Chip with simple program for Toy

[Kitchen Man]

On Fri, 24 Jun 2005 10:42:31 -0700, william coleman
<william.coleman@asu.edu> wrote:

Kruminilius W. wrote:

I'm trying to find the type of dial that you can turn in one direction or
the other without it stopping (unlike a potentiometer). I have no idea what
this part is called. Any thoughts?

Thanks.




It is called a pot without stops

I don't think so. The poster appears to be looking for a device that
indicates relative rotation, not absolute position. His example is of
a digital volume knob on an inexpensive stereo. He's got some good
answers on that - most intriguing to me is the electric motor
(generator, actually, since mechanical motion is producing current)
idea. I think using one of the cheaper optical encoders is probably
the simplest implementation.

--
Al Brennan

"If you only knew the magnificence of the 3, 6 and 9,
then you would have a key to the universe." Nicola Tesla

 

[Kitchen Man]

On Sun, 26 Jun 2005 14:17:33 +0100, Ben Stephens <bps500@york.ac.uk>
wrote:

Kruminilius W. wrote:
I'm trying to find the type of dial that you can turn in one direction or
the other without it stopping (unlike a potentiometer). I have no idea what
this part is called. Any thoughts?

Thanks.

What about those wheels in a balled computer mouse??? they do what you
want to do, but they are fragile.

Yep. Those are the optical encoders that have been discussed. I
think the OP wants a more rugged human interface - something to be
moved with the fingers?

--
Al Brennan

"If you only knew the magnificence of the 3, 6 and 9,
then you would have a key to the universe." Nicola Tesla

 

[John Fields]

On Mon, 27 Jun 2005 07:15:37 -0700, Kitchen Man <nannerbac@yahoo.com>
wrote:

The poster appears to be looking for a device that
indicates relative rotation, not absolute position. His example is of
a digital volume knob on an inexpensive stereo. He's got some good
answers on that - most intriguing to me is the electric motor
(generator, actually, since mechanical motion is producing current)
idea.

---
Unfortunately, it won't work since there's no way of determining
direction.

--
John Fields
Professional Circuit Designer

 

[Tom Biasi]

"Eiffel" <none@invalid.com> wrote in message
news:d96uhn$24jv$1@biggoron.nerim.net...

John Fields a écrit:

Could you briefly explain what "fu" is?

Just "Follow Up". Never thought that "fu" followed by a newsgroupname, an
explanation about the xposting, and a redirection of the contributions to
[newsgroupname] could be offensive. Sorry.

Eiffel

They're just jerking your chain a little bit.

 

[Doug McLaren]

In article <eThve.11588$lv.3909@newssvr33.news.prodigy.com>,
Steve Banks <none@here.isfine> wrote:

| >>3. How about some sort of foot pump like generator that you can pump the
| >>small lead battery back up, or possibly a lipo directly to save a few
| >>strokes, perhaps even while you are flying.
|
| >Not too sure if thats a serious suggestion?
|
| Just tryin to think outside the box. I didn't know if the car was anywhere
| nearby or if you were hiking somewhere remote and wanted a more self
| sufficient solution. It might not take that many pumps to charge a battery,
| lessee if I can take a quick guess:
|
| Lets say your setup can make two pounds of thrust for 600 seconds (10
| minutes), or 1200 lb. seconds.
|
| Lets also say you weigh 200lbs.
|
| that means for you to make the same output, you would have to put all your
| weight on the pedal for 1200/200 seconds or 6 SECONDS!

Your understanding of the idea of work (in the physics sense) is very
flawed.

Mere thrust (force) does not do work. Force * distance is what gives
you work (energy) -- you'd need to not only push on that pedal, but
actually move it. A long distance.

Assuming that you could push down on the pedal with 200 lbs of force,
if you moved it one foot that would produce 271 joules -- which would
give you one watt for 271 seconds or some combination thereof.

Assuming you had a 7Ah 12 volt battery (a popular size for field
boxes) and it was completely discharged, and you wanted to completely
charge it by pushing on your pedal with 200 lbs of force, you'd have
to push it 1116 feet. (This is assuming that everything is 100%
efficient, and that the voltage is exactly 12 volts and stays there,
of course -- assumptions that are not accurate. In the real world,
you'd have to push the pedal more to compensate for the losses and the
increase in voltage.)

That also means that this battery has enough power, fully charged, to
raise your body (if it weighs 200 lbs) 1116 feet in the air, assuming
all is 100% efficient.

It certainly is possible to charge batteries via muscle power, but a
lot more muscle is required than one might think.

I believe that a top athlete can produce about 1/2 horsepower (380
watts) for several minutes. Making the same (incorrect) assumptions
as before, this would mean that the athlete could charge that battery
in about 13 minutes, if the athlete could keep up that rate of work.

And this is just for a pretty small battery ...

To make this R/C related, your LiPo battery in your plane is probably
smaller than this field box battery. Perhaps 1/4th the size, So your
athlete could charge it in about three minutes, which is more
realistic, except that the battery couldn't tolerate that

--
Doug McLaren, dougmc@frenzy.com
When I die, I want to donate my body to science fiction.

 

[Grant Edwards]

On 2005-06-27, Doug McLaren <dougmc@frenzy.com> wrote:

that means for you to make the same output, you would have to
put all your weight on the pedal for 1200/200 seconds or 6
SECONDS!

Your understanding of the idea of work (in the physics sense)
is very flawed.

[...]

Assuming you had a 7Ah 12 volt battery (a popular size for
field boxes) and it was completely discharged, and you wanted
to completely charge it by pushing on your pedal with 200 lbs
of force, you'd have to push it 1116 feet. (This is assuming
that everything is 100% efficient, and that the voltage is
exactly 12 volts and stays there, of course -- assumptions
that are not accurate. In the real world, you'd have to push
the pedal more to compensate for the losses and the increase
in voltage.)

Yup. In the real world, you probably can't get much more than
50% efficiency in the generator, and the voltage is closer to
14V.

That translates into pushing with 200lb of force through a
distance of about 2600 feet (let's just say a half a mile).

I believe that a top athlete can produce about 1/2 horsepower
(380 watts) for several minutes. Making the same (incorrect)
assumptions as before, this would mean that the athlete could
charge that battery in about 13 minutes, if the athlete could
keep up that rate of work.

If I remember correctly from articles on human powered flight,
1/2HP (373W) was about the best you can get from a trained
cyclist for any decent period of time. IIRC, an average joe
probably can't produce more than 50-100W for more than a few
minutes.

7AH @ 14V == 352,800 Joules.

Assuming a generator efficiency of about 50% and somebody
pedalling at 100W, it would take about 7000 seconds
(352800/(100*0.5)). Roughly two hours.

--
Grant Edwards grante Yow! Yow! Maybe I should
at have asked for my Neutron
visi.com Bomb in PAISLEY--

 

[John Fields]

On 27 Jun 2005 13:12:19 -0700, "davidd31415" <davidd31415@yahoo.com>
wrote:

Hi,

I'm looking for a solution to water/dust proofing multi-pin electrical
connectors. A problem I have is that the grommet through which the
electrical wires will pass will need to be much smaller than the
connectors, and I'd like to be able to get the connectors in the box
without removing them from the wire. The box will need to be removed
on a regular basis as well (this will be used for testing purposes).

A design I've been envisioning is a a box which opens up like a
clamshell, and has two holes with half-grommets in each. When closed,
the box would bring the half-grommets together to form a whole grommet
again. This would allow one to open the box, lay a cable in the
grommet (instead of pushing it through as you would need to with a
regular grommet) and then close the box. The electrical connectors
could be in between.

I have not yet experimented with this so I am not sure if special
grommets would be required for this type of design.

The box will need to be waterproof; salt water will be sprayed on it
but it will not be submerged.

If anyone knows of such a design already in production out there, or
has another recommendation for a solution to this problem please post
it here!

These http://www.newmarpower.com/thrudex_boxes/thrudex_boxes.html would
be perfect, if one side was hinged.

I have been using di-electric grease compound until this time but dust
and sand ends up sticking to it and creating a big mess in the long
run.

---
If all you're looking to do is to waterproof a couple of made
connectors, why don't you just use some waterproof connectors in the
first place?

--
John Fields
Professional Circuit Designer

 

[Steve Banks]

"Doug McLaren" <dougmc@frenzy.com> wrote in message
news:kAYve.91708$PR6.19223@tornado.texas.rr.com...

In article <eThve.11588$lv.3909@newssvr33.news.prodigy.com>,
Steve Banks <none@here.isfine> wrote:

| >>3. How about some sort of foot pump like generator that you can pump
the
| >>small lead battery back up, or possibly a lipo directly to save a few
| >>strokes, perhaps even while you are flying.
|
| >Not too sure if thats a serious suggestion?
|
| Just tryin to think outside the box. I didn't know if the car was
anywhere
| nearby or if you were hiking somewhere remote and wanted a more self
| sufficient solution. It might not take that many pumps to charge a
battery,
| lessee if I can take a quick guess:
|
| Lets say your setup can make two pounds of thrust for 600 seconds (10
| minutes), or 1200 lb. seconds.
|
| Lets also say you weigh 200lbs.
|
| that means for you to make the same output, you would have to put all
your
| weight on the pedal for 1200/200 seconds or 6 SECONDS!

Your understanding of the idea of work (in the physics sense) is very
flawed bla bla bla.

I guess you missed the relationship between thrust and time. Hint: pretend
the airplane is just a fancy rocket. In this example it is putting out 2
lbs. of thrust, for 600 seconds, we are disregarding the weight of the
rocket.

Yes, you actually have to be moving the pedal with your entire weight for
those 6 seconds, but with a spring and proper gearing or leverage, you can
reproduce the same two pound push for 600 seconds.

Reaching for the work formula will not help you see this relationship
without a bit of algebra, and that woud be contrary to my stated goal of a
quick guess .

 

[John Savage]

junk1@davidbevan.co.uk writes:

Is SLA short for sealed lead acid? As thats what I have got now and I
cant seem to find them over 7Ah?

You haven't discovered the 'Cyclon' range then. I've been using a pair of
small 2.5AH Cyclons in a flashlight since about 1982 with no problems until
just last week when one of them developed an internal short. That's longevity
for you! You can get 2v cells in 25AH, so you'd need 6 of them. An advantage
of building using single cells is that if one goes bad you don't have to
replace the whole lot--just the one cell. Search: hawker cyclon lead
--
John Savage (my news address is not valid for email)

 

[Ban]

Continuum wrote:

What is the difference between hole mounted components and surface
mounted components?

Which is better?

There is no difference electrically, just size and package.
Since there is no difference, the question of better doesn't even arise. You
just take whatever is your preference.
--
ciao Ban
Bordighera, Italy

 

[Bob Myers]

"Continuum" <chaoscontinuum@gmail.com> wrote in message
news:1119928846.739039.105340@z14g2000cwz.googlegroups.com...

What is the difference between hole mounted components and surface
mounted components?

Which is better?

"Hole mounted" (i.e., "conventional lead") components have
long leads (attached pins or wires) which are intended to
go through plated holes in printed-circuit boards, and then
are soldered into place (either by hand or via various automated
methods, the most common/generic type being "wave"
soldering). "Surface-mount" components are just what the name
implies - rather than wirelike leads, these components generally
have short tabs with flat surfaces, or often simply metal "bumps"
on the sides/bottom of the package, which are placed against
conductive pads (without holes) on the surface of the PC board.
Such components are typically placed into position on the board
using automated "pick and place" equipment, held there via an
adhesive paste (often the same paste which carries the solder),
and soldered to the pads using various "reflow" methods
(in which the solder paste is melted - "reflowed" - in place,
AFTER being deposited, to create the soldered connection).

Which is better is too open-ended a question - better from
what perspective? SM components will often (if properly
placed and soldered) result in a more reliable product, are
certainly smaller (allowing more components per unit area),
and with little or no leads to speak of, SM components don't
have the problems which result from "parasitic" lead inductance
(such as lowered self-resonant frequencies for capacitors, etc.).
On the other hand, they're certainly more difficult for the hobbyist
to deal with (and even for pros to deal with, without the proper
equipment) in terms of construction, probing, and servicing of SM-
technology boards. Surface mount technology has become the
manufacturing method of choice for most high-volume electronics
production (which has had the side effect of making conventional
through-hole versions of some parts somewhat more difficult to find),
but may not be the best choice for low-volume production or for t
he hobbyist.

Bob M.

 

[John - KD5YI]

Kruminilius W. wrote:

I'm trying to find the type of dial that you can turn in one direction or
the other without it stopping (unlike a potentiometer). I have no idea what
this part is called. Any thoughts?

Thanks.

http://www.mouser.com/catalog/622/1050.pdf


Good luck,
John

 

[Doug McLaren]

In article <GS0we.458$4m3.43@newssvr19.news.prodigy.com>,
Steve Banks <none@here.isfine> wrote:

| "Doug McLaren" <dougmc@frenzy.com> wrote in message
| news:kAYve.91708$PR6.19223@tornado.texas.rr.com...
....
| > | that means for you to make the same output, you would have to
| > | put all your weight on the pedal for 1200/200 seconds or 6
| > | SECONDS!
| >
| > Your understanding of the idea of work (in the physics sense) is very
| > flawed bla bla bla.
|
| I guess you missed the relationship between thrust and time.

No, I understand the relationship. It's just not relevant to figuring
out how hard and long a person would have to work to recharge a
battery.

| Hint: pretend the airplane is just a fancy rocket. In this example
| it is putting out 2 lbs. of thrust, for 600 seconds, we are
| disregarding the weight of the rocket.

You are grossly mangling the physics involved and using that to come
up with absurd answers. This is not rocket science, an airplane is
not a fancy rocket, and when we're talking about charging a battery
with human power, there's no need to involve an airplane.

Hint: I have a degree in physics. I spent several years in school
studying this stuff, though this is _the_ basics, Physics 101. They
go over this stuff in the very first physics class in college, and
probably in high school too (I don't really remember what we covered
there.)

| Yes, you actually have to be moving the pedal with your entire weight for
| those 6 seconds, but with a spring and proper gearing or leverage, you can
| reproduce the same two pound push for 600 seconds.

So, what I really need to do is put this magic pedal under one of the
corners of my bed (so it's supporting much of my weight), and then
take a nap on it. Unlimited free energy! I could put one of these
pedals into an electric R/C plane, wrap it in rubber bands (so there's
a constant force on it) and use that to fly my electric plane across
the Atlantic. Look out Maynard! Look out laws of thermodynamics --
I've got a perpetual motion machine, in R/C form!

| Reaching for the work formula will not help you see this
| relationship without a bit of algebra, and that woud be contrary to
| my stated goal of a quick guess .

I didn't make a quick guess. I did all the math and algebra involved
(it's not very much, actually -- took a few minutes) and gave you
exact figures (and explicitly stated all the
assumptions/simplifcations I made.) Actually, I used the `units'
program to do most of the heavy lifting for me --

% units
2084 units, 71 prefixes, 32 nonlinear units
You have: 200 pounds-force * feet
You want: joules
* 271.16359

once I had that figure -- 200 pounds * 1 foot = 271 joules, the rest
was just simple algebra and arithmetic. Oh, you'll also need to know
that power = voltage * current (watts = volts * amps), one watt = one
joule/second, that there's 3600 seconds in an hour (to convert volts
and amp-hours to joules) and 746 watts in a horsepower (though somehow
I originally had it in my head that it was 760 watts. Not sure where
that mistake came from ....)

http://en.wikipedia.org/wiki/Mechanical_work may be of some assistance
if you still don't understand the physics involved. Go down to the
`Simpler formulae' part -- there's no need to use integrals to get a
good approximation of the human work needed.

--
Doug McLaren, dougmc@frenzy.com
Computers make it easy to do alot of things, but most of the things
that make it easier to do don't need to be done. -- Andy Rooney

 

[Dr Engelbert Buxbaum]

davidd31415 wrote:

So how would the sampling rate of an oscilloscope be related to what a
square wave ends up looking like on the scope? Is there a rule of
thumb for the sampling rate of an oscilloscope when sampling square
waves?

Shannons theorem says that in order to see a signal with frequency n,
the sampling rate needs to be at least 2*n. Practical application:
Humans hear frequencies of up to 20 kHz, so the sampling frequency of a
CD player needs to be at least 40 kHz. Frequencies higher than 20 KHz
are filtered out before sampling by a low pass, in order to prevent
artefacts (beating of sampling and signal frequency).

This also means that digital equipment can only truthfully represent
sine waves. Any other signal form contains high order harmonics, which
need to be cut of at some point. The closer a square wave gets to the
sampling rate, the more harmonics are removed, and the "rounder" the
signal will appear.

Thus how much higher than the signal frequency the sampling rate needs
to be depends on how good you want the signal form represented, but 10
times is good enough for most purposes.

 

[Steve Banks]

"Grant Edwards" <grante@visi.com> wrote in message
news:11c1a301164rn80@corp.supernews.com...

On 2005-06-28, Steve Banks <none@here.isfine> wrote:

| weight on the pedal for 1200/200 seconds or 6 SECONDS!

Your understanding of the idea of work (in the physics sense) is very
flawed bla bla bla.

I guess you missed the relationship between thrust and time.
Hint: pretend the airplane is just a fancy rocket. In this
example it is putting out 2 lbs. of thrust, for 600 seconds,
we are disregarding the weight of the rocket.

Yes, you actually have to be moving the pedal with your entire
weight for those 6 seconds, but with a spring and proper
gearing or leverage, you can reproduce the same two pound push
for 600 seconds.

Wrong.

Gears and levers do not provide a ratio between thrust/time and
thrust/time. The ratio provided by gearing/leverage provides
ratios of thrust/distance -- which is what was explained to you
using the formula for work.

Thanks, I am quite familiar with the formula. The point of the SPRING was
to store the energy that it could be dissapated as a two pound push for a
longer period.

Reaching for the work formula will not help you see this relationship

Because there is no such relationship.

What are you afraid of?!?

without a bit of algebra, and that woud be contrary to my
stated goal of a quick guess .

Quick, but not founded in real physics.

Ok, if you don't trust my instincts then look at Newtons second law and
pretend we are pushing on a large rock in space with our pedal. We have
known forces( 2 and 200lbs.) , assume a large mass, the times are known(600
and 6 seconds), the terminal velocities can be computed and will be the SAME
if we push for 600 seconds with 2 lbs or 6 seconds with 200 lbs. The only
trick here is in finding the right sized rock to limit how far you have to
move the pedal in those 6 seconds.

 

[Ban]

PeteS wrote:

There is no difference electrically, just size and package.

True at a basic level, but not at high frequencies. Bob noted lower
parasitics, and that's important in high speed interconnects (in fact,
even 0603 has high parasitics when presented with 2.5Gb/s pairs). For
really high speed stuff (up to 10Gb/s pairs) we're moving to 0201 and
even 01005 for things like series caps and pullups.

I agree with you that in a lot of cases, you choose what you prefer
(or need for the product), though.

Cheers

PeteS

Right, but 1GHz and up has never been done with through-hole parts. People
used stripline with special HF parts and it hasn't changed that much today.
And if the OP has to asked this question, he will not be into this kind of
specialized engineering, but maybe wants to make a white LED supply which
also pulses or is fading.
For the normal hobbyist level it is actually easier to work with reasonably
sized SMD parts. Very important are some intelligently applied design rules
(here lies often the beginners fault) which usually require to modify each
of the library parts. Hand-soldering requires specialized pads, keep-out
areas, you also need to modify the silk-screen, so when a beginner uses the
standard lib parts, he will place the components too close, the distance
between signal and ground-pour is too small etc.etc. This will lead to
frustration and failure in getting the circuit running. On the hobbyist
level a certain understanding is needed, so it might be better to get some
experience with soldering leaded parts...
--
ciao Ban
Bordighera, Italy

 

[Steve Banks]

"Doug McLaren" <dougmc@frenzy.com> wrote in message
news:lN5we.58517$j51.2511@tornado.texas.rr.com...

In article <GS0we.458$4m3.43@newssvr19.news.prodigy.com>,
Steve Banks <none@here.isfine> wrote:

| I guess you missed the relationship between thrust and time.

No, I understand the relationship. It's just not relevant to figuring
out how hard and long a person would have to work to recharge a
battery.

There are tons of losses to consider, nevertheless there is a direct
relationship between an airplanes battery capacity, and how hard and for how
long it can push against the air. If my guess was within an order of
magnitude then I'd call it a good guess.

| Hint: pretend the airplane is just a fancy rocket. In this example
| it is putting out 2 lbs. of thrust, for 600 seconds, we are
| disregarding the weight of the rocket.

You are grossly mangling the physics involved and using that to come
up with absurd answers. This is not rocket science, an airplane is
not a fancy rocket, and when we're talking about charging a battery
with human power, there's no need to involve an airplane.

I am not attempting to mangle physics, and I did mention that "there are
losses all over the place in this analogy" and I am well aware of what they
are though I choose not to ramble on and on about them.

Hint: I have a degree in physics. I spent several years in school
studying this stuff, though this is _the_ basics, Physics 101. They
go over this stuff in the very first physics class in college, and
probably in high school too (I don't really remember what we covered
there.)

Intelligence is no defense against stupidity

| Yes, you actually have to be moving the pedal with your entire weight
for
| those 6 seconds, but with a spring and proper gearing or leverage, you
can
| reproduce the same two pound push for 600 seconds.

So, what I really need to do is put this magic pedal under one of the
corners of my bed (so it's supporting much of my weight), and then
take a nap on it. Unlimited free energy! ...

Thats cute, I can imagine you doing a little dance showing off an imaginary
magic pedal to the room while you typed that.

| Reaching for the work formula will not help you see this
| relationship without a bit of algebra, and that woud be contrary to
| my stated goal of a quick guess .

I didn't make a quick guess.

That wasn't the point, and you grossly deviated when you made the comparison
of a ~1500mah 9.6 power pack to a 12V 7000ah battery. With a more correct
battery pack for the thrust and duration given it comes out to just under
120 strokes, or about a minute on the treadmill, which happens to be within
an order of magnitude of my 6 second guess. So thanks for lending validity
to my theory, which BTW didn't take me years of specialized college
coursework to realize.

 

[Ban]

sandeep wrote:

Hi to all ,
I want to know about the different BJT/FETamplifier configurations
about their working and which component is used in them and why?
And how they are used practically?
Thank you.

You want to know everything. Better inscribe for some EE classes at local
university. You then learn how to formulate questions and which books
(what's that?) to look in. Learn how to read schematics, the parts are all
mentioned there. You will find everything on the net too, just a little
persistance needed. Make it your hobby to build whatever amplifier you
meant. And do not want instant pleasure, that is possible with drugs only.
The way is arduous but when you have designed your first amp and it works
fine, the pleasure is all yours.
--
ciao Ban
Bordighera, Italy

 

[Tim Auton]

"Steve Banks" <none@here.isfine> wrote:
[snip]

That wasn't the point, and you grossly deviated when you made the comparison
of a ~1500mah 9.6 power pack to a 12V 7000ah battery. With a more correct
battery pack for the thrust and duration given it comes out to just under
120 strokes, or about a minute on the treadmill, which happens to be within
an order of magnitude of my 6 second guess. So thanks for lending validity
to my theory, which BTW didn't take me years of specialized college
coursework to realize.

Dude, you're funny.


Tim
--
Today's message was brought to you by Mary, Jane and a big number two.

 

[Ian Stirling]

In sci.chem.electrochem.battery Steve Banks <none@here.isfine> wrote:
<snip>

That wasn't the point, and you grossly deviated when you made the comparison
of a ~1500mah 9.6 power pack to a 12V 7000ah battery. With a more correct
battery pack for the thrust and duration given it comes out to just under
120 strokes, or about a minute on the treadmill, which happens to be within
an order of magnitude of my 6 second guess. So thanks for lending validity
to my theory, which BTW didn't take me years of specialized college
coursework to realize.

1.5Ah * 9.6V = ~15Wh.
To recharge in a minute (which will kill the cells) you'd need a power
of 15W*60 = 900W.
Say a efficiency of 80%, and you'r up to 1100W.
Trained athletes can't provide that much power, unless ground up and mixed
with oxidiser.