41 replies to this topic

[Ruud de la Rosa]

I have to make an Accelerometer for school, does anybody know how to make one electronical??

[Christiaan]

what accerelation will it be measuring? on what application? ie is it on a vehicle, is it a remote sensor, what?

[This message has been edited by Christiaan (edited 12-07-1999).]

[Ruud de la Rosa]

On a vehicle! I allready know that I can use 3 plates. The middle can move freely, and the other to are a condensator.

[Ursus]

I think you can use the fact that when the middle plate moves the capacitance will change (if you didn't know this already)

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Ursus
Trust me, send money.

[Yelnats]

A pendulum suspended on a semi rigid shaft with a strain gauge glued to it (Edmunds Scientific) and some rudimentry electronics will do the job.

[Ruud de la Rosa]

Ursus tell me more, what is the capacitance??
I've read you can measure Volts on the middle plate?

[This message has been edited by Ruud de la Rosa (edited 12-09-1999).]

[Ursus]

Forget what I said previously! If you put a constant voltage across the outer plates, the measured voltage at the middle plate will depend on where the plate is positioned. Do the maths yourself.


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Ursus
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[Ruud de la Rosa]

How do you know this? are there any WWW-pages about it?

[Ursus]

I did some thinking(and calculations), and came to that conclusion. But I could be wrong, I study electrical engineering but have steered away from the electrics so I these are not fresh in my mind. You didn't giv us a lot of info. At what level do you study BTW?

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Ursus
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[Ruud de la Rosa]

Hey, Ursus!
We're in the 6th Grade. Were almost 18 years old. And we are desperate for help!!

Cheers Ruud & co

[Ursus]

OK Ruud, I´ll tell you how I'm thinking. I'm assuming you know how to calculate votages across capacitors and stuff. You can look at the 3 plates as two capacitors, connected in serial, from one of the outer plates to the middle one and from the middle to the other outer plate. The The capacitances of these 2 "imaginary" capacitors depend on where the middle plate is located, but the overall capac. between the outer plates is not affected. You need to know that the capac. between 2 plates is proportional to 1/(dist. between plates). Verify this for yourselves! (you have to do something for yourselves, Ican't tell you everything. Damn, I sound like some teacher here.) You'll find that my previous statement is right (I hope, otherwise I guess I look liike a fool ).

Thats all for now. Class dismissed!

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Ursus
Trust me, send money.

[Ruud de la Rosa]

Thanks Ursus, If you come up with some more ideas tell them please. Ursus You're the best, TNX Ruud.

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GO RUBENS CRUSH SCUMMI
CHEERS RUUD

[Ursus]

It's a good thing you didn't use that sig before, or else I might not have been as inclined to help.


BTW. let me know if it works!
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Ursus
Trust me, send money.



[This message has been edited by Ursus (edited 12-14-1999).]

[Ruud de la Rosa]

It's me again, when you say 2 capacitators you mean |-|....|+/-|....|+| ???
BTW how can you measure the capacitance? I know how to calculate it but that's it.
Isn't the: Capacitance =X/dist. between plates. (X=a Constant)
Cheers Ruud

[Ursus]

You don't actually measure the capacitance, but the voltage at the middle plate. When the two capacitances change (as the middle plate moves)the voltage across them will also change, so measuring the voltage across one of them will indicate where the plate in positioned.

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Ursus
Trust me, send money.

[Ruud de la Rosa]

You can measur volt? I believed the only thing that changes is The elecktrical field, and so the number of Coulomb.
which formula can I use? Becaus eI found a couple of formula's like 1/Ctotal=1/Ca+1/Cb
and C=Q/U E=U/l C=capacitance Q=Coulomb's U=Volts E=Field...... l=distance between plates

C=(Eo x Er x A)/l
A= MxM=Msquare
Eo = Dielecktricum vacuum
Eo = Di-elecktricum of air in this case

Cheers Ruud

[Ursus]

The Coulombs stay the same (actually the equal in both capacitors)-> electric field stay the same -> U= E*d E=El.Field d=dist.

Your formulas i OK but it is your assumption that is incoorrect.

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Ursus
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[Ruud de la Rosa]

So you put For instance 12V on the first plate

......|..|......|......|......|...|
----|+.|......|-..+|......|.-.|.----|
|.....|..|......|......|......|...|.......|
|.....12V..........?....... ...0V........|
|___________________________|
..................+ / -
..................12 V
(the...dots don't mean anything, just there to replace the spaces)
How do you connect the sensors of the Volt meter to the capacitators??

[This message has been edited by Ruud de la Rosa (edited 12-17-1999).]

[This message has been edited by Ruud de la Rosa (edited 12-17-1999).]

[This message has been edited by Ruud de la Rosa (edited 12-17-1999).]

[This message has been edited by Ruud de la Rosa (edited 12-17-1999).]

[Ruud de la Rosa]

That drawing didn't came out as good as it should! You can copy this one and paste it in to you reply, that way you see it correct!
......|..|......|......|......|...|
------|+.|......|-....+|......|.-.|-------|
|.....|..|......|......|......|...|.......|
|.....12V..........?....... ...0V.........|
|_________________________________________|
..................+ / -
..................12 V .

[Ursus]

...................|-----(voltmeter)------|
...................|......................|
...................|......................|
......|..|......|..|...|......|...|.......|
------|+.|......|-....+|......|.-.|-------|
|.....|..|......|......|......|...|.......|
|.....12V..........?....... ...0V.........|
|_________________________________________|
..................+ / -
..................12 V .

This is how I was thinking. I.e connect from 0V plate/connector/ground to the middle plate.

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Ursus
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[Ruud de la Rosa]

Can you explain why the El. Field stays the same. I've read that if you decrease the distance, the distance is shorter so there is less resistance and the El. Field gains in strenght. E=U/l
U is the Voltage (12V) that stays 12V so if you decrease the distance E wil gain in strenght.

[Ursus]

That is if you have a constant voltage across a single capacitor. You have a constant voltage across the the *two* capacitors.
If you put two capacitors in series the *el. charge*(coloumbs) on both capacitors will be the same. -> El. field will be the same -> voltage will increase aross one cap. and decr. across the other if the plate moves.

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Ursus
Trust me, send money.

[Ruud de la Rosa]

Ahhhhhhhhhhh Thanks! Now were getting somewhere! You bring back a smile on my face I hope I won't need your help now!!

Cheers Ruud

[Ruud de la Rosa]

Just to be sure,
+|.................|+-|.................|-
<<................10 cm.................>>
.................10 Volts

So if the middel plate is 3 cm away from the (-) plate the voltmeter will indicate 3 Volts,
in the middle 5 Volts, 7cm from the (-)plate 7 volts?
Can i put the Volt sensor just anywhere on the middle plate??

[Ursus]

Yes and yes, the plate have the same potential everywhere.

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Ursus
Trust me, send money.

[Ruud de la Rosa]

Yes Yes Yes, Thanks a lot I finally got it!!!!!!

[Ruud de la Rosa]

One more Q: do I have to use plates that are the same size??? and the same thickness??

[Ruud de la Rosa]

Hey, Urus It didn't work!!!!!!!!
The volt meter has have an resistance that is unlimited!!!, We have an elecktrostatic one, but that one needs 6000V!!!!
We heard we can use bipolar, the frequency changes ?????
You can make it up to us by telling us how it works!!!!!!Please now we are really desperate!!!!!!!!!!!!!!! !!!!!!!!!!!!! !!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!

[Ursus]

Ooops, didn't think about the voltmeter .
(Why don´t you use 6000V??? High voltage is always fun ;))
What do you mean with "bipolar" and "the frequency changes"?


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Ursus
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[Ruud de la Rosa]

euhhh....alternatingcurrent A.C.??? The capacitator changes the frequency of the AC ??? Our teatcher told us that, but not how!!

[This message has been edited by Ruud de la Rosa (edited 12-23-1999).]

[Ruud de la Rosa]

??

[Ursus]

Hmmm... I'm a bit confused here. Maybe the capacitor can be connected to an oscillator in order to control it´s frequency. Or the capacitor is connected in serial with a resistor like a high-pass filter(the voltage across the resistor will change when the capacitor change). Maybe the capacitor can be connected to a toroid (or another inductive device) to create an oscillating circuit.

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Ursus
Trust me, send money.

[Ruud de la Rosa]

Tell me more about that oscillator thing, please.?

[Ursus]

An oscillator is often an integrated circuit who's frequency often is controlled by connecting a capacitor/resistor to it.

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Ursus
Trust me, send money.

[Ruud de la Rosa]

So if the capacitance changes then ....???

[Ursus]

...the frequency changes. If it is controlled by the capacitor, ofcourse

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Ursus
Trust me, send money.

[Ruud de la Rosa]

So how do I make it??

[Ruud de la Rosa]

?? Can you make me a diagram??Please??
BTW Happy 2000!!

[Williams]

How about this ? Get yourself a slide-potentiometer (say 10k), two resistors (say each 4.7k), two springs, a weight, a battery and a voltmeter.The strength of the springs and the size of the weight will depend how much acceleration you need to measure.

Connect the ends of the potentiometer to a battery. Connect the resistors in series across the battery. Connect the meter to the
slider terminal on the pot and to middle of the two resistors, on the 2v DC scale.

Mechanically attach the weight to the sider arm of the potentiometer, and attach the slider arm to two springs, which are fixed at each end:

O=========[]========O

<-- -->

O is fixed ends of springs.
= is the springs
[] is the potentiometer slider and
weight.

Now if you tip the whole thing vertically, that will be 1G of acceleration. I *think* if you tip it at 45 degrees, that will be 1/2 G of acceleration. You can use this method to calibrate it.

Now if you move the unit in the direction of the springs, you should should get acceleration reading on you meter, plus or minus depending on the direction.Moving it sideways should given you no measurement.

Using a center-calibrated analog +/- voltmeter might be the best way to display your result, instead of a digital meter. If you don't have a bipolar meter, you can use a unipolar meter if you unbalance
the two resistors to get a DC bias voltage (the voltage at zero Gs), or you can even use a second potentiometer in place of the resisitors to calibrate you zero-G point.

To measure larger or smaller G-forces, simply use a smaller or larger weight.

[Ruud de la Rosa]

Thanks williams but I'm not so good in English. Can you tell me what you measure with a slide potentiometer?? Amperes??

[Christiaan]

more like Potential Difference, ie voltage

[Christiaan]

more like Potential Difference, ie voltage, although my memory says a potentiometer is only a voltage divider