OT: Balancing Act

Balancing Act...

6-armed chandelier, globes evenly (60°, 12in radius) spaced

Hung it, leans like a @#$%^

Weighed each of the globes (accurate scales)...

2# 0.8oz 1# 15.6oz 1# 12.4oz 1# 10.0oz 1# 10.2oz 2# 3.6oz

Made in China, 9.6oz between lightest and heaviest :-(

What's the best order for closest balance ?:-) ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson
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Add weight to the light ones ?

Reply to
Rheilly Phoull

My first thought, but there's no place that wouldn't be visible.

Isn't there a nice tidy mathematical solution ?>:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

6 is a small number. Even 6 factorial is small enough for modern CPUs.
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Reply to
Hal Murray

6! = 720, but I doubt that it takes that many trials... due to the symmetry. ...Jim Thompson
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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

Trial and Error? Or average of adjacent globes? (Gn-1 + Gn + Gn+1)/3

15.6 6.666 0.8 8.8 10.0 7.73 12.4 10.86 10.2 8.73 3.6 9.8

Maybe move that 12.4 around.... Don't forget to number them;)

Cheers

Reply to
Martin Riddle

They're already tagged by weight ;-) ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

Is the chandelier level with no globes installed?

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Reply to
RosemontCrest

I'll bet it's faster overall to let the CPU crank through the duplicates than it is to write the code to skip them.

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Out of the box mode: Use christmas decorations to balance things.

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Reply to
Hal Murray

What the hell does "1#" and "2#" mean? Six globes, six weights. Ah, "lbs".

Fix (any) one of the weights at (0,1) -- call it W0. (radius is immaterial -- assume all weights are point sources equidistant from a center at (0,0) -- for easy math!)

For each permutation of the remaining 5, assign the weights to (0.866,0.5), (0.866,-0.5), (-0.866,0.5), (-0.866,-0.5), and (0,-1).

Compute the center of mass as ( sum (Xi * Wi), sum (Yi * Wi) ) = (X,Y) ignoring the constant scale factor (but don't forget W0 @ (0,1)!)

[Wi being weight of i-th globe, (Xi,Yi) being the above locations]

Score this as X^2 + Y^2 (no need to take sqrt) to determine it's "off-centerness"

Minimize "score" over the set of possible permutations.

[Actually, this does twice as many evaluations as needed but a no-brainer to code in this incarnation!]

High school programming assignment. (Have I missed something obvious?)

Maybe Sparky Larkin can impress you with his POWER BASIC prowess!

:-/

Reply to
Don Y

Probably ;-) ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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Reply to
Jim Thompson

Have any extra wheel weights? :)

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Reply to
Michael A. Terrell

Maybe, as a starter: D B A F C E

Reply to
Robert Baer

Or turn on the heavy ones to make them lighter??

Reply to
Robert Baer

yes, due to the symmetry (which is 12) you only need to try 60 combinations.

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Reply to
Jasen Betts

Hey, we can be *sure* he's got plenty of solder!

Best regards,

Bob Masta DAQARTA v7.40 Data AcQuisition And Real-Time Analysis

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Reply to
Bob Masta

Probably got answered already, but two ways...use octave and 'play' with formulas, or...wade in.

Just like packing a box, start with the largest and work your way to the smallest.

Ah, ha! I see the problem now. There's a vector torque in there. This requires a 2D distribution solution. So then think in terms of trying to 'uniformly' distribute the 'mass' over the 'area'. Thinnk largest to smallest single then in pairs, then in 'sheets' of 3 etc. Wow, tougher then it first seemed! Screw it, go buy bulbs all the same weight!

Reply to
RobertMacy

Jim,

Just went through it, can't get there from here, at least not without quite a bit of unbalance! Sorry.

Reply to
RobertMacy

You'll have to put an unobtrusive 'ring' wieght around each bulb. Each weight cosmetically the same, yet each weight has a 'different weight to make each 'bulb system' the same weight.

Reply to
RobertMacy

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