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Post by Ron Walker on Oct 21, 2022 10:28:03 GMT -7
Posted by: sgrijalva Mar 11 2021, 01:01 PM Grat job!. I have another procedure using truncated icosahedron´s dual figure wich is a Pentakis dodecahedron. The vertices of that figure correspond to the center of the faces of the truncated icosahedron. So searching on the web I found this: Pentakis Dodecahedron Vertex co-ordinates = Center face co-ordinates of truncated icosahedron C0 = 0.927050983124842272306880251548 = 3 * (sqrt(5) - 1) / 4 C1 = 1.33058699733550141141687582919 = 9 * (9 + sqrt(5)) / 76 C2 = 2.15293498667750705708437914596 = 9 * (7 + 5 * sqrt(5)) / 76 C3 = 2.427050983124842272306880251548 = 3 * (1 + sqrt(5)) / 4 V0 = ( 0.0, C0, C3) V1 = ( 0.0, C0, -C3) V2 = ( 0.0, -C0, C3) V3 = ( 0.0, -C0, -C3) V4 = ( C3, 0.0, C0) V5 = ( C3, 0.0, -C0) V6 = ( -C3, 0.0, C0) V7 = ( -C3, 0.0, -C0) V8 = ( C0, C3, 0.0) V9 = ( C0, -C3, 0.0) V10 = ( -C0, C3, 0.0) V11 = ( -C0, -C3, 0.0) V12 = ( C1, 0.0, C2) V13 = ( C1, 0.0, -C2) V14 = ( -C1, 0.0, C2) V15 = ( -C1, 0.0, -C2) V16 = ( C2, C1, 0.0) V17 = ( C2, -C1, 0.0) V18 = ( -C2, C1, 0.0) V19 = ( -C2, -C1, 0.0) V20 = ( 0.0, C2, C1) V21 = ( 0.0, C2, -C1) V22 = ( 0.0, -C2, C1) V23 = ( 0.0, -C2, -C1) V24 = ( 1.5, 1.5, 1.5) V25 = ( 1.5, 1.5, -1.5) V26 = ( 1.5, -1.5, 1.5) V27 = ( 1.5, -1.5, -1.5) V28 = (-1.5, 1.5, 1.5) V29 = (-1.5, 1.5, -1.5) V30 = (-1.5, -1.5, 1.5) V31 = (-1.5, -1.5, -1.5) Source: math.stackexchange.com/questions/3231265/how-do-i-calculate-the-coordinates-of-the-centers-of-the-faces-of-a-truncated-icmath.stackexchange.com/questions/3231265/how-do-i-calculate-the-coordinates-of-the-centers-of-the-faces-of-a-truncated-icSaúl.
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Post by Ron Walker on Oct 21, 2022 10:28:55 GMT -7
Posted by: Scott T Mar 12 2021, 12:04 AM Wow - that is fantastic information thank you Saul!
I have been struggling down a long muddy footpath and it turns out the proper mathematicians have already built a motorway direct to the destination!
It is tempting to be downhearted as the fast cars whizz by and get to the destination seemingly effortlessly - but actually this is great news. The aim is collectively to get to get to the destination. If there is a better way I am all for it.
I need to learn about Pentakis dodecahedrons!
That is the power of this wonderful forum - unexpected solutions and expertise can appear at any time from anywhere on the planet.
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Post by Ron Walker on Oct 21, 2022 10:29:17 GMT -7
Posted by: mrgare5050 Mar 13 2021, 04:42 AM Duke, you took the more scenic route perhaps - its the journey! (I'm in fortune cookie mode lol)
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Post by Ron Walker on Oct 21, 2022 10:29:38 GMT -7
Posted by: Ron Walker Mar 13 2021, 11:21 AM QUOTE(mrgare5050 @ Mar 13 2021, 04:42 AM) * Duke, you took the more scenic route perhaps - its the journey! (I'm in fortune cookie mode lol)
The journey might be fun and/or painful, but it is the end, when one arrives in OZ, that is certainly the most fun of all.
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Post by Ron Walker on Oct 21, 2022 10:30:09 GMT -7
Posted by: Scott T Mar 14 2021, 11:57 PM Pentakis dodecahedron
Many thanks to Saúl Grijalva for finding a more elegant means of determining the centre co-ordinates of the star plates. In the interests of visualising what is going on and expanding on Saúl’s post– the Pentakis dodecahedron is what you get if you add pentagonal based pyramids to every face of a dodecahedron.
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Post by Ron Walker on Oct 21, 2022 10:30:34 GMT -7
Posted by: Scott T Mar 14 2021, 11:59 PM As Saúl says, the co-ordinates of the vertices (corners) of the of the pentakis dodecahedron correspond to the centre points of the faces of a truncated icosahedron.
The eagle-eyed reader will have noticed two things:
1. That there was a typo on the numbers I posted for pentagon 4 – the centre co-ordinates should read 2.45, -3.96 & 0. I will edit these posts to correct them.
2. That the co-ordinates Saúl provided are different from those I calculated. There is no further error, it is just the two sets of coordinates relate to different diameter 'spheres' so can’t be compared very easily at a glance using cartesian co-ordinates. They also have a different frame of reference in terms of the arbitrary starting point (the direction face 1 is pointing). In the background I have plotted them all out graphically and compared them – they are indeed exactly the same.
One thing to be aware of (which will be important later): there are two types of vertex on the pentakis dodecahedron - one that connects 6 edges and one the connects 5 edges. Remember that although the picture of the dodecahedron looks relatively spherical, these vertices are in fact at slightly different distances from the centre (origin) of the form. In fact, this is for exactly the same reason we talked about much earlier on. The centres of the hexagonal and pentagonal faces of a truncated icosahedron are at different distances from the centre of the ‘sphere’.
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Post by Ron Walker on Oct 21, 2022 10:31:28 GMT -7
Posted by: Scott T Mar 15 2021, 05:19 AM Step 5.2 Plotting the stars on the same sphere: download your star catalogue
My basic assumption is that we will download an electronic star catalogue of some sort and we will have nothing more sophisticated than a spreadsheet to do all the subsequent maths for us. I happen to use Excel which I think can handle a maximum 50,000 rows of data but that is more than enough stars for an amateur planetarium. If not, you will need a different spreadsheet or perhaps need to write a code to do the maths.
Search for downloadable star catalogues (or catalogs) on line. Some catalogues contain way too many stars, some have all sorts of other unnecessary information which can be deleted. It is helpful if they are in comma separated variable format because that makes it easy to import into a spreadsheet. There may already be catalogues in Excel format.
I downloaded a free star catalogue based on the Higgs Yale Gliese catalogues. It is just a long list of 3 columns of numbers, each star is defined by its Right Ascension, Declination and magnitude and contains just short of 50,000 stars which is way too many for me.
You just need to strip the catalogue down to the bare essential columns of data: RA, Dec and magnitude.
Once you have an electronic list of stars you can really start to manipulate the information and prepare the data for the star plates. The easiest first step it to decide what your cut off magnitude is. Use the ‘sort’ function on the spreadsheet to put the stars in order of magnitude then delete any excessively faint stars (or bright stars if you are going to have individual projectors for the brightest stars).
I will freely share my spreadsheets with anyone who wants them (when I have finished them). When they contain thousands of stars the file size tends to be very big and can be bounced out by e-mail servers so I will do a second version with all of the calculations but only a couple of hundred ‘example’ stars to keep the file size small. Anyone can subsequently cut and paste the full star catalogue data into the correct columns.
We then need to look at the birds-eye view of the route again and have a think.
• We currently have our star plate centre co-ordinates specified in cartesian (x,y,z) co-ordinates. • A typical star catalogue lists star positions in Right Ascension (hours) and Declination (degrees)
If we want to plot our stars and centre points on the same sphere, we need to convert one co-ordinate system to another i.e. cartesian to RA and Dec or vice versa so that we are using the same system.
So how do we choose which system of co-ordinates to convert to? There is no right or wrong answer. You can choose either as long as everything ends up in the same format. However, there is a third factor which makes the decision easier.
Remember step 3 in the route map? We are eventually going to need to calculate the shortest distance between the centre of the star plate and the position of the star. Fortunately, there are ready made formulae out there on the internet to help us do this. While different formulae are available for the different co-ordinate systems, the most accessible ones tend to work in terms of a third co-ordinate system i.e. latitude and longitude; presumably because most practical problems involving finding distances between two points on a sphere are really geography problems concerned with finding the distance between two places on the surface of the earth.
So, there are other ways to go about this, but I have decided to convert everything (star plate centres and star positions) into latitude and longitude.
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Post by Ron Walker on Oct 21, 2022 10:32:02 GMT -7
Posted by: Scott T Mar 15 2021, 05:26 AM Notes on co-ordinate systems
Before converting our star positions and star plate centres to latitude and longitude a couple of helpful notes about co-ordinate systems:
1. Cartesian co-ordinates are simply distances along three perpendicular axes ( x,y and z) from a starting point (origin) and can be expressed in any suitable unit – inches, mm or just plain numbers. Cartesian co-ordinates are not particularly convenient for talking about positions on the surface of a small sphere, because any change in position on the surface, results in a change in the value of at least two of the axes.
2. Right Ascension and Declination. RA and Dec are traditionally used to define the position of stars on the celestial sphere they are nearly the celestial equivalents of latitude and longitude. Right Ascension divides the celestial sphere into 24 divisions (hours) around its equator (each hour is subdivided into 60 minutes and each minute subdivided into 60 seconds). Declination divides the elevation between the equator and the pole into +90 degrees for the north pole and -90 degrees for the south pole. Each degree can be subdivided into minutes and seconds. There are plenty of far better, more thorough explanations out there on the internet. An important complication to be aware of; right ascension and declination can be expressed in ‘full, longhand’ form or in decimal form. Remember not to mix them up.
If you need to convert longhand to decimal form:
Decimal Right Ascension = hours + (minutes/60) + (seconds/3600)
For example, if the right ascension is 2 hours, 30 minutes and 45 seconds, then the decimal form is:
2 + (30/60) + (45/3600) = 2.5125
The same sort of principal applies to converting degrees from longhand to decimal:
Decimal Degrees = degrees + (minutes/60) + (seconds/3600)
3. Longitude and latitude. These are traditionally used to define positions on the surface of the earth but can be applied to any sphere. There are some very clear descriptions are out there on the internet. Longitude divides the distance around the equator into 360 degrees (which as always can be subdivided into minutes and seconds). Latitude divides the arc from the equator to the north pole into 90 degrees. The south pole is at -90 degrees.
The same complication applies i.e. longitude and latitude can be expressed in degrees, minutes and seconds or in decimal form.
One further complication exists, just as it is possible to express any angle in degrees or radians, latitude and longitude can be expressed in degrees or radians. I tend to naturally think about angles in degrees, but we are going to use radians for three very good reasons:
a) The ready-made formulae for calculating distances between points on a sphere are expressed in radians b ) The ready-made formulae for converting cartesian co-ordinates to latitude and longitude are expressed in radians c) Trigonometric functions in Excel work in radians. If you want to work in degrees you have to add in an extra conversion.
I think in degrees but I like radians. Alien mathematicians would use radians (they are universal) whereas they would have to work out why we divide circles up into 360 degrees.
Converting the star catalogue Right Ascension and Declination into Latitude and Longitude
To convert the decimal form of Right Ascension and Declination into longitude and longitude in radians use the formula:
Longitude in radians = (Right Ascension ÷24)x2π
To convert declination (decimal form) into latitude in radians use the formula:
Latitude in radians = Declination * (π÷180)
While RA runs all the way around the sphere continuously from 0 to 24 hours, longitude divides the circle into +180 degrees heading east and -180 degrees in a westerly direction rather than counting from 0 to 360.
It is easy to apply a correction to the longitude using a spreadsheet or code using the ‘if’ function. If the longitude is greater than 180 degrees (or π radians), you simply subtract 360 (or 2π in radians).
The syntax for Excel is =IF (logical test, value if true, value if false). =IF(longitude<180,longitude, longitude-360) I will edit this bit in so that the story is in chronological order.
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Post by Ron Walker on Oct 21, 2022 10:32:47 GMT -7
Posted by: Ron Walker Mar 15 2021, 10:18 AM Great Scott...I mean, Great Stuff Scott!. This is the kind of stuff that I will need to read over several times to just begin to understand what has been put down here. Math has never been my strong suit so with you doing all of the work it will be easier for the rest of us. Forge on....but beware the ides of March.
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Post by Ron Walker on Oct 21, 2022 10:33:24 GMT -7
Posted by: Scott T Mar 15 2021, 12:31 PM QUOTE(Ron Walker @ Mar 15 2021, 05:18 PM) * Great Scott...I mean, Great Stuff Scott!. This is the kind of stuff that I will need to read over several times to just begin to understand what has been put down here. Math has never been my strong suit so with you doing all of the work it will be easier for the rest of us. Forge on....but beware the ides of March.
Thanks Ron. To be honest I have always been jealous of people to whom maths seems to come naturally- this stuff takes me ages and you don't see all of the stupid things I do on the way. It has taken me years to gradually change my slightly unhealthy outlook on maths. Too many people label themselves as either good or bad at maths and that's it. We really need to think in terms of being 'good enough' at the particular bit of maths we need to achieve the task we want to do.
I will of course send my final spreadsheet to anyone here who wants it when it is done - for use as is, further refinement or whatever.
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Post by Ron Walker on Oct 21, 2022 10:33:46 GMT -7
Posted by: Scott T Mar 15 2021, 12:38 PM Converting cartesian co-ordinates to latitude and longitude
The formula for converting cartesian co-ordinates to latitude and longitude is:
longitude = atan2 (y,x)
In this formula, y and x are two of the cartesian co-ordinates we worked so hard to find.
Atan2 is a function. Fortunately, it is built into Excel so you can just get the spreadsheet to calculate the value directly from the cells containing the y and x co-ordinates with no further bother.
If on the other hand you want to know what the atan2 function is doing, in full it is:
atan2 (y, x) = 2 atan (y/√(x²+y²)-x)
where atan is short for arctangent which is another way of saying the inverse tangent or tan-1.
The conversion for latitude is given by:
latitude = asin (z/R)
Where z is our z co-ordinate, asin is short for arcsin (also known as inverse sin or sin-1) and R is the radius of the sphere.
That is fine except no one has told us what the radius of our sphere is!
A pitfall to watch out for – do you remember back in section 5.1 when we talked about the circumradius of the truncated icosahedron being 4.956? Well it is not that simple! That is the radius for the corners not the centres. We are actually expecting the radius of the centres to be slightly less than 4.956 and different for pentagonal faces and hexagonal faces.
The radius of the sphere (distance from the origin to the point on the surface) can be relatively easily calculated using pythagoras in 3 dimensions and the cartesian co-ordinates (which sounds worse than it is). The formula is:
Radius = √ (x^2+y^2+z^2)
If you plug the numbers in, the Radius for the centres of the hexagonal faces comes out at 4.534568 and the radius for pentagonal face centres is 4.654877.
If you happen to be using the pentakis dodecahedron method the two radii are 2.598076 and 2.530927.
We have reached the top of the second hill on our journey. To recap, we have both the positions of the stars and the positions of the star plate centres converted to latitude and longitude so that they are easy to compare.
Time for another cup of tea.
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Post by Ron Walker on Oct 21, 2022 10:34:07 GMT -7
Posted by: Scott T Mar 20 2021, 01:15 AM Calculating the distance between two points on a sphere.
Fortunately, the problem of finding the distance between two points on a sphere is a very common geographical problem. The equation to calculate distance looks a bit scary but for practical purposes it only needs us to plug the formula correctly into a spreadsheet and take a little care over what the terms are. The formula is sometimes known as the haversine method.
For convenience the equation is broken down into three steps:
a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2) c = 2 * atan2( √a, √(1−a) ) d = R * c
Where φ represent latitude, and λ represent longitude. (φ1 is the latitude of the star and φ2 is the latitude of the star plate centre).
Noting that sin² (Δ φ /2) is shorthand for sin(Δ φ /2) x sin (Δ φ /2).
Δ φ means the difference between the two latitudes, (the difference between the star latitude and the star plate centre latitude)
Δ λ means the difference between the two longitudes.
R is the radius of the sphere.
Which radius should we use? The circumradius, the radius of the pentagon centres or the radius of the hexagon centres? The good news is that, now we have the both the centre points and star positions plotted as latitude and longitude, it does not really matter what radius we use as long as we pick one and stick to it for the rest of the calculations. The position on the surface of the sphere of a particular point of longitude and latitude does not change relative to other points on the sphere as the sphere gets bigger or smaller.
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Post by Ron Walker on Oct 21, 2022 10:34:27 GMT -7
Posted by: Scott T Mar 20 2021, 01:17 AM A forgotten step
In the section about converting between Right Ascension and longitude I forgot to mention one thing, While RA runs all the way around the sphere continuously from 0 to 24 hours, longitude divides the circle into +180 degrees heading east and -180 degrees in a westerly direction rather than counting from 0 to 360.
It is easy to apply a correction to the longitude using a spreadsheet or code using the ‘if’ function. If the longitude is greater than 180 degrees (or π radians), you simply subtract 360 (or 2π in radians).
The syntax for Excel is
=IF (logical test, value if true, value if false). =IF(longitude<180,longitude, longitude-360)
I will edit this bit in so that the story is in chronological order.
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Post by Ron Walker on Oct 21, 2022 10:34:46 GMT -7
Posted by: Scott T Mar 20 2021, 01:22 AM The last steps
We now have a list of stars and can calculated the distance from any given star to any star plate centre.
The penultimate step is to decide how big you want your star plates. The radius of the star plate should be big enough so that there is some overlap with the next plate but not too much.
You could be clever and calculate exact star plate radii -in fact, for a regular hexagon, the radius of the circle enclosing the hexagon is the same as the length of the edge of the hexagon OR using a spreadsheet you could just keep adjusting the star plate radius until it looks about right.
You could have large star plates that snugly enclose the hexagons and smaller plates for the pentagons of the standard icosahedron BUT here is the point where serendipity kicks in.
Earlier on we were wondering whether to use a standard truncated icosahedron or an isodistant truncated icosahedron. The great news is that although the two versions are composed of different shapes (hexagons vs ditrigons) the central point of each face is in the same position on the surface of the sphere regardless of type.
What does this mean? We can make all of the star plates the same size and divide the sky into 32 equal portions. We are effectively using the isodisdant version of the truncated icosahedron.
A star ‘belongs’ to a star plate if the distance from the centre of the star plate to the star is less than the radius of the star plate. In Excel the IF function makes it easy to generate a list of stars that belong to any given star plate.
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Post by Ron Walker on Oct 21, 2022 10:35:14 GMT -7
Posted by: Scott T Mar 20 2021, 01:29 AM Test Version of the spreadsheet
To test all the steps so far, I have plugged all the equations into a spreadsheet. To keep things manageable, I have only imported about 8000 stars to mag 6.5. Here is a snapshot of the spreadsheet. It obviously extends down for thousands of rows offscreen.
And a couple of initial test plots of some example star plates to demonstrate that the principle works!
These plots are very basic – the axis are latitude and longitude in radians. You will notice the distortion from perfectly circular star plates at the equator to elongated forms as you go towards the poles – this is not wrong - just a graphing artefact because plates near the poles cover more of longitude. I will do some better ones shortly.
Next steps will be more checking and some refinements including removing duplicate stars from the overlaps.
It looks like we are basically at the destination!
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