water analogy- a simple calculator

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

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

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

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

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.

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.

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.

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

Interesting, anyway.

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
Reply to
Phil Hobbs
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Right. Friction doesn't behave like resistance. All they have in common is that both can dissipate energy.

John

Reply to
John Larkin

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

Reply to
John Larkin

--
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
Reply to
John Fields

--
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."
Reply to
John Fields

:

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.

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

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

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

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.

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.

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!

George H.

t- Hide quoted text -

Reply to
George Herold

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

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

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

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

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.

(And for the viscosity fans, viscous drag goes like velocity squared, so it isn't like resistance either. 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).)

Progress can only be made against a resisting medium. ;)

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
Reply to
Phil Hobbs

--
"Parsing definitions"?

What's that supposed to mean?
Reply to
John Fields

An LC tank paralleled by current-limiting devices, like jfets maybe. That will produce linear decay at high amplitudes, but will go exponential at low level when the jfets get resistive. It certainly won't go into static friction mode at the end.

John

Reply to
John Larkin

In that case, alligators are like transformers.

John

Reply to
John Larkin

resistance.

On the whole, I'd prefer wrestling a transformer.

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
Reply to
Phil Hobbs

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.

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

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.

John

Reply to
John Larkin

resistance.

Have a fried transformer sandwich, on me!

John

Reply to
John Larkin

resistance.

I haven't fried a transformer in a looong time. So it would probably be a bit tough at this 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
Reply to
Phil Hobbs

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.

Reply to
krw

Now that you mention it, that's a pretty good analogy; but I wouldn't want to give the newbies the impression that the coil shape has anything to do with actual (electrical) inductance! :-)

And, on my side, it'd be a very lossy (i.e. low-Q) inductance indeed!

Thanks! Rich

Reply to
Rich Grise

No, John, that'd be more like a capacitor.

Sheesh! Will you get with the program here? ;-)

Cheers! Rich

Reply to
Rich Grise

George Herold:

No. Where? If it's some binary NG I can't see it.

Or like drag, that varies with the square of speed.

As I said, I'm not retired yet, so I must rake my brain to earn a decent living, not to perfect an introductory analogy that is already perfect for that scope.

Easy. Take a controller with a D/A that ouputs variable m

for(m = 0, i = 10; i > 0; i++) if(i % 1) m -= i; else m += i;

Reply to
F. Bertolazzi

Hmm. Yeah, that's be a fairly hard thing to analogize with the water pipe model, but I think everybody admits it's a very limited model, used just to introduce newbies to the concept of current flow.

Every single one of us was a newbie once, you know!

(some longer than others. )

Cheers! Rich

Reply to
Rich Grise

John Larkin:

That's why I love electronics: it's difficult, much more than software. Too bad that my love is seldom returned...

Reply to
F. Bertolazzi

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