Author Topic: EM Drive Developments Thread 1  (Read 765760 times)

Offline Ron Stahl

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Re: EM Drive Developments
« Reply #3420 on: 11/25/2014 05:37 PM »
Well perhaps we misunderstood each other.

I apparently mistook you to be championing this skeptical pseudo-scientific attitude that scoffs right out of the gate at anything dramatic or promising.  I was subject to this sort of sickness as a high school student when both my first and second year physics teachers told us on no uncertain terms that journeying outside our planetary system would never happen because the distances are too great, and the average man journeying off the planet would never happen because the expense was too great.  Of course it's since 1997 we've known both these assertions are wrong.

Now if you want to get Kip's view of this issue, I suggest look here:

http://time.com/3602525/christopher-nolan-physics-interstellar-kip-thorne/

time index around 6:20 and Kip's answer about 7:15.  The interviewer asks specifically if Kip is concerned about teaching people things like warp drive are possible, and what we find is Kip recommending humility, and noting that physicists think one way and later find they've had it all backward.  "You believe something and then discover you were wrong, every day."  He never answers the question directly.  I find that fascinating.

« Last Edit: 11/25/2014 05:39 PM by Ron Stahl »

Offline Rodal

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Re: EM Drive Developments
« Reply #3421 on: 11/25/2014 06:24 PM »
The purpose in posting Prof. Thorne's and Cramer's statements is to promote discussion: why and what are the reasons for which they decided to make these statements?.  The scientific attitude has always been to question everything.   Scientists are in search of the truth.  Neither Galileo, Einstein, Tsiolkowvsky, etc. were demoralized by discerning questions.  When Paul Davies brings up Hawking's cosmological protection principle (that would prevent the formation of wormholes), it should not be viewed as being a party pooper: it should be viewed as an intelligent question to search whether the presenter has a good answer for it.  If the presenter has a good answer, it benefits the whole audience.  It is to the benefit of the scientist to ask himself/herself, and to be asked by others, discerning questions. 

It was to the benefit of Amundsen to be well-prepared for the South Pole expedition, and it was to Scott's detriment to not be as well prepared and to have embarked without as much discerning planning.
« Last Edit: 11/25/2014 06:36 PM by Rodal »

Offline SteveKelsey

Re: EM Drive Developments
« Reply #3422 on: 11/25/2014 06:45 PM »
This might help the debate

http://thelifeofpsi.com/wp-content/uploads/2014/09/Hawking-1992.pdf

The conclusion has enough in it to reinforce the improbability of wormholes forming without new physics.
2001 is running a little late, but we are getting there.

Offline Rodal

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Re: EM Drive Developments
« Reply #3423 on: 11/25/2014 07:21 PM »
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
« Last Edit: 11/25/2014 07:39 PM by Rodal »

Offline aceshigh

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Re: EM Drive Developments
« Reply #3424 on: 11/25/2014 07:49 PM »
will a copy of this be forwarded to Dr White?

Online Mulletron

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Re: EM Drive Developments
« Reply #3425 on: 11/25/2014 07:50 PM »
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.
Quote
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.
Quote
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
Quote
"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
« Last Edit: 11/25/2014 09:49 PM by Mulletron »
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Offline Rodal

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Re: EM Drive Developments
« Reply #3426 on: 11/25/2014 08:13 PM »
will a copy of this be forwarded to Dr White?
I have let one of the senior members in his team know through an electronic message by other media (other than this forum).

Online Mulletron

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Re: EM Drive Developments
« Reply #3427 on: 11/25/2014 08:24 PM »
In a nutshell....Push while FM is heavy, pull while it is light. But on the flip side, the finite power supply that is exciting the FM is literally strapped down to the RM, reaction mass (the ship). So if the FM is gaining mass, then the power supply is losing mass by the same rationale.

This is very close but not quite right.  The thing that is losing mass when the fluctuating or active mass gains mass, is the rest of the universe.  The reaction mass is just the thing the active mass pushes off of.  The source of the gravinertial flux of the universe, is the mass of the universe itself.  So it is accurate to say, that when a gravinertial transistor like a Mach-Effect Thruster (MET) harvests momentum from the gravinertial field, it is causing the universe to accelerate in its expansion and hastens the arrow of time (entropy).  This is in fact why Tom Mayhoood, Woodward's grad student back in the 90's, put the sign on the lab door reading "Tomorrow's Momentum Today".  This is accurate to the physical theory.

Though Woodward has never weighed in on the issue so far as I'm aware, i would just note this could be an explanation for Dark Energy, or whatever force it is that is causing the observed acceleration in the expansion of the universe.   We've known since 1997, that instead of the universe expanding ever more slowly as it fights its own gravity as we'd expect, the universe is accelerating in its expansion, and this is the effect we ought to observe from LOTS of gravinertial harvesting from Mach-Effect devices.

Given your research, I want your opinion on this:

If a clever mechanical engineer built a MET assembly using strictly hydraulic actuators and fluids. Would it still work? This sounds weird I know, but there is a reason I'm asking you this. If I had a giant one of these hydraulic setups on a boat and a team of men working together to pump a hydraulic chamber with fluid (assuming it is on wheels or something so it'll actually move), push on it, then they withdraw the fluid from the chamber, and then they pull on it. And everything was timed perfectly. Would it still work? If so why? Where is the net force? If not why? What is the most fundamental thing here? Capacitors or mass fluctuations? This is a thought experiment.

Others please chime in.
« Last Edit: 11/25/2014 08:27 PM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline Ron Stahl

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Re: EM Drive Developments
« Reply #3428 on: 11/25/2014 08:27 PM »
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/0817638237

If 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.

Quote
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.

Quote
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.

Quote
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.

Quote
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.

Quote
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.

Quote
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.

Quote
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?

Quote
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.

Quote
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.
« Last Edit: 11/25/2014 08:40 PM by Ron Stahl »

Online Mulletron

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Re: EM Drive Developments
« Reply #3429 on: 11/25/2014 08:36 PM »
That first quote in your post isn't me, it is John. And there are some contradictions in your post.
« Last Edit: 11/25/2014 08:37 PM by Mulletron »
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Online Mulletron

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Re: EM Drive Developments
« Reply #3430 on: 11/25/2014 08:45 PM »
Fair enough. We'll pump hot liquid mercury in. Push on it. Pump the liquid through a heat exchanger on the way out. Then pull on it. Perfectly timed. How about now?

How is Woodward satisfying the requirement of the thing having to be accelerated in the first place?
For this thought experiment, we'll say it is already being accelerated by X means.

I don't see any way it can ever ever work.
« Last Edit: 11/25/2014 08:49 PM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline Ron Stahl

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Re: EM Drive Developments
« Reply #3431 on: 11/25/2014 09:12 PM »
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.
« Last Edit: 11/25/2014 09:15 PM by Ron Stahl »

Online Mulletron

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Re: EM Drive Developments
« Reply #3432 on: 11/25/2014 09:27 PM »
So what do you think about the thought experiment? Also....it seems, from your own words, the book, and the Oracle, the acceleration requirement must be relative to the "distant stars." So the acceleration felt by a body at rest here on Earth doesn't satisfy that requirement. What do you think?

This forum thread is the #3 Google search item when people search for EMdrive. The dialogue I'm starting here is an opportunity for you to sell your point.
« Last Edit: 11/25/2014 09:29 PM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline Ron Stahl

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Re: EM Drive Developments
« Reply #3433 on: 11/25/2014 09:41 PM »
I don't think pushing on hot mercury can have the high dE/dT necessary for dramatic effects.  If you want to dabble around at power densities you can (I never do that as my math is not trustworthy--I'm not an engineer) by looking at something like PZT with a k of about 1,000 that diminishes to 700 when clamped, or PMN with a k of 20,000.  Both these materials have an extension that is I think about 1 part in a thousand or 0.1% the thickness, and the frequency they operate on resonance is inversely proportional to thickness.  So proper neighborhood figures could be around 50 Khz, 2cm of these materials would have 200 micron extensions.  You can then get reasonable acceleration here:

http://www.spaceagecontrol.com/calcsinm.htm

I haven't done this in a long time and it is because I don't remember figures like these that I never became an engineer.  So don't trust my math.  Just saying, this is how you can get round figures and I think you'll see, the accelerations and power density of even lowly PZT is very high compared to what you're talking about.  PMN is much higher.  The power density of the special materials we intend to use is vastly higher again, since it has a k of 60,000 at 500 Mhz.

M-E generation scales linearly with frequency and quadratically with k.
« Last Edit: 11/25/2014 09:52 PM by Ron Stahl »

Offline Stormbringer

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Re: EM Drive Developments
« Reply #3434 on: 11/25/2014 10:32 PM »
I wonder if something like they use to investigate the dynamic casimr effect could be used?
I have read of several experimental set ups that use lasers to act as the moving casimir plate. One was that Egyptian girl that was making the satellite maneuver system and the other was someone using it to pull positrons and electrons out of the vacuum at one of the big physics labs.

But anyway if that would work the "mirror" moves at basically light speed or pretty near it.
« Last Edit: 11/25/2014 10:34 PM by Stormbringer »
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Offline 93143

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Re: EM Drive Developments
« Reply #3435 on: 11/25/2014 11:34 PM »
I don't see any way it can ever ever work.

I'm getting the strong impression that you still think the "mass fluctuations" are just the energy/matter being pumped back and forth in the device.

This is wrong.

Online Mulletron

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Re: EM Drive Developments
« Reply #3436 on: 11/25/2014 11:56 PM »
I don't see any way it can ever ever work.

I'm getting the strong impression that you still think the "mass fluctuations" are just the energy/matter being pumped back and forth in the device.

This is wrong.

Yeah you know, you're right. I'm not really addressing the problem here with the initial conditions of that thought experiment.

Quoting @Ron Stahl
Quote
This is very close but not quite right.  The thing that is losing mass when the fluctuating or active mass gains mass, is the rest of the universe.

The notion of how, or the validity of, the rest of the universe losing mass, so that the FM can gain mass, is non trivial. I dismissed it outright without even thinking about it.

Over to you @Ron Stahl

Edit: Page 73 of the book has a nice equation for the mass fluctuations, but nothing to support it other than a reference to the flux capacitor paper. Flux capacitors? Couldn't he have picked a better name that didn't scream gobbledygook? I am understanding why the "QVFers" have the funding. The PR job on Mach effects is just awful.

Here's the flux capacitor paper I found, this is the one referenced by the book :o
http://physics.fullerton.edu/~jimw/flux-cap.pdf

I mean I have an open mind, but dang already.

I see no resemblance.
« Last Edit: 11/26/2014 01:09 AM by Mulletron »
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Re: EM Drive Developments
« Reply #3437 on: 11/26/2014 02:15 AM »
I just can't mess with Flux Capacitors........and I want to go to space really really really bad.

That's where I'm leaving them. If I'm wrong about them.....Mea Culpa.
« Last Edit: 11/26/2014 02:26 AM by Mulletron »
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Offline 93143

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Re: EM Drive Developments
« Reply #3438 on: 11/26/2014 02:28 AM »
Flux capacitors? Couldn't he have picked a better name that didn't scream gobbledygook?

The name does more or less match the item.  And hey - if Woodward's ideas about wormholes are borne out too, it might actually end up being what makes time travel possible...

...

It's not the only time someone working on futuristic but potentially legitimate technology has done this.  Tom Ligon, of Polywell fame, once blew a fist-sized diode and recharged a bank of lead-acid batteries in an instant by mistakenly assuming a research machine called PXL-1 had undergone cathode poisoning, when what had really happened was that it had entered a low-loss mode and was behaving as a dynamic capacitor.

You guess what he decided to call the machine.
« Last Edit: 11/26/2014 02:34 AM by 93143 »

Offline ThinkerX

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Re: EM Drive Developments
« Reply #3439 on: 11/26/2014 03:11 AM »
Quote
I just can't mess with Flux Capacitors........and I want to go to space really really really bad.

That's where I'm leaving them. If I'm wrong about them.....Mea Culpa.

Until you change your mind in six or eight weeks. 

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