ago few months when I got into the topic of Ken Miura Robert Rose Lang had the courage to write an email, to show my poor resolution of the CP of this beautiful rose, which is already charted in the book of the annual convention of Origami USA 2007, as well as in the book of the 12th Convention JOAS , and his answer filled me with joy, not only allowed me to publish my poor doc but also gave me suggestions and good advice.
One of these suggestions was the Junior version of the Miura Ken Rose, described by him as the construction of a grid of horizontal divisions in 12/54, 25/54 and 39/54, each of which then divided to half, along with vertical divisions of 1 / 9 each. Then I was faced this issue of the divisions a little strange at first sight a little fanciful, because who in full possession of his mental faculties could be made up to half turn 54: D? The thing is not so obviously, if we consider that 6 / 54 are 1 / 9 and twice that make the first of the horizontal divisions suggested by Master Lang. Moreover, if we divide the paper in these units of 6 / 54 (or 1 / 9), 6 units and complete the 39/54 half of the bottom, and split one of those units to half do not sound so terrible.
So how consegumos 1 / 9 folding a sheet of paper? Quite a problem, if I can say. The search for an answer led me to Theorems Haga, beautifully described on the website of the Japan Origami Academic Society , by someone identified as Koshiro. The idea then now is to show how they work and how they base their results.
Right The First Theorem says something like: "If we take a corner of a square to a torque division mark on the opposite side, indicating an uneven division known in its adjacent side." To visualize take the simplest case of dividing a hand in two and take the opposite corner to it, and see that we get the other side.
As we see, gives an indication of 2 / 3 on the right side of this square. The explanation comes from the world of mathematics (what else?), The triangles are related to SAP and PBT, are said to be "like" or proportional, ie the second is like the first but proportionately larger, demonstration of this is the interior angles, there is a classic theorem of geometry (and here I use the word "classic" to excuse not to show it:)) which says that if a triangle has each of its sides perpendicular to the sides of other then the interior angles of both triangles are equal and therefore they are proportional, or similar. In this case it is clear that the SA side is perpendicular to PB (line segment AB), the AP side is perpendicular to BT and it's SP for PT.
Then SA = c * PB, AP = c * BT and SP = c * know PT and AP = PB = 1 / 2 and SA + SP = 1, we know how BT measured. If the triangles are proportional, AP / SA = BT / AP
PB / SA = BT / (1-AP) or (1 / 2) / SA = BT / (1 / 2) and BT = 1 / ( 4 * SA)
SA
how much better then? By Pythagoras we know that
and hence SA = (1-1/4) / 2 = (3 / 4) / 2 = 3 / 8
and BT = 1 / (4 * 3 / 8) = 1 / (3 / 2) = 2 / 3
However, in the general case we
generating the following table, which could be our workhorse when it comes to creating arbitrary divisions grids :
in a future entry may look at other ways exist for these divisions, known as the Second and Third Theorems Haga, if there is interest, of course: P
many greetings ...
One of these suggestions was the Junior version of the Miura Ken Rose, described by him as the construction of a grid of horizontal divisions in 12/54, 25/54 and 39/54, each of which then divided to half, along with vertical divisions of 1 / 9 each. Then I was faced this issue of the divisions a little strange at first sight a little fanciful, because who in full possession of his mental faculties could be made up to half turn 54: D? The thing is not so obviously, if we consider that 6 / 54 are 1 / 9 and twice that make the first of the horizontal divisions suggested by Master Lang. Moreover, if we divide the paper in these units of 6 / 54 (or 1 / 9), 6 units and complete the 39/54 half of the bottom, and split one of those units to half do not sound so terrible.
So how consegumos 1 / 9 folding a sheet of paper? Quite a problem, if I can say. The search for an answer led me to Theorems Haga, beautifully described on the website of the Japan Origami Academic Society , by someone identified as Koshiro. The idea then now is to show how they work and how they base their results.
Right The First Theorem says something like: "If we take a corner of a square to a torque division mark on the opposite side, indicating an uneven division known in its adjacent side." To visualize take the simplest case of dividing a hand in two and take the opposite corner to it, and see that we get the other side.
As we see, gives an indication of 2 / 3 on the right side of this square. The explanation comes from the world of mathematics (what else?), The triangles are related to SAP and PBT, are said to be "like" or proportional, ie the second is like the first but proportionately larger, demonstration of this is the interior angles, there is a classic theorem of geometry (and here I use the word "classic" to excuse not to show it:)) which says that if a triangle has each of its sides perpendicular to the sides of other then the interior angles of both triangles are equal and therefore they are proportional, or similar. In this case it is clear that the SA side is perpendicular to PB (line segment AB), the AP side is perpendicular to BT and it's SP for PT.
Then SA = c * PB, AP = c * BT and SP = c * know PT and AP = PB = 1 / 2 and SA + SP = 1, we know how BT measured. If the triangles are proportional, AP / SA = BT / AP
PB / SA = BT / (1-AP) or (1 / 2) / SA = BT / (1 / 2) and BT = 1 / ( 4 * SA)
SA
how much better then? By Pythagoras we know that
and hence SA = (1-1/4) / 2 = (3 / 4) / 2 = 3 / 8
and BT = 1 / (4 * 3 / 8) = 1 / (3 / 2) = 2 / 3
However, in the general case we
generating the following table, which could be our workhorse when it comes to creating arbitrary divisions grids :
in a future entry may look at other ways exist for these divisions, known as the Second and Third Theorems Haga, if there is interest, of course: P
many greetings ...
