Here’s something mildly interesting and - for me at least - puzzling:
The background is that I have numerous plastic tubes from some plastic shelves I broke down for moving house and have not yet reassembled. These tubes make a good test surface for binding knots, as they are smooth and slippery.
Particularly I have been using them for experimenting with the Gleipnir.
Now, seven tubes make a good bundle, with one in the middle surrounded by six, as seen in pic Tubes01.
On a whim, I drew out the middle tube and was mildly surprised that the structure didn’t collapse (Pic Tubes02).
Reflecting on this, I decided that since the middle tube had enforced ideal contact points for each of the surrounding tubes, they were pushing each other apart while the two Gleipnirs pushed them together.
(The string Gleipnir is there to keep the bundle together as I tie and re-tie the braided cord Gleipnir).
So, I’ve removed about 14% of the volume held by the string, but a gap remains because of a precarious arrangement of compression/tension. OK, but if I disturb that, it will collapse and the bindings will go slack.
Nope. All that happened was that the six tubes could be rearranged into various configurations without ever losing the tension on the bindings. For instance, three over three (Tubes03) or a pyramid of three + two + one (Tubes04).
What the heck is going on? How can I remove 14% of a bound volume without any reduction in the length of line to bind that volume?
I calculate the minimum length of string (excluding the binding) needed to encircle the bundles to be the same in each case, ( 6 + Pi ) times the diameter of the tubes.
These bundles follow some rules which are:
Each tube is the same diameter.
Each tube contacts the string exactly once.
Each tube is in contact with two other tubes that also contact the string.
The cross-sections of the bundled tubes in contact with the string form a continuous path and a complete revolution (with no doubling back onto the path).
These rules may need some refining. Also, the arrangement of the tubes need not have any symmetry (if one ignores the question of stability).
In general for N tubes in contact with the string:
There are N sections of string not in contact with the tubes. Each of these sections has a length of 1 tube diameter for a total of N times the tube diameter. There are N sections of string in contact with the tubes. The total length of these sections equals exactly 1 circumference of a tube, Pi times the tube diameter.
The minimum length of string needed to encircle the N-bundle will be ( N + Pi ) times the diameter of the tubes.
Expanding on your example:
One can hexagonally pack another 12 tubes around your hexagonal packing of 7 tubes. Around this new packing of 19 tubes, one can hexagonally pack another 18 tubes for a total of 37 tubes. The minimum length of string needed is ( 18 + Pi ) diameters.
Removing the interior 19 tubes allows one to rearrange the remaining 18 tubes into a 2 by 9 stacking of tubes with the same minimum length of string needed, ( 18 + Pi ) diameters. This repacking has eliminated about 51% of the tubes.
Note: The 6 by 1 arrangement mentioned by @KC has a minimum length of string needed equal to ( 10 + Pi ) diameters
I think that your sizing guage --the binding cord–
isn’t giving a perfectly accurate reading, which
requires equal tension. At least, I’ve no other way
to explain how the the bound volumes are equal
for 6-in-triangle = 6-in-rectangle.
I see that both have surrounded 6 circles;
and each has 6 “tangent-line” spaces between
pipes on the outer surface;
but only the rectangular one has similar
interior between-pipes spaces (think : insert
a rigid sheet/card down between pipes; there
will be an isosceles triangular space on eithe
side with those 4 pipes) which occurs 4 times;
and the triangular one has 4 slightly smaller
such spaces, as pipes intrude across the line.
OR am I missing that there is in the triangular
one a slightly greater arc to cover in turning
around its 3 “point” pipes (vs. the right-angle
turn but occurring 4 times for the rectangular one!?
I agree with AikBkj that the circumference of the hexagonal with 7 pipes is mathematically exactly equal to the 6 pipe triangle and the 6 pipe rectangle.
For what it is worth, I also agree that this result is very counter-intuitive at first. I would have bet that the circumference just happens to be similar enough that the difference might be explained by the cord stretching.
I think this is it exactly. In AikBkj’s images, this is illustrated with the green-striped segments. Focus on the first image (hexagonal arrangement) and the last (rectangular arrangement): the green-striped segments always form exactly 1 complete circle.
A whole different way to think about this: the rope only curves where it follows the pipes’ curvation. Since the rope produces a full circle the touched parts of the pipes also must make a full circle.
Perhaps some of the perplexing nature of the OP can be understood by considering a similar problem.
First, to help with the similarity, let’s consider that if the tubes are the same length, we are really talking about two dimensions, that is, the missing cross-sectional area. To rephrase the problem somewhat, what are the areas that can be enclosed within the same perimeter (length of string ignoring the knot)?
SIMILAR PROBLEM:
The problem: What is the maximum rectangular yard that can be “fenced-in” with a limited length of fencing, let’s say 40 meters of fencing?
Although those that have studied differential calculus will almost certainly recognize this problem, only arithmetic is required.
Given that the perimeter is 40 meters and that the length plus the width of the yard must be equal to half of the perimeter, the length plus width always equals 20 meters.
Some trials show that the Length, Width and Perimeter (in meters) and Area (in square meters) of the yard are:
More extensive trials and recognition of the symmetry present in this problem would seem to indicate that an equal length and width produces the largest rectangular fenced-in yard, that is, a square yard of 100 square meters.
In the more asymmetric cases, one sees that small changes in, let’s say, length, can produce a doubling or more of the width which in turn nearly doubles the area enclosed by the perimeter.
I have calculated the bound cross-sectional areas enclosed by the bundling string in the OP. The perimeters of the enclosing string are given in diameters and the enclosed areas are given in diameters squared. So, for example, if the actual tube diameter is three units, then the perimeters I give should be multiplied by three and the enclosed areas should be multiplied by 9 (3 squared).
Finding Bundle Area for a Regular Polygon Bundle, for example, the 6-in-Hexagon:
Typically, there are three types of areas to sum: wedges at corners (shaded green), rectangles between tube centers in contact with string (shaded gray) and an interior polygon (unshaded), see attached.
Green Wedges sum to one full circle: Area = (Pi/4) Diameter^2 = 0.7854 Diameter^2
Gray Rectangles (there are 6), each are (1/2) Diameter^2: Area = 3.0000 Diameter^2
Unshaded Interior Hexagon, side of 1 Diameter: Area = (3/2)*Sqrt(3) Diameter^2 = 2.5981 Diameter^2
