to the length of the hypotenuse, 90.94mm. The crease/joint line down the middle of the face gives us (s) the slant height, equal to the long leg of our right triangle, 74.25mm. Knowing a=105mm, we can find the perimeter (P), which is four times that, or 420mm. It is important for us to note that the perimeter is twice the width of an A4 sheet, for we will return to that relationship as we proceed.
n=4 (square base)
Step 5: Unfolding the Mystery – Part 3
A quick count of the total number of small right triangles per fold pattern per sheet, reveals a total of 32. Therefore, from the previous step, we know that only half are expressed on the surface, so the other half get swallowed into the internal structure of our model. This gives us the total surface area equal to one half the area of an A4 sheet.
By examining what is happening internally with the small right triangles that have been absorbed, we can find (h) height and (R) circumradius. Looking inside a half formed model (see photo) we can see that (h) height is equal to the short leg of our right triangle, 52.50, and that (R) circumradius is equal to the long leg, 74.25mm (one half of the diagonal through the square base). We now have only two unknown dimensions remaining, (r) inradius or apothem*, and the central angle (α). The central angle is simply 360 divided by the number of sides, 360/4=90⁰. To find the inradius (r) we can see that it divides the central angle in half creating a 45 45 90 isosceles triangle. Knowing one of its legs equals a/2, or 52.50, we see that the other leg, (r) must be the same.
Our complete description:
Some additional information of interest to us:
(n) x (h)=short length of sheet=210mm
(R) x 4=long length of sheet=297mm
Array: 4 by 4
Step 6: The Pentagonal Diyramid – When (h) = (a)
Having completed this analysis, we are in a position to find the fold pattern on a rectangular sheet which when formed and folded will take the form of a dipyramid whose base is a pentagon, with a height equal to the length of one of its sides. We will assume a perimeter (P) equal to 420mm as a given. This means the sides (a) are equal to 420/5, or 84mm. Height of the pyramid will be a/2, or 42mm, which we will assume to be the short dimension of our new small right triangle of the new fold pattern. This leap comes from the notion that the sheet width will always be one half the perimeter, 420/2=210, and that the short leg of the small triangle will always be 210/n (number of sides) which gives us the 42mm. In other words, for a pentagon, the sheet will be folded into five equal lengths of 42mm along this dimension.
To find the long length of that triangle, we must find (R), the circumradius of a regular pentagon whose sides are 84mm, per our internal look from the previous step. To do this we will need the central angle of an octagon, 360/5=72⁰. If we then show the inradius (r) bisecting that angle and a side, we can use that right triangle to find (R). We know right away two of its angles, 36⁰ and 90⁰, so the third must be 54⁰. We also know from above, that the short leg is 42mm. Now, taking advantage of the triangle calculator:
We can plug in three of those terms and find that r=57.81mm and R=71.46mm.
We now can fill in the following:
e=lateral edge (see below)
(n) x (h)=short length of sheet=5 x 42=210mm
(R) x 4=long length of sheet=285.84mm
Array: 4 by 5
It is not necessary to calculate (e) in order for us proceed, as it will always be the diagonal through our rectangle defined by h and R, but should one wish, it can be solved with the Pythagorean Theorem, a right triangle with legs 42 and 71.46mm gives us a hypotenuse equal to 82.89mm. In fact, it turns out that as soon as (n), (h) and (R) are known, as long as (P) is given, the sheet can easily be measured, marked, trimmed, scored, folded and formed.
Step 7: Prepping the Sheet
I’ve jumped a little ahead of myself, with regard to how to find the length of our sheet, and why it is always equal to 4 times R, thus creased in four equal divisions of that length. Again, an examination of our unfolded model of the square based dipyramid, brings us to this conclusion. We can see that along the long length, the top faces (red dots) are all contained in the two quarters of the sheet on either side of center, and that the bottom (yellow) are in the outside quarters. We know that each rectangle of the total array (16) is equal to one half the total number of faces of a square based dipyramid. Therefore, it is logical to assume that the pentagon pattern will require an array of twenty, twice the number of its 10 faces. We already know that we are dividing the short length by 5, therefore the long length must get divided by 4. We will see that top and bottom faces are distributed similarly, as well.
So we proceed by calculating the overall length of sheet: (R)x4=285.84 which for our purposes we will call 286. We can subtract that from 297 (the A4 dimension) to determine we must trim 11mm off the long dimension of our standard sheet.
The sheet then gets folded in four equal parts along the long dimension, using our standard techniques shown in (Un)Folding the Mysteries.
Dividing in fifths along the short dimension requires measuring and marking, before creasing.
Scoring and creasing the diagonals is more or less straight forward, using our standard technique. However, it is important to take note that for the pentagon (and all odd sided bases), the diagonal does not go corner to corner through the center of the sheet.
Step 8: Folding, Forming and Applying the Tape
Technique is the same, see (Un)Folding the Mysteries for details.
The Pentagonal Dipyramid is complete!
Step 9: In Conclusion
It turns out that for all regular dipyramids, when h=a, then s=R. Similarly, for all regular pyramids, when h=a/2, s=R. It is this relationship which allows us to find a unique rectangle that can fold in this manner.
For both pyramids and dipyramids, when the perimeter is constant, as the number of sides of the base (n) increases, (h) height and (a) length of side decreases. And as (n) approaches infinity, or the circle with circumference equal to (P), (R) approaches the radius of that circle, and (h) height approaches 0.
If my memory serves, this might be some sort of limit problem. Maybe someone out there can express it in that form for us.
Til next time, enjoy the folds!