Hi all

hehe  and yooouupps ,i am getting lost here

What is the specification of     "Q  " seems to be , as it is discribed as  the magnitude of charge  ??     what is it ?   , in respest of energy stored   ?

please answer

Laurent

Hmm woopy

Well it looks like we cant use air pressure unless we have 2 vessels one for the low pressure side and one for the high side. I dont think we can look at it this way. Not sure yet..  Been working on things today. Gota run to the store.

Mags

Hi Mag and P.99

Sorry to insist but i can 't make my mind on this

At  "Hyperphysics" they say that the energy stored in a cap can be assimilated to an air tank. So a capacitor is a condensor or as we name them in french a "condensateur".
And they speak of  Q  as the  "magnitude of charge"

And the stored energy seems to be different of this Q

so if somebody would be kind enough to put all this in simple understandable word or drawing ?

And to go on my stacked brain on this energy transfer, i redo some graph,

And in graph 3 you can see that the transfer pipe are the same as per graph 2. So the same resistance and heat lost ,    as the pipe dimension are the same   or do i miss something ?.

But aside those pipe resistance, we ADD the resistance of the inductor ((turbine  (inductor) in the pipe motorising the  flywheel)) .So the transfer is braked by the resistance of the pipe  (heat lost ) and further more by the "turbine inductor-flywheel resistance)

And the transfer of the stored energy ,which should loose 50 %  , due to pipe resistance not only does not decrease to 50%  but it increases , or   does simply improve to almost 98 %???

I need to learn much more on the inductor secret please help

good luck at all

Laurent

Woopy,

Do you have an understanding about inductive and capacitive reactances and why they dissipate no power?

Just remember; a capacitor or an inductor can store and release energy, whereas a resistor can do neither, it can only burn off energy. A piece of wire at low frequencies is purely resistive, so it burns off energy when connected between a source cap and a charge cap. When you replace that piece of wire with an inductor, no energy can be burned off so it stores it, then releases it. As inductors are not ideal, there will always be some loss of energy.

Have a look at this page, it may help.
http://www.allaboutcircuits.com/vol_2/chpt_11/1.html

.99

Woopy

Think like this. ;]

Have 2 air tanks to represent each capacitor.  When the tanks are at normal air pressure, consider the cap to have no charge. Here is an example.

When we do a cutoff (switch) before the source cap reaches 50%, and we have the inductor continue to pull from one side of the cap(lead or tank) to the other,  to complete the charge form the freewheeling. So after cutoff, the flywheel is decompressing one side of the cap(air tank) and compressing that to the other side.

With a single tank representing a cap, we cant visualize how a cap in circuit actually works.

Think of an AV plug. 1 side at a time gets compressed and the other decompressed, 1 side at a time.  So there can be the same situation with dual air tanks to represent the cap. ;]

I understand your trying to visualize it. ;]  Hope that helps.

Ive been looking into getting the flywheel rolling with HV impulse driven by a stout primary.  We envision a Tesla coil by this, but with the right rectifier and a particular cap and inductor size, the voltage on the sec dont have to be of great value loaded to charge the cap to usable more levels voltage.  It just seems very easy to pump a small cap to kilo volts for discharge into the primary. The secondary flywheel is of heavy gauge wire and many turns.
Just some things Im trying.

I have some materials coming this week for variable cap experiments. Im trying for some high capacitance. Will be able to see if it takes a lot of work to vary.  ;]

Mags