water analogy- a simple calculator

[F. Bertolazzi]

George Herold:

Got any data? It could be that the deceleration is constant, but that
we 'humans' like the deceleration to be less as we come to a stop.

If the car has ABS and you brake at high speed, it seems to barely slow
down. As the speed decreases the deceleration increases.

This is due to the fact that the brakes must dissipate the kinetic energy
"stored" in the running car without exceeding the tyres grip, energy that
is proportional to the square of speed.

 

[George Herold]

On Jan 12, 11:09 pm, Phil Hobbs
<pcdhSpamMeSensel...@electrooptical.net> wrote:

George Herold wrote:
On Jan 12, 5:15 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net>  wrote:
George Herold wrote:
On Jan 12, 3:27 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net>    wrote:
George Herold wrote:
On Jan 12, 7:16 am, John Fields<jfie...@austininstruments.com>      wrote:
On Tue, 11 Jan 2011 13:47:18 -0800, John Larkin

jjlar...@highNOTlandTHIStechnologyPART.com>      wrote:

Right. I can't think of an electrical analogy to friction.

---
Why would you think that resistance isn't analogous to friction?

---
JF

Did you look at the plots I posted John?  Friction causes a linear
decrease in amplitude, not exponential.

George H.

Cute demo.

I'm not really persuaded by the v**0 argument for frictional damping..
Long years of painstaking research in the field of yo-yo tricks has
convinced me that when you have a string sliding on a roller, once you
break it loose there's very little friction.

I suspect that if you put a load cell on the string, you'd find that the
actual retarding force was concentrated in narrow pulses near the peak
of each oscillation.  The work required to break the string loose is
pretty well constant, so you'd lose a fixed amount of energy per half
cycle.  The total energy is

The data is a lot better than the 'scope shots.  The static friction
acts only for an 'instant' when the rotor turns around.  I never
really looked closely, but there are no 'big' steps at the ends. If I
record some data I could try and fit the envelope... it should be
possible to determine the static amplitude loss at the turn around
points and constant loss through the rest of the motion.

There are actually two strings pressing on opposite sides of the
rotor.  And we can load them with differernt weights and change the
frictional force.

         I omega**2    k*theta**2
E =   ---------- + ------------
             2            2

where omega = d/dt(theta).  The average energy loss would be linear in
time, so

   ** It's the amplitude loss that is linear in time! **

I get to that point later.  Right now I'm doing the reciprocating yo-yo
case.  (Patience, Grasshopper.)

dE
--  = Qdot = I omega d(omega)/dt + k*theta*d(theta)/dt.
dt

At the extremes of motion, omega = 0, so if dE/dt over one cycle is some
constant B, then

As above amplitude loss for one cycle is constant. Not energy loss.

d(theta)/dt =  B/(k*theta)

so theta = (2B/k)*sqrt(t0-t),

where t0 is the time where the motion stops.  That's the case for car
brakes--you have to lighten up on the pedal as you slow down, to avoid
jerking to a stop.

Doesn't your car behave like that?  All the ones I've driven do.

Yeah sure that's how I drive a car.... If my car went back and forth
and I wanted to stop it, I'd apply the brakes at the turn around point
where the velocity is zero.... I guess I'm just saying I don't get the
(hic) analogy.

Deceleration of a braking car is due to sliding friction, but for a
given normal force on the brake pads, the deceleration isn't constant.
I'm not sure why that is, but it clearly shows that the coefficient of
friction model isn't always adequate as the speed goes to zero.

Got any data? It could be that the deceleration is constant, but that
we 'humans' like the deceleration to be less as we come to a stop.
Same goes when you start off. Unless you're a teenage boy with his
hot rod, you don't want to be slammed back into the seat with 1/2 g
right away, but perfer a gentle increase in acceleration.

Now we get to the coefficient-of-friction case, which is what your web
page assumes.

With the usual coefficient-of-friction approximation, i.e. your v**0
approach, the power consumed by the rotor in overcoming friction is

dE/dt =  omega Gamma,

where Gamma is the frictional torque.

At the peak velocity, theta = 0, so

d(omega)/dt = -Gamma/I,

and you get a linear decrease in the amplitude, as you say.

A colleague has worked out the math in detail.  I'd be happy to send
you the appropriate section of the manual.

If those were the whole story, I'd expect to see the envelope be convex,
i.e. with a linear slope at high amplitudes where the sliding friction
dominates, and a steeper slope at low amplitude where it's the stiction
that matters most.

I think the only point where static friction matters is right when it
stops.  At that point there is not enough torque in the spring to over
come the static friction and it stops.  And stops at a non-equilibrium
position.

But it stops twice per cycle, which is the point I was trying to make
above. Doesn't it?

Sure. If there was energy loss (from say the thread sticking?) right
at the turn around point wouldn't the amplitude have to undergo some
sort of 'step' change in amplitude?  The energies all in the spring at
that point.  I'm not sure what the thread is at one point we were
using 'spider' wire.  (use for fishing.)  It's rubbing against an 1"
diameter Al rod, There are two strings each touching for maybe 45
degrees of the circle.  We first tried a complete wrap of string and
that was much to much friction.  I think there's ~100 grams hanging on
each string.

You wouldn't get a noticeable discontinuity on each start because when
the stiction lets go, the rotor accelerates smoothly from 0.  There
might be something visible when it stops each time--another of those
square-root decelerations.

I'm confused... You want to model all the energy loss as occuring at
the turn around points.. where the rotor stops. (Is that correct?)
It's pretty easy to identify the energy in the system. We agree there
is some energy loss after each 1/2 cycle. So if the energy is only
lost at the turn around point... then there must be a decrease in the
amplitude at that point. It's all potential energy at that point.
(Geesh Phil you're a lot smarter than I am, I'm repeating myself
now.. I feel pretty silly writing down a potential energy equation
for you!)

Your plot's envelope is slightly concave, which looks like you have some
exponential behaviour in there someplace.

Yeah there is a little bit of 'concavity' to it.  I can't really say
if this is some velocity dependent damping... or perhaps something to
do with the friction.

The coefficient of friction approximation isn't that good, but it
certainly leads to a decay that's a lot closer to linear than to
exponential.

I was looking at some other data today.  The amplitude looks a lot
more linear.

Cool.  As I said, it's a great demo.  Nothing like a bit of data to
discourage pontification.

Thanks Phil. I hope it's alright if I continue to disagree with your
interpretation.

George H.

(And for the viscosity fans, viscous drag goes like velocity squared, so
it isn't like resistance either.
We 'attempt' to do that too.  I've never tried fitting the data, but
at least it's different than exponential.  We put two ~2 foot long
1/4"diam. aluminum tubes through the rotor with paddles one the
ends.  ... I was just thinking one way to test the v**2 depenedence
would be to drive it and look at the response.
If you take a boat with a displacement
hull, i.e. one that doesn't plane like a speedboat, then for any initial
velocity, its displacement goes like log(initial velocity * time).)

Interesting, anyway.

Say if you like Yo-yo's my son has a new model with a built in
clutch... (Or anti clutch... it grabs on when the angular velocity is
small.)  Making it 'sleep' is a piece of cake!

Progress can only be made against a resisting medium.

Yeah, feel free to 'resist' me any time.

Thanks,
grasshopper.



Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) nethttp://electrooptical.net- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

 

[George Herold]

On Jan 13, 12:26 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 20:05:30 -0800 (PST), George Herold





gher...@teachspin.com> wrote:
On Jan 12, 8:59 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 18:04:19 -0600, "k...@att.bizzzzzzzzzzzz"

k...@att.bizzzzzzzzzzzz> wrote:
On Wed, 12 Jan 2011 13:31:45 GMT, N0S...@daqarta.com (Bob Masta) wrote:

On Tue, 11 Jan 2011 13:41:39 -0800, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 11 Jan 2011 13:51:52 GMT, N0S...@daqarta.com (Bob Masta)
wrote:

Agreed in principle, but there can be serious debate over
the proper analogy even in this basic RLC case. Consider
that with a series connection, current is analogous to force
in that it is identical in all elements, while the voltage
drop across each element is analogous to relative velocity.
That doesn't fit well with the voltage-as-pressure concept.

Mechanical engineers use different analogies as needed, and
can make the math work out even when there isn't much
intuitive connection (to this poor EE, at least!).

Why do MEs need analogies? They can see and feel their stuff!

They need analogies so they know what equations to borrow
from the EEs!

When I started working for IBM in '74 I was surprised to see the heat transfer
folks down the hall were using "our" circuit simulators to do their work.

Ther was one thermal simulator, called Sauna, that was based on
connecting an array of dots in 3-space with (thermal) resistors in a
cubic matrix, and caps to ground, and then running a Spice-like sim.

I've used Spice to simulate thermal systems. This works pretty well:

ELECTRICAL THERMAL

1 amp 1 watt
1 farad 1 gram aluminum
1 volt 1 degree C
1 second 1 second
1 ohm 1 degC/watt

John- Hide quoted text -

- Show quoted text -

Yeah I do it with one resistor and one cap.. (Well I usually break the
resistor in half.)

When I have to explain thermal circuits to people I love to use the
'electronic' analogy.
(I'm not sure it helps them, but it's how I understand it.)  Of course
capacitors don't have phase changes.

Oh the Farad is equivalent to the heat capacity.  Joules/deg K.  (I
think that's right. RC is time, seconds in both 'units')

George H.

Unfortunately, most useful thermal systems are diffusive, unlike
lumped electrical circuits. And the diffusion is usually among nasty
3D shapes. So the modeling is ugly finite-element stuff, basically a
many-element 3D hairball of resistors and capacitors.

I guess I've never tried it on complex systems. Mostly I'm just
trying to guess what the time constant of the system will be.

Interestingly, there is no thermal equivalent of an inductor. So
thermal systems don't ring or oscillate, unless you sneak in some
nonlinear mechanical or transport phenomenon.

I wish I had some nice little thermal resistance calculator, even a 2D
one.

There must be some software company that would sell you such a thing.
Ever read the book "Hot air rises and Heat sinks" Not all that
technical, but an enjoyable read.

George H.

John- Hide quoted text -

- Show quoted text -

 

[John Fields]

On Wed, 12 Jan 2011 15:23:45 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 15:06:45 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 12 Jan 2011 09:02:27 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 06:16:37 -0600, John Fields
jfields@austininstruments.com> wrote:

On Tue, 11 Jan 2011 13:47:18 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Right. I can't think of an electrical analogy to friction.

---
Why would you think that resistance isn't analogous to friction?

---
JF

Grossly different behavior, explained elsewhere. Resistance is more
like low-Reynolds number viscoscity.

---
Why complicate things when all that's necessary for the analogy to be
valid is for the viscosity to be likened to resistance?

http://en.wikipedia.org/wiki/Viscosity

---

The only things that resistance and friction have in common is that
both are dissipative.

---
Then, since watts is watts and they both get hot from energy wasted
when work is being done on them, that's what causes them both to be
dissipative, and that's all that's necessary for the analogy to be
valid.


Valid to you, not to me. An analogy is valid to me if the systems
behave similarly, if graphs or waveforms in one domain look like ones
in the other domain. If you are happy with fuzzy poetical feelings,
fine, but they are not useful tools for teaching electronics. Worse
than useless, if the analogy encourages frank misunderstanding.

---
You're such a hypocritical bull shitter!

When you want to play fast and loose, like when you off-handedly make
the claim that latching relays have infinite gain, and you're
corrected, you try your damnedest to force your conveniently
reformulated definition of 'infinity' down everyone's throat, but when
someone else uses a perfectly good analogy to try to explain a
concept, you yell and scream and jump up and down and rail about the
incorrectness of the analogy, even though for the purpose at hand,
it's eminently suitable.
---

Cats "get hot from energy wasted when work is being done on them" so I
guess cats are good analogs of resistors.

---
That would be true if they were strapped across hot high-voltage
lines, but normally the resistor analogs would apply to some of their
innards.
---

A ringing LC tank rings forever, at ever-decreasing amplitude, and the
envelope decays exponentially. A mechanical oscillator damped by
mechanical friction doesn't decay exponentially and, at some point,
just stops.

Stuff like that is important to me. Maybe it's not to you.

---
It's important to me when it matters, but I don't think that an
analogy's failing to exibit second and third order effects of its
"target" matters.

I don't think you do either, and are just trying to win the argument
by throwing a lot of crap into the game.

---
JF

 

[George Herold]

On Jan 13, 2:10 am, Rich Grise <ri...@example.net.invalid> wrote:

George Herold wrote:

Oh the Farad is equivalent to the heat capacity.  Joules/deg K.  (I
think that's right. RC is time, seconds in both 'units')

Well, the Farad is the unit of capacity, which we geeks call
capacitance. ;-)

Cheers!
Rich

Oh it's about doing thermal things with an electrical analogy.

ELECTRICAL THERMAL

1 amp 1 watt
1 farad 1 gram aluminum
1 volt 1 degree C
1 second 1 second
1 ohm 1 degC/watt

There's John's list. If current -> watts of heat flow then the charge
is the amount of heat in Joules. And Q=CV so C = Q/V and heat
capacity is Joules/degree. If you haven't used this analogy I guess
it's confusing. But once you get it, you can trun thermal circuits
into electrical circuits. Which are just easier for me to grasp.

George H.

 

[George Herold]

On Jan 13, 12:34 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 18:40:42 -0800 (PST), George Herold





gher...@teachspin.com> wrote:
On Jan 12, 3:37 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 09:48:45 -0800 (PST), George Herold

ggher...@gmail.com> wrote:
On Jan 11, 4:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 11 Jan 2011 06:11:41 +0100, "F. Bertolazzi"

TOGLIe...@MAIUSCOLEtdd.it> wrote:
Michael A. Terrell:

John Larkin wrote:

OK, what's the electrical equivalent of the finger-on-the-hose thing?
Schematic, please.

A transformer: High flow at low pressure is transformed into low
flow at high pressure.

Humm. Maybe John is right.

HA!

A transformer could be a water-powered motor driving a pump. For
non-steady-state, a driven piston pushing another piston of a
different size.

Oh the piston thing works nicely for me as a transformer. Thanks.

It's easier to just explain the electricity. That way, you don't have
to explain two systems, when you're trying to teach one.

One interesting analogy is a synchronous buck switcher to a pair of
gears. That sort of works. You can apply conservation of energy in
both ideal cases and predict behavior without having to go into gory
detail.

The holy grail of automotive design is the continuously-variable
transmission. The electronic equivalent is easy. Eat your hearts out,
MEs!

 Did you ever see the variable Vee Belt drive on a snowmobile?
George H.

John

Yeah, and that's been used in some cars. I think it tends to be
unreliable at high horsepowers.

I can think of a mechanical equivalent to a synchronous switcher,
using clutches and torsion springs and flywheels and such. It would be
noisy and not very reliable.

My Audi has a 6-speed tranny with two parallel gear trains, 3 speeds
each, odd and even gears used sequentially, two clutches, no torque
converter. It's very efficient and shifts in something silly like 80
milliseconds. It's the equivalent of a tap-switched transformer.

John- Hide quoted text -

- Show quoted text -

My car is a 2002 Corolla.  But I do have a cool little ride'm lawn
mover.  You have two levers, that connect to the back wheels.  The
front wheels are like those on shopping carts and free turn where they
want.  The levers are like those on a tank.  The clutch mechanism is
''tangentail' spheres.  (I certainly can't describe it.)   But it's a
dang fast way to cut grass in summer.  (My son's been driving it since
before I'm free to mention.)

We don't have grass. I hate grass.



We've finally got a bit of snow here.  Two feet up the east coast and
we only get six inches, sigh.

There's only, officially, 48" of snow on the ground in Truckee. But
some of the mounds tossed up by the snow plows are easily 12' high.
And the year is young!

John- Hide quoted text -

- Show quoted text -

Everything was pretty much frozen here right around Xmass. We snow-
shoed and X-contry skied down the creek to Angle Falls.

http://www.topozone.com/map.asp?lon=-78.4347415&lat=42.6711722&datum=nad83

And then it all melted.

George H.

 

[John Larkin]

On Thu, 13 Jan 2011 06:50:51 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Jan 13, 12:26 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 20:05:30 -0800 (PST), George Herold





gher...@teachspin.com> wrote:
On Jan 12, 8:59 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 18:04:19 -0600, "k...@att.bizzzzzzzzzzzz"

k...@att.bizzzzzzzzzzzz> wrote:
On Wed, 12 Jan 2011 13:31:45 GMT, N0S...@daqarta.com (Bob Masta) wrote:

On Tue, 11 Jan 2011 13:41:39 -0800, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 11 Jan 2011 13:51:52 GMT, N0S...@daqarta.com (Bob Masta)
wrote:

Agreed in principle, but there can be serious debate over
the proper analogy even in this basic RLC case. Consider
that with a series connection, current is analogous to force
in that it is identical in all elements, while the voltage
drop across each element is analogous to relative velocity.
That doesn't fit well with the voltage-as-pressure concept.

Mechanical engineers use different analogies as needed, and
can make the math work out even when there isn't much
intuitive connection (to this poor EE, at least!).

Why do MEs need analogies? They can see and feel their stuff!

They need analogies so they know what equations to borrow
from the EEs!

When I started working for IBM in '74 I was surprised to see the heat transfer
folks down the hall were using "our" circuit simulators to do their work.

Ther was one thermal simulator, called Sauna, that was based on
connecting an array of dots in 3-space with (thermal) resistors in a
cubic matrix, and caps to ground, and then running a Spice-like sim.

I've used Spice to simulate thermal systems. This works pretty well:

ELECTRICAL THERMAL

1 amp 1 watt
1 farad 1 gram aluminum
1 volt 1 degree C
1 second 1 second
1 ohm 1 degC/watt

John- Hide quoted text -

- Show quoted text -

Yeah I do it with one resistor and one cap.. (Well I usually break the
resistor in half.)

When I have to explain thermal circuits to people I love to use the
'electronic' analogy.
(I'm not sure it helps them, but it's how I understand it.)  Of course
capacitors don't have phase changes.

Oh the Farad is equivalent to the heat capacity.  Joules/deg K.  (I
think that's right. RC is time, seconds in both 'units')

George H.

Unfortunately, most useful thermal systems are diffusive, unlike
lumped electrical circuits. And the diffusion is usually among nasty
3D shapes. So the modeling is ugly finite-element stuff, basically a
many-element 3D hairball of resistors and capacitors.

I guess I've never tried it on complex systems. Mostly I'm just
trying to guess what the time constant of the system will be.

Interestingly, there is no thermal equivalent of an inductor. So
thermal systems don't ring or oscillate, unless you sneak in some
nonlinear mechanical or transport phenomenon.

I wish I had some nice little thermal resistance calculator, even a 2D
one.

There must be some software company that would sell you such a thing.

$$$$$$$$$$$$$$$!!!

I need to play with Sonnet Lite, the free EM simulator. Maybe it can
be hijacked for 2D thermal analysis.

Ever read the book "Hot air rises and Heat sinks" Not all that
technical, but an enjoyable read.

I have it. Not impressed. Steinberg's "Cooling Techniques for
Electronic Equipment" is pretty good, but I wish he's stick to SI
units. He freely mixes inches, feet, BTUs, deg F, degC, calories,
joules, cm, meters, whatever.

John

 

[Phil Hobbs]

George Herold wrote:

On Jan 12, 11:09 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net> wrote:
George Herold wrote:
On Jan 12, 5:15 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net> wrote:
George Herold wrote:
On Jan 12, 3:27 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net> wrote:
George Herold wrote:
On Jan 12, 7:16 am, John Fields<jfie...@austininstruments.com> wrote:
On Tue, 11 Jan 2011 13:47:18 -0800, John Larkin

jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Right. I can't think of an electrical analogy to friction.

---
Why would you think that resistance isn't analogous to friction?

---
JF

Did you look at the plots I posted John? Friction causes a linear
decrease in amplitude, not exponential.

George H.

Cute demo.

I'm not really persuaded by the v**0 argument for frictional damping.
Long years of painstaking research in the field of yo-yo tricks has
convinced me that when you have a string sliding on a roller, once you
break it loose there's very little friction.

I suspect that if you put a load cell on the string, you'd find that the
actual retarding force was concentrated in narrow pulses near the peak
of each oscillation. The work required to break the string loose is
pretty well constant, so you'd lose a fixed amount of energy per half
cycle. The total energy is

The data is a lot better than the 'scope shots. The static friction
acts only for an 'instant' when the rotor turns around. I never
really looked closely, but there are no 'big' steps at the ends. If I
record some data I could try and fit the envelope... it should be
possible to determine the static amplitude loss at the turn around
points and constant loss through the rest of the motion.

There are actually two strings pressing on opposite sides of the
rotor. And we can load them with differernt weights and change the
frictional force.

I omega**2 k*theta**2
E = ---------- + ------------
2 2

where omega = d/dt(theta). The average energy loss would be linear in
time, so

** It's the amplitude loss that is linear in time! **

I get to that point later. Right now I'm doing the reciprocating yo-yo
case. (Patience, Grasshopper.)

dE
-- = Qdot = I omega d(omega)/dt + k*theta*d(theta)/dt.
dt

At the extremes of motion, omega = 0, so if dE/dt over one cycle is some
constant B, then

As above amplitude loss for one cycle is constant. Not energy loss.

d(theta)/dt = B/(k*theta)

so theta = (2B/k)*sqrt(t0-t),

where t0 is the time where the motion stops. That's the case for car
brakes--you have to lighten up on the pedal as you slow down, to avoid
jerking to a stop.

Doesn't your car behave like that? All the ones I've driven do.

Yeah sure that's how I drive a car.... If my car went back and forth
and I wanted to stop it, I'd apply the brakes at the turn around point
where the velocity is zero.... I guess I'm just saying I don't get the
(hic) analogy.

Deceleration of a braking car is due to sliding friction, but for a
given normal force on the brake pads, the deceleration isn't constant.
I'm not sure why that is, but it clearly shows that the coefficient of
friction model isn't always adequate as the speed goes to zero.

Got any data? It could be that the deceleration is constant, but that
we 'humans' like the deceleration to be less as we come to a stop.
Same goes when you start off. Unless you're a teenage boy with his
hot rod, you don't want to be slammed back into the seat with 1/2 g
right away, but perfer a gentle increase in acceleration.

I haven't stuck an accelerometer in the car, no. To be really
probative, it would need a measure of the pressure in the master
cylinder anyway, which is more difficult.

Try it on your way home from work--you'll come right off the back of the
seat if you don't ease off on the pedal.

Now we get to the coefficient-of-friction case, which is what your web
page assumes.

With the usual coefficient-of-friction approximation, i.e. your v**0
approach, the power consumed by the rotor in overcoming friction is

dE/dt = omega Gamma,

where Gamma is the frictional torque.

At the peak velocity, theta = 0, so

d(omega)/dt = -Gamma/I,

and you get a linear decrease in the amplitude, as you say.

A colleague has worked out the math in detail. I'd be happy to send
you the appropriate section of the manual.

If those were the whole story, I'd expect to see the envelope be convex,
i.e. with a linear slope at high amplitudes where the sliding friction
dominates, and a steeper slope at low amplitude where it's the stiction
that matters most.

I think the only point where static friction matters is right when it
stops. At that point there is not enough torque in the spring to over
come the static friction and it stops. And stops at a non-equilibrium
position.

But it stops twice per cycle, which is the point I was trying to make
above. Doesn't it?

Sure. If there was energy loss (from say the thread sticking?) right
at the turn around point wouldn't the amplitude have to undergo some
sort of 'step' change in amplitude? The energies all in the spring at
that point. I'm not sure what the thread is at one point we were
using 'spider' wire. (use for fishing.) It's rubbing against an 1"
diameter Al rod, There are two strings each touching for maybe 45
degrees of the circle. We first tried a complete wrap of string and
that was much to much friction. I think there's ~100 grams hanging on
each string.

You wouldn't get a noticeable discontinuity on each start because when
the stiction lets go, the rotor accelerates smoothly from 0. There
might be something visible when it stops each time--another of those
square-root decelerations.

I'm confused... You want to model all the energy loss as occuring at
the turn around points.. where the rotor stops. (Is that correct?)
It's pretty easy to identify the energy in the system. We agree there
is some energy loss after each 1/2 cycle. So if the energy is only
lost at the turn around point... then there must be a decrease in the
amplitude at that point. It's all potential energy at that point.

I'm not really a mechanical guy myself. (I've known some mechanical
guys who were pretty life-like, though.)

I'm assuming that the stiction grabs briefly at each stopping point,
then lets go after stretching out whatever little asperities have become
interlocked on the surface. That will slow down the acceleration in the
neighbourhood of the peaks, but they'll still appear smooth because the
stiction loss is all happening while the rotor is nearly stopped. It
would cause a slight change in the second derivative of the angle, i.e.
a minor flattening and asymmetry of the peak, but no jaggies.
Deforming the asperities should take roughly a constant force,
independent of amplitude, because (in my simple model) it's limited by
the yield stress of the materials. Similarly, the deformation should
occur over a distance determined by the size and shape of the asperities.

Both force and distance are constant in this model, so to leading order,
unsticking should take a constant amount of energy per cycle. That
leads to a square-root time dependence of amplitude, at least in cases
where stiction dominates.

As you point out, the energy is nearly all potential near the peaks, so
the peak shape has to change a bit to account for the reduced
acceleration during the sticking and unsticking. If it were
instantaneous, it wouldn't require any work, and so couldn't change the
shape of the curve.

Your plot's envelope is slightly concave, which looks like you have some
exponential behaviour in there someplace.

Yeah there is a little bit of 'concavity' to it. I can't really say
if this is some velocity dependent damping... or perhaps something to
do with the friction.

The coefficient of friction approximation isn't that good, but it
certainly leads to a decay that's a lot closer to linear than to
exponential.

I was looking at some other data today. The amplitude looks a lot
more linear.

Cool. As I said, it's a great demo. Nothing like a bit of data to
discourage pontification.

Thanks Phil. I hope it's alright if I continue to disagree with your
interpretation.

George H.

Of course! Surely nobody in this august company would be so ill-bred as
to confuse an argument with a quarrel.


Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

 

[John Larkin]

On Thu, 13 Jan 2011 08:08:47 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 12 Jan 2011 14:45:40 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 15:53:48 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 12 Jan 2011 09:04:52 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 13:25:14 +0100, "F. Bertolazzi"
TOGLIeset@MAIUSCOLEtdd.it> wrote:

John Fields:

On Tue, 11 Jan 2011 13:47:18 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Right. I can't think of an electrical analogy to friction.

---
Why would you think that resistance isn't analogous to friction?

Because he does not want to. ;-)

I don't want to because the behaviors are very different. You'd have
to actually understand friction behavior to see why.

---
Typical Larkinese intimidation attempt, to wit: "I'm betting that he
doesn't understand friction, so I'll intimate that he doesn't, and
I'll add the "actually" as a fillip (since I've puffed myself up as an
authority) to get him to shut the fuck up."
---

Neither static friction nor sliding friction behave like electrical resistance.

---
Nonsense.

"Stiction" is like a Zener and Teflon on Teflon is like charge flowing
through silver.


In that case, alligators are like transformers.

---
That's like saying that you don't understand analogy.

---
JF

Analogy is like popcorn.

John

 

[John Larkin]

On Thu, 13 Jan 2011 10:01:18 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 12 Jan 2011 15:23:45 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 15:06:45 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 12 Jan 2011 09:02:27 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 06:16:37 -0600, John Fields
jfields@austininstruments.com> wrote:

On Tue, 11 Jan 2011 13:47:18 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Right. I can't think of an electrical analogy to friction.

---
Why would you think that resistance isn't analogous to friction?

---
JF

Grossly different behavior, explained elsewhere. Resistance is more
like low-Reynolds number viscoscity.

---
Why complicate things when all that's necessary for the analogy to be
valid is for the viscosity to be likened to resistance?

http://en.wikipedia.org/wiki/Viscosity

---

The only things that resistance and friction have in common is that
both are dissipative.

---
Then, since watts is watts and they both get hot from energy wasted
when work is being done on them, that's what causes them both to be
dissipative, and that's all that's necessary for the analogy to be
valid.


Valid to you, not to me. An analogy is valid to me if the systems
behave similarly, if graphs or waveforms in one domain look like ones
in the other domain. If you are happy with fuzzy poetical feelings,
fine, but they are not useful tools for teaching electronics. Worse
than useless, if the analogy encourages frank misunderstanding.

---
You're such a hypocritical bull shitter!

When you want to play fast and loose, like when you off-handedly make
the claim that latching relays have infinite gain, and you're
corrected,

Corrected? Who did that? And who has obsessed over it for years?


you try your damnedest to force your conveniently

reformulated definition of 'infinity' down everyone's throat, but when
someone else uses a perfectly good analogy to try to explain a
concept, you yell and scream and jump up and down and rail about the
incorrectness of the analogy, even though for the purpose at hand,
it's eminently suitable.

I'm not forcing anything; don't want to, couldn't if I wanted. I said
that I don't like the water analogy because it's not accurate, and I
don't think it's helpful as a teaching tool. This is a discussion
group. You're the one escalating here.

I think analogies like this should have mathematical validity. You
don't. Enjoy.

John

 

[John Larkin]

On Thu, 13 Jan 2011 06:32:10 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Jan 12, 11:09 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net> wrote:
George Herold wrote:
On Jan 12, 5:15 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net>  wrote:
George Herold wrote:
On Jan 12, 3:27 pm, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net>    wrote:
George Herold wrote:
On Jan 12, 7:16 am, John Fields<jfie...@austininstruments.com>      wrote:
On Tue, 11 Jan 2011 13:47:18 -0800, John Larkin

jjlar...@highNOTlandTHIStechnologyPART.com>      wrote:

Right. I can't think of an electrical analogy to friction.

---
Why would you think that resistance isn't analogous to friction?

---
JF

Did you look at the plots I posted John?  Friction causes a linear
decrease in amplitude, not exponential.

George H.

Cute demo.

I'm not really persuaded by the v**0 argument for frictional damping.
Long years of painstaking research in the field of yo-yo tricks has
convinced me that when you have a string sliding on a roller, once you
break it loose there's very little friction.

I suspect that if you put a load cell on the string, you'd find that the
actual retarding force was concentrated in narrow pulses near the peak
of each oscillation.  The work required to break the string loose is
pretty well constant, so you'd lose a fixed amount of energy per half
cycle.  The total energy is

The data is a lot better than the 'scope shots.  The static friction
acts only for an 'instant' when the rotor turns around.  I never
really looked closely, but there are no 'big' steps at the ends. If I
record some data I could try and fit the envelope... it should be
possible to determine the static amplitude loss at the turn around
points and constant loss through the rest of the motion.

There are actually two strings pressing on opposite sides of the
rotor.  And we can load them with differernt weights and change the
frictional force.

         I omega**2    k*theta**2
E =   ---------- + ------------
             2            2

where omega = d/dt(theta).  The average energy loss would be linear in
time, so

   ** It's the amplitude loss that is linear in time! **

I get to that point later.  Right now I'm doing the reciprocating yo-yo
case.  (Patience, Grasshopper.)

dE
--  = Qdot = I omega d(omega)/dt + k*theta*d(theta)/dt.
dt

At the extremes of motion, omega = 0, so if dE/dt over one cycle is some
constant B, then

As above amplitude loss for one cycle is constant. Not energy loss.

d(theta)/dt =  B/(k*theta)

so theta = (2B/k)*sqrt(t0-t),

where t0 is the time where the motion stops.  That's the case for car
brakes--you have to lighten up on the pedal as you slow down, to avoid
jerking to a stop.

Doesn't your car behave like that?  All the ones I've driven do.

Yeah sure that's how I drive a car.... If my car went back and forth
and I wanted to stop it, I'd apply the brakes at the turn around point
where the velocity is zero.... I guess I'm just saying I don't get the
(hic) analogy.

Deceleration of a braking car is due to sliding friction, but for a
given normal force on the brake pads, the deceleration isn't constant.
I'm not sure why that is, but it clearly shows that the coefficient of
friction model isn't always adequate as the speed goes to zero.

Got any data? It could be that the deceleration is constant, but that
we 'humans' like the deceleration to be less as we come to a stop.
Same goes when you start off. Unless you're a teenage boy with his
hot rod, you don't want to be slammed back into the seat with 1/2 g
right away, but perfer a gentle increase in acceleration.

Cars definitely jerk as they come to a stop. I suspect it's the
elasticity of the suspension, all those struts and bushings, that
essentially put a spring between the brake housing and the frame. That
spring winds up during braking and, when the brakes finally go from
sliding to grabbing, there's a spring back.

John

 

[John Larkin]

On Thu, 13 Jan 2011 16:01:44 +0100, "F. Bertolazzi"
<TOGLIeset@MAIUSCOLEtdd.it> wrote:

George Herold:

Got any data? It could be that the deceleration is constant, but that
we 'humans' like the deceleration to be less as we come to a stop.

If the car has ABS and you brake at high speed, it seems to barely slow
down. As the speed decreases the deceleration increases.

This is due to the fact that the brakes must dissipate the kinetic energy
"stored" in the running car without exceeding the tyres grip, energy that
is proportional to the square of speed.

Ignoring brake heating issues, wouldn't the minimal-distance braking
strategy be almost constant deceleration, corresponding to constant
tire-road force, just short of sliding? Power dissipated would be
proportional to speed, but force would be constant.

John

 

[Phil Hobbs]

John Larkin wrote:

On Thu, 13 Jan 2011 16:01:44 +0100, "F. Bertolazzi"
TOGLIeset@MAIUSCOLEtdd.it> wrote:

George Herold:

Got any data? It could be that the deceleration is constant, but that
we 'humans' like the deceleration to be less as we come to a stop.

If the car has ABS and you brake at high speed, it seems to barely slow
down. As the speed decreases the deceleration increases.

This is due to the fact that the brakes must dissipate the kinetic energy
"stored" in the running car without exceeding the tyres grip, energy that
is proportional to the square of speed.

Ignoring brake heating issues, wouldn't the minimal-distance braking
strategy be almost constant deceleration, corresponding to constant
tire-road force, just short of sliding? Power dissipated would be
proportional to speed, but force would be constant.

John

It has nothing to do with tire adhesion, nor to suspension deflection.
The jerk happens when you're coming to a stop, regardless of how fast
you're doing it. And if the deceleration were constant, the suspension
deflection would be too, at least until after the stop.

Also, car suspensions are extremely stiff in the fore-and-aft direction,
because otherwise you couldn't control the car. The suspension
deflection is essentially all vertical, whereas the jerk on stopping is
definitely horizontal. You can watch the front suspension deflect due
to the resulting torque.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

 

[John Larkin]

On Thu, 13 Jan 2011 07:34:17 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Jan 13, 12:34 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 18:40:42 -0800 (PST), George Herold





gher...@teachspin.com> wrote:
On Jan 12, 3:37 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 09:48:45 -0800 (PST), George Herold

ggher...@gmail.com> wrote:
On Jan 11, 4:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 11 Jan 2011 06:11:41 +0100, "F. Bertolazzi"

TOGLIe...@MAIUSCOLEtdd.it> wrote:
Michael A. Terrell:

John Larkin wrote:

OK, what's the electrical equivalent of the finger-on-the-hose thing?
Schematic, please.

A transformer: High flow at low pressure is transformed into low
flow at high pressure.

Humm. Maybe John is right.

HA!

A transformer could be a water-powered motor driving a pump. For
non-steady-state, a driven piston pushing another piston of a
different size.

Oh the piston thing works nicely for me as a transformer. Thanks.

It's easier to just explain the electricity. That way, you don't have
to explain two systems, when you're trying to teach one.

One interesting analogy is a synchronous buck switcher to a pair of
gears. That sort of works. You can apply conservation of energy in
both ideal cases and predict behavior without having to go into gory
detail.

The holy grail of automotive design is the continuously-variable
transmission. The electronic equivalent is easy. Eat your hearts out,
MEs!

 Did you ever see the variable Vee Belt drive on a snowmobile?
George H.

John

Yeah, and that's been used in some cars. I think it tends to be
unreliable at high horsepowers.

I can think of a mechanical equivalent to a synchronous switcher,
using clutches and torsion springs and flywheels and such. It would be
noisy and not very reliable.

My Audi has a 6-speed tranny with two parallel gear trains, 3 speeds
each, odd and even gears used sequentially, two clutches, no torque
converter. It's very efficient and shifts in something silly like 80
milliseconds. It's the equivalent of a tap-switched transformer.

John- Hide quoted text -

- Show quoted text -

My car is a 2002 Corolla.  But I do have a cool little ride'm lawn
mover.  You have two levers, that connect to the back wheels.  The
front wheels are like those on shopping carts and free turn where they
want.  The levers are like those on a tank.  The clutch mechanism is
''tangentail' spheres.  (I certainly can't describe it.)   But it's a
dang fast way to cut grass in summer.  (My son's been driving it since
before I'm free to mention.)

We don't have grass. I hate grass.



We've finally got a bit of snow here.  Two feet up the east coast and
we only get six inches, sigh.

There's only, officially, 48" of snow on the ground in Truckee. But
some of the mounds tossed up by the snow plows are easily 12' high.
And the year is young!

John- Hide quoted text -

- Show quoted text -

Everything was pretty much frozen here right around Xmass. We snow-
shoed and X-contry skied down the creek to Angle Falls.

http://www.topozone.com/map.asp?lon=-78.4347415&lat=42.6711722&datum=nad83

That part of the country is really beautiful. The Brat went to
Cornell, and we got to explore a little. It's different from the mud
flats of Louisiana!

And then it all melted.

We skiied at Thanksgiving and Christmas, and I hear it's still
fabulous. It's been cold here, very cold (below 0F at night) in the
Sierras. So far, we're having a wet winter. California survives on
snowmelt.

John

 

[John Larkin]

On Thu, 13 Jan 2011 14:41:36 +0100, "F. Bertolazzi"
<TOGLIeset@MAIUSCOLEtdd.it> wrote:

John Larkin:

I can think of a mechanical equivalent to a synchronous switcher,
using clutches and torsion springs and flywheels and such. It would be
noisy and not very reliable.



The hydraulic analogy is also useful with switchers, to show that it is
possible to divide the content of one bucket in two buckets, so it is
possible to make a divide-by-two buck converter with two capacitors, a few
switches and no inductors. But the efficiency, given the fact that we
lowered the top half of one bucket to ground, wasting most of the energy,
would be lousy.

The electronic div-2 capacitor thing, a charge pump converter, can
approach 100% efficiency. I don't know if there's a hydraulic analog.

John

 

[Phil Hobbs]

John Larkin wrote:

On Thu, 13 Jan 2011 14:41:36 +0100, "F. Bertolazzi"
TOGLIeset@MAIUSCOLEtdd.it> wrote:

John Larkin:

I can think of a mechanical equivalent to a synchronous switcher,
using clutches and torsion springs and flywheels and such. It would be
noisy and not very reliable.



The hydraulic analogy is also useful with switchers, to show that it is
possible to divide the content of one bucket in two buckets, so it is
possible to make a divide-by-two buck converter with two capacitors, a few
switches and no inductors. But the efficiency, given the fact that we
lowered the top half of one bucket to ground, wasting most of the energy,
would be lousy.

The electronic div-2 capacitor thing, a charge pump converter, can
approach 100% efficiency. I don't know if there's a hydraulic analog.

John

Sure, just a pulley and two buckets. If the weights are the same, the
energy supplied by lowering one bucket halfway is enough to raise the
other bucket to the same point.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

 

[Phil Hobbs]

George Herold wrote:

On Jan 13, 11:36 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 13 Jan 2011 06:50:51 -0800 (PST), George Herold





gher...@teachspin.com> wrote:
On Jan 13, 12:26 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 20:05:30 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Jan 12, 8:59 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 18:04:19 -0600, "k...@att.bizzzzzzzzzzzz"

k...@att.bizzzzzzzzzzzz> wrote:
On Wed, 12 Jan 2011 13:31:45 GMT, N0S...@daqarta.com (Bob Masta) wrote:

On Tue, 11 Jan 2011 13:41:39 -0800, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 11 Jan 2011 13:51:52 GMT, N0S...@daqarta.com (Bob Masta)
wrote:

Agreed in principle, but there can be serious debate over
the proper analogy even in this basic RLC case. Consider
that with a series connection, current is analogous to force
in that it is identical in all elements, while the voltage
drop across each element is analogous to relative velocity.
That doesn't fit well with the voltage-as-pressure concept.

Mechanical engineers use different analogies as needed, and
can make the math work out even when there isn't much
intuitive connection (to this poor EE, at least!).

Why do MEs need analogies? They can see and feel their stuff!

They need analogies so they know what equations to borrow
from the EEs!

When I started working for IBM in '74 I was surprised to see the heat transfer
folks down the hall were using "our" circuit simulators to do their work.

Ther was one thermal simulator, called Sauna, that was based on
connecting an array of dots in 3-space with (thermal) resistors in a
cubic matrix, and caps to ground, and then running a Spice-like sim.

I've used Spice to simulate thermal systems. This works pretty well:

ELECTRICAL THERMAL

1 amp 1 watt
1 farad 1 gram aluminum
1 volt 1 degree C
1 second 1 second
1 ohm 1 degC/watt

John- Hide quoted text -

- Show quoted text -

Yeah I do it with one resistor and one cap.. (Well I usually break the
resistor in half.)

When I have to explain thermal circuits to people I love to use the
'electronic' analogy.
(I'm not sure it helps them, but it's how I understand it.) Of course
capacitors don't have phase changes.

Oh the Farad is equivalent to the heat capacity. Joules/deg K. (I
think that's right. RC is time, seconds in both 'units')

George H.

Unfortunately, most useful thermal systems are diffusive, unlike
lumped electrical circuits. And the diffusion is usually among nasty
3D shapes. So the modeling is ugly finite-element stuff, basically a
many-element 3D hairball of resistors and capacitors.

I guess I've never tried it on complex systems. Mostly I'm just
trying to guess what the time constant of the system will be.

Interestingly, there is no thermal equivalent of an inductor. So
thermal systems don't ring or oscillate, unless you sneak in some
nonlinear mechanical or transport phenomenon.

I wish I had some nice little thermal resistance calculator, even a 2D
one.

There must be some software company that would sell you such a thing.

$$$$$$$$$$$$$$$!!!

I need to play with Sonnet Lite, the free EM simulator. Maybe it can
be hijacked for 2D thermal analysis.

Ever read the book "Hot air rises and Heat sinks" Not all that
technical, but an enjoyable read.

I have it. Not impressed. Steinberg's "Cooling Techniques for
Electronic Equipment" is pretty good, but I wish he's stick to SI
units. He freely mixes inches, feet, BTUs, deg F, degC, calories,
joules, cm, meters, whatever.

John- Hide quoted text -

- Show quoted text -

Oh thanks I'll check out the reference. One thing I hate about
thermal stuff is all the different units. I can spend more time
making sure I've converted the units correctly than in doing the
calculation.

george H.

1 BTU is within a few percent of 1 kJ, which makes it easier.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

 

[George Herold]

On Jan 13, 11:36 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 13 Jan 2011 06:50:51 -0800 (PST), George Herold





gher...@teachspin.com> wrote:
On Jan 13, 12:26 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 20:05:30 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Jan 12, 8:59 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 18:04:19 -0600, "k...@att.bizzzzzzzzzzzz"

k...@att.bizzzzzzzzzzzz> wrote:
On Wed, 12 Jan 2011 13:31:45 GMT, N0S...@daqarta.com (Bob Masta) wrote:

On Tue, 11 Jan 2011 13:41:39 -0800, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 11 Jan 2011 13:51:52 GMT, N0S...@daqarta.com (Bob Masta)
wrote:

Agreed in principle, but there can be serious debate over
the proper analogy even in this basic RLC case. Consider
that with a series connection, current is analogous to force
in that it is identical in all elements, while the voltage
drop across each element is analogous to relative velocity.
That doesn't fit well with the voltage-as-pressure concept.

Mechanical engineers use different analogies as needed, and
can make the math work out even when there isn't much
intuitive connection (to this poor EE, at least!).

Why do MEs need analogies? They can see and feel their stuff!

They need analogies so they know what equations to borrow
from the EEs!

When I started working for IBM in '74 I was surprised to see the heat transfer
folks down the hall were using "our" circuit simulators to do their work.

Ther was one thermal simulator, called Sauna, that was based on
connecting an array of dots in 3-space with (thermal) resistors in a
cubic matrix, and caps to ground, and then running a Spice-like sim..

I've used Spice to simulate thermal systems. This works pretty well:

ELECTRICAL THERMAL

1 amp 1 watt
1 farad 1 gram aluminum
1 volt 1 degree C
1 second 1 second
1 ohm 1 degC/watt

John- Hide quoted text -

- Show quoted text -

Yeah I do it with one resistor and one cap.. (Well I usually break the
resistor in half.)

When I have to explain thermal circuits to people I love to use the
'electronic' analogy.
(I'm not sure it helps them, but it's how I understand it.) Of course
capacitors don't have phase changes.

Oh the Farad is equivalent to the heat capacity. Joules/deg K. (I
think that's right. RC is time, seconds in both 'units')

George H.

Unfortunately, most useful thermal systems are diffusive, unlike
lumped electrical circuits. And the diffusion is usually among nasty
3D shapes. So the modeling is ugly finite-element stuff, basically a
many-element 3D hairball of resistors and capacitors.

I guess I've never tried it on complex systems.  Mostly I'm just
trying to guess what the time constant of the system will be.

Interestingly, there is no thermal equivalent of an inductor. So
thermal systems don't ring or oscillate, unless you sneak in some
nonlinear mechanical or transport phenomenon.

I wish I had some nice little thermal resistance calculator, even a 2D
one.

There must be some software company that would sell you such a thing.

$$$$$$$$$$$$$$$!!!

I need to play with Sonnet Lite, the free EM simulator. Maybe it can
be hijacked for 2D thermal analysis.

Ever read the book "Hot air rises and Heat sinks"  Not all that
technical, but an enjoyable read.

I have it. Not impressed. Steinberg's "Cooling Techniques for
Electronic Equipment" is pretty good, but I wish he's stick to SI
units. He freely mixes inches, feet, BTUs, deg F, degC, calories,
joules, cm, meters, whatever.

John- Hide quoted text -

- Show quoted text -

Oh thanks I'll check out the reference. One thing I hate about
thermal stuff is all the different units. I can spend more time
making sure I've converted the units correctly than in doing the
calculation.

george H.

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 12 Jan 2011 21:26:09 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 20:05:30 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 12, 8:59 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 12 Jan 2011 18:04:19 -0600, "k...@att.bizzzzzzzzzzzz"





k...@att.bizzzzzzzzzzzz> wrote:
On Wed, 12 Jan 2011 13:31:45 GMT, N0S...@daqarta.com (Bob Masta) wrote:

On Tue, 11 Jan 2011 13:41:39 -0800, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 11 Jan 2011 13:51:52 GMT, N0S...@daqarta.com (Bob Masta)
wrote:

Agreed in principle, but there can be serious debate over
the proper analogy even in this basic RLC case.  Consider
that with a series connection, current is analogous to force
in that it is identical in all elements, while the voltage
drop across each element is analogous to relative velocity.
That doesn't fit well with the voltage-as-pressure concept.

Mechanical engineers use different analogies as needed, and
can make the math work out even when there isn't much
intuitive connection (to this poor EE, at least!).

Why do MEs need analogies? They can see and feel their stuff!

They need analogies so they know what equations to borrow
from the EEs!

When I started working for IBM in '74 I was surprised to see the heat transfer
folks down the hall were using "our" circuit simulators to do their work.

Ther was one thermal simulator, called Sauna, that was based on
connecting an array of dots in 3-space with (thermal) resistors in a
cubic matrix, and caps to ground, and then running a Spice-like sim.

I've used Spice to simulate thermal systems. This works pretty well:

      ELECTRICAL       THERMAL

       1 amp            1 watt
       1 farad          1 gram aluminum
       1 volt           1 degree C
       1 second         1 second
       1 ohm            1 degC/watt

John- Hide quoted text -

- Show quoted text -

Yeah I do it with one resistor and one cap.. (Well I usually break the
resistor in half.)

When I have to explain thermal circuits to people I love to use the
'electronic' analogy.
(I'm not sure it helps them, but it's how I understand it.) Of course
capacitors don't have phase changes.

Oh the Farad is equivalent to the heat capacity. Joules/deg K. (I
think that's right. RC is time, seconds in both 'units')

George H.

Unfortunately, most useful thermal systems are diffusive, unlike
lumped electrical circuits. And the diffusion is usually among nasty
3D shapes. So the modeling is ugly finite-element stuff, basically a
many-element 3D hairball of resistors and capacitors.

Yup, but the math still works the same way. Unlike AlwaysWrong, computers are
good at it.

Interestingly, there is no thermal equivalent of an inductor. So
thermal systems don't ring or oscillate, unless you sneak in some
nonlinear mechanical or transport phenomenon.

I wish I had some nice little thermal resistance calculator, even a 2D
one.

Maybe a fields simulator? Lumped analysis works fine, though.

 

[John Fields]

On Thu, 13 Jan 2011 08:55:17 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 13 Jan 2011 10:01:18 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 12 Jan 2011 15:23:45 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 15:06:45 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 12 Jan 2011 09:02:27 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 12 Jan 2011 06:16:37 -0600, John Fields
jfields@austininstruments.com> wrote:

On Tue, 11 Jan 2011 13:47:18 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Right. I can't think of an electrical analogy to friction.

---
Why would you think that resistance isn't analogous to friction?

---
JF

Grossly different behavior, explained elsewhere. Resistance is more
like low-Reynolds number viscoscity.

---
Why complicate things when all that's necessary for the analogy to be
valid is for the viscosity to be likened to resistance?

http://en.wikipedia.org/wiki/Viscosity

---

The only things that resistance and friction have in common is that
both are dissipative.

---
Then, since watts is watts and they both get hot from energy wasted
when work is being done on them, that's what causes them both to be
dissipative, and that's all that's necessary for the analogy to be
valid.


Valid to you, not to me. An analogy is valid to me if the systems
behave similarly, if graphs or waveforms in one domain look like ones
in the other domain. If you are happy with fuzzy poetical feelings,
fine, but they are not useful tools for teaching electronics. Worse
than useless, if the analogy encourages frank misunderstanding.

---
You're such a hypocritical bull shitter!

When you want to play fast and loose, like when you off-handedly make
the claim that latching relays have infinite gain, and you're
corrected,

Corrected? Who did that?

---
Everyone.
---

And who has obsessed over it for years?

---
You.
---

you try your damnedest to force your conveniently
reformulated definition of 'infinity' down everyone's throat, but when
someone else uses a perfectly good analogy to try to explain a
concept, you yell and scream and jump up and down and rail about the
incorrectness of the analogy, even though for the purpose at hand,
it's eminently suitable.

I'm not forcing anything; don't want to, couldn't if I wanted.

---
Which, then, belies your attempts.
---

I said that I don't like the water analogy because it's not accurate, and I
don't think it's helpful as a teaching tool.

---
More crap.

The reason you don't think it's helpful as a teaching tool isn't
because it's not accurate - since it's not required to be traceable to
NIST at the level at which it's used - it's because you've convinced
yourself that unless everything in the world is done the way you say
it should be done, it's being done improperly.
---

This is a discussion group.

---
As usual, a remarkable grasp of the obvious.
---

You're the one escalating here.

---
Yes, of course, but why would you say that unless, because I often
head you off at the pass when you make a mistake, or when you lie, and
refuse to kow-tow to you, you try to put escalation in a bad light and
make your position seem to be above reproach?

My point was, and always has been, that the water analogy is valid for
the formulation of basic concepts; for example, the one which sticks
in your craw, resistance.

For a constant input pressure and a decreasing pipe diameter, does the
flow of water not decrease?

Likewise, for a constant input voltage and a decreasing wire diameter
does the flow of charge not decrease?
---

I think analogies like this should have mathematical validity.

---
Which, of course, they do, whether or not you're capable of mapping
the transfer functions.
---

You don't. Enjoy.

---
A slap from an idiot is inconsequential.

---
JF