Tom
I applaud your experiments with the red model Stirling engine.
I too happen to have that very same model, so I have a "feel" for how much your wooden displacer is different from the metal one supplied, and just how long a time it takes to get the original version to absorb enough heat to begin to run properly.
I agree with your general point that the metal body (and the rather heavy metal displacer) may be having a lot of adverse influence over the efficiency of the engine exactly because it is storing and thereby wasting a lot of heat by constantly dumping it into the local environment.
Also I've noticed that the cold-end gets pretty warm! That doesn't seem very efficient.
Having said that, maybe it isn't as bad as it seems. The engine runs despite the cold end being surprisingly hot (it is still cooler than the hot end which gets really very hot). So the temperature gradient along the metal body - whilst seeming to be infuriatingly inefficient thermally - might actually be beneficial to the running of this particular type of motor (which is a desk top toy intended to amuse and entertain).
There is a general point here. The volume of working fluid in the model is tiny and is less than the volume of highly conductive metal used to contain it. It probably only works on a small temperature difference between input (hot) end and output (cold) end because no matter how hard you try to pump extra heat into the hot end it rapidly travels along the metal body towards the cold end. This has the effect of stuffing yet more heat into the metal work, raising the temperature of both hot and cold ends but keeping the temperature difference between them more or less the same, despite best efforts.
It just goes to show that unless you have a "good" design, it is actually quite hard to get more heat into one end of the working fluid only. Its a bit like trying to fill a teacup by pouring a bucketful of fluid into it - most of the fluid instantly goes to waste.
Which brings me to what is a "good" design? Obviously.... I don't know! But, I am intrigued by simple tubes with low thermal mass, low thermal conductivity and low specific heat capacity. Borosilicate/pyrex glass looks very promising in that regard. Even simple test tubes (commonly found in lots of designs these days) seem to offer lots of advantages over cast metal bodies with small internal volumes..........
Ted Warbrooke's Stirling 1: Question
Re: Ted Warbrooke's Stirling 1: Question
I've been familiar with Rijke tubes, for about as long as I've known about Stirling engines, I just don't buy the theory that there is anything more than a superficial resemblance. I don't think that there is any sound waves traveling, stationary or otherwise effecting the operation of a Stirling engine.
A rifle also has a long tube like a flute or a pipe organ and lo, when fired it makes a loud report! Do we conclude from that that the sound vibration of the gun is what propels the bullet? No, it is heat, and the expansion of a gas.
A Stirling engine and a rifle actually have more in common IMO than a Stirling engine and a Rijke tube.
A rifle also has a long tube like a flute or a pipe organ and lo, when fired it makes a loud report! Do we conclude from that that the sound vibration of the gun is what propels the bullet? No, it is heat, and the expansion of a gas.
A Stirling engine and a rifle actually have more in common IMO than a Stirling engine and a Rijke tube.
Re: Ted Warbrooke's Stirling 1: Question
No chance of that, (In My Opinion, of course, but I don't believe I could be wrong)
The only thing powering a Stirling engine is heat entering and causing the expansion of the working gas. Heat traveling through, escaping, migrating, producing a "gradient" anywhere else, are all, invariably detrimental to one degree or another. Again, to use the electrical analogy, these are short circuits of a sort. A heat gradient along the length of a pipe does not contribute to the expansion of a gas within the pipe, it is, rather, the effect of heat loss the effect of which is to drain power and reduce the expansion of the gas.
There is, unfortunately, a general acceptance that there is some inevitability the majority of heat put into a heat engine MUST go to waste.
It is presumed to be a "LAW"!
The second so-called LAW of thermodynamics. ENTROPY!
Unavoidable!
There is even a mathematical formula to calculate just exactly how much heat MUST, of necessity, go to waste.
It has a long standing history, the so-called "Carnot Limit".
The formula is: η = (Th - Tc) / Th * 100%
The maximum efficiency, or rather, the minimum inefficiency can be calculated quite easily here:
https://www.omnicalculator.com/physics/ ... efficiency
Or using any of the dozens of other online calculators available for the purpose.
All complete bollocks!
In all the history of the world, there has been no experimental support for any of this "Carnot efficiency" poppycock.
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Alphax
Re: Ted Warbrooke's Stirling 1: Question
Hi Tom
That is an interesting way of looking at it. I look at it differently. So I'll elaborate a little on how I see it at the moment:-
It seems to me that all small models (toys, really) are always going to be thermally very inefficient. Worse.... because they are small models and not laboratory instruments, we have no way of actually measuring the inefficiencies that we know are there. So we must guess that they are there and that they are "large". On that point we must agree, I think.
For my part, I accept that. In other words, as far as small models are concerned, I accept that they are very inefficient. But I also see that as positive and beneficial. Why? Because they work despite what must (presumably) be their very poor efficiency. So, for example, if someone had as their personal objective the task of improving the efficiency of a particular design (as used in a particular model), then there is scope to do so by definition - you can always get some improvement if you work at it. But only for that one particular model.
It is perhaps just a small point, but that is why I said that the temperature gradient and all the various (and large) thermal inefficiencies of that model are not "bad" things but actually can be viewed as "good" because it still works despite all the heat losses stacked against it.
I see that you prefer to focus on the fundamentals of what makes Stirling engines work and that you consider anything else (thermal gradients etc) as
undesirable wasting of heat. So lets look at that ........
That is an interesting way of looking at it. I look at it differently. So I'll elaborate a little on how I see it at the moment:-
It seems to me that all small models (toys, really) are always going to be thermally very inefficient. Worse.... because they are small models and not laboratory instruments, we have no way of actually measuring the inefficiencies that we know are there. So we must guess that they are there and that they are "large". On that point we must agree, I think.
For my part, I accept that. In other words, as far as small models are concerned, I accept that they are very inefficient. But I also see that as positive and beneficial. Why? Because they work despite what must (presumably) be their very poor efficiency. So, for example, if someone had as their personal objective the task of improving the efficiency of a particular design (as used in a particular model), then there is scope to do so by definition - you can always get some improvement if you work at it. But only for that one particular model.
It is perhaps just a small point, but that is why I said that the temperature gradient and all the various (and large) thermal inefficiencies of that model are not "bad" things but actually can be viewed as "good" because it still works despite all the heat losses stacked against it.
I see that you prefer to focus on the fundamentals of what makes Stirling engines work and that you consider anything else (thermal gradients etc) as
undesirable wasting of heat. So lets look at that ........
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Alphax
Re: Ted Warbrooke's Stirling 1: Question
Hi Tom
Can I explore what you said?
You said:-
The only thing powering a Stirling engine is heat entering and causing the expansion of the working gas.
That is roughly half right, but you have also got to consider the work output by the engine and the contraction of the working fluid. The problem is that the Stirling engine is a closed cycle system with a fixed quantity of working fluid. So it is true that heat enters the engine and causes the expansion (and heating) of the working gas - exactly as you say. But it is also true that having achieved its expansion then working fluid then has to contract in volume (and cool, having done work) to complete the cycle ready to pick up the next quantity of heat and repeat the cycle.
So, the working fluid itself experiences Volume changes and Temperature changes but does not change its Mass (which goes on unaltered from one cycle to the next).
During a sequence of cycles, the working fluid temperature changes through time from hot to cold to hot to cold and so on. The temperature changes are not step changes, they take time and are "smeared out" and smooth. They are gradients with respect to time (and also gradients with respect to position within the internal engine space because of volume changes). The gradients themselves within the working fluid are transients.
Therefore it is inevitable that the working fluid itself experiences transient thermal gradients during each cycle - even in a theoretical, ideal engine - because that is how it works.
Because the working fluid is in intimate contact with the walls of the engine it cannot avoid transferring some heat to those walls. That is the nub of the problem because the working fluid itself is experiencing continual changes of temperature during the completion of each cycle. So the walls of the engine will also experience changes of temperature. This also is inevitable. I think you recognise this too because you say:-
A heat gradient along the length of a pipe does not contribute to the expansion of a gas within the pipe, it is, rather, the effect of heat loss the effect of which is to drain power and reduce the expansion of the gas. Obviously, I agree with that statement.
Where things become interesting is that I sense that you believe that such losses are not inevitable and that they can somehow be avoided or reduced.
What I don't understand is how you think they might be avoided or reduced? I'd like too, though.
Incidentally, the function of the regenerator can be thought of as a step towards reducing such losses by trying to establish the "least worst" thermal gradients for the working fluid to pass through during each cycle.
Can I explore what you said?
You said:-
The only thing powering a Stirling engine is heat entering and causing the expansion of the working gas.
That is roughly half right, but you have also got to consider the work output by the engine and the contraction of the working fluid. The problem is that the Stirling engine is a closed cycle system with a fixed quantity of working fluid. So it is true that heat enters the engine and causes the expansion (and heating) of the working gas - exactly as you say. But it is also true that having achieved its expansion then working fluid then has to contract in volume (and cool, having done work) to complete the cycle ready to pick up the next quantity of heat and repeat the cycle.
So, the working fluid itself experiences Volume changes and Temperature changes but does not change its Mass (which goes on unaltered from one cycle to the next).
During a sequence of cycles, the working fluid temperature changes through time from hot to cold to hot to cold and so on. The temperature changes are not step changes, they take time and are "smeared out" and smooth. They are gradients with respect to time (and also gradients with respect to position within the internal engine space because of volume changes). The gradients themselves within the working fluid are transients.
Therefore it is inevitable that the working fluid itself experiences transient thermal gradients during each cycle - even in a theoretical, ideal engine - because that is how it works.
Because the working fluid is in intimate contact with the walls of the engine it cannot avoid transferring some heat to those walls. That is the nub of the problem because the working fluid itself is experiencing continual changes of temperature during the completion of each cycle. So the walls of the engine will also experience changes of temperature. This also is inevitable. I think you recognise this too because you say:-
A heat gradient along the length of a pipe does not contribute to the expansion of a gas within the pipe, it is, rather, the effect of heat loss the effect of which is to drain power and reduce the expansion of the gas. Obviously, I agree with that statement.
Where things become interesting is that I sense that you believe that such losses are not inevitable and that they can somehow be avoided or reduced.
What I don't understand is how you think they might be avoided or reduced? I'd like too, though.
Incidentally, the function of the regenerator can be thought of as a step towards reducing such losses by trying to establish the "least worst" thermal gradients for the working fluid to pass through during each cycle.
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Alphax
Re: Ted Warbrooke's Stirling 1: Question
Tom,
I'm unfamiliar with how to use the quote-function on this site - how do I avoid that horrible yellow stripe and get the neat fawn coloured box that you seem to manage so well?
I'm unfamiliar with how to use the quote-function on this site - how do I avoid that horrible yellow stripe and get the neat fawn coloured box that you seem to manage so well?
Re: Ted Warbrooke's Stirling 1: Question
I often resort to either copy/paste or just manually typing in the bb code
Code: Select all
[quote] Text here [/quote]
If just quoting the whole post, use the quote " button at the top right of the post being responded to to reply, instead of the "Reply" button.
Re: Ted Warbrooke's Stirling 1: Question
Must agree?It seems to me that all small models (toys, really) are always going to be thermally very inefficient. Worse.... because they are small models and not laboratory instruments, we have no way of actually measuring the inefficiencies that we know are there. So we must guess that they are there and that they are "large". On that point we must agree, I think.
No. Actually I emphatically contest all of that.
A small engine is not somehow automatically less efficient just because they are small
We certainly can measure efficiencies in a small engine as in any engine. If we do not "know" there is no reason to resort to guesswork, and there is no reason to assume any engine is efficient or inefficient without any actual measurement or experimental evidence
Well, looking on the bright side, there is room for making improvements, but an inefficiency being "beneficial"? Not as it stands IMO., I accept that they are very inefficient. But I also see that as positive...
.... and beneficial
My car blew a head gasket Looking on the bright side, it could be repaired, but the blown head gasket is not itself beneficial to the functioning of the engine. It may still limp along, blowing smoke and steam and backfiring, it could be worse!
That is the real crux of the issue, and has already been the subject of endless debate going back more than ten years:having achieved its expansion the working fluid then has to contract in volume (and cool, having done work) to complete the cycle ready to pick up the next quantity of heat and repeat the cycle.
viewtopic.php?f=1&t=478
Your sentence there, however is ambiguous.
Does the working fluid cool on its own accord, as a consequence of having expanded while doing work, (ie. Heat converted to work output) or does the expanded gas need to be cooled by the actual removal of the heat that was added, that is; cooled by being brought into contact with a sink?
The conservation of energy would suggest it must be one or the other but cannot be both simultaneously for any given quantity of heat.
If I add 1000 joules of heat to the engine, and the gas expands producing 1000 joules of work output, in what state does that leave the engine?
The heat has already been converted and has left the system as work output, so does not need to be removed to any "sink".
Heat sinks in heat engines are therefore unnecessary.
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Alphax
Re: Ted Warbrooke's Stirling 1: Question
Tom,
I am surprised.
Regarding whether size has any bearing on thermal efficiency, one of the things that matters is the surface area over which the working fluid has contact with the inside of the engine in relation to the volume of working fluid itself. This is a characteristic number called the working fluid surface-to-volume ratio (or SA:V) and it has dimensions of inverse length (L to the minus one). Length in this context means linear scale length (how big or how small the engine is in any direction). The surface area of the working fluid matters because it is the primary route by which Stirling engine working fluid "leaks" heat out into the environment before that heat can get the chance to do useful work elsewhere within the engine.
Consider two versions of exactly the same design but built to different sizes (one small one, one large one). The smaller one has a working fluid with a greater SA:V than the larger one (because surface area goes as L squared but volume goes as L cubed). Therefore if, whilst running, the engines are losing heat to the environment (which they all do), then the one with the largest SA:V factor working fluid (i.e. the smaller one) will lose more heat relative to the volume of its working fluid than the larger one will relative to its volume of working fluid.
Since loss of heat to the environment is one of the factors affecting engine efficiency, and since loss of heat is disproportionally (relative to volume of working fluid) worse in smaller motors, it follows that if the only variable being considered is size, then smaller motors are inherently less efficient than larger ones, all other factors being equal.
In other words, thermal efficiency scales adversely with decreasing size.
Another way of looking at this is to say that larger engines have a working fluid that is more easily able to retain heat within the engine to do useful work at the appropriate part of the cycle than smaller engines - thermal efficiency scales beneficially with increasing size.
I am surprised.
Regarding whether size has any bearing on thermal efficiency, one of the things that matters is the surface area over which the working fluid has contact with the inside of the engine in relation to the volume of working fluid itself. This is a characteristic number called the working fluid surface-to-volume ratio (or SA:V) and it has dimensions of inverse length (L to the minus one). Length in this context means linear scale length (how big or how small the engine is in any direction). The surface area of the working fluid matters because it is the primary route by which Stirling engine working fluid "leaks" heat out into the environment before that heat can get the chance to do useful work elsewhere within the engine.
Consider two versions of exactly the same design but built to different sizes (one small one, one large one). The smaller one has a working fluid with a greater SA:V than the larger one (because surface area goes as L squared but volume goes as L cubed). Therefore if, whilst running, the engines are losing heat to the environment (which they all do), then the one with the largest SA:V factor working fluid (i.e. the smaller one) will lose more heat relative to the volume of its working fluid than the larger one will relative to its volume of working fluid.
Since loss of heat to the environment is one of the factors affecting engine efficiency, and since loss of heat is disproportionally (relative to volume of working fluid) worse in smaller motors, it follows that if the only variable being considered is size, then smaller motors are inherently less efficient than larger ones, all other factors being equal.
In other words, thermal efficiency scales adversely with decreasing size.
Another way of looking at this is to say that larger engines have a working fluid that is more easily able to retain heat within the engine to do useful work at the appropriate part of the cycle than smaller engines - thermal efficiency scales beneficially with increasing size.
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Alphax
Re: Ted Warbrooke's Stirling 1: Question
So, Tom, can you give a convincing argument as to why you think that a small engine is not somehow automatically less efficient just because it is smaller than an otherwise identical, but larger, version?
I ask because I think that a smaller engine is automatically less efficient just because it is smaller than an otherwise identical, but larger, version. And I have given my arguments (previous post) to explain why I think that.
I am prepared to be shown I am wrong on this point, but need to understand the reason if that be the case. Thanks.
I ask because I think that a smaller engine is automatically less efficient just because it is smaller than an otherwise identical, but larger, version. And I have given my arguments (previous post) to explain why I think that.
I am prepared to be shown I am wrong on this point, but need to understand the reason if that be the case. Thanks.
Re: Ted Warbrooke's Stirling 1: Question
As far as the inevitability of heat transfer to the inner walls of the engine, I don't buy that such is really inevitable. I think it could be minimized or very nearly eliminated entirely, in a number of ways, and I'm thinking this would be desirable.
But the mindset for over a hundred years has been that some "gradient", cooling phase, heat loss, heat dissipation, "cold reservoir", "sink" or whatever is an absolute requirement that cannot and should not be avoided, so we invent cooling systems, fins, water jackets, long pipes, etc. to facilitate that necessary function.
That is, it seems, a difficult mind set to abandon, the idea is so deeply ingrained and accepted without question.
But, is there any reason a transfer tube carrying PRESSURE to a piston needs to be very long so as to dissipate all it's heat before getting to the piston?
The heat IS the pressure! That's just two different names, or ways of thinking about the same kinetic energy.
Leaking out the heat along the pipe is loosing pressure before it can get to and push the piston.
The pipe could be kept short as possible and be made of a non heat conducting/insulating material.
But instead, we do the opposite!
We would not imagine that there is any benefit to conducting electricity through an uninsulated wire that is partially grounded out along it's entire length, so why do we imagine that heat conducted through a pipe to deliver pressure to push a piston needs to lose heat along the way to produce a gradient?
It is like saying that not only is voltage drop and short circuiting of a wire unavoidable, but it is necessary and wires should be clad with a conductor instead of insulation in order to help facilitate the process. Slow running, inefficient motors are just a given. Short circuits are beneficial!
But the mindset for over a hundred years has been that some "gradient", cooling phase, heat loss, heat dissipation, "cold reservoir", "sink" or whatever is an absolute requirement that cannot and should not be avoided, so we invent cooling systems, fins, water jackets, long pipes, etc. to facilitate that necessary function.
That is, it seems, a difficult mind set to abandon, the idea is so deeply ingrained and accepted without question.
But, is there any reason a transfer tube carrying PRESSURE to a piston needs to be very long so as to dissipate all it's heat before getting to the piston?
The heat IS the pressure! That's just two different names, or ways of thinking about the same kinetic energy.
Leaking out the heat along the pipe is loosing pressure before it can get to and push the piston.
The pipe could be kept short as possible and be made of a non heat conducting/insulating material.
But instead, we do the opposite!
We would not imagine that there is any benefit to conducting electricity through an uninsulated wire that is partially grounded out along it's entire length, so why do we imagine that heat conducted through a pipe to deliver pressure to push a piston needs to lose heat along the way to produce a gradient?
It is like saying that not only is voltage drop and short circuiting of a wire unavoidable, but it is necessary and wires should be clad with a conductor instead of insulation in order to help facilitate the process. Slow running, inefficient motors are just a given. Short circuits are beneficial!
Re: Ted Warbrooke's Stirling 1: Question
I think your reasoning and mathematics analysis is flawed.Alphax wrote: ↑Sun Feb 06, 2022 10:51 am Tom,
I am surprised.
Regarding whether size has any bearing on thermal efficiency, one of the things that matters is the surface area over which the working fluid has contact with the inside of the engine in relation to the volume of working fluid itself. This is a characteristic number called the working fluid surface-to-volume ratio (or SA:V) and it has dimensions of inverse length (L to the minus one). Length in this context means linear scale length (how big or how small the engine is in any direction). The surface area of the working fluid matters because it is the primary route by which Stirling engine working fluid "leaks" heat out into the environment before that heat can get the chance to do useful work elsewhere within the engine.
Consider two versions of exactly the same design but built to different sizes (one small one, one large one). The smaller one has a working fluid with a greater SA:V than the larger one (because surface area goes as L squared but volume goes as L cubed). Therefore if, whilst running, the engines are losing heat to the environment (which they all do), then the one with the largest SA:V factor working fluid (i.e. the smaller one) will lose more heat relative to the volume of its working fluid than the larger one will relative to its volume of working fluid.
Since loss of heat to the environment is one of the factors affecting engine efficiency, and since loss of heat is disproportionally (relative to volume of working fluid) worse in smaller motors, it follows that if the only variable being considered is size, then smaller motors are inherently less efficient than larger ones, all other factors being equal.
In other words, thermal efficiency scales adversely with decreasing size.
Another way of looking at this is to say that larger engines have a working fluid that is more easily able to retain heat within the engine to do useful work at the appropriate part of the cycle than smaller engines - thermal efficiency scales beneficially with increasing size.
To simplify, take a cube. Sure, as you say: "surface area goes as L squared but volume goes as L cubed"
A cube however has six sides.
A large cube has exactly the same surface to volume ratio as a small cube. The same applies to a large or small engine of identical design
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Alphax
Re: Ted Warbrooke's Stirling 1: Question
Tom, I'm sorry, but that isn't correct at all.
Your statement that "A large cube has exactly the same surface to volume ratio as a small cube" is totally and utterly wrong. I'm sorry if that sounds confrontational, but there is no easy way to say it - sorry about that.
A worked example will prove that they do not have the same surface to volume ratio if they are of different sizes but the same shape. This fact holds true for ALL three dimensional objects, including engines.
Consider 2 cubes: Cube 1 is smaller than Cube 2.
CUBE 1: Has a side length of 5 cm. It has six identical sides, each is 5cm x 5cm and so the surface area = 6x5x5 cm squared = 150 sq cm.
CUBE 1: Its volume is therefore 5 x 5 x 5 = 125 cu cm.
CUBE 1: Its SA:V is therefore 150/125 = 1.2 cm to power -1
CUBE 2: Has a side length of 10 cm. It has six identical sides, each is 10cm x 10cm and so the surface area = 6x10x10 cm squared = 600 sq cm.
CUBE 2: Its volume is therefore 10 x 10 x 10 = 1000 cu cm.
CUBE 2: Its SA:V is therefore 600/1000 = 0.6 cm to power -1
SO..... firstly, two cubes of different sizes DO NOT have the same SA:V.
Secondly the SA:V for the smaller cube (Cube 1) is larger than for the larger cube (Cube 2) - it is 1.2 compared to 0.6.
Thirdly, heat would therefore be lost more quickly from Cube 1 in relation to its volume than Cube 2 in relation to its volume.
The analysis and maths is not flawed, it is correct as presented. But of course it is only one argument in considering the question (of whether smaller engines are inherently less thermally efficient).
HOWEVER.... what I was hoping for was for me to try to understand your belief that a smaller engine isn't less efficient just because it is smaller. Can you explain what you mean?
Your statement that "A large cube has exactly the same surface to volume ratio as a small cube" is totally and utterly wrong. I'm sorry if that sounds confrontational, but there is no easy way to say it - sorry about that.
A worked example will prove that they do not have the same surface to volume ratio if they are of different sizes but the same shape. This fact holds true for ALL three dimensional objects, including engines.
Consider 2 cubes: Cube 1 is smaller than Cube 2.
CUBE 1: Has a side length of 5 cm. It has six identical sides, each is 5cm x 5cm and so the surface area = 6x5x5 cm squared = 150 sq cm.
CUBE 1: Its volume is therefore 5 x 5 x 5 = 125 cu cm.
CUBE 1: Its SA:V is therefore 150/125 = 1.2 cm to power -1
CUBE 2: Has a side length of 10 cm. It has six identical sides, each is 10cm x 10cm and so the surface area = 6x10x10 cm squared = 600 sq cm.
CUBE 2: Its volume is therefore 10 x 10 x 10 = 1000 cu cm.
CUBE 2: Its SA:V is therefore 600/1000 = 0.6 cm to power -1
SO..... firstly, two cubes of different sizes DO NOT have the same SA:V.
Secondly the SA:V for the smaller cube (Cube 1) is larger than for the larger cube (Cube 2) - it is 1.2 compared to 0.6.
Thirdly, heat would therefore be lost more quickly from Cube 1 in relation to its volume than Cube 2 in relation to its volume.
The analysis and maths is not flawed, it is correct as presented. But of course it is only one argument in considering the question (of whether smaller engines are inherently less thermally efficient).
HOWEVER.... what I was hoping for was for me to try to understand your belief that a smaller engine isn't less efficient just because it is smaller. Can you explain what you mean?
Re: Ted Warbrooke's Stirling 1: Question
If I take a stick and cut it to the length of the side of a cube, call that s for stick, the surface area is
s2 x 6 or 6(s2)
The internal area is s3 or "stick length" cubed.
So the ratio is 6(s2) : s3
And will always be 6(s2) : s3 regardless of how big or small the cube may be relative to anything else.
s2 x 6 or 6(s2)
The internal area is s3 or "stick length" cubed.
So the ratio is 6(s2) : s3
And will always be 6(s2) : s3 regardless of how big or small the cube may be relative to anything else.
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Alphax
Re: Ted Warbrooke's Stirling 1: Question
No....Tom, that isn't right. I think you may have a "blind spot" regarding ratios of variables with different exponents (powers of 10).
The algebraic value of your expression is always constant - as you correctly point out - simply because it is a true algebraic expression for any cube of any size. But the numerical value of the ratio as soon as you replace the term 's' in the expression with a real number to represent the size of stick (cube) ALWAYS results in a different value in the ratio equal to 6(s2):s3 when a different value of 's' is used. That is in the nature of algebraic expressions which have different exponents (2 and 3 in this case) in the denominator and numerator. The expression itself is the same but different values of 'S' give different values of the ratio because of the different powers of 10 in the numerator and denominator. Sorry, but that is maths.
Just calculate the numerical value for 6(s2):s3 using s=5 and then compare it with s=10. You get two different outcomes (1.2 and 0.6), despite the fact that the algebraic expression used (as you correctly derive) remains the same.
I'm sorry I can't make it any plainer, but why not ask someone to explain it to you?
Alternatively, look up any on-line surface area to volume calculator for basic 3D shapes and input numbers for 2 different sized cubes (or indeed spheres, cones etc) and you'll find that you've somehow missed the point here . Just one at random you can use (there are others) is here:-
https://www.omnicalculator.com/math/sur ... lume-ratio
I am truly sorry if I seemed a little confrontational on this point but it isn't really open to any interpretation at all - your algebraic formula (R=6(s2):s3) is PERFECTLY CORRECT for cubes, but for some reason you are unable to substitute two different values for 's' (such as s= 5 then s=10) and see that the two resulting ratios are in fact very different to each other (by a factor of 2 in this case), even though the algebraic expression which you have derived is correct and perfectly applicable in both the case where a cube has 5cm side and another cube has 10 cm side (or indeed ANY size at all!!). Different values of 's' give different answers using the same expression 6(s2):s3.
As I say, it seems that on this instance you are struggling with the arithmetic but not the algebra!
Beyond that I don't think I can help! But hopefully the penny will drop in the morning and you'll suddenly "get it" and reach the well known and well understood result that different sized cubes have different surface-area to volume ratios. I'm so sorry to have to put that to you, I really am and take no pleasure in trying to explain it to you, but I'm afraid there is not any doubt.
Thanks!
The algebraic value of your expression is always constant - as you correctly point out - simply because it is a true algebraic expression for any cube of any size. But the numerical value of the ratio as soon as you replace the term 's' in the expression with a real number to represent the size of stick (cube) ALWAYS results in a different value in the ratio equal to 6(s2):s3 when a different value of 's' is used. That is in the nature of algebraic expressions which have different exponents (2 and 3 in this case) in the denominator and numerator. The expression itself is the same but different values of 'S' give different values of the ratio because of the different powers of 10 in the numerator and denominator. Sorry, but that is maths.
Just calculate the numerical value for 6(s2):s3 using s=5 and then compare it with s=10. You get two different outcomes (1.2 and 0.6), despite the fact that the algebraic expression used (as you correctly derive) remains the same.
I'm sorry I can't make it any plainer, but why not ask someone to explain it to you?
Alternatively, look up any on-line surface area to volume calculator for basic 3D shapes and input numbers for 2 different sized cubes (or indeed spheres, cones etc) and you'll find that you've somehow missed the point here . Just one at random you can use (there are others) is here:-
https://www.omnicalculator.com/math/sur ... lume-ratio
I am truly sorry if I seemed a little confrontational on this point but it isn't really open to any interpretation at all - your algebraic formula (R=6(s2):s3) is PERFECTLY CORRECT for cubes, but for some reason you are unable to substitute two different values for 's' (such as s= 5 then s=10) and see that the two resulting ratios are in fact very different to each other (by a factor of 2 in this case), even though the algebraic expression which you have derived is correct and perfectly applicable in both the case where a cube has 5cm side and another cube has 10 cm side (or indeed ANY size at all!!). Different values of 's' give different answers using the same expression 6(s2):s3.
As I say, it seems that on this instance you are struggling with the arithmetic but not the algebra!
Beyond that I don't think I can help! But hopefully the penny will drop in the morning and you'll suddenly "get it" and reach the well known and well understood result that different sized cubes have different surface-area to volume ratios. I'm so sorry to have to put that to you, I really am and take no pleasure in trying to explain it to you, but I'm afraid there is not any doubt.
Thanks!