Author Topic: Tube Design The Equal Mass Theory  (Read 2523 times)

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Tube Design The Equal Mass Theory
« on: January 21, 2011, 00:07:01 am »
last revision 1 21 11
last revision 1 30 11 Photo0068.jpg observation on OT8

TUBE DESIGN The Equal Mass Theory
You have to ask the right question. Was Stanley
 trying to match the mass or the weight of the inner and outer tubes in the demonstration cell? Were the tubes acting like some tuning fork with acoustically equal legs with modification by slots
to account for differences in mass or area of the inner and outer tubes?
 
In an effort to get back to basics, let's take a look at the theory that the notches on the Meyer Demonstration Cell were made to adjust the weight
of the outer Tubes to equal that of the inner Tubes.
 
The basic method is to calculate the mass of the tubes by using known
and reasonable measurement of the tubes.
 
In the published diagram of the Water Fuel Cell (Demonstration Unit)
Exhibit E2: Tubular Cluster-Array on page 28 of the International Independent Test-Evaluation Report page 28, the following measurements are given:
 
Outer Tube   0.75 inches in diameter,  wall thickness 0.035 inches
Inner Tube   0.50 inches in diameter,  Wall thickness 0.049 inches
Tubes material T304 Stainless
 
Length of the tubes not provided on this page, however it has been reported to be 18 inches long for the outer tube and 19 inches for the inner tube.
 
Now for the math:
 Goal :determine volume of metal in inner and outer tubes and thus relative weights

The volume of a cylinder is given by Pi times r(squared) time L where
Pi is about 3.14, r equals the radius of the cylinder and L equals the length of the cylinder.  Think of a tube as a cylinder of metal with a smaller tube or air within it.
 
Volume of inner tube
 Pi times( 0.25 squared) equals the area of the end of the inner cylinder.
 Pi times( (0.25  minus 0.049) squared) equals the area of the cylinder of
air  within the inner tube.
Thus 0.19635 is the area of the end of the inner cylinder
 and  0.1269 is the area of the end of the cylinder of air
Thus 0.06945 is the area the end of the tube
If one multiplies this by 19 then the cubic inches of stainless steel in the
inner tube is 1.3196 cubic inches.
 
Volume of the Outer Tube
 
In a similar calculation for the outer tube:
Pi times (0.375 squared) equals the area of the end of the outer tube.
Pi times(( 0.375 minus 0.035 ) squared) equals the area of the cylinder of air within the outer tube.
Thus .44178 is the area of the end of the outer tube
and .3632 is the area of the end of the air cyclinder within the larger tube. Subtracting , one gets .0786 for the area of the outer tube.
Multiplying by 18 yields a volume of stainless steel metal of 1.4148
cubic inches.
 
So now comparing weights of the 19/18 configuration, the inner tube is only 1.3196/1.4148 or 93.2%
the weight of the outer tube. Also the amount of metal cut out for slots is negligible in terms of weight of the outer tube. (See slot correction below)
----------------
Observation on Photo0068.jpg
Outer tube OT8 clearly has a different cell wall thickness (on the order of 0.040-0.049.)  The inner tube IT8 is appears to be shorter. Was Stan trying to adjust mass ratio  to compensate for the now heavier OT, or because the smaller gap meant a change in area ratio? Only seven of the nine tube sets have the IT longer
than the OT. One of the exceptions is OT8. And it is the one that has a thicker wall. Coincidence?.
----------------------------------------------
So, now let's see if these calculations have any basis in reality.
If the cross-sectional area of the .5 x .049 tube is .06945
and the cross-sectional area of the .75 x.035 tube is .0786
.06945/.0786 =.8835

Another Check on the math

Now let's compare the calculated .06945 and .0786 figures
For a given length of tube the ratios of the calculated weights/foot should check out against the real world calculation
So I'm going to onlinemetals to use their calculator.
 
 http://www.onlinemetals.com/calculator.cfm
 
Plugging in Material Stainless
                 Alloy    304
                 Round

Size .5    Length 10 feet  Wt is 6.7588
Size .402 Length 10 feet  Wt is 4.3690
 
Calculated weight                   2.3898
 
Size .75  Length 10 feet    Wt is 15.2074
Size .68  Length 10 feet    Wt is 12.501
 
Calculated weight                   2.7064
 
  2.3898/2.7064 =  .8830 using onlinemetals calculator
So the ratios of the weight per unit of length of tube essentially match
(Slight difference due to differing values of pi or rounding)
regardless of the method.
 
So if there is no handy table of the particular tube set you are designing just calculate the weight of a bar or outside diameter
then calculate the weight of a bar that is the same diameter minus twice the wall size, and you will get the weight of the tube.
 I like the online metals calculator because it has a variety of stainless types in their data base.

The value of the tubes have about 12% difference in weight and the notches can only modify the weight by .6 to 1.2%.(if the area of the notches is between .25 and .5 inches) Why would Stan put them in the tubes unless they were important?
 
Milling, cutting or filing stainless tubes is time consuming and difficult so therefore they were there for a reason. Acoustics!
not mass not area, imho
 
Conclusion
  By looking at the various videos of the demonstration cell, it is
clear that the notches do not constitute even 10%of the surface area of the outer tube(3.1416X .75 x18) = 42.4 pi x d x l=a  about 4.2 square inches. (.25 sq inches/ for notch/38.45 sq in inside area of outside tube =0.0065 or less than 1% correction or if you assume .5 sq inches then 1.2% correction) also  (difference in wall size neglible .049 - 0.035 for wt calc).
Therefore the weight correction provided by notches is less than 1.2%
 The notches would have to be either multiple per tube or very wide and long. I think the area matching and weight matching theories are not as strong as the acoustic match theory.Go to any of the tube acoustic sites and small differences in length show big changes in frequency for open ended pipes or chimes.

To give credence to this take a look at the stress that Scott Cramton puts on acoustic tuning and also view the size of the notches on the hhoiseparator cells.
 
Cramton WFC Replication 022
or
http://www.youtube.com/user/HHOiseparator#p/a/u/0/FJzrMG__EGI

The tube lengths are comparable to Stan's Demonstration Cell but the wall thicknesses  of both tubes in the Cramton replication appear to be is much greater (perhaps Schedule 40).  The outer tube also has a thicker wall in the hhoiseparator Cramton replication and appears to be the same tube wall thickness at the inner tube, whereas in the Meyer Demonstration unit , the amount of material that is removed by the notching is much smaller (despite the fact that it's wall thickness is thinner than the inner tube)(on the order of .25-.5 square inches).  One would expect that the 2 sq inches of notch is not enough compensate for the greater weight of the outer tube to make the inner and outer tubes equal weights because the  tubes appear to the the same wall thickness( and length). And in the Meyer Demonstration Unit over 3 square inches would need to be notched out to balance out the weight of the inner and outer tubes. Because the inner tube of the Cramton replication is so much heaver relative to the outer tube an even greater amount of surface area (which you see) would need to be removed. However, the notches are comparable in length which argues for the acoustic match or tuning theory.
See Cramton area,mass,acoustic thread
 
Comparison of area

Further refinement: Correction for notch size in Meyer Cell
 
Let's say the notches are  1 inch by 1/4 inch, therefore the area of the outer tube needs to be reduced by .25 inches.
 
Area of cylinder inside of (outer tube)  3.1416 x .68 x18 = 38.45             Area of notch estimate .25 sq/in                          -0.25
 
            Corrected              area                                     38.30
 
Area of cyclinder(inner tube) 3.1416 x .5 x 18      25.274
  Ratio of area    1.52
 Clearly the ratio of  amount of area of the tube overlap is not close to a 1:1
 
Conclusion
The weight or mass and the areas of the inner and outer tubes are not greatly influenced by notches but the acoustics are!!-------------------------------------------------------------------

Thanks for the link Donaldwfc, the site you suggested is a faster calculation!!
 http://www.speedymetals.com/
« Last Edit: November 08, 2011, 13:22:35 pm by jim miller »

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Re: Tube Design The Equal Mass Theory
« Reply #1 on: January 21, 2011, 01:48:05 am »
yes


if you go to buy these tubes online at www.speedymetals.com and put in the dimensions it tell you how much each piece weighs, and they are extremely close


factor in the mounting brackets and cut a notch in the outer tube and you can match them


equal mass means the voltage pulls on the tubes the same amount, same resistance, same number of electrons


i don't understand why so many people buy huge thick tubes, you can use thin tubes, they are cheaper too, 35 thousandths is pretty thin, and the inner tube is smaller, so it's thicker to give the same mass