last revision 1 21 11
last revision 1 30 11 Photo0068.jpg observation on OT8
TUBE DESIGN The Equal Mass TheoryYou 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)
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Observation on Photo0068.jpgOuter 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?.
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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 mathNow 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 022or
http://www.youtube.com/user/HHOiseparator#p/a/u/0/FJzrMG__EGIThe 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 ConclusionThe 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/
