| Author |
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Panther
Guest
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 Posted:
Fri Dec 02, 2005 10:35 pm Post subject:
Re: using current to measure a time |
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Thanks but I think the poster above you's suggestion is simple enough to do.
"Nick." <me@here.com> wrote in message
news:pan.2005.12.02.07.06.33.92246@here.com...
| Quote: | On Thu, 01 Dec 2005 21:32:18 +0000, Panther wrote:
How I could use a potential divider to set a pulse when the circuit
breaks?
You cannnot. A potential divider divides potential.
Or a more complicated circuit design?
Some simple designs (e.g. a resister and a capacitor) have already been
suggested but are too complex (?)
If you really must use the stopwatch you mentioned what you really need is
an edge-triggered monostable. You may get away with a simple inverter,
depending on the input logic of the stopwatch. Either way, you will
probably run into problems with "switchbounce".
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Panther
Guest
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| Posted:
Fri Dec 02, 2005 10:37 pm Post subject:
Re: using current to measure a time |
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| Quote: | Use a high-impedance voltmeter, like a DVM.
Record the voltage while the balls are apart. It will very probably be
very close to zero. You can assure this by momentarily shorting the
capacitor terminals.
Drop the ball. While the balls are in contact, current will flow and
start to charge the cap through the resistor, with a time constant
of T = RC.
The voltage across the capacitor at the exact moment that the balls
separate will tell you T by using that exponential equation that I
can't remember now, but since you're in school you should look it up
anyway.
|
Thank you!!!!!!!!!!!!!! This looks very simple so I will give it a go. But
could you please explain what the time constant T = RC means? I've no idea.
So you're saying, for example, if I were to do the practical now and record
the voltages, I would be able to work out the time section using some
formula? IE I don't need the equipment? As that would very handy.
Thanks |
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Rich Grise
Guest
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| Posted:
Sat Dec 03, 2005 1:35 am Post subject:
Re: using current to measure a time |
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On Fri, 02 Dec 2005 22:37:55 +0000, Panther wrote:
| Quote: | Use a high-impedance voltmeter, like a DVM.
Record the voltage while the balls are apart. It will very probably be
very close to zero. You can assure this by momentarily shorting the
capacitor terminals.
Drop the ball. While the balls are in contact, current will flow and
start to charge the cap through the resistor, with a time constant
of T = RC.
The voltage across the capacitor at the exact moment that the balls
separate will tell you T by using that exponential equation that I
can't remember now, but since you're in school you should look it up
anyway.
Thank you!!!!!!!!!!!!!! This looks very simple so I will give it a go. But
could you please explain what the time constant T = RC means? I've no idea.
So you're saying, for example, if I were to do the practical now and record
the voltages, I would be able to work out the time section using some
formula? IE I don't need the equipment? As that would very handy.
|
T = RC is the "time constant". It's the amount of time it takes for the
charge to reach 63% (or something) of its final value, R is in ohms, and
C is in farads. With T on the horizontal axis, the cap voltage rises at
some exponential rate...
Maybe one of these sites will be more helpful:
http://www.google.com/search?q=%22time+constant%22
Good Luck!
Rich
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John Fields
Guest
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| Posted:
Sat Dec 03, 2005 1:35 am Post subject:
Re: using current to measure a time |
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On Fri, 2 Dec 2005 22:35:20 +0000 (UTC), "Panther" <black@cat.com>
wrote:
| Quote: | Thanks but I think the poster above you's suggestion is simple enough to do.
|
---
It seems to me that by your not cross-posting, by your off-handed
rejection of several solutions which have been presented to you,
and by not using in-line references, you're not really looking for a
solution to your "problem", but instead are more interested in
playing games.
Am I wrong?
--
John Fields
Professional Circuit Designer |
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John Fields
Guest
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| Posted:
Sat Dec 03, 2005 1:35 am Post subject:
Re: using current to measure a time |
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On Fri, 2 Dec 2005 22:37:55 +0000 (UTC), "Panther" <black@cat.com>
wrote:
| Quote: |
Use a high-impedance voltmeter, like a DVM.
Record the voltage while the balls are apart. It will very probably be
very close to zero. You can assure this by momentarily shorting the
capacitor terminals.
Drop the ball. While the balls are in contact, current will flow and
start to charge the cap through the resistor, with a time constant
of T = RC.
The voltage across the capacitor at the exact moment that the balls
separate will tell you T by using that exponential equation that I
can't remember now, but since you're in school you should look it up
anyway.
Thank you!!!!!!!!!!!!!! This looks very simple so I will give it a go. But
could you please explain what the time constant T = RC means? I've no idea.
So you're saying, for example, if I were to do the practical now and record
the voltages, I would be able to work out the time section using some
formula? IE I don't need the equipment? As that would very handy.
|
---
How do you propose to do the practical without the equipment?
--
John Fields
Professional Circuit Designer |
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Andy Baxter
Guest
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| Posted:
Sat Dec 03, 2005 7:37 am Post subject:
Re: using current to measure a time |
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John Fields said:
| Quote: | On Fri, 2 Dec 2005 22:35:20 +0000 (UTC), "Panther" <black@cat.com
wrote:
Thanks but I think the poster above you's suggestion is simple enough to do.
---
It seems to me that by your not cross-posting, by your off-handed
rejection of several solutions which have been presented to you,
and by not using in-line references, you're not really looking for a
solution to your "problem", but instead are more interested in
playing games.
|
I think he's been given a practical assignment in school physics /
electronics, but isn't sure how to do it, and doesn't really want to do
the experiment anyway.
To Panther - if you don't do the experiment, you're cheating yourself. The
whole of science is based on trying out your ideas against the real world
by testing them to see if they work, so if you don't do that bit, you're
not really learning science.
--
http://www.niftybits.ukfsn.org/
remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with [html] or [attachment] in the subject line. |
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Panther
Guest
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| Posted:
Sat Dec 03, 2005 10:04 am Post subject:
Re: using current to measure a time |
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| Quote: | To Panther - if you don't do the experiment, you're cheating yourself. The
whole of science is based on trying out your ideas against the real world
by testing them to see if they work, so if you don't do that bit, you're
not really learning science.
|
Nononono
I am doing the practical it's just that i have ONE WEEK to do it in and if I
just measure the voltages then I won't have time to work out the time, so if
I could record the voltages and do the calculations later using some formula
it would bebetter. |
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Panther
Guest
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| Posted:
Sat Dec 03, 2005 10:05 am Post subject:
Re: using current to measure a time |
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| Quote: | It seems to me that by your not cross-posting, by your off-handed
rejection of several solutions which have been presented to you,
and by not using in-line references, you're not really looking for a
solution to your "problem", but instead are more interested in
playing games.
Am I wrong?
|
I'm sorry but I cannot do those complicated solutions, I am poor at
practicals/electronics. Simpler circuits are better. |
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Andy Baxter
Guest
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| Posted:
Sat Dec 03, 2005 5:35 pm Post subject:
Re: using current to measure a time |
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Panther said:
| Quote: | To Panther - if you don't do the experiment, you're cheating yourself. The
whole of science is based on trying out your ideas against the real world
by testing them to see if they work, so if you don't do that bit, you're
not really learning science.
Nononono
I am doing the practical it's just that i have ONE WEEK to do it in and if I
just measure the voltages then I won't have time to work out the time, so if
I could record the voltages and do the calculations later using some formula
it would bebetter.
|
It shouldn't take you long to work out the times - it's a simple formula.
Look up 'capacitor', 'exponential decay', and 'time constant' on google.
--
http://www.niftybits.ukfsn.org/
remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with [html] or [attachment] in the subject line. |
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Rich Grise
Guest
|
| Posted:
Sun Dec 04, 2005 1:35 am Post subject:
Re: using current to measure a time |
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On Sat, 03 Dec 2005 10:05:49 +0000, Panther wrote:
| Quote: | It seems to me that by your not cross-posting, by your off-handed
rejection of several solutions which have been presented to you,
and by not using in-line references, you're not really looking for a
solution to your "problem", but instead are more interested in
playing games.
Am I wrong?
I'm sorry but I cannot do those complicated solutions, I am poor at
practicals/electronics. Simpler circuits are better.
|
Then your best advice is to go back to school, but read the textbooks
this time. There are no simpler answers to your question if you refuse
to become educated in the methods and background involved.
The ball is in your court.
Good Luck!
Rich |
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Rich Grise
Guest
|
| Posted:
Sun Dec 04, 2005 1:35 am Post subject:
Re: using current to measure a time |
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On Sat, 03 Dec 2005 10:04:05 +0000, Panther wrote:
| Quote: | To Panther - if you don't do the experiment, you're cheating yourself. The
whole of science is based on trying out your ideas against the real world
by testing them to see if they work, so if you don't do that bit, you're
not really learning science.
Nononono
I am doing the practical it's just that i have ONE WEEK to do it in and if I
just measure the voltages then I won't have time to work out the time, so if
I could record the voltages and do the calculations later using some formula
it would bebetter.
|
Well, record the damn voltages then!
Set up your thing, with a DC power supply and series resistor to the
swinging ball. Connect a capacitor from the stationary ball to ground.
Measure the voltage at the capacitor. Any ordinary DVM should be able
to do that - in the US, they're US$9.95. Write this voltage down. It
doesn't matter what it is, just write it down. Swing the ball. Measure the
voltage at the capacitor, QUICKLY! Write that number down.
Record the value of the resistor and capacitor in step 1, and use
the equation for a charging capacitor to find out how long they were
touching.
If I gave you any more detailed of an answer than that, I would be
doing your labwork for you, which is even worse than doing your
homework for you.
Get up off your dead butt and learn something.
If you only have a week left, and haven't by now learned the material,
then maybe you should get into politics or something.
Good Luck!
Rich |
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Panther
Guest
|
| Posted:
Sun Dec 04, 2005 9:40 am Post subject:
Re: using current to measure a time |
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OK I will do that thanks.
But you should really be blaming the school, not me for the lack of time. We
were given the assignments about a week ago, and expected to think of
something in a few days. I get practically no help from the teacher anyway.
I had to wait for ONE HOUR FOURTY FIVE MINUTES to get help for FIVE minutes,
because the teacher has to run out somewhere. I hope the next year
coursework involves no practical.
"Rich Grise" <rich@example.net> wrote in message
news:pan.2005.12.03.21.41.53.723040@example.net...
| Quote: | On Sat, 03 Dec 2005 10:04:05 +0000, Panther wrote:
To Panther - if you don't do the experiment, you're cheating yourself.
The
whole of science is based on trying out your ideas against the real
world
by testing them to see if they work, so if you don't do that bit, you're
not really learning science.
Nononono
I am doing the practical it's just that i have ONE WEEK to do it in and
if I
just measure the voltages then I won't have time to work out the time, so
if
I could record the voltages and do the calculations later using some
formula
it would bebetter.
Well, record the damn voltages then!
Set up your thing, with a DC power supply and series resistor to the
swinging ball. Connect a capacitor from the stationary ball to ground.
Measure the voltage at the capacitor. Any ordinary DVM should be able
to do that - in the US, they're US$9.95. Write this voltage down. It
doesn't matter what it is, just write it down. Swing the ball. Measure the
voltage at the capacitor, QUICKLY! Write that number down.
Record the value of the resistor and capacitor in step 1, and use
the equation for a charging capacitor to find out how long they were
touching.
If I gave you any more detailed of an answer than that, I would be
doing your labwork for you, which is even worse than doing your
homework for you.
Get up off your dead butt and learn something.
If you only have a week left, and haven't by now learned the material,
then maybe you should get into politics or something.
Good Luck!
Rich
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Andy Baxter
Guest
|
| Posted:
Sun Dec 04, 2005 4:36 pm Post subject:
Re: using current to measure a time |
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Panther said:
| Quote: | OK I will do that thanks.
But you should really be blaming the school, not me for the lack of time. We
were given the assignments about a week ago, and expected to think of
something in a few days. I get practically no help from the teacher anyway.
I had to wait for ONE HOUR FOURTY FIVE MINUTES to get help for FIVE minutes,
because the teacher has to run out somewhere. I hope the next year
coursework involves no practical.
|
Good luck with it,
andy.
--
http://www.niftybits.ukfsn.org/
remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with [html] or [attachment] in the subject line. |
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