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Overunity Machines Forum



Trawoeer Power Pyramid Version 12 - Electrical output from a homemade pyramid

Started by hartiberlin, June 28, 2011, 04:05:30 PM

Previous topic - Next topic

0 Members and 2 Guests are viewing this topic.

k4zep

Hi Diego, Netputne, All,

I stand absolutely corrected!  I was totally wrong on how to find the middle of the pyramid!
Diego, you are right.  Memory is a terrible thing to trust.  You guys are on the ball!

Respectfully,
Ben K4ZEP

Pascuser

@K4ZEP

No problem; and my memory is more bad than yours I think.
About calculation to place correctly the reactor, there is a small notice to do: it depends on your pyramid construction. The calculation that I have done is for an ideal pyramid with edges in wire.

But you have edges made with 20mm tubes. And depending on the way you built your pyramid the same computation works or not. For my pyramid, where all edges are 50cm and equal, and the peak on the square base are real peaks from edges coming of the apex, the computation is correct.

But now the gegyx question has most importance, not only because he will build the 1 meter pyramid for me because he helps me in my next building scale 1:1; but because depending on dimensions this modifies the positioning of the 1/3 point for reactor.

If edges from the apex were 1 meter laid on the square base, then you should only have to add the tube thickness of 2cm to the 23.6cm and the right position is 25.6cm.
But if these edges from the apex are shorter so to have a 1 meter length from the prolongation of the edges on the base of the square base, then not only you add 2cm but you must substract something. So maybe 25cm is a correct placement for Thomas pyramid (23.6+2-?=25).

But for YOUR pyramid, if it is built slightly differently from the way Thomas did (where your measured dimensions of edges) the position will be modified. I can compute for you the exact position of the reactor placement if you give the dimensions of all edges of the pyramid, in an exact way, because the imprecision of how it is built gives an imprecision varying from many centimeters in the placement of the center of the reactor.

If Thomas gave the exact answer to Gegyx of what are his lengths for his edges, I can calculate too his exact placement to confirm that 25cm is the correct height for his reactor's center. The calculation is only a way of doing exactly what Thomas said doing with wires coming from the center of each triangle gypsum sides with a hole in the center. You can avoid this construction and know the exact point he has with calculation; and once again I offer to do the calculation if you give the dimensions.

k4zep

Quote from: Pascuser on July 28, 2011, 02:56:23 AM
@K4ZEP

No problem; and my memory is more bad than yours I think.
About calculation to place correctly the reactor, there is a small notice to do: it depends on your pyramid construction. The calculation that I have done is for an ideal pyramid with edges in wire.

But you have edges made with 20mm tubes. And depending on the way you built your pyramid the same computation works or not. For my pyramid, where all edges are 50cm and equal, and the peak on the square base are real peaks from edges coming of the apex, the computation is correct.

But now the gegyx question has most importance, not only because he will build the 1 meter pyramid for me because he helps me in my next building scale 1:1; but because depending on dimensions this modifies the positioning of the 1/3 point for reactor.

If edges from the apex were 1 meter laid on the square base, then you should only have to add the tube thickness of 2cm to the 23.6cm and the right position is 25.6cm.
But if these edges from the apex are shorter so to have a 1 meter length from the prolongation of the edges on the base of the square base, then not only you add 2cm but you must substract something. So maybe 25cm is a correct placement for Thomas pyramid (23.6+2-?=25).

But for YOUR pyramid, if it is built slightly differently from the way Thomas did (where your measured dimensions of edges) the position will be modified. I can compute for you the exact position of the reactor placement if you give the dimensions of all edges of the pyramid, in an exact way, because the imprecision of how it is built gives an imprecision varying from many centimeters in the placement of the center of the reactor.

If Thomas gave the exact answer to Gegyx of what are his lengths for his edges, I can calculate too his exact placement to confirm that 25cm is the correct height for his reactor's center. The calculation is only a way of doing exactly what Thomas said doing with wires coming from the center of each triangle gypsum sides with a hole in the center. You can avoid this construction and know the exact point he has with calculation; and once again I offer to do the calculation if you give the dimensions.

Good Morning Pascusser,

Thanks, In my pointed conversations with Thomas about the correct size of the pyramid (as I don't want to do this twice), it seemed that the absolute dimensions of the pyramid are not as critical as the building dimensions of the reactor/materials and its placement.  As the inside area/dimensions are what are doing the work, reflecting the ground energy, I presume that (the center of the gypsum) is the absolute dimension you measure for centering and and mounting point inside the gypsum.  The metal does not reflect the energy (not confirmed didn't think to ask), only the gypsum. Remember he showed how to measure for the center the reactor using ONLY pieces of gypsum.   That is all the absolute information I can give you. 

I suspect the "tuning" by adjusting/clipping the length of the wires on the non-connected ends of the Capacitor/inductor are what make up for the small variances in the overall system.  The metal would appear to be only a ground plane to bleed away excess charge (when connected to a active ground) that builds up in a working pyramid and NOT a part of the active energy producing system.

Respectfully
Ben K4ZEP


neptune

@Pascuser . Back in the year2006/7 I believe that Thomas said that the dimensions given [ base 1 meter square , diagonals 1 meter] were the internal dimentions of the inside of the plasterboard [gypse] shell . If this is true , would this give a "focus" [centre of gravity] at exactly 25 cms from the floor ?

k4zep

Quote from: neptune on July 28, 2011, 08:02:15 AM
@Pascuser . Back in the year2006/7 I believe that Thomas said that the dimensions given [ base 1 meter square , diagonals 1 meter] were the internal dimentions of the inside of the plasterboard [gypse] shell . If this is true , would this give a "focus" [centre of gravity] at exactly 25 cms from the floor ?

Neptune,

You couldn't have said more in less words if you tried, EXCELLENT.

Respectfully
Ben K4ZEP