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Post by Ron Walker on Oct 21, 2022 9:59:18 GMT -7
Posted by: sgrijalva Mar 9 2021, 10:08 AM Wow !. Excellent information, congratulations !. Now I am trying to build a new projector with the shape of a truncated icosahedron, not optical because it is out of my economic scope but a pinhole type as was my initial idea but I ended up building one in the shape of a dodecahedron 20 years ago. I have some information so far but yours will be very helpful. Thank you for your contribution, I will be pending for the next ones.
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Post by Ron Walker on Oct 21, 2022 10:04:27 GMT -7
Posted by: Ron Walker Mar 9 2021, 02:07 PM "This hobby is a disease", says it all.
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Post by Ron Walker on Oct 21, 2022 10:05:21 GMT -7
Posted by: Scott T Mar 9 2021, 02:25 PM Thanks sgrijalva - hope this continues to be useful.
Part 5 How do you sort the stars from a star catalogue onto the correct star plate?
To my mind this is one of the key problems in making an optical projector.
I took a very rough and ready approach for my first experimental projector but for this version I want a computer to ‘do it properly’. I don’t mind admitting this problem has had me scratching my head for a long time. How do you allocate each star from a star map or catalogue to the correct face of the truncated icosahedron? I am sure proper mathematicians or geometry specialists can solve this efficiently and quickly with their wizardry (as indeed I think Paul Bourke and Jamie Hicks have done on their thread).
There is a way to do it with simple individual steps. It needs google, a spreadsheet and some perseverance. I am going to illustrate every step here. It will take a few posts.
Here is the basic ‘route map’. We need to visualise the basic plan before we delve into the nitty gritty detail of how to actually to do it.
1. The starting point is to plot the central point of each of the 32 faces of the truncated icosahedron onto an imaginary sphere.
2. We then plot all of the stars from a star catalogue onto the surface of the same sphere.
3. For any given star plate, we will calculate the shortest distance between the centre of the star plate and an individual star (see diagram)
4. The star and star plate ‘belong together’ if the shortest distance from the centre of the plate to the star (calculated in step 3) is less than the radius of the star plate.
5. Repeat steps 3 and 4 until you have done it for every star and each of the 32 star plates.
6. Remember to eliminate any duplicate stars that ‘belong’ to more than one star plate (because they are positioned in the slight overlaps between faces).
All of these steps could be done by hand (but it would be horrendously time consuming). A computer code or spreadsheet will carry out the individual calculations nearly instantaneously. To be continued.
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Post by Ron Walker on Oct 21, 2022 10:07:13 GMT -7
Posted by: Ron Walker Mar 9 2021, 02:55 PM QUOTE(Scott T @ Mar 9 2021, 02:25 PM) * Thanks sgrijalva - hope this continues to be useful.
Part 5 How do you sort the stars from a star catalogue onto the correct star plate?
To my mind this is one of the key problems in making an optical projector.
I took a very rough and ready approach for my first experimental projector but for this version I want a computer to ‘do it properly’. I don’t mind admitting this problem has had me scratching my head for a long time. How do you allocate each star from a star map or catalogue to the correct face of the truncated icosahedron? I am sure proper mathematicians or geometry specialists can solve this efficiently and quickly with their wizardry (as indeed I think Paul Bourke and Jamie Hicks have done on their thread).
There is a way to do it with simple individual steps. It needs google, a spreadsheet and some perseverance. I am going to illustrate every step here. It will take a few posts.
Here is the basic ‘route map’. We need to visualise the basic plan before we delve into the nitty gritty detail of how to actually to do it.
1. The starting point is to plot the central point of each of the 32 faces of the truncated icosahedron onto an imaginary sphere.
2. We then plot all of the stars from a star catalogue onto the surface of the same sphere.
3. For any given star plate, we will calculate the shortest distance between the centre of the star plate and an individual star (see diagram)
4. The star and star plate ‘belong together’ if the shortest distance from the centre of the plate to the star (calculated in step 3) is less than the radius of the star plate.
5. Repeat steps 3 and 4 until you have done it for every star and each of the 32 star plates.
6. Remember to eliminate any duplicate stars that ‘belong’ to more than one star plate (because they are positioned in the slight overlaps between faces).
All of these steps could be done by hand (but it would be horrendously time consuming). A computer code or spreadsheet will carry out the individual calculations nearly instantaneously. To be continued.
Just thinking out loud but one wonders if the entire circle of stars is actually smaller then the corresponding five or six sided shape. The proper expansion to the screen (overlap) would be dependent on the focal length of the particular lenses used. Perhaps this is why commercial projectors are limited to very little change in dome diameter before one would need to change the fl of the lens. Using a zoom lens would make the projector more of a universal instrument for many size domes. But then again commercial star plates are adjusted and drilled according to dome size. Technically at a particular fl a lens will expand coverage as a dome gets bigger so a particular fl should work for any size dome. Then one wonders if star plates are made for a particular size dome because while the projector remains the same size, the offset from the center to the end of the dumbbell to the projected image on the dome must change position on the plate as the projection angle changes for a given star as the dome changes size.
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Post by Ron Walker on Oct 21, 2022 10:07:43 GMT -7
Posted by: Scott T Mar 10 2021, 11:37 AM Hi Ron - absolutely, I think Jamie Hicks stroke of genius on his thread was to plan to use zoom lenses. There is a downside in terms of light absorption with more optical elements, cost and limitations on the size of the star plate if you use standard photographic zoom lenses but it is a really, really good way to go.
I will get to the sensitivity of the design to dome diameter, focal length etc a bit later- but I am going to press on with sorting the stars onto the right star plate.
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Post by Ron Walker on Oct 21, 2022 10:08:06 GMT -7
Posted by: Scott T Mar 10 2021, 11:40 AM 5.1 The co-ordinates that define the centres of the 32 star plates The birds-eye view of the route to matching stars to their star plates is fairly simple and clear, but when you actually set off on the journey it quickly becomes apparent that the road is not paved with smooth tarmac all the way. You need to take some twisty paths to keep going toward the destination. Our starting point is to find the co-ordinates of the centres of the star plates. A google search quickly throws up some formulae in Wikipedia which can provide the co-ordinates of the vertices (corners) of a standard truncated icosahedron. At time of writing, I could not find formulae for the co-ordinates of an isodistant truncated icosahedron so we are going to work on a standard truncated icosahedron for a while (but we will return to the isodistant form later). Similarly, I could not find an equivalent formula that directly calculates the co-ordinates of the centres but we will calculate the positions of the vertices first, then use those to get to the centres.
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Post by Ron Walker on Oct 21, 2022 10:08:58 GMT -7
Posted by: Scott T Mar 10 2021, 11:45 AM This was my first wobbly moment. It’s not too hard to use the penultimate equation to work out that the value of the squiggle -phi (ψ) is 1.61 and that 3 ψ is 4.85 and so on. I have to admit I didn’t really understand what is meant by an ‘even permutation’ and did not feel confident to get the values of all 60 vertices correct even after trying to digest Wikipedia’s permutation entry. Fortunately, when the road looked like it had hit an impassable ravine google provides a handy footbridge to cross the obstacle. A google search for ‘coordinates of vertices of polyhedra’ brought up a full list of all 60 vertices! It was a bit like having the answers at the back of the maths book – the questions look so much easier when you have the answers to look at! The beginning of the list looks like this: 0, 1, 4.854101966249685 0, 1, -4.854101966249685 0, -1, 4.854101966249685 0, -1, -4.854101966249685 and goes on for a full 60 lines (I will give a full list in a more useful format soon). Cartesian co-ordinates are simply distances along the x,y and z axes from a starting point (origin) and can be expressed in any suitable unit – inches, mm or just unitless numbers. We need to remember the last line from the Wikipedia section - that these co-ordinates apply to a truncated icosahedron with an edge length of 2 i.e. the length of the edge of each pentagon or hexagon is 2. The circumradius is 4.956 (also mentioned in the last line of the Wikipedia section) and is the radius of our imaginary sphere which touches every corner of the truncated icosahedron. Rather than just blindly trust the internet, I plugged these co-ordinates into gnuplot- which is a free graph plotting programme from the internet – at the moment, Excel (which is my usual spreadsheet) does not support plotting 3 dimensional x,y,z scatterplots. gnuplot is not really user friendly and takes a little effort to get used to but before too long it showed that the co-ordinates are indeed from the corners of a truncated icosahedron.
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Post by Ron Walker on Oct 21, 2022 10:09:21 GMT -7
Posted by: Scott T Mar 10 2021, 12:34 PM Finding the centres Knowing the co-ordinates of the vertices is a really good step, but we really want to know the co-ordinates of the centres of each of the faces. If there is a ready-made list of these values out there on the internet I could not find it. Fortunately, the co-ordinates of the centre of a regular hexagon or pentagon can be calculated by simply taking the average value of the co-ordinates of its corners! So, all we need to do is work out which co-ordinates in the list of 60 vertices belong to which face of the icosahedron. Unfortunately, the list of co-ordinates is a fairly random list. The co-ordinates are not grouped together conveniently into the faces of the truncated icosahedron. You may have the sort of mathematical brain that can look at that big long list of numbers and work it out in your head which co-ordinate belongs to which hexagon (or pentagon). I am afraid I do not - so I needed a simple way of doing it. There was nothing else for it – I borrowed my son’s football, drew on an equator and meridian and started sorting through the list of co-ordinates by writing them on the ball. This helped me to group the co-ordinates together. Here is a picture. I had to work fast in case he wanted his ball back!
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Post by Ron Walker on Oct 21, 2022 10:10:01 GMT -7
Posted by: Scott T Mar 10 2021, 12:36 PM One more thing to note. There is no universal systematic way of giving each face a name or number – I simply called them 1 to 32 but in no real order. It does not really matter as long as you know which face you are talking about at the end. Here is the order I used:
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Post by Ron Walker on Oct 21, 2022 10:21:17 GMT -7
Posted by: Scott T Mar 10 2021, 12:44 PM So finally, after going through all that here is complete list of vertices grouped into 32 faces. The top left value for Pentagon 4 now correctly reads +1 not -1 as originally posted
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Post by Ron Walker on Oct 21, 2022 10:22:38 GMT -7
Posted by: Scott T Mar 10 2021, 12:47 PM And here is how you calculate the centre point for each face - you just take the average value of x,y and z of the co-ordinates of the hexagon or pentagon:
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Post by Ron Walker on Oct 21, 2022 10:25:30 GMT -7
Posted by: Scott T Mar 10 2021, 12:49 PM and here is the important bit - the list of the co-ordinates of the centre points of each face. Edited post to give the list sorted into hexagons and pentagons- a bit more useful. Post re-edited to correct the typo on pentagon 4
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Post by Ron Walker on Oct 21, 2022 10:26:03 GMT -7
Posted by: Scott T Mar 10 2021, 12:52 PM Phew- we have reached the top of the first hill! Time for a cup of tea. If you have made it this far there are only two more big hills to climb and they are not quite as steep as this one.
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Post by Ron Walker on Oct 21, 2022 10:27:11 GMT -7
Posted by: moonmagic Mar 10 2021, 06:35 PM QUOTE(Scott T @ Mar 10 2021, 02:52 PM) * Phew- we have reached the top of the first hill! Time for a cup of tea. If you have made it this far there are only two more big hills to climb and they are not quite as steep as this one.
Well, Let me firstly admit much of this is now beyond me, but I did read it all while struggling with comprehension. (Your explanations are strong, but my mind is not) I had my chuckle moment when you listed step#5 ( 5. Repeat steps 3 and 4 until you have done it for every star and each of the 32 star plates.)
About then, I had FULL justification for my lore of finding and procuring old optical projectors. It is certainly easier than designing and making your own! (humor intended)
It is humbling for sure to see your and other's work. If I attempted to make my own projector it would look as if a five year-old had been given a wood block, a little scrap of metal, a short section of wire, a crayon, and a piece of duct tape with which to secure the pieces together.
This is truly an AMBITIOUS project and I wish you the best as you continue and I will be following along as you post.
From this side of the pond's vantage point, I also enjoyed a quick laugh with the mention of your son's football. BEFORE I saw the photograph where you had plotted onto YOUR football, I had this quick vision of the shape of OUR American football. Then it occurred to me: IF someone would use that shape, compute the configurations for the stars WE would generate the financial support for MANY MORE NEW optical planetariums in this country. You see, there is ALWAYS money to support American football especially at College/University levels, but little money can ever be found for the promotion of the sciences.
Cheers! mm
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Post by Ron Walker on Oct 21, 2022 10:27:30 GMT -7
Posted by: Ron Walker Mar 10 2021, 06:37 PM For someone who has trouble with math you're doing a remarkably fine job.
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