Lets beat up Carnot

[Tom Booth]

This is how, right or wrong, I imagine the gas in a hot air engine drives the piston out and how the piston returns.

Inside the engine there is a pool of gas molecules.

Relatively few molecules are heated at any one time.

If say, only one molecule were heated (among the billions and billions inside the engine) the piston might move a tiny bit but would return as soon as one outside molecule collided with the other side of the piston.

To make any significant movement of the piston MANY molecules of the "working fluid" would need to be heated simultaneously before the outside molecules had a chance to drive the piston back again.

Many gas molecules inside and outside the engine have no participation in these exchanges but remain near equilibrium.

When a heated molecule strikes the piston and the piston moves the energy of the molecule is transfered to the piston and the molecule falls back into the "pool".

There is always a tendency for the "system" to return to equilibrium and for the piston to return to its original starting position after any such impact on either side because there is a fixed number of molecules trapped inside the engine. If the piston is "left" to one side or the other of its starting, or equilibrium position it will return due to subsequent random impacts on both sides.

If a very large number of molecules of the working gas are heated the piston can be propelled outward with some momentum while those molecules participating in this impact fall back into the pool. There can then be an imbalance created which leaves the piston far from the equilibrium position and the outside impacts then momentarily far outnumber the inside and the piston is driven back by impacts from the outside.

Remember most of the gas in the "pool" are effectively "lying on the ground" as Matt put it. Or may as well be, because whatever motion they have is equally matched by molecules on the outside so that the forces cancel.

[Fool]

Sorry this took so long. I had to find my 2014 Lap top. Chisel off many layers of rock, fossils and dinosaur bones. Man they pile up fast. Dust it off, Charge it up. Start up and stoke the steam boiler. Allow time for it to come up to temperature and pressure. Boot strap load the operating system. Find the old files. Start the spread sheet and word processer. Okay so some of that is artistic license, storytelling. LOL

I promised to comment on this thread with some mathematics, so here it goes. Some of this is corrections, minor, and some is branching out to a more detailed pondering, some is just grammar and convention.

1 Joule = 0.737562 Ft-Lbs (Not Ft/Lbs)
Round able to 0.74
1 Joule is 1 Newton of force used to move an object 1 meter of distance. AKA: Energy.

Torque is in Lbs-Ft (Pounds times Feet), often colloquially spoken: Foot Pounds. Not the same as Energy. Torque is Newtons, or Pounds times a perpendicular distance, length of lever arm, Meter or Foot, and can be applied to something without motion. Again, not Energy, Force times a moved Distance. Yes, Force times lever arm no motion necessary. AKA twisting force.
There are two Heat Capacities, Cv and Cp, at STP, using SI (System International), Meters, Kilograms, Seconds, (I convert everything to SI.) :
Cv = 718 to J/kg Constant Volume Heat addition.
Cp = 1005 J/kg Constant Pressure Heat addition.
They change very little over the pressures and temperatures given for this example.

Displacer Diameter = 0.3048 m = 1 foot x 12 x 2.54 / 100
Cylinder effective length = 0.0127 = (1 inch - 0.5 inch) x 2.54 / 100
Effective volume = 0.000926667 m^3 = 56.54866776 in^3 (Close enough to yours.)

Density of air = 1.225 kg/m^3 at STP
Mass of air = 0.001135167 kg = Density x Volume = 0.000926667 x 1.225
Moles of air per kilogram = 0.0289 Moles/kg
Moles of air = 0.039279122 = 0.0289 x 0.001135167

T1 STP = 293.15 K = 20 C
Given starting temperature.
T2 = 373.15 K = 100 C
Given ending temperature T hot.
Delta-T = 80 K and C = T2 - T1

P1 = STP = C 101325 Pa (Pascals) (N/m^2) = 14.6959494 psi
The given starting, Standard Temperature and Pressure, STP.

Calculations:
P2 = 128976.373 Pa = P1 x T2 / T1 = 18.70644216 psi
Delta pressure = 27651.37302 Pa = 4.010492758 psi
This is at V1.

Heat added to get to T2, P2:
This would be Heat Capacity times Delta T times mass. Since there are two Heat Capacities Cv and Cp, the correct one must be justified. Since the volume of this is not changing and the pressure is, we choose Cv for constant volume (isochoric) heat addition.

Qh = 65.2039713 Joules = 718 x 80 x 0.001135167

That number is different from yours because you used Cp. If I use Cp I get 91.26 Joules. That is different from your 96 Joules because of rounding differences. So we are comparable here. Cv is smaller than Cp because temperature rises with increased pressure. Since Cv must account for increased temperature from pressure increases, it takes less heat energy/Joules, than for Cp, to get to the same temperature.

Now to calculate the PV factors, sometimes called Pressure Volume Energy. It assumes a Delta-V of total V, or Vtotal, and zero pressure change, also called constant pressure process. In a real adiabatic expansion the pressure will drop, but we are just calculating it here. It would be easier to understand seeing a graph of this.

P1V1 = 93.894497 Joules = 101325 Pa x 0.000926667 m^3
P2V1 = 119.5181 Joules = 128976.373 Pa x 0.000926667 M^3
Constant volume, increased pressure from temperature.

PV energy difference from added heat,
P2V1 – P1V1 = 25.623605 Joules

PV heat change divided by heat in:
25.623605 / 65.2039713 = 0.392976139 or about 39%

Carnot for this is:
80 / 373.15 = 0.214390996 or about 21%

At this point it will be noted that the change in PV energy doesn’t equal the Heat added. The percent loss isn’t the same as expected by the Carnot Theorem either. However, the Carnot theorem is for a complete cycle, not a single stroke or for a PV energy equivalence. However a constant volume and pressure CALCULATION also has zero energy out, so the overall “energy out” efficiency of this process is also ZERO.

I think the opening post confused total energy at T2 with “Delta Energy” from Delta Pressure 4 psi, or change in PV energy factor. And confusion of Cv for Cp. It also didn’t recognize that real Energy/Work out was zero. Efficiency is real work out over real work in, not Energy Stored.

This concludes the corrections to the opening post.

I would like to explore a little further and look at some maximum work out from a half cycle, or single expansion stroke. The actual real work available will be less, than the following, because we will be ignoring mechanical, and other, losses.

Let us try expanding, adiabatically with work (isentropically), until the volume P3 = P1, or it is back down to the atmospheric/beginning pressure.

Expanding back down to atmospheric will give the maximum available work out of the working gas for an isochoric process, because at the point where P3 equals P1, the piston has reached its maximum velocity and kinetic energy. Expanding any further will require pulling a vacuum with associated work input and a decrease in piston velocity, i.e., less maximum energy to output as work. I’ve use the equations from the Wikipedia page on Adiabatic Processes for the following:

https://en.m.wikipedia.org/wiki/Adiabatic_process

Single stroke P2 back to P1 = P3:
Gamma = Cp/Cv = 1005 / 718 = 1.399721448
Alpha = 1/(Cp/Cv-1) 1/(1005/718-1) = 2.50174216
A good check of Gama at this point is that:
Gama also = (5+2)/5 = (Degrees of Freedom + 2) / Degrees of freedom
Degrees of freedom = 5 for an Ideal diatomic gas like oxygen and nitrogen.

W = 20.29956414 Joules = -Alpha x n x R x T2 x
{(P3/P2)^([Gama-1]/Gama) – 1}

Where:
W = Work
n = number of moles = 0.039279122
R = Gas constant for air = 8.314
W = -2.50174216 x 0.039279122 x 8.314 x 373.15 x
((101325/128976.373)^((1.399721448-1)/1.399721448) -1)
W = 20.29956414 Joules

Comparing the isentropic work with the Carnot efficiency:

20.29956414/65.2039713 = 0.311324046 or about 31%.
That is still not as low as the Carnot efficiency of about 21.4%

At this point it should be noted that the “perfect” efficiency of the one half cycle Adiabatic/Isentropic process is less than 100%, yet it isn’t as low as the Carnot Efficiency. That is because the Carnot efficiency theorem needs a return stroke to complete the cycle. The return stroke if Isentropic will need 20.29956414 Joules of energy input of work and will get back to P2 and V1 and T2. Not P1, V1 and T1. This leaves zero energy for output as work. However, 65.2 Joules of heat would need to be rejected to reduce the pressure and temperature back to initial conditions. A graph would also help to visualize this.

V3 = V1 x (P1/P2)^(-1/Gamma) = 0.000926667 x
(101325/128976.373)^(-1/1.399721448)
V3 = 0.001101009 m^3
This is larger than V1meaning it has expanded, and is now back to P1.

T3 is determined by Pressure ratio times T2

T3 = T2 x (P1/P2)^([Gama-1]/Gama) = 373.15 x
(0.000926667 / 128976.373)^([ 1.399721448-1]/ 1.399721448)
T3 = 348.3030909 K
Note: This is significantly lower than T2 but not back to T1. It can be reflected by the need for there to be a higher temperature than T1 to support the larger volume of V3.

It also means to return to T1, at this volume, heat must be rejected, or stored in the regenerator. The Carnot process can also be used by putting in work, for isentropic adiabatic expansion/vacuum, to create a lower pressure than P1/atmospheric, and a lower temperature back down to T1. The work will be returned on the compression stroke because the internal pressure is lower than the external atmosphere pressure. AKA adiabatic bounce, work in equals work out, total zero for a complete cycle.

It will buy nothing unless the return stroke is done at T1, constant temperature heat rejection. Then the return stroke will require less than the extra expansion work delivered. And the adiabatic, “Carnot” compression back to T1 and P1 initial conditions will require input of work further slowing the piston.

My conclusion from this exercise is that the Carnot Theorem is verified, or at least not disproven.

Disclaimer: I’ve checked and rechecked the above calculations and have made every attempt to assure their correctness. If errors are found, please check for them, and I’m informed, if possible I’ll repost a correction. Thanks for allowing my posting here. My apologies for being so long, tedious, and boring. I intend it to be useful and informative.

[Goofy]

Hi Fool,

Not to offend your calculations, but we have to agree on the definitions.

Quote:
"P1 = STP = C 101325 Pa (Pascals) (N/m^2) = 14.6959494 psi
The given starting, Standard Temperature and Pressure, STP."
(on 273,15 K)

Regarding STP from WIKI : https://en.wikipedia.org/wiki/Standard_ ... %201%20bar).

So your definition is from before 1982, using 1 atm instead of 1 bar

By the way the Molar weight of air is 28,9 gram/mol, not 28,9 mol/gram.

In STP, 1 mole of any gas will occupy a volume of 22.4 L and contains 6 x 10^23 molecules.

I haven´t been through it all yet, but interesting . . . .

BR

[VincentG]

Thanks Fool for a very well thought out and detailed response. When making the original post I found it very difficult to gather all of the information in a single source with no deviations. A text book may have been helpful here.

I found it most interesting that the formula for static force and torque over time(as how a vehicle is measured) are the same formula but taken as different quantities and generally recommended to not be interchanged.

I was not suggesting that the pressure from heating the gas alone is work. Im suggesting that the rapid oscillation of pressure(force) can be utilized to produce work in an unorthodox way, perhaps yet unknown, to myself included. And that the pressure can still result from far less gas than would be traditionally required to act on a piston.

Something like acoustic levitation comes to mind among other things......Something that would harness the kinetic energy of the gas molecules themselves.

[Fool]

Goofy, AKA BR, Brilliant! Right! Thanks for the corrections.

For the record I learned about STP before 1982, and even then learned it has different definitions. It seems like every time I look it up, it's different. LOL. I was kind of defining it for my specific post, not as an authority. Good point, and thanks for the history. It's best to be standard. Thanks again.

VincentG, yes physics is a lot more baffling than it should be. I guess that is why they keep teaching it over and over to every new generation.

It seems when comparing rotary physics with straight line physics several baffling anomalies rear up.

The thing I found interesting is that the heat and or temperature didn't completely disappear during an adiabatic expansion. ??? Go figure? Unfortunately heat must be rejected to complete the cycle.

Using kinematic impulses, pressure or sonic waves, acoustic levitation, whatever it's called, won't break the iron grip of Carnot. It may push harder on the piston, but for a shorter length of time. When a pulse first leaves it must travel a distance before impacting the piston. It hasn't pushed on the piston for that travel time. It has a short period while it's bouncing off the piston. It then travels back to where it came from causing the opposite of pushing. The upshot is the piston accelerates the same total amount, or less, as for a constant pressure model .

If it has any impact to the system, it will be a worsening from the Carnot cycle, a lowering of efficiency. Furthermore, it seems like it's easy to forget that acoustic levitation requires a very large input of energy to make a few tiny drops of liquid levitate. Kind of like a helicopter.

I think, too much effort is spent on efficiency.

Your goal of getting more heat in, and lower output temperature/pressure is a much better one. Maximize pressure/temperature difference and heat input. Power to weight/size will depend mostly on pressure-differential and volume-change.

Work equals P•V.

P depends on T. Note: Those are best thought of as Delta, Differential, or changes in, quantities. The piston won't move without a pressure differential. A pressure differential implies a temperature differential, at times. Especially in a heat or "temperature engine".

Heat exchanger improvements seem to me to be the area we should concentrate. That and materials. Tom Booth seems to spend a good share of his time in those two areas.

Carry on, doing well.

[Tom Booth]

"1 Joule is 1 Newton of force used to move an object 1 meter of distance. AKA: Energy."

This statement is incorrect.

[VincentG]

The thing I found interesting is that the heat and or temperature didn't completely disappear during an adiabatic expansion. ??? Go figure? Unfortunately heat must be rejected to complete the cycle.

This is the common misconception with these engines. While yes, heat must be rejected to complete the cycle, work can also be produced directly from this rejection of heat and high potential state. I'll submit the flame licker engine as a prime example. It runs purely on adiabatic cooling from essentially atmospheric pressure at elevated temperature.

In the Stirling engine, at the end of this vacuum cycle, the engine is in another high potential state, that is top dead center at minimum internal volume with no compression work negative to get there. The air is then heated and the piston driven to maximum volume to complete the cycle. Again, with no expansion work negative.

[Tom Booth]

The thing I found interesting is that the heat and or temperature didn't completely disappear during an adiabatic expansion. ??? Go figure? Unfortunately heat must be rejected to complete the cycle.

This is the common misconception with these engines. While yes, heat must be rejected to complete the cycle, work can also be produced directly from this rejection of heat and high potential state. I'll submit the flame licker engine as a prime example. It runs purely on adiabatic cooling from essentially atmospheric pressure at elevated temperature.

In the Stirling engine, at the end of this vacuum cycle, the engine is in another high potential state, that is top dead center at minimum internal volume with no compression work negative to get there. The air is then heated and the piston driven to maximum volume to complete the cycle. Again, with no expansion work negative.

While I basically agree with your point here, though maybe not entirely as your assessment does not appear to take into account cooling due to work output, which I think is significant, I'm interested in the purely mathematical proof or argument presented.

You present a sound logical argument, but that does not directly address the mathematics.

This statement:

"1 Joule is 1 Newton of force used to move an object 1 meter of distance. AKA: Energy."

I think, is wrong and also incomplete. The Joule, I think, is at the heart of the mathematics. If there is a flaw in the understanding, definition or interpretation of this basic mathematical unit this could have a cascading effect on the entire mathematical derivation.

I would submit that it does not take 1 Joule to simply "move" ANY object 1 meter.

1 Newton is not defined as the movement of just ANY "object" a distance of one meter. There are constraints on the definition which I think we can assume carry over to the definition of the Joule as 1 Newton-meter.

First of all the "object" is 1 kilogram.

Secondly the object is not simply moved but "accelerated" from zero to one meter in one second.

Once in motion an object remains in motion unless influenced by some opposing forces.

An additional joule will accelerate the 1 kilogram object to two meters per second. After which it will continue at that speed without additional energy input.



In other words, it does not require 1 Joule to move an object one meter. One Joule could move an object an infinite distance. It would then be "carrying" one Joule of energy as it continued to travel, forever, until it impacts another object transferring that unit of energy to the other object.

Until that Joule is given up to some opposing force (baring friction etc.) it does not require ANY additional energy to maintain a motion of 1 meter per second or whatever the velocity.

It will COST 1 Joule of energy, transfered, to decelerate the 1 kilogram object traveling at 1 meter per second back down to zero.

I do not know if this redefinition has any bearing on the mathematics that follows as I've only just started reading the first few paragraphs, but considering the Joule is such a basic mathematical unit and it seems that even this very basic element is in question, what follows, as far as it may depend on or utilize the Joule as a mathematical unit of energy...

Well, as I said, it could have a cascading effect, one error leading to another.

I don't yet know of that is the case, but something that perhaps should be kept in mind, guarded against or possibly corrected for.

Unless I'm wrong in my understanding of the relationship between a Joule and a Newton-meter. (?) Which may be, the subject seems a bit of a muddle.

[VincentG]

Tom as far as I can find the standard definition of a joule has no time constraint. Just 1 Newton applied to 1kg over a distance of 1 meter. Time may be calculated I assume. Though some sources don't specify 1kg as the subject of the force.

[Tom Booth]

Tom as far as I can find the standard definition of a joule has no time constraint. Just 1 Newton applied to 1kg over a distance of 1 meter. Time may be calculated I assume. Though some sources don't specify 1kg as the subject of the force.

I'm not saying that a Joule as a unit of energy is time constrained. Quite the opposite. An object may "carry" a Joule of "kinetic energy" indefinitely.

The definition of a Newton is:

The newton (symbol: N) is the unit of force in the International System of Units (SI). It is defined as ... the force which gives a mass of 1 kilogram an acceleration of 1 metre per second per second.

The point being that if a Joule equals the energy to move an object one Newton-meter that does not, I don't think, means that it takes one Joule to move an object each and every meter. That is, it does not take 2 joules to move the object 2 meters or three joules to move the object three meters.

A meter is not the same thing as a Newton meter. The Joule of energy is the energy needed for the production of one Newton-meter or the energy required to accelerate 1 kilogram to a velocity of 1 meter per second.

The time constraint is implied or inherited from the definition of a Newton, but it is not "attached" to the joule as a unit of energy, it is just part of the definition of what constitutes a joule. -> the energy required to produce one Newton meter, to accelerate one kilogram to a speed of 1 meter per second, not the energy required to "move" just ANY object "one meter".

That implies that one Joule of energy can move a bull dozer or an entire mountain or a galaxy one meter the same as it could move one kilogram one meter, and that another Joule would be required to move the object again one additional meter.

Once in motion an object continues in motion. The Joule becomes the kinetic energy of the object "forever" until transfered to some other object.

The statement: "1 Joule is 1 Newton of force used to move an object 1 meter of distance" then gives a false impression of the mechanics of motion, as if every meter of motion of any object whatsoever "uses" one Joule of energy in the production of the motion of ANY object whatsoever a distance of one meter.

That, I think, is transparently absurd, as if the same quantity of energy is required to move a grain of sand as to move a continent a distance of one meter.

[Tom Booth]

1 Joule = 0.737562 Ft-Lbs (Not Ft/Lbs)
Round able to 0.74
1 Joule is 1 Newton of force used to move an object 1 meter of distance. AKA: Energy.

Torque is in Lbs-Ft (Pounds times Feet), often colloquially spoken: Foot Pounds. Not the same as Energy. Torque is Newtons, or Pounds times a perpendicular distance, length of lever arm, Meter or Foot, and can be applied to something without motion. Again, not Energy, Force times a moved Distance. Yes, Force times lever arm no motion necessary. AKA twisting force.

This (the above quoted passage) is, all in all, a confused, mind numbing jumble of nonsense impossible to untangle or make any sense of.

Well, maybe not "impossible", but I'm struggling.

The main problem is it makes no clear distinction between "Newton-meters" in the context of torque and "Newton-meters" in the context of energy.

Maybe that is the intention, to make the distinction between the two different concepts clear, but doesn't really succeed, IMO.


Prior it is stated: "Some of this is corrections, minor, and some is branching out to a more detailed pondering, some is just grammar and convention."

Taken as a "correction" or clarification of terminologies, the paragraph begins to make sense.

Is this to "correct" a previous misunderstanding (by VincentG ?) if so, it's not clear what is being corrected or clarified, if that is the intention.

It seems self contradictory, confusing the usage of the term "Newton" as used in different contexts: torque vs. energy (joules), but perhaps is intended to untangle a previous misunderstanding?

Sorry, I haven't read the whole thread and still don't know the relevance if any to the mathematics that follows.

I'm taking the recent post in isolation as it seems to have just recently dropped out of the blue. I guess I need to go back and read the whole thread from the beginning.

Anyway, the subject of "Newtons" in relation to "joules" requires (further?) clarification I think. What the term "Newton" or "Newton-meter" means in the two different contexts, and how this does or doesn't relate to joules (energy). But, I'm probably not the person for that job as I'm not entirely clear about it and am, at this point, just trying to sort it out myself.

Online sources I'm finding to be of little help as they seem to repeat the same common errors (or simplifications?) Or is there something I'm not getting?

This video (Link below) for example is helpful, but I can't help but ask, what if the grocery cart being pushed by 1 Joule 1 meter is full of bricks?

Can one Joule overcome the inertia of a 1 kilogram cart just the same as a 10 ton cement truck full of cement?

Does this grocery cart weigh just 1 kilogram? Can that be assumed since the subject is the Newton meter and that is part of what defines a Newton? ("the force required to impart an acceleration of one meter per second per second to a mass of one kilogram") at least in this context (heat, work, energy, joules etc.)

Is the grocery cart just being "moved" one meter, or is it being "accelerated" to the velocity of 1 meter per (whatever).

Does the actual weight or mass of the cart or other object not matter?


https://youtu.be/76EvyvRdd6c?si=RuwaMta47njAyMke


Such "simplified" (?) explanations only tend to add to the confusion.

I can't imagine that 1 joule of energy can actually move ANY object 1 meter to make 1 "Newton-meter" as some supposedly well defined SI unit without any additional reference to gravity, weight, actual mass, friction, resistance, inertia, time, velocity or something.

Nearly all the online tutorials I've found just gloss over the subject.

As I stated previously, I can only assume these restricting or defining specific details are "inherited" so that it can be "assumed" by the use of the term "Newton" in the term Newton-meter aka "joule" that the cart in the video must actually have a mass of 1 kilogram and that it is not "moved" but rather "accelerated" and after being "moved" one meter by one joule would continue moving based on Newton's 1st law of motion.

The 1 joule of energy is applied to the cart to overcome its inertia, setting it in motion, accelerating it to a velocity of 1 meter per (some time interval depending on the mass?)

Unfortunately, if "fools" explanation is wrong or unclear or incomplete, the explanations that can be found online are mostly just as bad of worse and do little or nothing to clarify anything...

At least what I've been able to find so far.

I BELIEVE however, that this "Joule" in the context of the Newton-meter, in terms of heat and energy and work, has to do with overcoming inertia so as to accelerate a mass (of 1 kilogram) to some velocity, (1 meter per/second) not just "move" any object a distance of 1 meter.

What makes it worse, in terms of clarifying things, is that this usage of the term Newton-meter is "rare" and/or "discouraged" so as to "avoid confusion", so the subject is not often addressed in depth and just remains a muddle.

[Goofy]

Again definition on Joule from Wiki:

"A result of this similarity is that the SI unit for torque is the newton-metre, which works out algebraically to have the same dimensions as the joule, but they are not interchangeable. The General Conference on Weights and Measures has given the unit of energy the name joule, but has not given the unit of torque any special name, hence it is simply the newton-metre (N⋅m) – a compound name derived from its constituent parts.[26] The use of newton-metres for torque but joules for energy is helpful to avoid misunderstandings and miscommunication."

Again : "but they are not interchangeable."

As we are discussing HEAT, it is perhaps better to use the "old" Caloric definition (Thermochemical) :

1 calorie is defined as the amount of heat needed to raise the temperature of one liter of water by one degree Celsius (or one kelvin)

This gives (me at least) a better relation between energy and heat. But as we have moved away from Calories to Joules, that is :

1 Calorie is he amount of energy equal to exactly 4.184 J

Keep this in mind, instead of some mishmash with torque . . .

BR
(Best Regards)

;-)

[Tom Booth]

This is better:

https://youtu.be/cPgjSeDciQ8?si=sb5uhXQLdEv2Mqc7

Clearly, with less friction, the weight would continue traveling right off the table, or with the same 1 Newton of force applied, travel many meters.

An object, such as a levitating superconductor on a magnetic track or an object in space could travel an unlimited distance. Does that make more Newton-meters or increase the force in joules?

I think, obviously not.

Friction can be reduced, even on earth, to near zero. So one Joule/Newton-meter of force could accelerate an object so that it moves much more than just one meter. The 1 Joule of force has been converted into kinetic energy, which can then be transfered to other objects..

After 1 meter, an additional joule will accelerate the 1 kilogram to 2 meters per second.

After 2 meters another joule will accelerate the object to 3 meters per second.

The original force is not lost with distance but accumulates with each additional joule, (of course, less friction, but that can be minimized or even very nearly eliminated altogether)

This looks good if you have 1/2 hour to kill.

https://youtu.be/QAsnpR6bqQo?si=I0kmh4HXnbbM2jw_

Very basic and thorough, though I've only watched the first 5 minutes.

[Tom Booth]

This next part is probably worth a view also

https://youtu.be/5T_O6Zt5Xt4?si=rzz70E8TiiXvdR3P

The main takeaway being the mass or inertia of the object matters.

Another interesting point is the sum of multiple forces (or vectors) either augment or cancel each other.

This helps make clear what I was driving at in the "Aligning heat 'vectors'" thread :

viewtopic.php?t=5556

In the Stirling engine we have multiple forces we are trying to align to maximize acceleration or RPM rather than having the forces working against each other and canceling out.

[Tom Booth]

...

1 Joule = .74 Ft/Lbs - Now traditionally a Joule is not used to quantify a static force, but that's what the internet says the conversion is, and we will need to measure power as a static force for this example. Perhaps this is where this whole example falls apart, ...

There is a kind of segue in the Wikipedia article about the Newton-meter:

The unit is also used less commonly as a unit of work, or energy, in which case it is equivalent to the more common and standard SI unit of energy, the joule.[2] In this usage the metre term represents the distance travelled or displacement in the direction of the force, and not the perpendicular distance from a fulcrum as it does when used to express torque. This usage is generally discouraged,[3] since it can lead to confusion as to whether a given quantity expressed in newton-metres is a torque or a quantity of energy. However, since torque represents energy transferred or expended per angle of revolution, one newton-metre of torque is equivalent to one joule per radian.

Second-ish paragraph down:

https://en.m.wikipedia.org/wiki/Newton-metre

A radian is about, (a little less than) 1/6 of the circumference of a circle. The distance of the radius wrapped around the circumference.

Could this perhaps be a means of relating joules to rotational force, or just make for additional confusion?