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 Post subject: A segmented bowl...
PostPosted: Mon May 24, 2010 2:17 am 
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...is made up from rings of 32 segments. The segments have a cross section of ¾ x ¾ inch. Each segment will be 1inch long over its longest side. what is the largest bowl (external dimension) that could be turned using these rings?

Ralph :wink:

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PostPosted: Mon May 24, 2010 9:05 am 
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I'm going to say "five inches" max.
With 32 segments at 1" length (longest side) it figures out to be almost 5-1/8" circumference but, if you turn it to round and loose the corners of the segments it will reduce the diameter.
No, I'm not really that smart, I used a book called "Smoleys Four Combined Tables" to calcuate the diameter from my "assumed" circum. and guessed! :lol: :lol: :lol: :lol: :lol: :lol:

Rog

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PostPosted: Mon May 24, 2010 12:54 pm 
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Bump! :wink:

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PostPosted: Mon May 24, 2010 2:08 pm 
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A 32" circumference means a ~10.19" diameter. Taking the corners off while bringing down to round would reduce it a bit so I am going to say 10" diameter for the finished bowl.

Bob


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PostPosted: Mon May 24, 2010 2:19 pm 
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Radius = 5.077 inches.

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PostPosted: Mon May 24, 2010 4:03 pm 
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Hey Winston,

How did you come up with that? I've been playing with it, but I can't make it work without assuming a benchmark maximum of r=5.093.
32/π = 10.1859
r = 5.093

360/32 = 11.5º

Then I took the Cos of the bisected angle, assuming the middle of the chord to be the low spot.
11.5/2 = 5.75º

So, rCos 5.75 = 5.0674

The problem with my method is that the benchmark r of 5.093 is really and artifact.

Tom

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PostPosted: Mon May 24, 2010 6:16 pm 
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I did it with a measurement on a CAD drawing.

Using straight math, I get the same answer:

Half-angle = 360 / 32 / 2 = 5.625º

0.5 / tan (5.625) = 5.0766....

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PostPosted: Tue May 25, 2010 9:25 am 
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Thanks Winston,

I knew I was doing something wrong.

Tom

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PostPosted: Tue May 25, 2010 4:19 pm 
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OOOPPPPSS!! :oops: :oops:
I obviously got the wrong answer!! I quoted radius instead of diameter. I know that pi is 3.141 so the diameter whould have been closer to 10 than 5 !!DOH!!!
Must have gotten on the wrong page in my Smoley's!! :oops: :oops:
Anyway Ralph, who won the coveted "Soggy Points"? And lets get on to the next puzzle.

Rog

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 Post subject:
PostPosted: Tue May 25, 2010 6:24 pm 
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Looks to me as if Winston gets the soggy point today!

Ralph :wink:

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 Post subject:
PostPosted: Wed May 26, 2010 8:26 am 
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I think my head just exploded.....

and to think I always daydreamed in math class thinking "this stuff will NEVER come in handy in the practical world"

Lawrence


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