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Accelerometer?
#1
Posted 06 December 1999 - 23:19
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#2
Posted 07 December 1999 - 16:41
[This message has been edited by Christiaan (edited 12-07-1999).]
#3
Posted 09 December 1999 - 00:13
#4
Posted 09 December 1999 - 00:43
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Ursus
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#5
Posted 09 December 1999 - 21:56
#6
Posted 09 December 1999 - 22:43
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).]
#7
Posted 10 December 1999 - 00:40
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Ursus
Trust me, send money.
#8
Posted 10 December 1999 - 05:18
#9
Posted 10 December 1999 - 05:35
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Ursus
Trust me, send money.
#10
Posted 10 December 1999 - 22:59
We're in the 6th Grade. Were almost 18 years old. And we are desperate for help!!
Cheers Ruud & co
#11
Posted 10 December 1999 - 23:19
![:)](https://bb2.autosport.com/public/style_emoticons/default/smile.gif)
Thats all for now. Class dismissed!
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Ursus
Trust me, send money.
#12
Posted 11 December 1999 - 19:30
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GO RUBENS CRUSH SCUMMI
CHEERS RUUD
#13
Posted 13 December 1999 - 19:29
![;)](https://bb2.autosport.com/public/style_emoticons/default/wink.gif)
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).]
#14
Posted 17 December 1999 - 01:04
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
#15
Posted 17 December 1999 - 05:58
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Ursus
Trust me, send money.
#16
Posted 17 December 1999 - 06:24
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
#17
Posted 17 December 1999 - 06:46
Your formulas i OK but it is your assumption that is incoorrect.
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Ursus
Trust me, send money.
#18
Posted 17 December 1999 - 18:08
......|..|......|......|......|...|
----|+.|......|-..+|......|.-.|.----|
|.....|..|......|......|......|...|.......|
|.....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).]
#19
Posted 17 December 1999 - 18:21
......|..|......|......|......|...|
------|+.|......|-....+|......|.-.|-------|
|.....|..|......|......|......|...|.......|
|.....12V..........?....... ...0V.........|
|_________________________________________|
..................+ / -
..................12 V .
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#20
Posted 17 December 1999 - 18:44
...................|-----(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
Trust me, send money.
#21
Posted 17 December 1999 - 08:07
U is the Voltage (12V) that stays 12V so if you decrease the distance E wil gain in strenght.
#22
Posted 17 December 1999 - 08:27
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.
#23
Posted 17 December 1999 - 08:31
![:)](https://bb2.autosport.com/public/style_emoticons/default/smile.gif)
Cheers Ruud
#24
Posted 22 December 1999 - 03:51
+|.................|+-|.................|-
<<................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??
#25
Posted 22 December 1999 - 04:50
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Ursus
Trust me, send money.
#26
Posted 22 December 1999 - 04:56
#27
Posted 22 December 1999 - 05:02
#28
Posted 22 December 1999 - 19:52
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!!!!!!!!!!!!!!! !!!!!!!!!!!!! !!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!
#29
Posted 22 December 1999 - 22:31
![:o](https://bb2.autosport.com/public/style_emoticons/default/redface.gif)
(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
Trust me, send money.
#30
Posted 23 December 1999 - 21:19
[This message has been edited by Ruud de la Rosa (edited 12-23-1999).]
#31
Posted 26 December 1999 - 21:50
#32
Posted 27 December 1999 - 21:33
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Ursus
Trust me, send money.
#33
Posted 28 December 1999 - 00:42
#34
Posted 28 December 1999 - 15:50
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Ursus
Trust me, send money.
#35
Posted 29 December 1999 - 19:37
#36
Posted 29 December 1999 - 21:32
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Ursus
Trust me, send money.
#37
Posted 30 December 1999 - 01:34
#38
Posted 02 January 2000 - 22:57
BTW Happy 2000!!
#39
Posted 03 January 2000 - 10:46
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.
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#40
Posted 03 January 2000 - 19:34
#41
Posted 04 January 2000 - 22:26
#42
Posted 04 January 2000 - 22:31