pennies with a hole

[green]

Was reading ITMD, The Concentric Ring Experiment page 248. The experiment used an IB VLF detector to compare results. Was looking at some data I posted in the past using a PI (IB coil) to compare. One of the charts was labeled wrong, constant current instead of constant rate. Including corrected charts. The penny with the hole has a higher signal at start of decay. The copper penny with hole charts almost the same as zinc penny. The third chart includes the return signal when Tx is on. Tx on slope close to Tx off slope after slope is straight line decay linear X log Y. Calculated frequency for the targets, (1/2pi)/TC. Is that correct. Would the detector frequency effect which signal was higher, coin or ring when using a VLF detector? If yes, is there way to change the test using a PI that would cause the penny to have a higher signal than the ring?

Just noticed penny has higher signal than ring when Tx on. Does that mean penny should be higher than ring with a VLF?

[WM6]

Was reading ITMD,

Just noticed penny has higher signal than ring when Tx on. Does that mean penny should be higher than ring with a VLF?

Thanks for all your testing efforts.

For me, most interesting in your last presentation, is the fact, that on given timing, (say 35usec
with zinc penny) no matter of hole or not, there is no distinguishable difference exposed:

[Ferric Toes]

On a PI, the starting amplitude for the decay will always be higher for objects with a low conductivity. This is because the eddy current tries to maintain the internal field in the object just as it was prior to switch off. (or ON in the case of a CC circuit). For lower conductivity (higher resistance) the voltage must be higher to give the initial current necessary to do this. An extreme example is the TX coil itself, when you break the circuit at the end of each pulse, you get several hundred volts as the coil tries to maintain its current, but the decay, via the damping resistor, is very fast compared to the targets one wants to detect.

Drilling a hole in a coin lowers the conductivity, hence the result in the graph. Enlarge the hole still further and the starting amplitude will be higher still.

Eric.