Building Sundials
Designed by math professor Hugo Michnik (Germany) and published in April of 1922, the bifilar sundial has the interesting variant with respect to other sundial that the time is indicated by the intersection of the shadows of two wires (Nomon) independent or beam placed horizontally, perpendicular to each other and parallel to the sundial is horizontal type.
Although the original design has the horizontal sundial and the rods perpendicular to each other and aligned with the cardinal points, it can be developed for any inclination of sundials and arrangement of the hours indicator rods.
Unfortunately, there is more information about the inventor and we can not offer a better tribute to his invention.
One of the remarkable qualities of the bifilar sundial is the possibility of developing the layout of the hours so that these are equiangular in the horizontal plane, ie separated from each other by 15 °, in the same way as the sundial Equatorial Quadrant. Remember that clock hour angles Azimuthal Quadrant in Horizontal or Quadrant sundial Horizontal developed and are not uniformly distributed radially, hence the ease of construction of the bifilar sundial clock.
This publication will attempt to develop a bifilar sundial Horizontal Quadrant as a starting point and use the information previously calculated base for the development of sundial Azimuthal Quadrant in Horizontal made a previous delivery.
A property that makes the clock Sun bifilar particularly interesting is that according to the vertical separation between the rods fail to get different patterns radial to show hours on the sundial, the most attractive arrangement of the rods that allow us to obtain a radial pattern of hours over uniformly around a center and the angle of separation between time and time is of 15 º, separation time identical to that of the Sun in daytime running across the sky.
To understand the functioning of the clock begin to recall the working principle of Azimuth sundial, because essentially you are trying to determine the length of the shadow cast on the basis of angle of the Sun with respect to the horizontal and its direction is determined by the angle of the Sun with respect to the meridian.
A long vertical Nomon "AG" is illuminated by the Sun, "S" (Figure 1), projecting on the floor with a length "L" in the shadow of the gnomon, which depends on the height "a" of the Sun "S" on the horizon, oriented at an azimuth "Z" direction to the South Pole.
The height of the sun is a function of local time where the gnomon the latitude of the place and the declination of the sun as the day on which this by observation.
Having at hand these parameters, the height "a" of the Sun can be found by solving the equation of spherical trigonometry used for the development of Azimuth sundial:
Sen (a) = sin (L) x sin (d) + cos (L) x cos (d) x cos (H)
FIGURE 1
Where:
l = latitude of location in degrees.
d = Declination of the Sun for the day when you are determining the height Solar.
H = the hour angle in degrees and counted from noon.
a = The angle of the Sun's altitude with respect the horizon in degrees.
azimuth angle of the sun is a function of the declination of the sun to the time of measurement, the height of the sun above the horizon and the hour angle (hour) we can calculate it by solving the equation of spherical trigonometry as follows:
Cos Z = [Sen (l) x Sen (a) - Sen (d)] ÷ (Cos (l) x Cos (a))
Where "d", "a" and "l" have the same meaning the above formula and "Z" is the azimuth angle of the sun for that instant.
The shadow length is determined by solving the following trigonometric equation:
L = AG ÷
Tan (a)
Where:
L = Length of the Shadow.
AG = length of the gnomon.
a = height of the sun in degrees.
The time clocks azimuth is then defined by the length of the shadow and the sun azimuth angle
FIGURE 2
If we place a horizontal rod aligned with the North-South height "Hn", his shadow in the same way that the vertical shadow of the gnomon is cast by the sun according to the Sun's altitude and its azimuth as we can see in Figure 2. The shadow point "V" of the rod is projected in the horizontal plane under "W" to a length L (L = Hn ÷ Tan (a)) starting from the point "P" line "VP" perpendicular to the horizontal plane and angle "Z" which is the azimuth of the Sun . The result is the projection of the shadow of the rod in the horizontal plane and parallel to the North-South (The rod is parallel to the north-south line in the vertical plane) off at a distance "X" axis North South.
To determine the distance "X" which is the shadow of the rod with respect to the North-South, we must make the same calculations we did to determine the length of the shadow of Nomon vertical, in this case find the length "L" line "PW", the length of the line "PW" is required for the calculations. After finding this value "L" We find the value of "X" according to the azimuth angle of the Sun "Z" using plane trigonometry.
X = L x Sen (Z)
If the rod horizontal turn 90 degrees so that it is parallel with the east-west the Shadow of the same projects shifted to a value "Y" East line West as shown in Figure 3.
calculations "a", "L" and "Z" for this case are identical to the previous one, just change the direction of the rod. Given the length "L" Line segment "PW" to any point on the rod, we can calculate the offset "Y" with respect to the reference based East-West azimuth "Z".
Y = L x Cos (Z)
The length "L" for East-West oriented rod is determined by: L = Hs ÷ Tan (a).
FIGURE 3 Figure 4 shows the case of two vertically aligned points but separated at a distance.
FIGURE 4 The two points are projected onto the horizontal plane along parallel paths according to the height "a" of the Sun, so that the projection (shadow) of them are on the line "PQ" keeping azimuth "Z". If the lowest point (G) moves down, its shadow (in blue) is near the intersection of the vertical line to the intersection of the lines North-South and East-West under "P", increasing the distance "L" between the two shadows (decreases the value of Y). If the point "G" increases, the distance between the projections decreases and when both points are at the same height, only have a projection in the same place as the projection of the upper red "J" and have the same case the gnomon vertical Azimuthal clock.
clarify this point is good, because in the clock-wire rods are separated from each other as points in Figure 4 and the vertical separation between them defines the pattern of the track of the hours on the sundial.
Figure 5 shows the case where the two rods are overlapping, perpendicular and aligned to the cardinal points, lines North-South and East-West.
FIGURE 5
The top rod (green) aligned with the axis North-South "is at a height" Hn "(PJ) of horizontal. Sunlight casts the shadow of this rod parallel to the north-south axis and displaced a distance "X" as shown in Figure 2. Bottom Rod (Blue) is perpendicular to the top and is at a height "Hs" (PG) horizontal plane and its shadow is shifted a distance "Y" of the east-west as shown in Figure 3. Both shadows intersect at point "O" (Figure 5). If we look carefully, we see that the intersection of the shadows is not on the projection line "PQ" although the projection of the points "J" and "G" that are aligned with the vertical passing through the point "P" when projected onto the projection line "PQ."
Point "O" is then defined by the distance "X" and "Y" in the horizontal plane.
We consider that in the horizontal plane have a Cartesian coordinate system where the point "P" is the origin, the x-axis is the east-west and the y-axis is North-South line.
Point "P" in the horizontal plane is the vertical projection of the points "J" and "G" that is, if we look from the horizontal plane vertically above the two rods, these apparently intersect at point "P" of horizontal rods and lines do about North-South and East-West and for this particular we can imagine the Cartesian coordinate system in the horizontal plane. The vertical line formed by the points "J", "G" and "P" is remarkable and use it as a reference for determining the point "X" and "Y" point "O" on our Cartesian plane. The point "O" is the solar time indication of the bifilar sundial.
For a certain time of day point "O" is separated from the north-south axis by the value X1 and the east-west axis of Y1, which implies that this point "O" has the coordinates (X1, Y1). After a time, the shadows have moved in the horizontal plane copying the movement of the sun in the sky, moving the intersection point O to point O ', which is defined on the Cartesian plane by the coordinates (X2, Y2 ), as shown in Figure 6.
FIGURE 6
We can assume that the point "O" is part of a straight line (green) does not necessarily pass through source "P" the coordinates and defined by the equation of the lines:
Y1 = K1 x X1 + b
Where "K1" is the slope of the line and "b" to the point where this line intersects the y-axis (north-south) as illustrated in Figure 7.
FIGURE 7
slope "K1" of the line is associated with the hour angle, which runs from noon, which for our Cartesian plane for the y-axis north South. So the slope of the line takes the expression:
K1 = Tan (90 - h1)
But Tan (90-h1) is equal to COTAN (h1) which in turn is:
K1 = 1 ÷ Tan (h1)
Where "h1" is the hour angle in degrees. To clarify, at 1:00 PM, the angle "h1" would have a value of 15 °, for 2:00 PM "h1" would be worth 30 and so on, but measured with respect to the North-South, while the slope angle is determined from the East-West.
Applying the same reasoning, the point O 'also form part of a line (in blue) have in common with the previous point of intersection with the north-south axis, ie point "B" in order to obtain a radial distribution of hours, the position of O 'is defined by the time "h2." The sweep angle "h1" to "h2" should be 15 degrees in one hour, and we are looking for a regular distribution of time on the clock dial bifilar Horizontal. Mean, we are concerned that the separation between straight line and is 15 degrees to get a pattern equiangular of hours in the horizontal quadrant, so the lines that we build must be related to the angle of 15 degrees for each hour spent.
The general equation for the second line is:
Y2 = K2 x X2 + b
Its slope can be determined by:
K2 = Tan (90 - h2)
That is:
K2 = 1 ÷ Tan (h2)
recall the other hand, the value of "X" was
the expression X = L x Sen (Z)
Y "L":
L = Hn ÷ Tan (a)
The value of "Y" is obtained from:
Y = L x Cos (Z)
being "L" to the rod East-West:
L = Hs ÷ Tan (a)
Substituting for the equation of the first line we have:
X1 = Hn x Sen (Z1) ÷ Tan (a1)
Y1 = Hs x Cos (Z1) ÷ tan (a1) K1 = 1 ÷
Tan (h1)
Hs x Cos (Z1) ÷ tan (a1) = Hn x Sen (Z1) ÷ [As (a1) x Tan (h1)] + b ... (1)
For the second line:
X2 = Hn x Sen (Z2) ÷ tan (a2)
Y2 = Hs x Cos (Z2) ÷ tan (a2) K2 = 1
÷ Tan (h2)
Hs x Cos (Z2) ÷ tan (a2) = Hn x Sen (Z2) ÷ [Tan (a2) x Tan (h2)] + b ... (2)
not forget that " Z1 "and" a1 "is the azimuth of the sun and the height thereof for the time" h1 "... and ..." Z2 "and" a2 "are the azimuth and the height of the sun for hours "h2." "Hn" is in both cases the height of the rod as North-South oriented and "Hs" the height at which the rod is oriented east-west.
If we look closely, what we have is a system of equations with two unknowns, the value of "b" and the value of "Hs." Choose "Hs" as a variable that allows us to obtain separate times regularly in the clock face by adjusting the value of "Y". With the value of "Hn" to be fixed, will determine the size of the quadrant.
Subtracting equation (1) to equation (2) to eliminate "B" we get:
Found the value of "Hs" or stick height indicator east-west when we find the value of "b", you just have to enter the value of "Hs "in any of the above equations and solve for" b ".
To reinforce perform a calculation with the following information:
Hn = 100 mm (height rod North-South)
d = -4.22 ° (declination of the Sun for 01/10/1909)
L = 8.27 ° (latitude for the sundial, Puerto Ordaz, Venezuela)
h1 = 15 º (11 AM)
h2 = 30 º (10 AM)
calculate the height "a" of the Sun using the formula set out at the beginning of the post.
a1 = 70.511 º. A2 = 57.579 º
We find the azimuth for each hour.
Z1 = Z2 = 50.685 ° 68.446 °
Now we can find
Hs Hs = 14.3 mm (Rounding)
b = -99 mm (Rounding)
As for Azimuthal clock, we can simplify calculations through the application of a spreadsheet to verify compliance with the values \u200b\u200bof "b", "Hs" and "Hn" for any time and day of the year, taking the 12 M and 6 AM and PM to be cases individuals.
FIGURE 10
With the value of "Hn" we can get an idea of \u200b\u200bthe size of our quadrant, calculating the value of "X" to 7 AM and 5 PM, for if we do the calculations for the 6 am gives us a negative value is meaningless.
To find the maximum length of the quadrant, do the calculation for the first day of January, the sun is almost at its maximum declination south with a value of -23.01 °, are the days with elongated shadows June and shorter shadows in the case of this latitude of 8.27 °. In the Southern Cone are reversed values.
The greatest value of "X" with a height of North-South rod 100 mm, which corresponds to January is 495 mm. The value of "Y" enables us to determine the width of our clock, which according to the calculations of Figure 10 is about 51 mm. This narrowing is due to the next thing we're on the equator and near the quadrant in which they are the dipstick.
With the value of "b" and the length of the quadrant, we can make a template for the sundial bifilar sundial, playing with the direction of the projection of the shadows so that the shadow of the media do not hide interception of the shadows of the rods indicator of the time. For this reason, the support rod "NS" should not be on the point of convergence (b) of the lines indicating the hours of the quadrant.
For convenience, I used a computer design program, but this activity can do with a little patience with pencil, ruler and compass.
Due to the length of my rod indicator (390 mm) Solar Quadrant is confined to a square of 450 mm by 450 mm to the media of the rods are on dial. Figure 11 shows the appearance of the template.
FIGURE 11
We must caution quadrant indicate the position of the Solar stand bases indicator rod during assembly to ensure that the same clock are in the correct position. In Figure 11, the cross marks indicate the location of the supports for the rods.
Although specialty stores materials for the realization of models we can find metal rods or plastic the diameter you want and of good length, for the clock as we are making use indicator wire rods a couple of coat hangers as shown in the figure below.
FIGURE 12
Wire to use the chop with pliers or a hook and shear corresponds to the part where they hang straight down "pants", it has a diameter of 2.2 mm with a length of 390 mm as shown in Figure 13. Obviously these wires
not be perfectly straight, but the most righteous use to find and / or trying to straighten them as best we can. Unrighteousness induce errors in the time display.
FIGURE 13
As annotated wires used should not be too long to prevent curling under its own weight as it would produce an error Construction on the clock.
bases like the dial of this watch we will make them from cardboard construction of 2 mm thick and must be shaped to hold the wire by itself and focus, the best solution is to do them so that the area of \u200b\u200bcontact or support wire is triangular as shown in Figure 14.
FIGURE 14
With regard to the height of the stand, we must consider that the calculations are made for the rod axis without considering the diameter of the same. Do not take this into account would be to introduce another small factor that would influence the timestamp, however we can ignore this detail if our wire is thin and the employee of 2 mm or take into account if we like the geometric developments to determine the "V" support for the wire in terms of the height Hn and / or Hs
Figure 15 shows a plan for the template I will use wire to support the "NS" and height 100 mm (measured from the center of the wire diameter), in this case the angle chosen for " V "bracket is 45 degrees, but can be any angle and base width is 30 mm and above in the peaks of the" V "10 mm.
FIGURE 15 Figure 16 corresponds to the level of support for the establishment of the rod "EO" with a height of 14.4 mm rod.
FIGURE 16
so marked asymmetry between the two carriers is due to the latitude of the place, which for this design is 8.27 º N.
expected appearance of the clock once projected bifilar assembly is shown in Figure 17.
FIGURE 17
The following photos show us the practical realization of the bifilar sundial. Template
quadrant Solar.
Gluing sundial template on the cardboard base of 2 mm thick.
cardboard cut excess.
media templates glued to the cardboard base of 2mm.
Stands cut.
supports placing the rods on the dial. Note that the brackets are on the "cross" indicator.
bifilar Watch face armed.
With the clock assembled, we went to a clear place and properly oriented.
Solar Time 8:30 a.m.
Note that the time is indicated by the intersection of the shadows, 8:30 AM (ET Solar).
Polar watch and clock-wire.
We see that our solar clock works perfectly and it was expected according to theoretical investigation above.
Analysis by Professor Hugo Michnik simplify the formulas used in this blog, so that the separation between the rods to obtain a regular graduation of hours on the dial Solar meets the formula Hs = Hn x Sen (L) and distance "b" to the equation b = Hn x Cos (L). If we compare the results obtained for Hs b, we see that are the same according to the simplified formulas.
With this realization, we have unveiled the "secret" of the bifilar sundial at the same time it shows the ingenuity of man to explore all possibilities for each of their technical inventions, not limited to a single design or model of device but trying and exhausting all possibilities. In reality it is an act of perseverance, concern, curiosity and inventiveness that is what has allowed the Self People arrive at the current technological development, technology and science.
Puerto Ordaz in December 2009
