Am thinking of building a small dome prototype that consists of identical prefab pentagon panels to be constructed in shop and moved to location to be secured together. Panels consist of 5 identical 2x’s with compound angles fitting the perimeter of the pentagon and sheathing added. I had the compound angles necessary at one time but the years have gone by and now they are misplaced. Any math folk out there that can answer this quest? Thanks.
Half of good living is staying out of bad situations.
The other…proper application of risk.
Replies
I think the information you need is here
http://www.nas.com/~kunkel/dodec/dodec.htm
but not in exactly the format you want. It will require some algebraic massaging. I Googled for dodecahedron, and this was the first hit. I've seen several discussions of compound angles here, and also heard that there is one on Joe Fusco's site.
oh lord help me.Half of good living is staying out of bad situations.
The other...proper application of risk.
Back in the 80s I went to a highschool geometry teacher with the same questions about the compound angles dealing with a dome of a hexagon/pentagon configuration. He said to come back in a little later. When I returned he told be the exacting compound angles for the hex/pent dome couldn't be done but gave me the single compound angle needed for the all pentagon formula. Stupid me probably put it in a special place for safe keeping and now it's gone.
Been too long to try to jump back into the math stuff. When looking for the paper with the angles on it I came across a bunch of algebra and geometry stuff I was working on at one time and those chicken scratches were all greek to me.Half of good living is staying out of bad situations.
The other...proper application of risk.
How about this? I didn't check the math on the dihedral angle or the central angle, just copied what was on the referenced web site. They may be wrong.
From the first link I gave you, http://www.nas.com/~kunkel/dodec/dodec.htm
The dihedral angle, the interior angle between two adjacent faces = 116.565 degrees.
From http://kjmaclean.com/Geometry/dodecahedron.html
The central angle, the angle between two radii of the circumscribed sphere passing through the opposite ends of the same edge = 41.810 degrees.
The end angle, the angle between an edge and a radius of the circumscribed sphere passing through the end of the edge = (180 - CA) / 2 = 69.095 degrees.
The face angle, the angle between two adjacent edges = ((5 - 2) x 180) / 5 = 108 degrees.
So the compound angle at the ends of the 5 identical 2x's making the perimeter of a pentagon panel will be a 41.810 & 69.095 degrees?Half of good living is staying out of bad situations.
The other...proper application of risk.
The 41.810 is an angle down in the middle of the dodecahedron. It is only there to allow you to calculate the end angle.
Picture a compound miter saw, with a 2x up against the fence. The upper surface is the surface that will show on the outside of the dodecahedron, then the surface against the fence is the surface that will be face to face with the corresponding 2x in the next pentagon over.
The angle you want to see on the top and bottom surfaces is 54 degrees, half of 108 degree face angle.
The angle you want to see on the front and back surfaces is the end angle, 69.095 degrees.
The bevel angle you want to see along the full length of the top of each 2x is 58.282 degrees, one half of the dihedral angle.
I hope this doesn't read as condescending. That's not my intent at all. It's just that the forum isn't close enough to real time, so I can't stop and check periodically to see if I left you behind, so I start trying to pack in every possible detail. Let my know if I've still left you behind.
It's proper to condescend to men of low estate. My math whizzes went after I finally learned how to program my VCR and now I want nothing to do with detail anymore. I'm trying to simplify life in an increasingly complex world.
Tonight I'll rip some lengths and start cutting for a model prototype as a trial run to see how it fits. If it's successful I'll post a pic before I begin on the storage unit. Thanks.Half of good living is staying out of bad situations.
The other...proper application of risk.
Unc- How many of the pentagon panels would it take to make a 1/2 sphere?Half of good living is staying out of bad situations.
The other...proper application of risk.
You can't build any closed, partial shape with complete panels. Take a look at the first picture here http://kjmaclean.com/Geometry/dodecahedron.html
In this orientation, face up, you can slice through the gray plane and get a ~3/8 sphere with one pentagon and five trapezoids.
Or you can slice through the teal plane and get a ~5/8 sphere with six pentagons and five triangles.
Or you can slice exactly between the two planes and get a 1/2 sphere with one pentagon, five hexagons and five triangles.
Face up is not the only possible orientation. You can also choose point up, or edge up, or something irregular. But there's no way you can slice it in half and come away with only complete panels.
Thanks for your time here Unc.
It was probably 20 years or so back I talked to that math teacher. Can't really recall what he said for sure. Sounds like my error in reception. Did manage to find my old paperwork for the hex/pent configuration. I've got a pic somewhere of the old dome shed I'll post when I find it to give you a chuckle. Thanks again. Half of good living is staying out of bad situations.
The other...proper application of risk.
Doesn't address the math- Unc's done a great job of that already- but here is a link with more info than you probably ever wanted to know about Buckminster Fuller's little geodesic dome idea. What a genius.
http://www.insite.com.br/rodrigo/bucky/buckyball.txt
Kevin Halliburton
"I am not a noun — I seem to be a verb." -R. Buckminster Fuller-
Edited 2/12/2003 10:44:30 AM ET by wrecked angle
Ha! He certainly was the man, wasn't he. I remember reading something of his once and he had this long, long paragraph describing something full of 5 syllable words and I got to looking at it and discovered it was one sentence. Roar!
Ol' boy sure woulda given Greenspan a run for his money.Half of good living is staying out of bad situations.
The other...proper application of risk.
Will this approach to math help at all?
Hey -- that's not so bad! At least in 2010 students wil be learning how to do real calculations again, even though in Spanish. (BTW, it should be "produccion", not "production).
Unc-Please forgive this latenight scrawl but is there a formula for finding '?' by using only 'a' or would 'b' be necessary also.
All heavy lines are equal to each other as all thinner lines are equal also.
There's no way to find it with just A, and here's why. The only things we know for sure are the length of A, the fact that A is less than B, and the fact that the central angle of A plus the central angle of B = 72 degrees. But given those three facts, I can draw an infinite number of decagons, from B = 1.0001 x A to B = 10,000,000 x A and beyond, each of which would have a different diameter.
Sorry.
But if you'll tell me the length of one edge of a full pentagon, and the length of A, I can probably calculate B, and from that calculate the diameter.
Edited 3/11/2003 2:07:51 AM ET by Uncle Dunc
Thanks unc-
a= 47"
b= 94"
Can a formula to arrive at ? be derived using a & b dimensions?
I couldn't think of a formula, which is not to say there isn't one. I had to find the distance from the center of the decagon to the ends of a and b, which also entailed finding the angle at the center between the two ends of a or b. With that data is was not difficult to calculate the distance from the center to the midpoints of a and b. Since each a side faces a b side across the decagon, the distance across the decagon is just the sum of the distances from the center to the midpoints of the two edge lengths.
center to midpoint of a:112.31
center to midpoint of b:104.67
sum: 216.98
It sounds like your dome is quite a lot bigger than I though it would be, from the picture of the pile of panels you posted.
Thanks dunc- so ? is looking like 18'1".
Depending on how well this one goes up if the next one has a=70.5" and b=141" then could I deduce that ?=325.47
Yep. As long as a is exactly half of b, the same proportions will hold.
So we are saying that d and e are not equal distances and it is as the image portrays from the center point?
That's right. I've been assuming that all ten points lie on a circle. If that's true, then the distance from the center to the midpoints of the short sides has to be longer than the distance from the center to the midpoints of the long sides. That may be easier to visualize if you draw the circle that passes through all ten points.
Because you drew it that way, I've also been assuming that it's a convex polygon, meaning it doesn't zig-zag in and out, and that it is five-way symmetric. If both of those are true, then all ten points are constrained to lie on a circle. It can't be any other shape.
Ya, I see the lengths affecting the distance. Not sure of the definition of the convex polygon. These are all symmetrical flat panels I believe making the base design in question 5way symmetrical , the basic structure containing 10 hexagons, 5 half hexagons and 6 pentagons.
The only place that has an ever slight concave are the 5 'b' lengths which compose the bottom sill plates of the 5 half hexagons of whose tops are mounted directly beneath the in-leaning pentagons and the lower sides of the bottom hexagons. This base plate of the half hex ends up sitting slightly inside the plane of the same half hex top side.
The 5 pentagons in turn mount their lower sides to the upper sides of the bottom hexagons, and the top sides of the pentagons each mounting to the bottom sides of the top hexagons, the top hexagons upper sides each mounting to an adjoining top hexagon upper side completing the circle. The top plate of those top hexagons forming a same size pentagon.
The whole structure leans in on itself except for the bottom half hexagons.
Well, why didn't you just say it looks like half of a soccer ball? :)
I assume you're putting a door in one of the bottom hexes? When are you going to bolt it together?
well now that I've got the deck plan :O)
Going to put it two block high off the ground on a deck.
Yep door in a hex. Kind of testing the waters for a larger one if this goes well. Using 6 foot lengths instead of 4 foot would make around a 27 foot dia. and still be able to handle the panels.
Uncle Dunc-
I was wondering if I might pester you again.
In the previous posts on this thread describing the pent-hex configuration, each pentagon panel is made from 5 identical sticks with the same compound angle on the ends of each. The hexagon panel composed of it's own compound angle also.
In the pic here is a dome constructed of what seems to be all hexagons instead of the hex/pent structure.
Is it possible to produce an all hexagon panel dome style building as what is shown here using the same design strategy of identical compound angles on the end of each identical stick as found in the afore mentioned hex/pent design?
Thus the skeleton frame of this all hexagon panel dome would be made using all identical sticks with the same compound angles on each.
If so, is there a formula used to decipher what would be the compound angle on the ends of those hexagon sticks, and what would that compound angle be?
I much appreciate your responses earlier in this thread. Thanks
Wow! That was a year ago already? Time flies, etc.
Did you get that one bolted up? Got pictures? Already posted the pictures and I just forgot?
>> Is it possible to produce an all hexagon panel dome style building as what is shown here
>> using the same design strategy of identical compound angles on the end of each identical stick ...
No. To prove it to yourself, draw a honeycomb pattern on a piece of paper and try to shape it into a dome. Can't be done without stretching or shrinking some of the edges. You can make a cylinder with regular hexagons, but not a sphere.
Look at the horizontal strut at the right center edge of the picture. The left end of the strut meets the right point of a hex. (Duh!) Follow across that hex to the opposite point, and the strut coming out of that point is nowhere near horizontal. Even if each strut is perfectly straight, the trend line through them is curved, and the curve is much sharper than the circumference of the dome. The only way that curve can happen is if the sides of each hex are not all the same length. Extended far enough, that distortion is going to squeeze some hex until one of its sides disappears, and it comes out as a pent instead of a hex.
Well ya, I see that now. Just jumped the gun in hopes of a new adventure.
The panels for the previous dome structure are awaiting in a shed. Was out of state working for 7 unexpected months and blew my summer.
I have the lumber for the floor decking frame which will set atop about a dozen of these 2ft high 2ft dia fiberglass cylinder type housings.
I stand there looking at all this stuff and wonder what I was thinking and then turn around and think about chasing another one. ROAR!
I saw a book in a bookstore once titled 'Diary of a Mad Builder'. Sure wish I'd have bought that piece of literature;o).
Hey Uncle,
It's me again.
Got a question.
The dome floor shape as shown with the five 4ft and five 8ft sides equally distributed as the are.
Using 2x8's around the perimeter, is 36 degrees the common angle cut on the end of each of the 4 and 8 ft boards or would the two different lengths alter it?
Also, is Merlot pronounced "mer-low"?
If the sides were all the same length, the end angle would be 72 degrees, not 36. Draw a line from each point to the center of the circle and you get ten triangles. The central angle of each triangle is 360 degrees divided by 10 = 36 degrees. The sum of the three angles of a triangle is 180, so 180 minus 36 = 144, divided by two for the two equal angles = 72 degrees.
The central angle of the 4' sides is 23.64 degrees, and the end angle is (180 - 23.64) / 2 = 78.18 degrees. The central angle of the 8' sides is 48.36 degrees, and the end angle is (180 - 48.36) / 2 = 65.82 degrees. These angles will be valid for any size platform as long as the long edge is exactly twice as long as the short edge.
>> Also, is Merlot pronounced "mer-low"?
That would be my guess, but my French is a good bit weaker than my trig.
Sounds like I ought to just break down and go learn some trig.
Thanks.
Uncle Dunc-
If I wanted to enlarge the decking perimeter making the 4ft lengths 48 1/4 inches would that mean lengthing the 8ft lengths to 96 1/2 inches to keep the angle cuts in sync?
Theoretically, yes. But given the limited precision of the tools you're likely to be using to lay out your cuts, I don't think you need to worry about it.
You will need to change the layout of the panels that intersect with the 48-1/4" edge, but again, I don't think you'll need to change the angles that you cut on the panel edges, just widen that edge of the panel by 1/4".
Edited 5/9/2004 5:32 pm ET by Uncle Dunc
Well, actually the lengths were 47" and 94" and I fear the precision of the base on a deck like this up on pylons. There's little room for tweaking once the hexes get tied up together.
The whole concept is an 'I'll make it fit' deal so anything I can do to help help smooth the operation ahead of time I try to add.
If I mis-calculated it'll be either a headache and time or a bonfire.
Trying to cover the sill seams with the 4x8 sheeting for strength as in the diagram. You think I can get away with 5/8osb instead of 3/4?
>> You think I can get away with 5/8 osb instead of 3/4?
Yeah, I think you can get away with it in the sense that no point load will break through it. I have no idea how much it will flex when you walk on it. What are you planning for joists?
I went back and read through the thread to see if you ever said what you were going to use the dome for, and missed it if it was there. If I were building it for shop, I'd want a pretty damn stout floor, like joists 16" OC and 1-1/4" T&G plywood, or two layers of something thinner glued and screwed together with all the joints offset. Or maybe not, when it came time to actually go down to the lumberyard and write the big check.
If it's going to be a storage building, 3/8" might be adequate.
>> Sounds like I ought to just break down and go learn some trig.
The amount of trig you'd have to learn is really not all that much.
1) There's a mnemonic for remembering the trig functions of a right triangle, SOH CAH TOA. If that's not memorable enough, maybe you could remember it as "suck a toe." For each of the two acute angles, where opposite is the length of the side across from the angle and adjacent is the length of the side next to the angle that is not the hypotenuse,
sine = opposite / hypotnuse
cosine = adjacent / hypotnuse
tangent = opposite / adjacent
I got through your dome using the tangent almost exclusively, maybe used the sine a few times.
2) A squared + B squared = C squared, where A and B are the lengths of the two short sides of a right triangle, and C is the length of the hypotenuse.
3) The sum of the interior angles of any triangle is 180 degrees.
Of course you do have to remember enough algebra to manipulate the equations. Also, a scientific calculator is much easier than lugging around a book of trig tables.
Dunc-
It'll end up a storage shed with wood storage sliding underneath. It's kind of a test dome to see if I can pull it off without a lot of problems.
Previously done two 14ft dia domes of same configuration using 35 inch sides for the panels. Both are just riding on a single layer of 4inch flackblock, no mortar, vapor barrier with a drymix of peagravel and cement, tamped.
The idea is if this is successful I think the same format could be used with larger prefab panels on a block stemwall with slab or riding on a shallow pressure treated foundation for an economical structure built easy when ready to shell out the green and do something right. Shooting for 28ft dia. cabin or garage.
This dome has PT 2x8s for the deck framing, longest span at 10ft, on center 24".
The reason the question of 5/8 or 3/4 is I've managed to keep the cost down on this pretty good. Bulk buy of PT 2x8 lumber from a bigbox cast off 'cause of bark showing and such, something like 25% of original cost, the 7/16 osb from back in the glory days of $4.50 a sheet and the framing lumber from a pallet of pallet building material a buddy sold me cheap. Stuff adds up fast and it is just a shed.
And still got to get the roofing. I'll post some pics as it all progresses. Thanks for the data.
Also, is Merlot pronounced "mer-low"?
Exactement! Vous l'avez bien dite! Alors, débouchez la bouteille, et nous allons arroser ça...!Dinosaur
'Y-a-tu de la justice dans ce maudit monde?
Exactement! Vous l'avez bien dite! Alors, débouchez la bouteille, et nous allons arroser ça...!
Well ya, I think that probably sounds something like what I said when I saw the price of 3/4" t&g osb board today. Local lumber yard $34.99! fer cryin' out loud.
The place also has the last remaining supply of 80 CCA pressure treated 10ft 2x8s at $9.99 a board. Got me to thinking.
15 bucks more an area sheet and I could have 2x flooring. Hmmm.
ROAR!!!
If you said what I wrote, you probably convinced the lumber yard their price wasn't high enough, LOL!
Exactement! Vous l'avez bien dite! Alors, débouchez la bouteille, et nous allons arroser ça...! Translated, that means: Exactly! You said it well! Right, now--uncork that bottle; let's drink to that!
Dinosaur
'Y-a-tu de la justice dans ce maudit monde?
rez;
Uncle dunc has his numbers right but don't forget that if you cut the lumber on a miter saw it reads 0 degrees for a 90 degree cut ... so miter saw settings for the respective cuts would be 11.82 degrees and 24.18 degrees (90 degrees - 78.18 degrees = 11.82 degrees and 90 degrees - 65.82 degrees = 24.18 degrees). Most table saw miter attachments read similarly too.
...arising from the bowels of the earth.
He's crazy I tell you, crazy!
you gonna shoot the SOB who dumped all them washer tubs in yer yard? Or are they dryer drums? Can't quite tell from here.
View Image
Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations.
sludge pump housings from a local factory.
PEEE ---EEEWWW...wash yer hands before ya touch them beans.
View Image
Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations.
A little late, but you might find this interesting.
http://www.page.sannet.ne.jp/jun_m/tenkaizu_dl/index-eng.html
I can see it now.
Hinged houses of the future.
Portable dome structures delivered to your site and erected in one hour.
Let's see, I think I bought a pile of piano hinges at an auction a while back...now where'd I put those things. ;o)
Dunc- Thanks for the help with the math.
Things came out pretty much on the money. I owe ya one.
Glory be! It fits. Math really works.
Lucky guess....
Life is not a journey to the grave with the intention of arriving safely in a pretty and well preserved body, but rather to skid in broadside, thoroughly used up, totally worn out, and loudly proclaiming.... WOW!!! What a Ride!
Math really works.
Ya, well, now that you mention it.
I misplaced the paper I made with all the base layout dimensions on it.
Had measured and recorded the points for the location of the hexes and halfhexes taken off the decking frame of the first so when I did the next one it would be easy street and I can't recall the process I used on the first.
Remember this diagram I'd made off my CAD program...View Image
On this floor diagram the thin lines of the perimeter are 47" and the thick lines are 97" in length. The red dots are the known staked points for a hexpanel and a halfhex.
The shed will rest on railroad ties and what I need to do is place the ties down before the dualhex panel units are placed on them.
Yet my lightening fast brain has failed to come up with an easy solution to find the measurements needed to locate the remaining RRtie locations for the hexpanels.
Might you be up to the challenge of the math problem so I can get the show on the road once the snow melts?
Cheers
Parolee # 53804
All ten points lie on a circle. The radius of the circle is 114.74". Make yourself a giant trammel. If you don't know where the center is, swing an arc from each of your four known points. They should all intersect in one point, or at least a fairly small quadrangle. From the known center, draw a circle. It should pass through each of your four known points. Then make two smaller trammels, 47" and 94", and alternate lengths around the circle starting from your four known points.I calculated the radius of the circle from the drawing. The distance from the center of the circle to the midpoint of the shorter edge is 112.31". The line from the center meets the edge at a right angle, so the midpoint of the edge, the end of the the edge, and the center of the circle form a right triangle. The hypotenuse of that triangle is the radius of the circle.radius = square root ((112.31 * 112.31) + (23.5 * 23.5))
= square root (12613.5361 + 552.25)
= square root (13165.7861)
= 114.74 (rounded)Just as a cross check, I repeated the calculation with the longer edge and got the same value, when rounded to two decimal places.
Beats a giant protractor :o)
Thanks
Parolee # 53804
>> Beats a giant protractor :o)That was my thought, too.>> ThanksGlad I could help.
Try visiting
http://www.outwater.com
http://www.shelterpub.com/_shelter/domebuilder%27s_blues.html Try this site lots of info not sure if it has the formula your looking for but interesting reading...
It was interresting .
Tim Mooney
And can be true depending on the builder.
Rez,
Any experience using 3d CAD solid modeling software and have access to it? Just like a calculator has replaced pen and paper, 3d CAD has also replaced what it is think you are trying to do/calculate. Where I work we use EDS Unigraphics. It's pretty high end stuff, but it's impossible to make a mistake once the part is modeled. Anyway there are alot of cheap (free?) programs out there that can do what you need.
Jon
I go brain dead after 5minutes anymore and for a onetime thru problem the learning curve is too great. But thanks.