Measuring Capacitive Load Cells?

I Have found this very interesting

and I have  invited some of the skilled engineers to
assist us on this forum .

I hope they can help , save the environment with us
in there spare time please.

http://www.loadstarsensors.com/why-build-capacitive-load-cells.html

Dan
www.securesupplies.biz

Quote from securesupplies on September 14th, 2013, 08:45 PM

and I have  invited some of the skilled engineers to
assist us on this forum .
Dan
www.securesupplies.biz

Wow that would be awesome :) could really take things to a whole new level.

Quote from securesupplies on September 14th, 2013, 08:45 PM

http://www.loadstarsensors.com/why-build-capacitive-load-cells.html

A capacitive strain gauge aay.  Interesting idea and it would be far easier to connect to digital circuits.  The only problem I see with it is if the RC circuit is at too low of a frequency, you won't see the stress point show up in the waveform.  So for very high-speed events, it may or may not work quite as well as a conventional strain gauge.

What do we use now to measure capactive load?

Quote from Dog-One on September 15th, 2013, 08:39 AM

Quote from securesupplies on September 14th, 2013, 08:45 PM

http://www.loadstarsensors.com/why-build-capacitive-load-cells.html

A capacitive strain gauge aay.  Interesting idea and it would be far easier to connect to digital circuits.  The only problem I see with it is if the RC circuit is at too low of a frequency, you won't see the stress point show up in the waveform.  So for very high-speed events, it may or may not work quite as well as a conventional strain gauge.

Quote from securesupplies on September 15th, 2013, 09:30 AM

What do we use now to measure capactive load?

I have no idea.  Your link refers to mechanical load and stress sensors.

I can't say I know what a "capacitive load" even is.  I'd have to see the definition.  I do OTOH know what an inductive load is.

Quote from Dog-One on September 15th, 2013, 09:53 AM

I can't say I know what a "capacitive load" even is.  I'd have to see the definition.  I do OTOH know what an inductive load is.

Capacitive loads are the same as inductive loads, only in cap. loads the current is 90 degrees ahead of the voltage as opposed to ind. loads where the voltage is 90 deg. ahead of the current.
If you have a motor for instance and measure the current to it, you would then get the apparent current, which contains both the active and the, in this case, reactive  inductive load(s).
The active and reactive loads are added in the Pythagoras triangle, A and B, and the resulting power, C, is then the apparent power.
If you were to connect a capacitor parallel to the motor which has the same (capacitive) reactive load as the motor's inductive reactive load, then the 2 cancel eachother out, leaving only the active current which the motor is putting out as it's spinning.
That's the short version btw
http://en.wikipedia.org/wiki/Power_factor

Quote from Lynx on September 15th, 2013, 10:08 AM

Capacitive loads are the same as inductive loads, only in cap. loads the current is 90 degrees ahead of the voltage as opposed to ind. loads where the voltage is 90 deg. ahead of the current.

Voltage being 90 degrees ahead of current...  No problem, I fully understand that.

Current being 90 degrees ahead of voltage...  Big problem, not possible.  You need a voltage potential to induce current flow.  I'm thinking what you have instead is voltage being 270 degrees ahead of current.  If I'm wrong here so be it, because I can't understand it any other way--my pee brain does not compute.  The only exception would be if current has mass and inertia and will keep flowing well after the voltage differential has dissipated.  That's a whole different ball game there.  Even still, something had to get the current flowing in the first place.

Quote from Dog-One on September 15th, 2013, 11:00 AM

Quote from Lynx on September 15th, 2013, 10:08 AM

Capacitive loads are the same as inductive loads, only in cap. loads the current is 90 degrees ahead of the voltage as opposed to ind. loads where the voltage is 90 deg. ahead of the current.

Voltage being 90 degrees ahead of current...  No problem, I fully understand that.

Current being 90 degrees ahead of voltage...  Big problem, not possible.  You need a voltage potential to induce current flow.  I'm thinking what you have instead is voltage being 270 degrees ahead of current.  If I'm wrong here so be it, because I can't understand it any other way--my pee brain does not compute.

Haha
Well, that's the laws of physics for you right there
Look at it this way: When a capacitor is completely discharged and you apply a voltage to it, the very first blink of an eye the capacitor voltage is zero and the current going to it is ridiculously high, so in that case the current comes first, sort of speaking.
You could buy that in the case of an inductive reactance that the voltage comes first, then the current, well that's also due to the intrinsic physical properties of inductive reactances if you will.
To start with, the instance you apply a voltage to one end of an inductor the voltage immediately comes out the other end of it, but due to the inductive reluctance to change, by nature it starts out restricting the current, only to adapt to status quo and slowly but surely let the current increase to whatever it's feeding.
Here, don't take my word for it, check it out for yourself, http://www.eeweb.com/blog/andrew_carter/ac-inductive-and-capacitive-reactance

Quote from Lynx on September 15th, 2013, 11:57 AM

Look at it this way: When a capacitor is completely discharged and you apply a voltage to it, the very first blink of an eye the capacitor voltage is zero and the current going to it is ridiculously high, so in that case the current comes first, sort of speaking.

I'm not convinced.  Sounds like a measuring error to me, because a voltage of zero is no voltage and the reason it would be no voltage is because you have connected a direct short by way of a discharged capacitor across the potential.  Disconnect the capacitor and I can assure you there is a voltage there if in fact you can charge the capacitor.

Voltage ~ Pressure
Current ~ Flow Rate

Starting from steady state, no pressure, no flow.  Any questions?   :P

Quote from Dog-One on September 15th, 2013, 12:31 PM

Quote from Lynx on September 15th, 2013, 11:57 AM

Look at it this way: When a capacitor is completely discharged and you apply a voltage to it, the very first blink of an eye the capacitor voltage is zero and the current going to it is ridiculously high, so in that case the current comes first, sort of speaking.

I'm not convinced.  Sounds like a measuring error to me, because a voltage of zero is no voltage and the reason it would be no voltage is because you have connected a direct short by way of a discharged capacitor across the potential.  Disconnect the capacitor and I can assure you there is a voltage there if in fact you can charge the capacitor.

Voltage ~ Pressure
Current ~ Flow Rate

Starting from steady state, no pressure, no flow.  Any questions?   :P

Ok, you win

Quote from securesupplies on September 15th, 2013, 09:30 AM

What do we use now to measure capactive load?

Here's what I would do:
I'd start out by measuring the resistance over whatever it's I'm out to get the capacitance on, that would be the R bit of it all, provided it's a circuit where the R and the C are attached to eachother in parallel.
Then I would subject the circuit to a low voltage AC, from a standard 60 Hz transformer, and then measure the RMS current going through it.
The current would be the RMS value of the applied voltage divided by the impedance, Z, of the circuit and Z would in turn be made out of the sum of the R and XC according to the formula Z squared = R squared + XC squared, where XC = 1/(2*pi*f*C).
So if you divide the applied voltage with the measured current you first get the impedance, then you subtract the resistance from the impedance using XC = the square root of (Z squared - R squared).
The frequency, f, is 60 Hz, so a little reverse engineering would leave you with
C (the capacitance) = 1/(XC*2*pi*f) = 1/(XC (from the above)*2*3.1415*60)
That's the short, awkward, version of it, here's the more sophisticated version, http://www.electronics-tutorials.ws/accircuits/AC-capacitance.html

to measure WFC capacitance and resistance you also can use the following cheap tool:

http://open-source-energy.org/?tid=469&pid=6965#pid6965