Quote from: Rodal on 10/15/2014 08:41 pmThose thinking that the results are an experimental artifact should seek artifact explanations that are consistent with the experimental data being linearly dependent on just these three variables. Yes dr Rodal, working on that.Better to keep a stiff upper lip, hey !

Those thinking that the results are an experimental artifact should seek artifact explanations that are consistent with the experimental data being linearly dependent on just these three variables.

7 data points is sparse to conduct statistical analysis

but you do a great job.

Though I wonder if it wouldn't be more appropriate to conduct the regressions in log log plane, as the low values dispersions tend to be squashed by the low absolute levels, while their relative dispersion around the (linear) predicted values seem more natural to me : log(experimental/predicted) or equivalently log(experimental)-log(predicted) (then squared as for the least square regression). A linearly scaling formula that predicts 1.0µN for a 1.1µN measure has as much error than predicting 1N instead of actual 1.1N. Maybe this is already the case in your R^2 results ?

QuotedF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).Which boils down to dF = (PQ/f)*((1/w_big)-(1/w_small)).P - PowerQ - Quality factorf - Drive frequencyw_big - diameter of the big endW_small - diameter of the small end.My goodness, where is the Unruh radiation, the Hubble horizon, the Casimir effect or any other strange factors?The only way MiHsC enters the picture is because it led Prof. M to the above equation.As it stands the equation can be written as dF = [ Stored power/w_big - Stored power/w_small ] / fwhere stored power = Q * Power. Does that mean anything helpful?

dF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).

It is an argument against thermal explanations (stored power instead of loss power)

The geometrical factor (that came from McCulloch's theory) needs to be explained by magnetic, artifact or any other explanation

QuoteIt is an argument against thermal explanations (stored power instead of loss power)Perhaps, but it is exactly the same equation as always, just arranged differently.

Quote from: frobnicat on 10/15/2014 09:46 pm...Please re-write or withdraw your comment about the Chinese.

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Quote from: aero on 10/15/2014 09:58 pmQuotedF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).Which boils down to dF = (PQ/f)*((1/w_big)-(1/w_small)).P - PowerQ - Quality factorf - Drive frequencyw_big - diameter of the big endW_small - diameter of the small end.My goodness, where is the Unruh radiation, the Hubble horizon, the Casimir effect or any other strange factors?The only way MiHsC enters the picture is because it led Prof. M to the above equation.As it stands the equation can be written as dF = [ Stored power/w_big - Stored power/w_small ] / fwhere stored power = Q * Power. Does that mean anything helpful?dF = (PQ/f)*((1/w_big)-(1/w_small)) = [ Stored power/w_big - Stored power/w_small ] / fwhere stored power = Q * Power.

...Thank you for the log log plots. Agree that in log log everything looks too good... don't know if it holds for some eminent statistician (which I am not) but I find the log log more convincing as we have very few diversity if the 3 Brady et al entries get squashed. Save the outlier, the other two still look good relative to much lower magnitudes overall.Still wondering what to do with ranges in the tabulated data...

Quote from: aero on 10/15/2014 10:53 pmQuote from: aero on 10/15/2014 09:58 pmQuotedF = (PQ/c)*((L/w_big)-(L/w_small)) = (PQ/f)*((1/w_big)-(1/w_small)).Which boils down to dF = (PQ/f)*((1/w_big)-(1/w_small)).P - PowerQ - Quality factorf - Drive frequencyw_big - diameter of the big endW_small - diameter of the small end.My goodness, where is the Unruh radiation, the Hubble horizon, the Casimir effect or any other strange factors?The only way MiHsC enters the picture is because it led Prof. M to the above equation.As it stands the equation can be written as dF = [ Stored power/w_big - Stored power/w_small ] / fwhere stored power = Q * Power. Does that mean anything helpful?dF = (PQ/f)*((1/w_big)-(1/w_small)) = [ Stored power/w_big - Stored power/w_small ] / fwhere stored power = Q * Power.OK, but this is telling a huge amount of information.Why should the stored power give you the force?It should not if the force is a thermal artifact.It should not if the force is an artifact due to losses in the medium.It does leave the door open for the force being due to being an artifact due to resonance.

OK, let's try to narrow this down.dF = (PQ/c)*((L/w_big)-(L/w_small)) L= 6.6, 9.0 and 9.9 inches w_small = 6.6 incheswhat do you want for w_big ?

Quote from: aero on 10/16/2014 12:00 amQuoteOK, let's try to narrow this down.dF = (PQ/c)*((L/w_big)-(L/w_small)) L= 6.6, 9.0 and 9.9 inches w_small = 6.6 incheswhat do you want for w_big ?Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.OK, I'm using metric units.So I will useL=0.16764, 0.2286, 0.25146 m with the same numbers previously used for w_big and w_smallQUESTION: Do you want me to use this L's only for NASA Eagleworks or also for Shawyer and China?Or do you want to think what values you want for them UK and China?[Please assume that the Chinese used the same dimensions as the larger Shawyer device, which makes sense from their text and also because Chinese used 1 KW]

QuoteOK, let's try to narrow this down.dF = (PQ/c)*((L/w_big)-(L/w_small)) L= 6.6, 9.0 and 9.9 inches w_small = 6.6 incheswhat do you want for w_big ?Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.

Quote from: Rodal on 10/16/2014 12:06 amQuote from: aero on 10/16/2014 12:00 amQuoteOK, let's try to narrow this down.dF = (PQ/c)*((L/w_big)-(L/w_small)) L= 6.6, 9.0 and 9.9 inches w_small = 6.6 incheswhat do you want for w_big ?Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.OK, I'm using metric units.So I will useL=0.16764, 0.2286, 0.25146 m with the same numbers previously used for w_big and w_smallQUESTION: Do you want me to use this L's only for NASA Eagleworks or also for Shawyer and China?Or do you want to think what values you want for them UK and China?[Please assume that the Chinese used the same dimensions as the larger Shawyer device, which makes sense from their text and also because Chinese used 1 KW]These L's are the dimensions of the NASA Eagleworks device. I don't know the dimensions of the Shawyer device. If you do, then use them consistently. That is, small diameter, Cavity length and Large diameter.

In McCulloch's quantised inertia the Unruh waves are allowed only if they fit exactly within the Hubble horizon (or within a local Rindler horizon). For the formula to apply the EM Drive cavity walls must act like a horizon.

Overdue emphasis on physical explanations rather than mathematical examination of the data is misplaced, particularly with anomalous results of early experimental results with high uncertainty bars. An objective, cool, mathematical viewpoint (rather than passionate subjective beliefs) is called for.

McCulloch's simple formula, without any fudge factors, and with a minimum of parameters, does a much better job at predicting the experimental results than anything else presented so far...

4) So far, all the experimental data variation in the US (NASA Eagleworks, including the statistical outlier), the UK and China can be explained solely in terms of just three variables: A) (1/DiameterOfSmallBase-1/DiameterOfBigBase) B) Q (resonance quality factor) C) Power Input

Quote from: aero on 10/16/2014 12:31 amQuote from: Rodal on 10/16/2014 12:06 amQuote from: aero on 10/16/2014 12:00 amQuoteOK, let's try to narrow this down.dF = (PQ/c)*((L/w_big)-(L/w_small)) L= 6.6, 9.0 and 9.9 inches w_small = 6.6 incheswhat do you want for w_big ?Use the same w_small and w_big that you have been using. My measurement of w_big =9.9 inches. Those two values, w_big and w_small come directly from Prof. M's formula, don't change them. The length that is uncertain in my mind is L.OK, I'm using metric units.So I will useL=0.16764, 0.2286, 0.25146 m with the same numbers previously used for w_big and w_smallQUESTION: Do you want me to use this L's only for NASA Eagleworks or also for Shawyer and China?Or do you want to think what values you want for them UK and China?[Please assume that the Chinese used the same dimensions as the larger Shawyer device, which makes sense from their text and also because Chinese used 1 KW]These L's are the dimensions of the NASA Eagleworks device. I don't know the dimensions of the Shawyer device. If you do, then use them consistently. That is, small diameter, Cavity length and Large diameter.We might as well do the best job we can at the outset rather than re-visit later on.Here is a link to Shawyer's paper: http://www.emdrive.com/IAC-08-C4-4-7.pdfplease let me know what L dimensions you want to use for the smaller and the larger Shawyer devicesThanks

Seeking clarity here.Quote4) So far, all the experimental data variation in the US (NASA Eagleworks, including the statistical outlier), the UK and China can be explained solely in terms of just three variables: A) (1/DiameterOfSmallBase-1/DiameterOfBigBase) B) Q (resonance quality factor) C) Power Input So...if I am following this correctly, the logical thing for the research teams to do would be to run thousands (?) of tests in which these three points differ somewhat, ideally in a vacuum. Aka...different 'truncated cone sizes', different frequencies (?), degrees of power input, that sort of thing, right? And then a much clearer picture as to what is going on should emerge. Is that a fair assessment?

Main experimental need is to be able to tune precisely to the (initially unknown) frequency that gives highest amplitude. And then for the device to stay tuned to that frequency and not deviate from it while very high resonance with very low damping is achieved. This was very difficult for researchers to accomplish on a reliable basis. Once that is accomplished the need is then for NASA to make EM Drive that runs at 1000 watts (like in the UK and China) instead of 20 watts to produce much larger forces.

Secondary need is to explain precisely what is the physical effect that is producing the thrust measurements.