Talk:Analogical models

Note: The discussion below was originally on the Impedance matching Talk article, however it was moved here because this seems a more appropriate topic for Analogical models. Sholto Maud 02:34, 5 May 2006 (UTC)

Engines, road wheels and gearboxes

Would anyone like to comment on my assertion that a gearbox is a mechanical impedance matching device between the engine and the road wheels?--Light current 21:25, 2 May 2006 (UTC)

Interesting...I just posted a question on the maximum power theorem talk asking whether the maximum power theorem applies when one converts from one energy form to another. It seems to apply when you resrict the energy form to electrical energy, but there is some question at the Maximum_power_theorem article whether this can be generalised so as to apply when one converts from the electrical form of energy to the mechanical form of energy. Is this related to your assertion? Sholto Maud 02:33, 3 May 2006 (UTC)

Yes. You are very perceptive! 8-) I believe Max power transfer theorem can be applied to most systems of energy conversion. Maybe that page needs expanding. But best to discuss mechanical implications here first.--Light current 02:39, 3 May 2006 (UTC)

I raised this issue previously in another form at thermodynamics talk. User:PAR replied with:

"I don't think gear changing is a form of impedance matching. Usually you have something like P = Fe Ve = Fw Vw where Fe and Fw are the force of the engine and the wheel respectively and Ve and Vw are the velocities of the engine and the wheel respectively, and P is power. The values of Fe and Ve are rather fixed, but you want to vary Fw and Vw according to your needs, and thats what the gears do. When going up hill, or accelerating, you want a large Fw and you settle for a small Vw, but when cruising, you want high Vw and only need a small Fw. I'm not sure about the other cases." PAR 14:34, 21 December 2005 (UTC)

My agenda is a curiosity with how generally the theorem can be applied. I definately think this needs to be expanded (along with the math, circuit diagram analogies and logical proofs) Sholto Maud 02:45, 3 May 2006 (UTC)

Well if you can get hold of Dynamical Analogies by Harry F. Olson (pub Van Nostrand 2nd ed 1958), you will see that any gearbox with an input and output shaft can be considered as a mechanical(rotational) impedance transformer. If the load cannot accept all the power transmitted from the prime mover (engine), what happens? I think PAR is not very well informed on this.--Light current 03:03, 3 May 2006 (UTC)

I'M not very well informed on this!! Could it be that one of, or a combination of, velocity, movement, inertia result when load power 'cannot accept' (is less than?, "<") source/mover power? Thanks for the ref, I'll look it up, and contact PAR. Sholto Maud 03:16, 3 May 2006 (UTC)

I haven't yet got hold of the Olson ref, but curiosity is getting the better of me... if the implication of Olson's work is that the maximum power theorem applies within mechanical systems, what happens if we have a system which converts accross power forms, from electrical energy to mechanical energy? Does maximum power apply? THere is some contention at Maximum_power_theorem suggesting that conversion from electrical to mechanical (dynamo) can be 90% efficient. I assume this to mean that maximum power transfer from electrical form of energy to mechancial form of energy occurs at 90% efficiency and not the 50% efficiency reported in electronic circuits. Do you know if this is this correct? Sholto Maud 03:29, 3 May 2006 (UTC)

Thats a very interesting question. As far as I know, he doesnt deal specifically with Max power tranfer theroem, but Ill see if anything can be inferred. I will have to scan the book. I think the term 'available power' is going to come into the argument here and Im going to have to look it up!

BTW rotational power is (ang vel*torque). How does torque vary with ang vel in a motor or engine and is this important to your question? Also rotational resistance is I/t (or is it dI/dt) (where I is the moment of inertia) 8-)--Light current 04:01, 3 May 2006 (UTC)

..very important. So from torque article..

• Energy = torque*angular displacement (radians)
• Power = torque*angular velocity ("rotational speed", radians per sec.)

...so to graph the maximum power transfer efficiency in this instance it seems that we need to determine the analogs for load and internal resistance such that max pow trans eff = PL/Pint+PL. I'll be very interested to know what your scan of Olson turns up. Sholto Maud 06:53, 3 May 2006 (UTC)

Rotational resistance

From Olsons table of equivalents, emf is analogous to torque and current is analogous to ang vel. Also rotational resistance is analogous to electrical resistance. So rot resistance r_R isgiven by

r_R =f_R/(dθ/dt)

or r= Torque/speed or the slope of the torque/speed curve if its non linear. One question is whether any energy can be dissipated/stored in this so called rotational resistance! I think not. Also for a constant resistance it is plain that torque must be proportional to(ang)speed. This is not usually the case in engines/motors methinks. --Light current 13:25, 3 May 2006 (UTC)

NB. Please see my later post regarding the possible error of assumption in this argument--Light current 06:48, 6 May 2006 (UTC)

Assuming no losses in the gearbox (which we can), it then follows that the torque-speed product (ie power) at the input and output must be equal. If the load torque increases, speed will decrease for constant power and vice versa. So with no load,(engine idling) there can be no torque and the engine resistance is zero. With full load (engine nearly stalled) there can be lots of torque but very low speed (high o/p resistance). So it seems the load torque actually determines the resistance of the engine/motor! Its seems also therefore that the load and source resistances are automatically matched thro' the gearbox. Does this mean max power transfer? Or something else? You always get max power transfer in the gearbox system itself but the gear ratio will determine at what resistance your are running the source. Is there an optimum value for this resistance?. For real engines/motors and loads I would say yes. For imaginary (perfect) sources and loads maybe not.--Light current 14:05, 3 May 2006 (UTC)

In conclusion I would say that:

• in an ideal gearbox, you get perfect matching between an ideal engine and ideal load with no losses occuring

the output resistance of the engine is automaticaly adjusted to be the same as the input to the gearbox. the output resistance of the gearbox is necessarily the load resistance (theyre connected) no power is lost in these resistances as they are not energy disspating (but only notional) components. The maximum power transfer theorem does NOT apply to rotating mechanical systems.

Gearboxes can match the load to the engine at any speed/torque thus enabling the engine to run at max effy (or output resistance)--Light current 15:51, 3 May 2006 (UTC)

In response to Light current, I think maybe you are right, but...

I can get my car from zero to sixty mph using only fourth gear (I have a stick shift). I have to be very careful to not fully engage the clutch (THATS where the impedance match is important) and it takes a long time to get up to speed, but it can be done. Again, the reason for using different gear ratios is to adjust your force/speed ratio (actually torque/speed ratio). When you want to accelerate, you want a large acceleration (a). That means you want a large force (F) since F=ma. That means you want low gear.

Its like the electrical impedance problem. In simple terms, you have a motor that delivers a certain torque, like the voltage in the electrical case. You have friction in the motor which "impedes" or resists that force, like the resistor in the electrical case. And then you have rotation (angular velocity) which maxes out at some value, like the current in the electrical case. The power dissipated is the angular velocity times the torque. If you put a load on the motor by engaging the clutch, some power will be transmitted through the clutch. The maximum power transmitted will be when the two loads (internal and external) match. If you want acceleration, you drop to low gear. Now the external load is that you are basically pushing against the inertia of the car. That's what you have connected to the other side of the clutch. Its not friction, so there is no maxing out of the velocity, velocity just keeps increasing, giving you what you want - acceleration. If you don't want acceleration, but rather just a steady velocity, then you want high gear - now the load is friction, not inertia, giving a "terminal" velocity. That terminal velocity is higher for higher gears, but the engine always feeds to the same load at the clutch.

When I said that changing gears is not a form of impedance matching, I was concentrating on the fact that the impedance match is at the clutch, not the gearbox. The gearbox is an impedance transformer which supplies the proper load to the clutch, depending on what you want out of the vehicle - acceleration, low speed, or high speed. PAR 19:16, 3 May 2006 (UTC)

Ah well I think the clutch is a lossy component that creates heat (hence burning your clutch out) and what I have said will not apply there (probably). Also my previous analysis was based on ideal (lossless) motors, gearboxes etc. If you bring in lossy components, maybe you can get an impedanace match , this time with energy loss. I wouldnt like to say ATM 8-)--Light current 23:18, 3 May 2006 (UTC)

Fantastic work people. Thank you PAR & Light current for your congenial collaborative contributions. I do not claim authority in this area and would like to clarify and summarise what you have both uncovered in a future post, and relate it back to the previous dicussion if possible. Congratulations, I feel this is truly wonderful work!! THankyou, and please continue. :) Sholto Maud 02:27, 4 May 2006 (UTC)

I dont think we have bottomed it all out yet, but its a start and its shown up at least one of my misconceptions. And of course there are acoustical systems (loudspeakers) also.... 8-)--Light current 02:53, 4 May 2006 (UTC)

It would be beneficial to have Olson's table of equivalents posted in a table on Wikipedia. I'm not sure if this would be appropriate for the impedance matching page. It seems more suited to a new page on analogical models, with a subheading on formal analogies and table of equivalents. Sholto Maud 03:24, 4 May 2006 (UTC)

Yes I think you are right. Im not sure how to do a table though! If someone could do a blank table 16 cols wide and 12 rows deep I could fill it in.--Light current 03:31, 4 May 2006 (UTC)

Scrappy Analogical_models article created with table 16 cols wide and 12 rows deep. Hopefully there is enough to go on in the code for how to fill out the table. Sholto Maud 04:11, 4 May 2006 (UTC)

Its ok. I ve copied a table and Im modifying it. That page Analogical_models is in a terrible state!. Whoops ! sorry I didnt realise you had just created it!! But thats the one that this info should go on. While Im there Ill try to tidy it.--Light current 04:14, 4 May 2006 (UTC)

Pseudo summary

1. emf analogous to torque = f_R
2. current analogous to angular velocity = dθ/dt

• 2.1 is angular velocity also = to ɯ*t = 2* $\pi$*f = 360*f ?

3. rotational resistance analogous to electrical resistance = r_R =f_R/(dθ/dt)
3. Power = Voltage*Current = Current*Current*Resistance = angular velocity*dθ/dt*f_R/(dθ/dt)
4. Power = torque*angular velocity
hence

5. torque*angular velocity = dθ/dt*dθ/dt*f_R/(dθ/dt)

And so torque = angular velocity*f_R/_(dθ/dt) = emf

Here is a table that gives maximum power in resistance matching. It might be beneficial to construct the analagous table using rotational power.

Effect of Connecting Various Load Resistances to a Practical EMF Generator

Load Resistance (Ohms)	Internal Resistance (Ohms)	Total Resistance (Ohms) RT	Current (amps) I	Load Voltage (Volts) VL	Load Power (Watts) PL	Internal Power (Watts) Pint	Total Power (Watts) PT	Power Efficiency %
RL(oad)	Rint(ernal)	RT(otal)= RL + Rint	= E/RT	= I(RL)	= I^2(RL)	= I^2(Rint)	= PL+Pint	= PL/PT
0	6	6	20	0	0	2400	2400	0 %
2	6	8	15	30	450	1350	1800	25%
3	6	9	13.3	40	533	1067	1600	33.3%
6	6	12	10	60	600	600	1200	50%
12	6	18	6.7	80	533	267	800	66.7%
18	6	24	5	90	450	150	600	75%
infinite	6	infinite	0	120	0	0	0	100%

Analagous table in rotational systems (needs experiemental data and definition of load rotational resistance)

Load Rotational Resistance	Internal Rotational Resistance	Total Resistance	angular velocity = dθ/dt	Load torque = fRL	Load Rotational Power PRL	Internal Rotational Power PRint	Total Rotational Power PRT	Rotational Power Efficiency %
rRL	rRint	rRT = rRL + rRint	= fR/rRT	= fR/rRT* rRL	= (dθ/dt)^2 * rRL	= (dθ/dt)^2 * rRint	= PRL+PRint	= PRL/PRT
0	6	6	20	0	0	2400	2400	0 %
2	6	8	15	30	450	1350	1800	25%
3	6	9	13.3	40	533	1067	1600	33.3%
6	6	12	10	60	600	600	1200	50%
12	6	18	6.7	80	533	267	800	66.7%
18	6	24	5	90	450	150	600	75%
infinite	6	infinite	0	120	0	0	0	100%

Sholto Maud 11:07, 4 May 2006 (UTC)

Load rotational resistance and internal rotational resistance

Lightcurrent, I'm not literate enough to find the analogs for load rotational resistance and internal rotational resistance. Does Olson specify load rotational resistance and internal rotational resistance in his book? These values are important for electrical resistance matching for maximum electrical load power, so I assume they are important for determining max rotational load power Sholto Maud 21:57, 5 May 2006 (UTC)

Well, all it says is , rotational resistance is equivalent to electrical resistance, and has dimensions of I/t, where I is the moment of inertia. Load rotational resistance is the same thing as internal rotational resistance in the book. I think what youre talking about is a counter torque caused by friction etc at the wheel rims, athough it is shown as a wheel with a sort of braking pad applied to it! Mystery!??--Light current 22:15, 5 May 2006 (UTC)

Ah just found a quotation:

Mechanical rotational resistance is represented by a flywheel with a sliding friction brake which causes dissipation.

Here we have a problem with my previous analysis where I determined that rotational resistance was not dissipative! Oh dear -- must do more thinking!--Light current 22:58, 5 May 2006 (UTC)

I'm not sure what the exact meaning of "rotational resistance" is, except from the above discussion. If it is equivalent to electrical resistance, then yes, it must have dimensions of I/t, because of the differential equation in the section below. Each term must have the same dimensions, and for rotation the differential equation is:

$\tau =I{\frac {d^{2}\theta }{dt^{2}}}+R{\frac {d\theta }{dt}}+G\theta$

where $\tau$ is the torque, I  is the moment of inertia, R  is the rotational resistance, and G  is a potential energy, like a watchspring, or something. θ is the angle of rotation and t  is time. If we say that angles are dimensionless, then the first term on the right has dimensions $I/t^{2}$ and the second term has dimensions $R/t$ and since these must have the same dimensions, you can see that the dimensions of R are I/t. The middle term is always dissipative, whether its a rotational differential equation, electrical, whatever. PAR 04:57, 6 May 2006 (UTC)

I think rotational resistance can be simply described as 'restance to rotation' its nothing more than that. Your equation above would explain my problem. Torque can cause dissipation or not depending upon the elements of the system. My very simple expalnation conveniently ignored the first and the third terms in the general equation (which Im sure is correct having just checked it in Olson). So my argument above was considering a system with no moment of inertia, and no rotational compliance (ie stiff coupling) THis is equivalent to an electrical system with no capacitance or inductance. However, to have a rotating system where you ignore the angular momentum is plainly silly! So I think my previous argument is not necessarily correct.

OTOH, in the steady state, I think we can indeed ignore the first and third terms in a mechanical rotating system. The o/p resistance of the engine should be very high but not infinite in order to obtain o/p torque at any desired speed. The engine must be speed controlled. That is, it must be more like a current source than voltage source. Now for maximum power to be transmitted to the load, the load resistance should be equal to the engine resistance. This can be achieved by means of the gearbox. THerfore the max power transfer theorem also applies to mechanical rotating systems. (I think). All this just goes to show that Im not a mechanical engineer doesnt it? 8-(--Light current 07:24, 6 May 2006 (UTC)

Empirical data should verify or falsify the hypothesis that max power transfer theorem applies in mechanical rotating systems. If internal electrical resistance is the electrical resistance of the power source, then wouldn't internal rotational resistance be the resistance to rotation of the combustion (dissipative/lossy) engine? Sholto Maud 09:36, 6 May 2006 (UTC)

I dont know. 8-( I mean what is the physical cuase of such a resistance to rotation. Is it the fact that the engine's stiff, or the fact that it will only go at the speed determined by the rate of fuel injection. If the latter, how can it be a real dissipative resistance? All I'd say is that a constant speed engine/motor is analogous to a current source and in a current source there is power dissipation equal to the current times the voltage across it. This power is not dissipated in a resistOR as such, but in a resistANCE. A current source has a variable resistance depending on how much current it has to supply and at what voltage. Does, also, an IC engine have such a variable resistance inside? --Light current 17:10, 6 May 2006 (UTC)

Hmmm. (We might need an expert - I'm only stabbing in the dark). The Resistivity#Temperature_dependence article says, "In general, electrical resistivity of metals increases with temperature." So if the IC engine is made of some kind of metal, then its specific (electric) resistance (resistivity) will vary with temperature, so it seems to have a variable (electric) resistance inside. But are we 360 degrees (one rotation:) ) back at the problem of maximum power in the conversion different energy types? I say this because it seems that rotational resistance is expressed not in the form of rotational energy, but heat energy?! Sholto Maud 00:41, 7 May 2006 (UTC)

Yes but an IC engine is not electrical. It may be something to do with thermodynamics. But I still think Ic engines could have a variable o/p resistance to go at different speeds with the same torque (or vice versa). How this 'resistance' is varied, Im not sure. Yes we need a mechanical engineer! 8-|--Light current 02:11, 7 May 2006 (UTC)

Agreed. IC not electrical. Not sure if they have variable resistance but it 'feels' like it should. Lets see if we can recruit a mechanical engineer. If the generality of the maximum power theorem can be demonstrated (with associated proof) in mechanical and other systems, it will provide support for the related project of attempting to render the theorem more strongly as a maximum power principle. I am curious to know if this is the case or not. Without wanting to dissipate matters, proofs and demonstrations of maximum power transfer holding in many systems might be well suited to the maximum power principle discussion page. Sholto Maud 11:52, 7 May 2006 (UTC)

...where I determined that rotational resistance was not dissipative...

Say I'm looking at an automobile, and I see that it is connected to a gearbox and a output shaft that exits the engine compartment. With the right measurement instrument mounted to the output shaft (torque sensor), I can measure the mechanical energy leaving the engine compartment. That energy is, of course, the torque (the "rotational resistance") times the angle turned. The "rotational resistance" is the torque divided by the angular velocity.

But I don't think it's possible to determine whether that resistance is dissipative or not, without checking out what's on the other side of the firewall.

• Perhaps on the other side of the firewall, the car is up on jacks and someone is applying the brakes. In that case, the rotational resistance is dissipative, all the energy going to heat up the brake pads.
• Perhaps on the other side of the firewall, the car is traveling at constant high speed along a level highway. In that case, the rotational resistance is dissipative, all the energy going to wind resistance (slightly warming up the air) and rolling resistance (warming up the rubber in the wheels).
• Perhaps on the other side of the firewall, the car crawling at constant low speed up a steep mountain road. In that case, the rotational resistance is not dissipative -- all the energy is being converted to gravitational potential energy, which (in theory) can be losslessly converted back to some other form of energy.
• Perhaps on the other side of the firewall, the car is accelerating from a slow start along a level road. In that case, the rotational resistance is not dissipative -- all the energy is being converted to kinetic energy, which (in theory) can be losslessly converted back to some other form of energy.

My point is that just because something absorbs power from a power source ("looks resistive", "looks like a load"), doesn't necessarily mean it gets hot. Perhaps it really is dissipative, and turns all that energy into heat. Or perhaps it is not (entirely) dissipative, and it converts that energy into some other form -- either immediately sending that energy into some other system, or storing that energy in some form. (There is always a small amount of inefficiency, turning some of that energy into heat -- but that inefficiency can be much, much less than the 50% that I and too many others mis-understood as the maximum efficiency implied by the maximum power transfer theorem).

Also -- the energy that goes out is almost entirely dependent on the load on the other side of the flywheel, in the same way that the energy that comes out of a wall socket or battery is almost entirely dependent on the load connected to it. When no load is attached, no power comes out, the torque or current is zero.

However, without looking at the other side of the firewall, I *can* tell the difference between

• energy flowing out -- "looks like a load", torque applied by motor is in the same direction as the shaft rotation, "torque divided by angular velocity is positive"

vs.

• energy flowing in -- "looks like a power source", torque applied by motor is in the opposite direction of shaft rotation, "torque divided by angular velocity is negative".

Energy may be flowing in for several reasons:

• Perhaps on the other side of the firewall, is some other engine or starter motor, pushing the flywheel of the motor faster.
• Perhaps on the other side of the firewall, the car is traveling at constant low speed down a mountain road -- downshifting.
• Perhaps on the other side of the firewall, the car is decelerating from highway speeds along the highway exit.

Again, I can't tell the difference between these "looks like a power source" without looking outside the engine compartment. But I certainly can tell the difference between a "external power source" vs. an "external load".

I'm pretty sure the maximum power principle does not apply to high-power muscle cars -- the engine can supply far more power than the wheels can accept, so there is no need to extract the maximum possible power from the engine. (Pushing the pedal to the metal causes the wheels to "peel out" and lose traction, getting to the finish line too late). But perhaps it can be applied to underpowered cars for the short time they are straining to accelerate to highway speeds on the entrance ramp.

--68.0.124.33 (talk) 21:35, 26 January 2008 (UTC)

Dynamical analogies table

Many physical processes are described by a second-order linear differential equation:

$F=m{\frac {d^{2}x}{dt^{2}}}+r{\frac {dx}{dt}}+gx$

where F is a generalized force, m is a generalized mass, r is a generalized friction coefficient, and g is a generalized potential.

	Mechanical rectilineal	Mechanical Rotational	Electrical	Acoustical
x	Linear distance	Angular distance	Charge
$dx/dt$	Linear velocity	Angular velocity	Current
$F$	Force	Torque	Voltage
$r$	Friction	Friction	Resistance
$m$	Mass	Moment of inertia	Inductance
$g$			1/Capacitance

Energy is the general force times the general distance, while power is the general force times the general velocity.

• Also added this table to the Analogical models page, making it even more scrappy. But youse can work out what is best...must go. Sholto Maud 04:13, 4 May 2006 (UTC)
• table in the end is in really bad shape posted an external link at the end of the article with the whole table completed. I don't understand how to write tables in wikipedia. hopefully someone that does can complete it. —Preceding unsigned comment added by 129.171.180.27 (talk) 04:01, 26 February 2009 (UTC)

Structural vs functional analogs: two formal analog definitions

Perhaps a good issue for the article:

• Structural analog: same "network structure" and of Lumped elements. (if the internal parameters are not ajusted, they may be not functional analog).
• Functional analog: same black box, that is, same input/output behaviour. (example Thévenin equivalent circuits).

They are different things, but both define formal "analog". --Krauss (talk) 12:11, 18 December 2014 (UTC)

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The first figure is inconsistent

The use of current/voltage compared to motion/force is inconsistent. The consistent model for inertia is inductance and the analog of a spring is a capacitor. Then modeling electrical current as motion & force with voltage is consistent. Michael McGinnis (talk) 04:16, 25 March 2019 (UTC)

It's a perfectly valid model called the mobility analogy It is widely used and has certain advantages. There are many possible models, there is no one correct analogy. The choice depends on the parameters of the problem one is trying to address. SpinningSpark 13:28, 25 March 2019 (UTC)

Hydraulic Analogy strange

The text of the Hydraulic Analogy does not match the current linked page, which is about electrical-hydraulic analogy only. In any case, it is not clear that there is an analogy between hydraulics and maths: the hydraulics obeys the maths: what it is describing is a computer not a simulator, I guess. (Yes I know there are analog computers, but still.) Rick Jelliffe (talk) 04:17, 18 June 2023 (UTC)