The Equal Area Theory
Modified 26 1 2011 added more Crampton info
Modified 30 1 2011 reconsidered with new length info
Ask the right question. Are the tubes in the demonstration cell being matched for Equal Area to match Capacitor Plate size or are they being matched for acoustic reasons??
Meyers Demonstration Cell
The Meyer Demo Cell had an outer tube diameterof 0.75
The Meyer Demo Cell had an inner tube diameter of 0.5
The wall thickness of the outer tube was 0.035 inches
The wall thickness of the inner tube was 0.049 inches
The gap between the tubes was .18 inches
The estimated size of the slots is .25 x 1 inch
The length of the outer tube has been given as 18 inches
The length of the inner tube has been given as 19 inches
The length of the tubes may be as short as 13 and 5/16ths
New info released by Dynodon
The material used was 304 stainless
see calc
.75- (2x.035) equals inner diameter of .68 inches Therefore .68-.5 =0.18 inch per inch of height of tube is the amount of material to be removed if equal area to be maintained. If the outer tube was 18 inches long ,then 18 squared/100 = 3.24 sq/in. This is the area of the notch that would be needed to maintain equal areas. See the Meyer Estate videos. If Stan was going for equal surface concept. the notch cut out is not even close to 3.24 square inches. (or maybe Stan did not make enough notches)
-----------------------------------------
Revised calc based on 13.583 inches
13.583 x 18/100 = 2.94 square inches
Assume .25 x1 sq inch /slot, 2 slots= .5 sq inch area match now closer
------------------------------------------------
)Therefore acoustic tuning by notches is more likely than surface area match.(see Lawton comment on notches) This is simlar to tuning an organ pipe and Stan being a God-fearing man probably knew this from being in church and seeing the church organ........
Crampton Cell
The Cramton cell had an outer tube diameter of 0.75 inches
The Cramton cell had an inner tube diameter of 0.5inches
The wall thickness of the outer tube was Calculable
The wall thickness of the inner tube is unk. presumes same as outer pipe
The gap between tubes was 1.2 mm
The estimated dimensions of the slots was 13 x 19 mm
The length of the outer tube after tuning was 445 mm
The length of the inner tube was 18 inches or 450mm
The material used was Stainless Steel 316L
Clamps and attached Z brackets and are factored in and adustment are made in the outer tube length to account fot this increased mass which affect the outer pips audio frequency in air. He gives a link to free audio tuning software
Total Area of outer surface of inside tube
CxPiXL 0.5(3.1416)(18) = 28.27 sq inches
Total Area of inner surface of outer tube assuming 1. mm gap
Volume of metal for outer tube was cubic inches
Volume of metal for inner tube was cubic inches
Volume of metal for outer tube with slots cubic inches
Volume of metal for inner tube overlapping slotted tube
What is the ratio of mass between inner and outer tubes?
What is the ratio of mass between inner and outer tube overlapped
What is the ratio of mass between inner and outer tubes after adjusting for slots?
Is the the amount of mass removed for the slots sufficient to match up the masses of the the two tubes?
How big would the slots have to be to match the tubes in terms of mass?
If the bottom of the slot is the new acoustic length, how many slots and what dimensions will remove sufficient mass to match the mass of the inner tube? Use multiple slots to remove sufficient mass to match the tube set acoustically
( seeLawton comment regarding length of slot0
Let L =length of slot let W = the width of the slot or sigma the
all the slots assume same depth. Then L times the W (% of circumference) x weight for 1 inch of tube. is the cubic inches of metal removed for a slot angle/360 So now use the new value
of L....... ( this is getting complex ..................lets say that half the tube is removed as a slot 180 degrees and is 2 two inches deep
then the weight removed will be the same cutting off 1 inch of tube. if the slot was 1/3 of the circumferece of a one inch tube
120 degrees and 1 inch deep the mass removed would be 1/3 the mass of a 1 inch section of tube. Therefore if you want to make the masses match you need to solve an equation that removes sufficient metal with notches to make the tubes equal in mass and acoustically the same.or VARY THE WALL THICKNESS, DIAMETER, AND LENGTH to maintain the proper gap, area overlap and weight per tube allowing for small notches or slots
to make small adjustments. Lawton "it has been suggested that the depth of the slot.." All of the above discussion assumes that the bottom of the slot is the new acoustic length. Actually if that were the case then cutting off one inch of a tube would remove the same mass as a 180 egree slot 2 inches deep but the acoustics will be different.
