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Segmented Sphere

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Working on a large prototype segmented sphere made from pine 2x materials, each ring segment is 1.5 x 1.5 in dimension starting out with an approx. 19-inch diameter ring. Is there a magic ratio or formula I can use to determine the length of the next outer ring as I am progressing from the largest ring working my way down to the next smaller ring in the stack?
I usually make a full scale drawing and make measurements for each stack and determine the segment lengths from those measurements. I was trying to wrap my head around this geometry but I am a little distracted with multiple issues this week. With the segment dimension being square I would think there would be a ratio or multiplier to use to calculate the progressively shorter segment lengths? Old timers is setting in this week. A circle is very uniform? I am starting with (12) 15 degree 5-inch long segment for the largest ring, the next ring stack is about 4-7/8 inches long, the next stack is progressively shorter as you work your way out to the last ring.
Is there a ratio or multiplier that can be used against the first segment length to determine the next segment length?
 

Dennis J Gooding

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Your choice of terms is a bit confusing can you clarify a bit? I assume that since your are using "2x" materials that the ring thickness is to be 1.5 inches for all rings. I don't understand what you mean by "each ring segment is 1.5 x 1.5 in dimension.... " . Are you not using wedge-shaped segments?
 
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Dennis,

Yes, wedge shaped as in 12 pieces per ring cut at 15 degree, so I will have longer segments around the equator of the sphere. The ring of segments at the equator are 5-inches long, so each progressive ring segment will be shorter, I was wondering if there was a ratio, factor, multiplier that you can use against the 5-inch length to determine the length of the segment for the next progressive segment ring stack. I have this project drawn to scale and have the needed measurements, but was wanting to know if there was a quick way to calculate the lengths on other sized spheres without drawing it to scale and taking measurements for each ring stack diameter.
 

Bill Boehme

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Dennis,

Yes, wedge shaped as in 12 pieces per ring cut at 15 degree, so I will have longer segments around the equator of the sphere. The ring of segments at the equator are 5-inches long, so each progressive ring segment will be shorter, I was wondering if there was a ratio, factor, multiplier that you can use against the 5-inch length to determine the length of the segment for the next progressive segment ring stack. I have this project drawn to scale and have the needed measurements, but was wanting to know if there was a quick way to calculate the lengths on other sized spheres without drawing it to scale and taking measurements for each ring stack diameter.

Is five inches the diameter of the ring?

If I understand what you are asking, the problem I see with trying to scale the pieces based on knowing the dimensions for one particular size sphere is that it would require planing the thickness of the rings in proportion to the size of the sphere. How about software for segmented turning? It will give you a cut list with precise dimensions (saves time), you can do 3D visualization, and play what-if with different configurations (wastes time).
 
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Bill,

5 inches is the length of the largest set of (12) segments at 15-degrees making a 19 inch diameter ring. Staying with the 1-1/2 inch thick segment stacks was causing me some grief as the width of the rings need to increase the farther out you go from the equator. I am trying to stay busy working on a project to keep my mind occupied, I have a funeral to attend on Monday and these types of projects keep me from dwelling on other issues.
 
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I finished this prototype a day ago to work through the process for making a large piece for a customer, the shape will be a little different.

vase.jpg
 
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Update : Did rough calculation and don't think you stock will be wide enough (inside to outside diameter) to make a sphere. That is as you get further away from the equator, the radius for the ring will change more than the width of the stock. The calculation below would be for the second ring. You need to just use .75 to get the first.
-------------
It is some simple math -- ha -- not explained very well and the picture probable is much help. You just need to find the side of the the triangle that is formed from the center to the outermost point of your new ring. This will be the radius of the the circle and from that you can find the segment size for you wedge.
A is the top out side of your equator ring.

B is the top outside of your next ring.

C is the center of your equator ring ( .75 thickness)

F is the top outside of your third ring.

These are the points of triangles formed from the center to the outer ring. The height

of the unknown side of triangle is the radius of your new ring and from that you can find the approximate wedge length to cut.

A -C, B-C, F-C are the radius of your sphere and is 9.5

If I understand this correctly you want to find the segment size to make each ring B and F

You need to find the radius of B-D and F-G etc. and from the radius the segment length.

B-D is square root of ( B-C squared minus C-D squared)

C-D is .75 + 1.5 ( ½ of the stock thickness [center ring] and the 1.5 of the new ring.

C-D is 9.229 exact

The circumference is of the new ring is 2*pi() *C-D. = 57.991

now each wedge you cut should be 57.991/12 == 4.832

Continue in like manner for rest.

Good luck.
This picture is useless. It intent is to show a cross section of the sphere with each layer moving to center.

I 's are the thickness of you ring (1.5)


F G
I
I
I

B D
I
I
I
A
I
I ------------------------- -------------------------------------- C
I
 
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Thank You Bob,

You are correct, the width of each progressive stack going away from the equator needs to be wider to provide enough material for turning.
 

Dennis J Gooding

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Mike,
Yes, there is a simple way to calculate the outside diameter of any given ring. If R is the radius of the sphere (half the diameter), h is the distance outward from the center of the sphere to the base of the ring in question then the required diameter d of the ring is given by

D = 2 x squareroot [ R-squared minus h-squared]

(I would give a nickel for an equation editor to use here.)



If you were making a 19-inch diameter sphere using 1.5 inch layers, you would need 13 layers and the diameters of the successive disks, measured from the center outward, would be:

19.00, 18.94, 18.46, 17.45, 15.84, 13.37, 9.42
 
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Here’s another way to calculate the width of the rings.
Take a piece of big graph paper and put down a center line. Draw the shape of the piece on the left using the the size of the ring you want finished say 3/4 thick when turned so you have the outline of the shape, outside and inside line. Figure out how thick each layer is and mark that line to the center. Ex, if your rings are going to be 1 inch tall you would have block 1 in high from the bottom to the top.

Now you have the thickness laid out so back to the outline of the vase. For each segment add about 1/2 in to each side of the line so each segment is the width you plan plus 1 inch. This is how wide each segment will be. Now measure from the center line to the outside of the line that has the overage and that’s how big you need to make the ring. Ex if the center to outside is 3 inches then the total ring is 6 inches. Measure for each ring and you have how big the rings need to be made.

Now you can make all of the rings since you have the width. I make everything at 15 degrees so I need 12 pieces per ring.

I can post a photo of my plans of someone want to see it. It’s a pretty easy way to do this and there’s no complicated math.
 
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Dennis,

That is pretty close to what I came up with, the only issue was the segment ring width had to change the further away from the equator I got.

Don,

That is the process I usually use when making a segmented project, with a sphere being uniform and revolving around 360 degrees I was hoping
there was a quick and easy factor to use when determining segment lengths on each progressive stack. It only takes a few minutes to draw the project
up on paper and take measurements for each stack diameter and determine the segment lengths.
 

Bill Boehme

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(I would give a nickel for an equation editor to use here.)

If you are using Windows don't forget about the character map app. As long as you use one of the eight fonts that are in the forum editor I believe that it should display correctly. The default Verdana typeface family has some commonly used math symbols such as square root (√), numeric superscripts (so that you can write things like x² + y²), superscript lowercase "n", numeric subscripts, Greek alphabet, degrees symbol (°, you can also type this from the keyboard by holding down the ALT key while typing 0176 on the numeric keypad ... not the numbers on the main keyboard), ⅛ through ⅞, delta ∆, infinity ∞, summation ∑, integral ∫, differential ∂, plus-minus ±, relational symbols (> < ≤ ≥ ≠ ≡). I have an app for my Wacom Intuos tablet that enables me to use the pen to write a complex equation and it will convert it to a ready-to-publish equation (PDF graphic format).
 
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Mike, I do not want to complicate your life but...
You can vary the thickness of each layer to maintain the ratio of height to width of the segments normal to the surface of the turned sphere. As you get closer to the top or bottom, the layers would get progressively thinner. Not easy but would look cool.
Jerry
 

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I was wondering if it couldn't be done like building an igloo. Of course figuring out the compound miter for each layer would be challenging.
 
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I was wondering if it couldn't be done like building an igloo. Of course figuring out the compound miter for each layer would be challenging.

John,
The thought had crossed my mind and then I woke up and reality slapped me along side the head. :D
I would assume each and every segment would be the same size with the same angle cuts on all four sides.
That would certainly require precision angle cuts and a jig to secure the piece for cutting and sanding to attain accurate angles.
 
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I was wondering if it couldn't be done like building an igloo. Of course figuring out the compound miter for each layer would be challenging.
Why did you have to bring that up!@#$%^&*()_+
I got to thinking about how someone would accomplish that and decided that each layer would need to be angled on each side of the width of the strip of wood, then to cut the individual pieces you could clamp the piece to the fence and use the same angle as doing flat segments I think or maybe not. This could drive a person crazy its to bad that wood isn't as easy to fit together as snow.
 

Dennis J Gooding

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If you are using Windows don't forget about the character map app. As long as you use one of the eight fonts that are in the forum editor I believe that it should display correctly. The default Verdana typeface family has some commonly used math symbols such as square root (√), numeric superscripts (so that you can write things like x² + y²), superscript lowercase "n", numeric subscripts, Greek alphabet, degrees symbol (°, you can also type this from the keyboard by holding down the ALT key while typing 0176 on the numeric keypad ... not the numbers on the main keyboard), ⅛ through ⅞, delta ∆, infinity ∞, summation ∑, integral ∫, differential ∂, plus-minus ±, relational symbols (> < ≤ ≥ ≠ ≡). I have an app for my Wacom Intuos tablet that enables me to use the pen to write a complex equation and it will convert it to a ready-to-publish equation (PDF graphic format).

Bill, if type a posting using the MS Word equation editor and the Verdana font can I import the .docx file directly to this site or would I have to convert the equations to an image file and import the result separately from the text?
 

Dennis J Gooding

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Well shucks, I guess I just answered my own question. Looking back, I find that I posted an article on a simple aid for turning spheres in the tutorials and tips forum that included equations with the text.
 
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Dennis,

Somewhere between 1/2 and 3/4 of an inch depending on how well I stacked the rings together.
Think about it ! The largest ring in the center and then one of two following would work at say 3/4" but as you progress in the circle the segments must be wider. The best way would be to draw 2 circles to represent the inside and out side then draw parallel lines to represent the layers and determine the width of the segments and the diameter for each layer.
 
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Don,

I quickly figured out on the large sphere that the further out away from the equator you went the wider/deeper each segment ring needed to be to provide enough wood to make the curve of the radius. You have plenty of room around the equator and the first couple of rings heading north and south, after that you need to add about 1/2 an inch of width/depth to each segment ring so you have a little fudge factor to work with.
 

Dennis J Gooding

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Mike, did you see my recent post in the Tutorials and Tips Forum "Calculation of Ring Sizes for Segmented Spheres"? It should answer your questions.
 
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