NASA'S MICROWAVE PROPELLANT-LESS THRUSTER ANOMALOUS RESULTS:
CONSIDERATION OF A THERMO-MECHANICAL EFFECT
J. J. Rodal, Ph.D.
ABSTRACT:
It has been argued that the anomalous results found by NASA's Brady et.al for microwave cavities (that supposedly act as a propellant-less thruster) cannot be due to thermal effects because a) the temperature increase would need to amply exceed several degrees C to be explained by thermal effects and b) thermal effects take place too slowly (minutes) and cannot explain the impulsive response of the thrust pendulum exhibiting a rise to full amplitude in half the pendulum's period (rise to full amplitude in little over 2 seconds).
These (analytically unsupported) arguments are invalidated here: a thermo-mechanical effect (thermal buckling) is shown that occurs in less than 1 second (for the copper thickness that has been argued as employed for the microwave cavity), with a temperature increase of a degree C or less and that results in forces of the same magnitude (microNewtons) as reportedly measured by NASA.
1. CALCULATION OF TEMPERATURE INCREASE
Considering the copper thickness at the big circular flat base of Brady et,al.'s truncated cone to be thermally insulated (by the printed circuit board material) at the surface z=0 and be subject to heat (energy per unit time, per unit area) "heatFlux" at the surface z=thickness, the complete transient solution for the temperature increase is (where "time" is time in seconds) [Carslaw, H. S., and J. C. Jaeger]:
deltaT=(heatFlux*thickness/thermalConductivity)*( (time/fourierTime) + ((1/2)*((z/thickness)^2) - (1/6)) + temperatureSum
where:
temperatureSum=(heatFlux*thickness/thermalConductivity)*(-(2/(Pi^2))*NSum[(((-1)^n)/(n^2))*(Exp[- (time/fourierTime)*((n*Pi)^2)])*Cos[n*Pi*(z /thickness)], {n, 1, Infinity}])
fourierTime = (thickness^2)/thermalDiffusivity
thermalDiffusivity = thermalConductivity/(density*heatcapacityperunitmass)
For copper, we have the following material properties:
density=8940 kg/(m^3);
thermalConductivity=390 (J/s)/(m*degC);
heatcapacityperunitmass=385 J/(kg*degC);
therefore:
thermalDiffusivity = 0.00011331 m^2/s
fourierTime = 8825.38 thickness^2 s /m^2
So we see that steady-state conditions occur very fast due to the very small thickness.
For example, for thickness = (0.025 inch) * (25.4/1000 m/inch) = 0.635 mm
fourierTime = 0.00355862 s
Therefore the temperatureSum term is negligible for time responses exceeding milliseconds. The term ((1/2)*((z/thickness)^2) - (1/6)) is also negligible in comparison with (time/fourierTime), so essentially we are left with
deltaT ~ (heatFlux*thickness/thermalConductivity)*(time/fourierTime)
~ heatFlux*time / (density*heatcapacityperunitmass*thickness)
For copper,
density*heatcapacityperunitmass = heatcapacityperunitvolume = 8940*385 (* J/((m^3) * degC)*)
= (3441900 J)/(degC m^3)
hence,
deltaT ~ (heatFlux* time)/(3441900 thickness) degC m^3/ J
where heatFlux has units of W/m^2, thickness in meters and time in seconds.
Calculation of the Heat Flux:
For the transverse electric mode TE012 (p. 18, Table 2. Tapered Cavity Testing: Summary of Results) of the "Anomalous Thrust..." paper by Brady et.al., the Input Power was 2.6 Watts.
The input power gets converted into heat (by eddy-currents from the magnetic field) and since for the transverse electric mode TE012 only the axial magnetic field is non-zero in contact with the big diameter base (the small end was insulated by a polymer dielectric), the heat flux is:
heatFlux = InputPower /FluxedArea = 2.6 W /FluxedArea
where the FluxedArea is:
(Pi/4)*( BigDiameter^2 ) / (heatedDiameterRatio^2)
where
heatedDiameterRatio = BigDiameter / DiameterOfAreaExperiencingHeatFlux
accounts for the fact that the magnetic flux in mode TE012 contacts only a fraction of the entire circular areas at the ends of the truncated cone.
heatFlux = InputPower * (heatedDiameterRatio^2) / ( (Pi/4)*( BigDiameter^2 ) )
and substituting this into the expression for deltaT
deltaT = heatFlux*time/(density*heatcapacityperunitmass*thickness)
=InputPower*(heatedDiameterRatio^2)*time /
( (Pi/4)*(BigDiameter^2)*density*heatcapacityperunitmass*thickness )
For example, the TE012 truncated cone input power and using the copper material properties:
InputPower=2.6 J/s
density=8940 kg/(m^3);
heatcapacityperunitmass=385 J/(kg*degC);
deltaT =InputPower*(heatedDiameterRatio^2)*time /
( (Pi/4)*(BigDiameter^2)*density*heatcapacityperunitmass*thickness )
=(( heatedDiameterRatio^2) * time)/( 1.03972*10^6 * (BigDiameter^2) * thickness) degC m^3 / s
2. CALCULATION OF TEMPERATURE AND TIME AT WHICH BUCKLING OCCURS
The thermal (membrane) stress of a restrained plate produced by a temperature difference "deltaT" is simply:
sigma = (ElasticModulus/(1- poissonRatio))*alpha*deltaT
where alpha is the coefficient of thermal expansion, ElasticModulus is the modulus of elasticity and poissonRatio is the Poisson's ratio of the plate's material. (See for example Noda et.al p.414 or Roark, top of page.583, Nr. 2).
From Timoshenko (p. 391 Eq. 9.16) or Roark (Table 35, Nr. 11, p.554, referring to Timoshenko's solution), the buckling (membrane) stress of a simply supported circular plate is:
sigma = 0.35*((thickness/plateRadius)^2)*(ElasticModulus/(1-poissonRatio^2))
and since plateRadius = BigDiameter/2
sigma = 1.4*((thickness/ BigDiameter)^2)*(ElasticModulus/(1-poissonRatio^2))
Therefore, equating both expressions the temperature difference that will produce buckling of the circular plate is:
bucklingdeltaT =((thickness/ BigDiameter)^2)*( 1.4/( alpha *(1+poissonRatio)))
For copper:
alpha = 17*10^(-6) 1/degC
poissonRatio=0.3
therefore, for a circular copper plate, the temperature difference that will produce buckling is only related to the square of the ratio of the thickness to the diameter of the circular plate as follows:
bucklingdeltaT =((thickness/ BigDiameter)^2)*63348.4 degC
For example, for
thickness = (0.025 inch) * (25.4/1000 m/inch) = 0.635 mm
BigDiameter =0.2793 m = 10.996 in (aero's estimate)
bucklingdeltaT =0.33 degC
So, very low temperature differences between the plate (and the rest of the truncated cone) are required to buckle it. The thinner the plate, the lower the temperature difference (between the plate and the rest of the truncated cone) that is required to buckle it.
Equating the expression for the deltaT required for buckling with the deltaT expression obtained at the end of section 1, we have
InputPower*(heatedDiameterRatio^2)*time/ ((Pi/4)*(BigDiameter^2)*density*heatcapacityperunitmass*thickness )
=((thickness/ BigDiameter)^2)*( 1.4/( alpha *(1+poissonRatio)))
therefore the time at which buckling occurs is:
bucklingtime=1.4*(Pi/4)*density*heatcapacityperunitmass*(thickness^3)
/ (InputPower*(heatedDiameterRatio^2)*alpha*(1+poissonRatio)))
For the following (copper) material properties
density=8940 kg/(m^3);
heatcapacityperunitmass=385 J/(kg*degC);
alpha = 17*10^(-6) 1/degC
poissonRatio=0.3
we get the following time at which buckling occurs:
bucklingtime =(5553.17*thickness) ^3 /(heatedDiameterRatio^2 InputPower) J/m^3
and for the InputPower=2.6 J/s
bucklingtime =(4038.47*thickness) ^3 /heatedDiameterRatio^2 s/m^3
For example, for
thickness = (0.020 inch) * (25.4/1000 m/inch) = 0.508 mm
heatedDiameterRatio =4
bucklingtime = 0.54 seconds
3. CALCULATION OF BUCKLING AND POST-BUCKLING DISPLACEMENT
The originally flat, circular plate, simply supported at its edges, under in-plane stress, buckles into a stress-free spherical shape.
Denote by xbar and ybar the horizontal rectangular cartesian coordinates and by zbar the vertical cartesian coordinates of the spherical buckled and postbuckled state centered at the origin of these coordinates, such that
xbar^2 + ybar^2 + zbar^2 = R^2
where R, a function of time, R(time), is the radius of curvature of the buckled and postbuckled shape.
Define a new set of rectangular cartesian coordinates with the origin vertically displaced upwards such that w(x,y) is the vertical coordinate displacement of the buckled shape with respect to the original flat configuration:
x=xbar
y=ybar
w(x,y) = zbar - (R - wmax)
Such that
w(0,0) = wmax
and the boundary conditions:
w(BigDiameter/2,0)=0
w(0,BigDiameter/2)=0
Then,
x^2 + y^2 + (w(x,y) - wmax + R)^2 = R^2
w(x,y) = wmax + Sqrt[ R^2 - x^2 - y^2 ] - R
w(x,y) = wmax + R (Sqrt[ 1 - (x/R)^2 -( y/R)^2 ] - 1)
which satisfies w(0,0) = wmax identically. While the other two equalities give
w(BigDiameter/2,0)= wmax + R (Sqrt[ 1 - ((BigDiameter/2)/R)^2 ] - 1) = 0
w(0,BigDiameter/2)= wmax + R (Sqrt[ 1 - ((BigDiameter/2)/R)^2 ] - 1) = 0
giving:
wmax = R (1 - Sqrt[ 1 - ((BigDiameter/2)/R)^2 ] )
and hence
w(x,y) = R (Sqrt[ 1 - (x/R)^2 -( y/R)^2 ] - Sqrt[ 1 - ((BigDiameter/2)/R)^2 ])
The thermal strain is simply
epsilonT = alpha *deltaT
where alpha is the coefficient of thermal expansion and deltaT the temperature difference. In the stress-free buckled configuration, this strain must be equal to the change in length divided by the original length:
epsilonT =( theta*R - (BigDiameter/2)) / (BigDiameter/2)
= alpha *deltaT
where theta is the angle, measured at the origin of the xbar, ybar, zbar coodinated system, measured between the vertical coordinate zbar and the simply supported ends. Therefore, this angle theta is:
theta = (1+ alpha*deltaT) * (BigDiameter/(2*R))
Also, from the definition of the angle theta, we know:
Sin[theta] = (BigDiameter/(2*R))
Which gives the following transcendental equation for the radius of curvature R of the buckled and post-buckled shape:
Sin[(1+ alpha*deltaT) * (BigDiameter/(2*R))] = (BigDiameter/(2*R))
A solution of this transcendental equation for arbitrarily large deformations would involve Elliptic functions (as in the Elastica solution), but since the coefficient of thermal expansion of copper is very small (alpha = 17 *10^-6 1/degC) and the temperature differences involved in this problem are small (deltaT ~ a few degrees C), it is known that
(1+ alpha*deltaT) ~ 1
such that
Sin[(1+ alpha*deltaT) * (BigDiameter/(2*R))] ~ (BigDiameter/(2*R))
Therefore we can use perturbation solution of the transcendental equation, by expanding the sine of theta as follows:
[(1+ alpha*deltaT) * (BigDiameter/(2*R))] - ( [(1+ alpha*deltaT) * (BigDiameter/(2*R))]^3)/3! + ... = (BigDiameter/(2*R))
giving:
alpha*deltaT - ( (BigDiameter/(2*R))^2) *( (1+alpha*deltaT)^3) /6 = 0
and solving for R:
R = (BigDiameter/2) * ((1+alpha*deltaT)^(3/2)) / Sqrt[6*alpha*deltaT]
= (BigDiameter/2) / Sqrt[6*alpha*deltaT]
Therefore
w(x,y) = ( (BigDiameter/2) / Sqrt[6*alpha*deltaT] ) * (Sqrt[ 1 - 6*alpha*deltaT *(x/((BigDiameter/2))^2 -6*alpha*deltaT *( y/((BigDiameter/2))^2 ] - Sqrt[ 1 - 6*alpha*deltaT ])
Expanding the square root terms, since (1+ alpha*deltaT) ~ 1, gives
w(x,y) = ((BigDiameter/4) * Sqrt[6*alpha*deltaT] ) * (1 - (x/(BigDiameter/2))^2 - ( y/(BigDiameter/2))^2)
therefore the maximum displacement of the buckled and postbuckled shape, occurring at the center of the circular plate (x=y=0) is given by:
w(0,0) = wmax = ((BigDiameter/4) * Sqrt[6*alpha*deltaT] )
and we check again, that the boundary conditions w(BigDiameter/2,0)=0 and w(0,BigDiameter/2)=0 are satisfied by this expression.
Recalling the previously derived expression for the buckling deltaT:
bucklingdeltaT =((thickness/ BigDiameter)^2)*( 1.4/( alpha *(1+poissonRatio)))
it follows that the buckling displacement at the center of the plate is:
wmaxBuckling=((BigDiameter/4)*Sqrt[6*alpha*(((thickness/BigDiameter)^2)*(1.4/
(alpha*(1+poissonRatio))))] )
wmaxBuckling = 0.724569* thickness / Sqrt[1+poissonRatio]
wBuckling(x,y)=0.724569*thickness*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2)/ Sqrt[1+poissonRatio]
and for poissonRatio = 0.3
wmaxBuckling = 0.635489 * thickness
wBuckling(x,y) = 0.635489 *thickness*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2)
So, the buckling displacement at the center of the plate is only a function of the thickness of the circular plate: it does not depend on the diameter of the plate, the coefficient of thermal expansion, or the temperature difference.
Now I derive the postbuckling displacement at the center of the plate, which is a function of time. In the previously derived expression:
w(0,0) = wmax = ((BigDiameter/4) * Sqrt[6*alpha*deltaT] )
if we substitute the previously derived expression for deltaT
deltaT =InputPower*(heatedDiameterRatio^2)*time /
( (Pi/4)*(BigDiameter^2)*density*heatcapacityperunitmass*thickness )
wmax = ((BigDiameter/4) * Sqrt[6*alpha*( InputPower*(heatedDiameterRatio^2)*time /
( (Pi/4)*(BigDiameter^2)*density*heatcapacityperunitmass*thickness ))] )
wmax=heatedDiameterRatio*Sqrt[((3/(2*Pi))*alpha*InputPower*time)/
(density*heatcapacityperunitmass*thickness)]
w(x,y,time)= heatedDiameterRatio*Sqrt[((3/(2*Pi))*alpha*InputPower*time)/
(density*heatcapacityperunitmass*thickness)]
*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2)
For the following (copper) material properties and InputPower:
density=8940 kg/(m^3);
heatcapacityperunitmass=385 J/(kg*degC);
alpha = 17*10^(-6) 1/degC;
InputPower=2.6 J/s;
this give the postbuckled displacement as a function of time, plate thickness and heated diameter ratio:(for input power=2.6 watts)
wmax=heatedDiameterRatio*Sqrt[time)/thickness]/403847. m*( (m/s)^(1/2))
w(x,y,time)=heatedDiameterRatio*Sqrt[time)/thickness]*(1/403847)
*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2)
4. CALCULATION OF POST-BUCKLING SPEED, ACCELERATION AND INERTIAL FORCE RESULTANT
Similarly we can compute the partial derivatives of the postbuckled displacement with respect to time: the speed and the acceleration:
dw/dt=(1/2) heatedDiameterRatio*Sqrt[((3/(2*Pi))*alpha*InputPower)/
(density*heatcapacityperunitmass*thickness*time)]
*(1-*x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2)
d^2w/dt^2=(-1/4) heatedDiameterRatio*Sqrt[((3/(2*Pi))*alpha*InputPower)/
(density*heatcapacityperunitmass*thickness*(time^3))]
*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2)
which for
density=8940 kg/(m^3);
heatcapacityperunitmass=385 J/(kg*degC);
alpha = 17*10^(-6) 1/degC;
InputPower=2.6 J/s;
gives
dw/dt=(heatedDiameterRatio/(807695. *Sqrt[time*thickness]))
*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2) m/s (s*m)^(1/2)
d^2w/dt^2= - (heatedDiameterRatio/((1.61539*10^6) *Sqrt[thickness*time^3]))
*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2) m/s^2 ((s^3)*m)^(1/2)
The postbuckling speed decreases with time, inversely proportionally to the square root of time. The postbuckling acceleration decreases with time as the inverse of time^(3/2).
To compute the inertial force resultant, we need to integrate the acceleration across the whole surface of the circular plate. To do this is most convenient to express the acceleration in polar coordinates r and phi (where r is the radial in-plane polar coordinate measured from the center of the plate and phi is the in-plane azimuthal polar angle) instead of the rectangular coordinates, using the transformation x=r*Cos[phi] and y=r*Sin[phi]:
d^2w/dt^2=(-1/4) heatedDiameterRatio*Sqrt[((3/(2*Pi))*alpha*InputPower)/
(density*heatcapacityperunitmass*thickness*(time^3))]
*(1-(r*Cos[phi]/(BigDiameter/2))^2-(r*Sin[phi]/(BigDiameter/2))^2)
m/s^2 ((s^3)*m)^(1/2)
= (-1/4) heatedDiameterRatio*Sqrt[((3/(2*Pi))*alpha*InputPower)/
(density*heatcapacityperunitmass*thickness*(time^3))]
*(1-(r/(BigDiameter/2))^2) m/s^2 ((s^3)*m)^(1/2)
inertialReaction=density*thickness*Integrate[r*d^2w/dt^2,{r,0 BigDiameter/2},{phi,0,2*Pi}}]
=(-1/4) heatedDiameterRatio*Sqrt[((3/(2*Pi))*alpha*InputPower)/
(density*heatcapacityperunitmass*thickness*(time^3))]
*density* thickness *( BigDiameter^2)*Pi/8
=(-1/32)*heatedDiameterRatio*(BigDiameter^2)
*Sqrt[((3Pi/2)*density*alpha*InputPower* thickness)/
(heatcapacityperunitmass *(time^3))]
Essentially, due to the boundary conditions (no out-of-plane deflection at the edges of the circular plate) the simply-supported circular plate has half the inertia as if the whole plate accelerated with the same acceleration of the center of the plate.
For
density=8940 kg/(m^3);
heatcapacityperunitmass=385 J/(kg*degC);
alpha = 17*10^(-6) 1/degC;
InputPower=2.6 J/s;
inertialReaction=-(((BigDiameter^2)*heatedDiameterRatio*Sqrt[thickness /( time^3)])/ 460.129)
N (s^(3/2) m^(-5/2) )
The inertial reaction force at buckling is obtained by replacing the expression for the bucklingtime:
bucklingtime=1.4*(Pi/4)*density*heatcapacityperunitmass*(thickness^3)
/ (InputPower*(heatedDiameterRatio^2)*alpha*(1+poissonRatio)))
inertialReaction = - (((alpha^2) (InputPower^2) (BigDiameter^2) (heatedDiameterRatio^4)
Sqrt[((1 +poissonRatio)^3)] )/(16.9964 *density*(heatcapacityperunitmass^2)*thickness^4))
and for
density=8940 kg/(m^3);
heatcapacityperunitmass=385 J/(kg*degC);
alpha = 17*10^(-6) 1/degC;
InputPower=2.6 J/s;
poissonRatio=0.3;
inertialReaction = - (BigDiameter^2)* (heatedDiameterRatio/( 1669.99*thickness) )^4 microN m^2
The inertial reaction force is a very nonlinear function of the plate thickness (to the fourth power ! )
5. EXAMPLES: THICKNESS, BUCKLING TIME AND INERTIAL FORCE RESULTANT
For:
BigDiameter =0.2793 m = 10.996 in (aero's estimate)
inertialReaction = - (heatedDiameterRatio/( 3159.94*thickness) )^4 microN m^4
So, for example we can compute the following table:
thickness (in)/ (mm) Buckling Time (sec) heatedDiameterRatio Buckling reaction Force (microNewtons)
0.027/ 0.6858 0.590120 6 -58.7626
0.023/ 0.5842 0.525285 5 -53.8171
0.018/ 0.4572 0.393413 4 -58.7626
0.014/ 0.3556 0.329074 3 -50.8071
0.009/ 0.2286 0.196707 2 -58.7626
0.0045/ 0.1143 0.0983534 1 -58.7626
For:
BigDiameter =0.397 m = 15.63 in (Fornaro's estimate)
inertialReaction = - (heatedDiameterRatio/( 2650.44*thickness) )^4 microN m^4
So, for example we can compute the following table:
thickness (in)/ (mm) Buckling Time (sec) heatedDiameterRatio Buckling reaction Force (microNewtons)
0.033/ 0.8382 1.07743 6 -53.2034
0.027/ 0.6858 0.849773 5 -57.2553
0.022/ 0.5588 0.71829 4 -53.2034
0.016/ 0.4064 0.491212 3 -60.1722
0.011/ 0.2794 0.359145 2 -53.2034
0.0055/0.1397 0.179572 1 -53.2034
6. CONCLUSIONS
I have shown that a thermo-mechanical effect (thermal buckling of the base of the truncated cone) can account for some of the "anomalous" results reported by NASA's Brady et.al. I have shown that the buckling time is under 1 second for copper thicknesses under 0.84 mm (33 thousands of an inch) and just 2.6 watt power input. I have shown that the buckling temperature increase required is of the order of 1 deg C or less. I have shown that thermal buckling can produce a sudden output response.
I have shown that the calculated buckling forces agree with the measured force (55.4 microNewtons). The buckling force is a very strong function of plate thickness (to the fourth power), to prevent thermal buckling from occurring it suffices to have a thicker copper sheet (1/8 inch or thicker would completely prevent this thermal buckling under these input powers).
This thermal buckling effect does not depend at all on air as a conducting medium; it will take place in a complete vacuum as well, since the axial magnetic field in the transverse electric mode TE012 results in heating of the copper by producing eddy currents on it.
Thermal buckling of a thin copper sheet produces extremely small reaction forces (microNewtons) and as such it is the kind of effect that is usually disregarded in experiments. It is of possible concern here due to the experimental methodology of using very small power inputs (2.6 watts in mode TE012) to measure very small forces in the torsional pendulum.
7. APPENDIX
Cotterell and Parkes (based on Cotterell's Ph.D. thesis at the University of Cambridge) correctly point out that the distribution of the heat flux "is not significant in the problem" of thermal buckling of a circular plate, whether the heating takes place uniformly over the whole circular plate or is concentrated in a central region. Cotterell chose a distribution with a heatedDiameterRatio =1/0.3=3.333 instead of the heatedDiameterRatio=1 analyzed by Noda et.al. The fact that the exact distribution is not significant for the deltaT that will produce buckling or for the buckling displacement follows from equilibrium: the membrane stress (=E*alpha*deltaT) force resultant (the integral of the membrane stress through the thickness) is is reacted at the simply supported edges (that constrain the in-plane displacement). The membrane force resultant is uniform and it is equal in the polar radial and angular (azimuthal) directions. If only a central area is heated, the membrane stress is still equilibrated throughout. If the plate has uniform thickness and isotropic material properties, the strain in the non heated area prior to buckling is the same as in the heated area.
The fact that the solution satisfies that these buckling variables are independent of the heated area distribution is shown by the fact that these variables are indeed independent of the heatedDiameterRatio:
bucklingdeltaT =((thickness/ BigDiameter)^2)*( 1.4/( alpha *(1+poissonRatio)))
wBuckling(x,y)=0.724569*thickness*(1-(x/(BigDiameter/2))^2-(y/(BigDiameter/2))^2)
/ Sqrt[1+poissonRatio]
8. REFERENCES
Brady, D, White, H., March, P., Lawrence, J., and Davies, F., Anomalous Thrust Production from an RF Test Device Measured on a Low-Thrust Torsion Pendulum, 50th AIAA/ASME/SAE/ASEE Joint Propulsion Conference, Propulsion and Energy Forum, July 28-30, 2014, Cleveland, OH
Carslaw, H. S., and J. C. Jaeger, Conduction of Heat in Solids, Oxford University Press; 2nd edition (April 10, 1986), ISBN-10: 0198533683
Cotterell, B., and Parkes, E. W., Thermal Buckling of Circular Plates, (United Kingdom's) Aeronautical Research Council, Ministry Of Aviation, Reports and Memoranda No. 3245, September, 1960
Noda, N., R. Hetnarskj, Y. Tanigawa, Thermal Stresses, CRC Press; 2nd edition (October 27, 2002), ISBN-10: 1560329718
Roark, R. J., and W.C.Young, Formulas for Stress and Strain, McGraw-Hill Book Company; 5th edition (February 1976) ISBN-10: 0070530319
Timoshenko, S. P., and J.M. Gere, Theory of Elastic Stability, McGraw-Hill; 2nd edition (1961), ISBN-10: 0070647496
I remember that Mach and Einstein ended up being at odds. Mach never accepted Einstein's theory. Check for yourself.
Alto. You make it sound like there was a scientific feud between 'em. There was not.
Einstein could not integrate Mach's principle in his work, but Einstein also realized that his own work was incomplete regarding the definition of inertia.
Sorry. An immaterial bit of background history regarding Woodward's humility regarding Mach, but still an immaterial bit that adds nothing of substance to the pragmatic implementation of Woodwards's work.
In short. Who cares?
The point is, and history shows, that Mach didn't accept Einstein's theory. The reason Mach didn't accept Einstein's ideas is because Einstein didn't accept Machian inertia. Einstein stuck with Newton......
Now to the application of this (and why it is important), and the logical paradox that follows from saying that nothing with the Woodward effect violates Einstein's theories.
Quoting @Ron Stahl: And no, there is nothing in Woodward's work that is inconsistent with Einstein.
Yet Mach's principle is central to the Mach Effect. Mach rejected Einstein. Therefore Mach effects are
inconsistent with Einstein. Einstein operated on Newtonian inertia, which was inherent to matter itself.
Mach effects don't operate using Newtonian inertia.
So Woodward is going his own way. This is no real new insight.
Indeed Woodward acknowledges his split from Einstein several times in his book:
Here's some language from the book, which I have excerpted to show Woodward's disdain for Einstein, in favor of pure Machian ideology, (history and countless experiment shows this is a bad move).
Preface XVII: Woodward commenting on Einstein, "Guided by his version of the Equivalence
principle and what he later called Mach’s principle, he also ignored the standard
techniques of field theory of his day."
Pages 18 & 22 Woodward uses the terminology "
so-called Einstein Equivalence Principle". Showing in my view not complete acceptance.
Yet on page 123, He acknowledges EEP as correct: "The reason why the Equivalence Principle is important in this case is that it
asserts that the active gravitational, passive gravitational, and inertial masses of an object are
the same. So, if you vary one of the masses, the other masses change, too. If this aspect of the
Equivalence Principle is correct (
and it is), then it is almost trivial to show that mass
variation has serious propulsive advantages."
He acknowledges Einstein is correct in every way, but except how inertia works.Here's why this is import.
One of the major pitfalls in science, and indeed here on this forum, (whereby picking one theory vs another) is
the problem of black and white thinking.
It is clear that neither Newtonian inertia, nor Machian inertia fit the bill. So the truth must be somewhere in the middle. As I have quoted Feynman as saying many times before, "All mass is interaction." The genius of Feynman should not be overlooked, nor should his observations. Now please bear with me while I humbly expand on what the Great Feynman said, "All mass (including inertial mass) is (all) interaction." The true origin of inertial mass is interaction with all fields, near and far. Dr. McCulloch seems to acknowledge this with his theory of MiHsC, whereby he postulates at the edge of galaxies, where gravitational interaction is low and accelerations are low, inertia may in fact behave very differently.
Most things in this world operate on a spectrum.
So should we.
In closing, it makes no sense to dismiss a theory (or pick a horse) based on such black/white thinking. If you want to kill a theory, find paradoxes, as I tried with Mach effect thrusters a couple pages back. It makes no sense to dismiss Mach's ideas on inertia, or Newton, or Haisch&Rueda, or other "QVers." Dollars to donuts, they are all correct. Instead, formulate a theory (or at least your own understanding) on your own that takes into account ALL available information. Not favored information. Open your mind; find the truth.
The reason I keep saying that METs, MLTs, Qthrusters and all that other stuff are related is because of the above reasoning. You can say they're Machian, Quantum Vaccumian, or whatever. Both ideas share the same continuum.
We could save the world further delays by really trying to suss out the similarities of the all these thrusters. Let me start. Electromagnetic interaction (pulsed and RF) with dielectrics supposedly interacting with distant matter and/or the Quantum Vacuum, causing motion.
We need to unify those ideas.
How about putting a MET in a tapered frustum resonant cavity and see what happens?
Op ed complete.
Next subject:
Quoting @Ron Stahl
"Tomorrow's Momentum Today"
Firstly thank you for making that idea clear on this forum. That really intrigued me, because while I've been trying to study the QV as much as possible lately, I saw this video by some professor who said that he and other believe that Quantum Fluctuations (particle pairs) are briefly taking energy from the very immediate future (picoseconds) holding onto it for a short period of time (picoseconds) and then return that energy to the past when they annihilate.
I'm telling you, we're talking past each other on these supposedly opposing thruster paradigms, they are two faces of the same coin my friends.
Edited some spelling
Alto. You make it sound like there was a scientific feud between 'em. There was not.
Actually there was. I suggest you read the essays devoted to this here:
http://www.amazon.com/Machs-Principle-Newtons-Quantum-Einstein/dp/0817638237If you don't want to pay for it, get it on interlibrary loan. It's the single best source of info about the historical issues between Einstein and Mach.
Einstein could not integrate Mach's principle in his work, but Einstein also realized that his own work was incomplete regarding the definition of inertia.
It is anyone's guess what Einstein might have accomplished if he hadn't had the falling out with Mach, but certainly Einstein did not take a stand concerning the origin or inertia, save to coin the term "Mach's Principle" and he did base much of GR on those notions. Saying he built Mach's Principle into GR would be overstating the issue. GR is completely compatible with Mach's Principle, but not reliant upon it.
The reason Mach didn't accept Einstein's ideas is because Einstein didn't accept Machian inertia. Einstein stuck with Newton......
Neither of these statements are true. You just are shooting from the hip in ignorance here.
Yet Mach's principal is central to the Mach Effect. Mach rejected Einstein. Therefore Mach effects are
inconsistent with Einstein.
You're having a terrible time trying to form logical syllogisms and really, I dunno what to recommend accept to say, this above is historically, factually and logically wrong. Mach effects are not inconsistent with Einstein. That's just silly and preposterous.
Einstein operated on Newtonian inertia, which was inherent to matter itself.
No, we just agreed that Einstein suspended judgement about inertia, and he did. You're straining at stuff and making claims with no reason to suppose you might be correct. Read the book above so you don't make these mistakes. Barbour makes all this quite clear.
So Woodward is going his own way.
No, he's not. You do not understand Einstein, you do not understand Mach and you do not understand Woodward. I suggest you actually READ these folks before making any more claims like this.
Indeed Woodward acknowledges his split from Einstein several times in his book:
Never. He would never. Not once. Show me where I'm wrong here. I am telling you, never would Woodward deny Einstein is correct. You are making this stuff up and need to start reading for comprehension.
Pages 18 & 22 Woodward uses the terminology "so-called Einstein Equivalence Principle". Showing in my view not complete acceptance.
You're fabricating evidence to support your conclusion. Woodward would never agree to what you're assuming in order to get your conclusion. You do realize when you put words in other people's mouths like this, you run dangerously close to liable?
Yet on page 123, He acknowledges EEP as correct: "The reason why the Equivalence Principle is important in this case is that it
asserts that the active gravitational, passive gravitational, and inertial masses of an object are
the same. So, if you vary one of the masses, the other masses change, too. If this aspect of the
Equivalence Principle is correct (and it is), then it is almost trivial to show that mass
variation has serious propulsive advantages."
He acknowledges Einstein is correct in every way, but except how inertia works.
Einstein never took a stand about the origin of inertia. He liked Mach's explanation, but after their falling out he realized he did not need to rely upon Mach's Principle to get GR.
One of the major pitfalls in science, and indeed here on this forum, (whereby picking one theory vs another) is the problem of black and white thinking.
Oh for cryin' out loud, don't you dare accuse me of being an adolescent. Black and white thinking is a characteristically adolescent trait and anyone who has had ad psych knows this. Your thinking is utterly clouded by your lack of familiarity with the real source materials. You can't draw conclusions because you're operating from ignorance. I suggest you read those sources and stop making claims about people's views that you are for the most part, unfamiliar with.
PZT is the simplest and cheapest way to test theory. The stuff is very cheap on Ebay, has fairly large dE/dT and under certain conditions can generate millions of gees accelerations. Also, PZT can provide 1w piezomechanical action to generate a 2w Mach Effect, and 2w electrostrictive action to rectify that M-E into useful force, both from the same signal; so this is simpler than having to provide 2 separate signals to the thruster. The down side is the phase angle between the piezo and electrostrictive is locked by the material, so you don't have control over it.
For mastery purposes, using a material with one or the other of these electromechanical actions would be preferable. Almost all materials are electrostrictive in some measure, but not all are piezoactive, so choosing a strong electrostrictor with no piezo action affords this opportunity to demonstrate mastery over the phase angle in the lab. You can for instance, thrust in one direction with 90* phase angle, in the opposite direction at 270* and have no thrust at 000 and 180*; all with the same power into the device. This is a great way to do a demonstration, which is what Woodward is currently after with the PMN, IIUC.
