## Resistance

Discussions on introductory science topics. Ask simple or beginner questions and expect clear and level answers.

### Re: Resistance

I guess that, had I thought about it, I might have said that an electric circuit powered by a multi-cell battery, with one cell sulfated, would come close to the pipe type circuit.

But, yes, pipes are a poor analogy unless you do a whole bunch of finagling with the elements. Truth is, most analogies don't work because voltage is energy per Coulomb, and that is a case not seen elsewhere in ordinary life, or at least I think. I learned electric circuits long before I learned anything about fluids, so I had the experience of using electricity as an analogy to understand hydraulics and gas dynamics. I actually found that helped.

### Re: Resistance

CanadysPeak » April 27th, 2014, 1:47 am wrote:Truth is, most analogies don't work because voltage is energy per Coulomb, and that is a case not seen elsewhere in ordinary life, or at least I think.

Actually, Canadys, any "potential" is Energy per unit.
Pressure is energy per unit volume.
Chemical potential (diffusional), i.e. (RT/F).log(concentration), is energy per particle, and for ions in solution it works exactly as electric potential (see Nernst equation).
Height from ground is energy per mass unit (given gravity).

And for ideal systems - i.e. neglecting frictions, particle-particle interactions, considering stationary conditions (no accelerations), etc - all these yield flow equations similar to Ohm's law.

Dave's conclusion above about all resistance being "not additive, the smallest aperture (Highest Resistance) is virtually the only controlling factor" is probably true for pipes, because the resistance along the pipe is much smaller (totally irrelevant) with respect to the resistance of the smaller aperture.
If you put a 1 kOhm resistance in series with a 1 MOhm, you don't see any additivity, even though that's electric stuff, unless you have a very sensitive gauge. And if you add 20 feet copper wire you also do not see any difference.
However, when you consider COMPARABLE resistances, they are additive in hydraulic systems as well, and there is no reason why they would not be.

neuro
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### Re: Resistance

neuro » Sun Apr 27, 2014 7:23 am wrote:
CanadysPeak » April 27th, 2014, 1:47 am wrote:Truth is, most analogies don't work because voltage is energy per Coulomb, and that is a case not seen elsewhere in ordinary life, or at least I think.

Actually, Canadys, any "potential" is Energy per unit.
Pressure is energy per unit volume.
Chemical potential (diffusional), i.e. (RT/F).log(concentration), is energy per particle, and for ions in solution it works exactly as electric potential (see Nernst equation).
Height from ground is energy per mass unit (given gravity).

And for ideal systems - i.e. neglecting frictions, particle-particle interactions, considering stationary conditions (no accelerations), etc - all these yield flow equations similar to Ohm's law.

Dave's conclusion above about all resistance being "not additive, the smallest aperture (Highest Resistance) is virtually the only controlling factor" is probably true for pipes, because the resistance along the pipe is much smaller (totally irrelevant) with respect to the resistance of the smaller aperture.
If you put a 1 kOhm resistance in series with a 1 MOhm, you don't see any additivity, even though that's electric stuff, unless you have a very sensitive gauge. And if you add 20 feet copper wire you also do not see any difference.
However, when you consider COMPARABLE resistances, they are additive in hydraulic systems as well, and there is no reason why they would not be.

Well, I'll have to think a bit about all this, but gravitational potential energy is not per kg.

And, I do understand the comparable problem. I was being a bit disingenuous when I posed that situation, and you've caught me up. Nonetheless, I'll try to zero in on what I think to be the nub.

In electricity, it seems important (at least to me) that one understand voltage as being defined by the gradient of the electric field, and it be path independent. If I want to try to create the closest analogy to those two criteria that I can think of, I need to avoid the situation where I get pressure drop as a function of length (which then is path dependent). Thus, I dislike pipes. Moreover, I do prefer analogies that can actually be demonstrated, and closed pipe flow has to be really finagled in order to get it to look like Ohms Law. You can do it, but you have to be somewhat less than candid.

I'm not saying you're wrong. I'm simply saying that conductance (or admittance, if you will) is the conceptually sound way of looking at current. I have never seen the water flow in pipe analogy be helpful to anyone in understanding electricity. To me, electricity seems so easy that I can't imagine needing help to conceptualize it.

### Re: Resistance

CanadysPeak » April 27th, 2014, 1:33 pm wrote:In electricity, it seems important (at least to me) that one understand voltage as being defined by the gradient of the electric field, and it be path independent.

I feel terribly pedantic, but though the field is not path-dependent, current flow is, if the path is resistive.

Again in hydraulic terms, you can consider a linear pressure gradient (from a basin at 100 feet height to sea level). Now, consider a straight, large pipe or a winding, narrow pipe along the same linear pressure gradient: water flow in the two pipes will differ only due to the different specific resistance (per meter in the constant pressure gradient, i.e. per meter of height, not per meter of pipe length).

Just to try and share my "enthusiasm" for the analogy, this is how I try to explain resistive-capacitive behavior of a cell membrane to my students (as you may know, bology and medicine students hate electricity and anything which sounds mathematical or physical [or possbly this only apply to Italian students])

To me, electricity seems so easy that I can't imagine needing help to conceptualize it.

Do you really mean that? even for capacitive and inductive aspects?

neuro
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### Re: Resistance

Neuro,

I will answer the bulk of your reply a bit later; I must run out just now. But, capacitive and inductive effects seem so simple. Perhaps I missed something, cause I do know those puzzle a lot of students.

I will venture, in a quick glance, that you may have hit on the key to using the pipe analogy, i.e., separating the velocity head and the friction head. More later.

### Re: Resistance

Hi neuro,

neuro wrote:If you put a 1 kOhm resistance in series with a 1 MOhm, you don't see any additivity, even though that's electric stuff, unless you have a very sensitive gauge.

You were doing fine till this statement. Electrical resistance is extremely additive. Any half decent digital ohm meter will easily see the RT sum of 1Meg and 1K. I have a 8 decade resistance box that allows you to dial up any precise resistance up to 8 digits of accuracy. I can dial 1234567.8 ohms if I wish.

This additive aspect is very critical for accurate ADC and DAC functions used in almost any signal analog/digital processing.

My point earlier was simply that because water can't be compressed (more or less) the pressure is distributed evenly in my example (C) for chambers X,Y,Z. It doesn't require a pressure gradient to flow through the individual chambers. Water is pushed through (at the Speed of Stick) to replace missing water and the main pressure gradient will only be found at the extreme small and local aperture between Inside and Outside the System. If you radially widen that exit aperture then the next smallest aperture becomes dominant in a series pipe situation.

This is demonstrated by there being no real difference between Tank (B) and Tank (C) setup in actual flow over time.

Best to all
Dave :^)

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### Re: Resistance

I'm coming out like a nit-picking pedant, and that's not fun, but somebody has to point this stuff out. If, in order to teach electricity, you use a water flow analogy where you have to say that water is incompressible, what do we do if a student hears that and believes it?

One of the saving graces is that most students pay no attention to what we say to them. Anyway, as I said, it's no damn fun being the wet blanket; use the analogy.

### Re: Resistance

CanadysPeak wrote:One of the saving graces is that most students pay no attention to what we say to them.

Your cynicism is only surpassed by your experience of the real world, my friend. Where would we be without education, which is only marginally better than its alternative? Heaven forbid we should allow the hoi polloi to do their own thinking.

Regards Leo
Obvious Leo

### Re: Resistance

Dave_Oblad » April 27th, 2014, 9:49 pm wrote:My point earlier was simply that because water can't be compressed (more or less) the pressure is distributed evenly in my example (C) for chambers X,Y,Z. It doesn't require a pressure gradient to flow through the individual chambers. Water is pushed through (at the Speed of Stick) to replace missing water and the main pressure gradient will only be found at the extreme small and local aperture between Inside and Outside the System. If you radially widen that exit aperture then the next smallest aperture becomes dominant in a series pipe situation.

Why do you keep saying there is no gradient?
If no pressure gradient exists between two points ther will be no water flow.
What would move preferentially each water molecule in a direction rather than the opposite one?

There is "no" gradient where there is virtually "no" resistance.
You are talking about a system which can be exmplified by a series of pieces of copper wire separated by finite resistors (orders of magnitude greater than copper wire), and say that the system is "dominated" by the smallest aperture: it would be the same in the electrical equivalent, provided you do not have a 6-digits Ohmeter or Amperometer.

But then, also in the hydraulic system, if you had a 6-digits manometer you would see the pressure drop across each aperture, and even the pressure gradient along the pipe. [actually you don't need the manometer to tell you there is a pressure gradient: the fact itself that water moves means that there are asymmetrical forces acting on each water molecule, i.e. there is a pressure gradient]

You cannot talk about "dominating" factors in hydraulics and "6-digit precision measurements" in electricity. Either we talk about first order approximation in both systems or we talk about precise behavior. If we choose the latter perspective the hydraulic analogy certainly does breaks down, but that, I believe, is the limit of any "analogy". If t were the same it would be the same, not an analogy....

By the way:
I may well be wrong, but the fact that water is "nominally incompressible" is a good analogy to Kirkoff's law, isn't it?
You cannot see a current flow from A to B unless exactly the same current leaves B to flow somewhere else...

neuro
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### Re: Resistance

Obvious Leo » Sun Apr 27, 2014 11:37 pm wrote:
CanadysPeak wrote:One of the saving graces is that most students pay no attention to what we say to them.

Your cynicism is only surpassed by your experience of the real world, my friend. Where would we be without education, which is only marginally better than its alternative? Heaven forbid we should allow the hoi polloi to do their own thinking.

Regards Leo

Leo,

Here's the thing that bothers me. It's not just this particular analogy, but I see it time after time in science. Let's say you tell a student that he can model equivalent resistance by putting two hoses in series or in parallel. Now, suppose that student actually likes science and learning, so he goes home and tries this. Now, nobody owns a bunch of precision manometers, so the student is gonna probably figure out to measure what he can, i.e. current flow rate.

So, he hooks a 50 foot, 3/4" hose to the faucet and fills a five gallon bucket while timing the process. let's just say he gets 28 seconds.

So, he gets another 50 foot, 3/4" hose and hooks it onto the first, and repeats the experiment, thinking he'll now get 56 seconds. Sumnabitch, it takes 33 seconds! He figures he messed up, so he repeats the test, but he gets 37 seconds, 31 seconds, and so on. Never close to 56 seconds.

Well, as long as he's got the hoses out, he tries the parallel test. He snakes one around the corner to the other faucet, so now he's got two 50 foot, 3/4" hoses hooked to their respective faucets, and he thinks this should take 14 seconds. Well, it's not so bad as the first trial. this time, it takes 18 seconds. He tries again, getting 16 seconds, and 18 seconds again.

Now, you and I know why this works this way, but the student has only managed to learn to never, ever believe anything he is told in school. And, he still doesn't understand about equivalent resistance. By the way, I put equivalent resistance in italics because that's what gets cut in half, not resistance.

Cynical? Moi?

### Re: Resistance

I'm rather fond of analogies as an educative tool, even if they aren't strictly speaking precise. They force you to look at a question in a slightly different way and actually think about what's going on instead of simply learning a rule and accepting it. As DragonFly would say, "neurons that fire together wire together", and you might regard the process of thinking through an analogy as one of opening up different neural pathways. This is neuro's business and not mine but you get what I mean. You never know when these different neural pathways might come in handy, so analogies enhance our lateral thinking in a more general kind of way.

As a non-engineer I've found this thread interesting because the exchanges between blokes who clearly know ten times more than I do on the subject have literally forced me to think about resistance in electrical circuits in a different way, and as a result I now definitely have a deeper understanding of a concept which I felt I already understood. To be honest I've never given the matter a thought since my high school days when I was tinkering with ham radio projects from "Electronics for boys", because my interests later took me in a different direction, but I'm reassured to discover that even a decrepit old fart like me is not such an old dog that he can't learn a new trick.

Regards Leo
Obvious Leo

### Re: Resistance

Hi all,

neuro wrote:Why do you keep saying there is no gradient? If no pressure gradient exists between two points there will be no water flow.

Wrong: In a closed pipe, water pushes water through the pipe. The gradient only needs to be located at the exit. This was my "Speed of Stick" reference. If you push on one end of the stick, the same pressure will appear at the other end of the stick. (given a stiff stick..lol)

I decided to seek out an Expert and have exchanged a few emails. He has also visited this thread. He had this to say:

Donald E. Simanek, Emeritus Prof. of Physics, Lock Haven University of Pennsylvania wrote:In your example of three constricted pipes, I ask this: If you had one system with three narrow pipe segments of length x butted together in series (equivalent to one narrow pipe of length 3x), and another system with only one segment of length x (the systems otherwise identical), would they have the same flow rate? If there's no significant dissipative processes like laminar flow and viscosity, the answer should be "yes" for the pressure difference between reservoir and output is the same in both systems. Now if the three segments were separated as in your diagram, should that matter? No.

There are of course minor issues in water flow through pipe such as:
Water Viscosity becomes an issue in extremely narrow pipes.
Flow resistance along the pipe interior surface.
Various eddy currents based on flow Geometries.
Atmosphere saturation in water allows some compressibility.

But I have said the Aperture(s) are all I care about. So make the connecting pipes Teflon. Connecting Pipes are large enough to ignore viscosity. De-saturate the air out of the water and ignore eddy currents as they are symmetrical between identical chambers (more or less).

This leave two major players in water flow. Static Pressure and Dynamic Pressure. Static pressure is like the atmospheric pressure on every square inch of your body. Dynamic pressure is like holding you hand out the window of a moving car and feeling the air pressure push your hand.

Now:
In example Tank (C) no water flow leaves the Tank that doesn't also leave the aperture exit after chamber (Z). Thus the Flow through all 4 restrictions must be identical in volume. Thus the Dynamic Pressure (Flow) in each chamber of (X,Y,Z) are all identical.

The Static Pressure in all three chambers will be identical as set by the pressure difference from the Tank to the Exit. There is nothing blocking Static Pressure between the 3 chambers.

Therefore..
The Static and Dynamic pressure in all three chambers will be identical.
There is no additive Resistance from Tank to chamber (X), (X) to (Y), and (Y) to (Z).

Thus: The identical restrictions in example (C) are not Additive.
Thus: Example Tank(B) is identical to Example Tank(C).

If the Resistance to flow along the interior walls of the restrictions is bothering you, then replace them with extremely thin barriers such as a plastic playing card with a 1/2 inch hole in the centers in all 4 positions of example Tank(C) and the exit in example Tank(B).

Sorry so long getting back here, but wanted to consult an Expert before going further.

Best wishes,
Dave :^)

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### Re: Resistance

Dave_Oblad wrote:Atmosphere saturation in water allows some compressibility.

Dave. I realise you're eliminating this variable from your experimental scenario but could you elaborate on what this actually means?

Regards Leo
Obvious Leo

### Re: Resistance

Hi Leo,

Sure.. we pump air into an Aquarium so fish can breath. The water absorbs the air. When we boil water, the air expands and boils out of solution. Air is compressible and would thus add compressibility to the water. Water, without atmospheric impurity, is virtually non-compressible. But given the scale of my examples, would play almost no role in pressure distribution within the Compartments of X,Y,Z in my Tank(C) example.

Neuro is not completely wrong, but his examples come from the human body, which is pressure regulated with arteries that have elasticity, among other complex attributes. My example is extremely rigid and super simple. The Gradient Neuro seeks is there in my example, but is located in a very small portion at the very end of the last exit aperture. It is Not a Stepped Gradient through each chamber.

Put another way, a pressure meter inserted into all three cells of (X,Y,Z) would all read the same. Which brings us full circle to the statement that Series Resistance in water pipes is not Additive. Thus series water pipes fail in the analogy for electrical series resistance.

Hope this helps,

Dave :^)
Last edited by Dave_Oblad on April 30th, 2014, 8:32 pm, edited 1 time in total.

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### Re: Resistance

It does indeed. I just need to think of aerated water and it all becomes crystal clear. The more gaseous particles that the water contains the more compressible it is because it forces the gas out. I should have been able to work that out for myself, Dave, but thanks for your indulgence.

Regards Leo
Obvious Leo

### Re: Resistance

The matter is that facts count.

Ordinary water is compressible. Just because we have been told in school that it is not doesn't alter the facts. If water were truly incompressible, most of the oil refineries in the world would close and nobody would be able to build large naval ships, or even some tanks for that matter. When you see 6" tungsten cut like a topiary, that's a black swan event.

Water flows downhill. There are a number of downhills possible. We can use height, velocity, density, but downhill it is. We call that downhill a pressure gradient. Zero pressure gradient is called a pond. If we were to posit that water flows just fine without a pressure gradient, then how would we make an analogy to Ohms law where a voltage gradient results in charge flow?

### Re: Resistance

While you are technically correct.. water is generally considered non-compressible.

Wiki wrote:The bulk modulus of water is 2.2 GPa. The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.

For the pressure levels we are dealing with in my examples, it's fine to consider water as incompressible.

Water downhill is not a pressure gradient exactly. A pressure gradient is what I would expect to find in a swimming pool, the deeper I go. Each additional foot deeper I descend, the greater the pressure becomes. This would be a true Pressure Gradient.

From the two statements above.. suppose I was to glue a bathroom scale to a wall and hand you a 10 foot steel rod. I ask you to hold one end of the rod and place the other end of the rod against the now vertically inclined scale and push on the rod. I asked you to apply 10 pounds of pressure at your end of the rod, which is horizontal to the ground and the pressure plate of the scale it perpendicular to the rod, so you can see the weight of the pressure you are applying. You apply 10 pounds of pressure to the rod and see the scale read 10 pounds. Now... where is the pressure gradient in this example? Do you believe you have to apply 20 pounds of pressure on your end of the rod.. to see the scale show 10 pounds?

Of course not. There is NO gradient along that Rod. This is the same as the water in the horizontal pipe.

The Gradient only exists at the physical interface between the pressure of the water in a pipe and the pressure outside the pipe (atmospheric pressure). If the exit aperture from the water pipe is a lot smaller than the diameter of the water pipe, then virtually all the Gradient is located in that very small area around the aperture. You won't see a pressure gradient along a horizontal pipe.

However.. I suppose one could make a model of series resistance if we inclined the pipe. Then I would have no choice but agree we would see a pressure gradient down the length of the water pipe, because now it is an inclined tank (per se). In this case extra pipe would add extra resistance in terms of pressure and the added pipe sections would be additive. So as a model, perhaps this works ok, sort of. (except for the back-wash effect.. but hey.. no model is perfect...lol)

Regards,
Dave :^)

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### Re: Resistance

While you are technically correct.. water is generally considered non-compressible.

Wiki wrote:The bulk modulus of water is 2.2 GPa. The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.

For the pressure levels we are dealing with in my examples, it's fine to consider water as incompressible.

Water downhill is not a pressure gradient exactly. A pressure gradient is what I would expect to find in a swimming pool, the deeper I go. Each additional foot deeper I descend, the greater the pressure becomes. This would be a true Pressure Gradient.

From the two statements above.. suppose I was to glue a bathroom scale to a wall and hand you a 10 foot steel rod. I ask you to hold one end of the rod and place the other end of the rod against the now vertically inclined scale and push on the rod. I asked you to apply 10 pounds of pressure at your end of the rod, which is horizontal to the ground and the pressure plate of the scale it perpendicular to the rod, so you can see the weight of the pressure you are applying. You apply 10 pounds of pressure to the rod and see the scale read 10 pounds. Now... where is the pressure gradient in this example? Do you believe you have to apply 20 pounds of pressure on your end of the rod.. to see the scale show 10 pounds?

Of course not. There is NO gradient along that Rod. This is the same as the water in the horizontal pipe.

The Gradient only exists at the physical interface between the pressure of the water in a pipe and the pressure outside the pipe (atmospheric pressure). If the exit aperture from the water pipe is a lot smaller than the diameter of the water pipe, then virtually all the Gradient is located in that very small area around the aperture. You won't see a pressure gradient along a horizontal pipe.

However.. I suppose one could make a model of series resistance if we inclined the pipe. Then I would have no choice but agree we would see a pressure gradient down the length of the water pipe, because now it is an inclined tank (per se). In this case extra pipe would add extra resistance in terms of pressure and the added pipe sections would be additive. So as a model, perhaps this works ok, sort of. (except for the back-wash effect.. but hey.. no model is perfect...lol)

Regards,
Dave :^)

With the steel rod, a steel rod isolated like that looks like a superconductor. Suppose we put the steel rod in a tightly fitting sleeve so that there is some friction between the sleeve and the rod. Now, let's have the rod push the scale at a constant velocity (this is gedanken, so we can arrange that). You may measure 10 pounds on the scale, but you will have to push with 10,3 pounds, for example. It is the two conditions of motion and friction that necessitate a gradient.

Wike may call water incompressible, and that's fine. If a banker or a streetcar conductor believes that, all is good. But, if an engineer believes that we're in deep dodo.

### Re: Resistance

Ok.. technically everything is compressible.. even Diamond. But I made my point and an expert agreed.. so that's good enough for me. Anyway, while we are on the subject, here is an interesting problem.

But first some back-story. After High school I went to Devry Institute to get a BS Degree in Electronics. I did real well 1st semester.. straight "A"s. I was TA in almost every class, helping slower students. College was keeping me from be drafted as my number was up. 2nd Semester was going ok until I butted heads with my analytical Math Professor.

He was teaching methods to solve complex problems but I kept looking for and finding short-cuts. I always got the right answer, but seldom used the right method. But almost none of the other classmates ever even got the right answer, despite their using the correct methods. His grading was:
A) Right Method and Right Answer
B) Right Method but Wrong Answer
C) Wrong Method but Right Answer
D) Wrong Method and Wrong Answer

So I coasted along with a B- average grade. Then one day some professor offered this problem to the school. It was posted in every classroom, anybody could try to solve it. So I noted the problem, saw how to solve it and solved it.. in about 2 minutes. I wanted to be first to solve it, so I was working on it when the Professor noticed me doing it. He said: "Haw! You solve that and I'll give you an "A" for the course." So I doubled my efforts and did all the Math long-hand, no slide rule. Figured if I didn't solve it first, I may still win the contest by accuracy. At the end of class I raced to the Office and submitted my answer. Turns out I was first after all. Then was told we had 1 month to solve it... Crap!

So about 50 students and 10 Professors (including my nemesis) submitted Entries. Can you guess who won? Me of course. I got a moment in the spot-light and a small article in the local newspaper. My big mistake was flaunting my victory to my Math Professor. He said to me privately, that he would see to it I would never graduate from that school. He used every trick he could to reduce my grade. In the end, about half of the class failed, including me. I had a B- Average but required a solid B.

I complained to the Dean, but he said sorry, not his place to second guess the Professor. I confronted the Professor and asked about his promise.. he said tough, I refused to learn the correct methodology. I pointed out that his reasoning was irrational. Would he rather trust his life to someone that always used the correct Method but never got the right answer or me, who always got the right answer, regardless of Method. That just made him madder and I got expelled. Just as well, I would never be able to pass his course with him in charge, no matter how far I bent over. So without a Degree and no College Deferment, I got drafted and sent to Vietnam.. almost got killed.. but made it home safe. Pursued Electronics anyway, from the bottom via OJT. Wasn't easy, but Business cared more about Productivity than Honors, once you get past the hiring prejudices.

A few years later, when I had made Senior R&D Engineer, I had the opportunity to select new-hires. None of the students from Devry (from that time) could solve some rather Basic Math Problems. But they knew the Dot Pitch in an Epson LA30 Printer. Go Figure?

So, here is the Problem that caused me so much grief:

All values in Ohms. Solve for RT.

Catch ya later my friend....

Best Regards,
Dave :^)

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### Re: Resistance

Forgot to mention.. not only was I first to hand in the answer.. mine was the only correct answer submitted.

Meaning I beat my Math Professor in his own field. That was rather key to his negative treatment of me. He was the first Academic Bully I had ever met. I've been bullied all my life. Physical in School, Academic in College, and Rank in Military and Business. I felt compelled to prove myself to everyone since that Professor.

So, perhaps he did me a favor.

But I changed when my wife passed away in 2003. It broke me spiritually. Now I am just trying to coast to the end. I am done fighting. Although this place has awakened some of that competitive spirit again.. so I am kinda liking it here.

Best wishes ol man,
Dave :^)

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### Re: Resistance

Ok.. technically everything is compressible.. even Diamond. But I made my point and an expert agreed.. so that's good enough for me. Anyway, while we are on the subject, here is an interesting problem.

But first some back-story. After High school I went to Devry Institute to get a BS Degree in Electronics. I did real well 1st semester.. straight "A"s. I was TA in almost every class, helping slower students. College was keeping me from be drafted as my number was up. 2nd Semester was going ok until I butted heads with my analytical Math Professor.

He was teaching methods to solve complex problems but I kept looking for and finding short-cuts. I always got the right answer, but seldom used the right method. But almost none of the other classmates ever even got the right answer, despite their using the correct methods. His grading was:
A) Right Method and Right Answer
B) Right Method but Wrong Answer
C) Wrong Method but Right Answer
D) Wrong Method and Wrong Answer

So I coasted along with a B- average grade. Then one day some professor offered this problem to the school. It was posted in every classroom, anybody could try to solve it. So I noted the problem, saw how to solve it and solved it.. in about 2 minutes. I wanted to be first to solve it, so I was working on it when the Professor noticed me doing it. He said: "Haw! You solve that and I'll give you an "A" for the course." So I doubled my efforts and did all the Math long-hand, no slide rule. Figured if I didn't solve it first, I may still win the contest by accuracy. At the end of class I raced to the Office and submitted my answer. Turns out I was first after all. Then was told we had 1 month to solve it... Crap!

So about 50 students and 10 Professors (including my nemesis) submitted Entries. Can you guess who won? Me of course. I got a moment in the spot-light and a small article in the local newspaper. My big mistake was flaunting my victory to my Math Professor. He said to me privately, that he would see to it I would never graduate from that school. He used every trick he could to reduce my grade. In the end, about half of the class failed, including me. I had a B- Average but required a solid B.

I complained to the Dean, but he said sorry, not his place to second guess the Professor. I confronted the Professor and asked about his promise.. he said tough, I refused to learn the correct methodology. I pointed out that his reasoning was irrational. Would he rather trust his life to someone that always used the correct Method but never got the right answer or me, who always got the right answer, regardless of Method. That just made him madder and I got expelled. Just as well, I would never be able to pass his course with him in charge, no matter how far I bent over. So without a Degree and no College Deferment, I got drafted and sent to Vietnam.. almost got killed.. but made it home safe. Pursued Electronics anyway, from the bottom via OJT. Wasn't easy, but Business cared more about Productivity than Honors, once you get past the hiring prejudices.

A few years later, when I had made Senior R&D Engineer, I had the opportunity to select new-hires. None of the students from Devry (from that time) could solve some rather Basic Math Problems. But they knew the Dot Pitch in an Epson LA30 Printer. Go Figure?

So, here is the Problem that caused me so much grief:

Cube.jpg

Catch ya later my friend....

Best Regards,
Dave :^)

So, I just tried it, guessing that you must have used some sort of intuitive approach rather than equations. So I didn't write out any equations, but just pretended to be a squad of electrons marching up to the junction and then imagining how many could go each way in the same time (oh, I turned it sideways because my glasses are crooked - old farts common complaint). I got 6/5 Ohm (I wasted a lot of time thinking up a creative way to do it; the addition took 2 or 3 seconds). Later, if I ever find a pencil, I'll draw a highway interchange and try the same thing, or maybe some water pipes (I do think this is a good place to use water pipes). Or maybe Pac-Men? Good problem. I have no idea if my answer is right - don't really care - cause I'm more interested in how to think about problems like this.

By the way, I must share a similar story in which I was hoist on my own petard. I used to teach Physics labs (yesterday was my last - turned in my keys - too damn old for Facebook and clickers). I tried to find questions and puzzles to get people thinking. So I made a cylindrical arrangement of small capacitors, apparently with lots of interconnections, all held on a 3D plastic frame. I asked the equivalent capacitance (not as a a class requirement, but just as a puzzle for good students). Most people took about 30 to 60 minutes to write out all the equations and solve. One student took about 30 seconds, yet got the right answer. Turns out he has no 3D visualization skills, so he misread what was there, solved it completely wrong, but still got the answer.

Reminds me of the old boy down in Alabama back in the 60s. He had a tv repair shop back when they still fixed tv sets. The guy was completely illiterate - couldn't even read the word SAMS on a schematic, but he had a good sense of color, so he would look for components whose color didn't seem right, change them out, and most of the time it worked. Since heat was the biggest culprit back then, I sort of believe him, but. . .?

I enjoy arguing with people I respect. Little of what I say makes any real difference in the larger picture, but it keeps my mind active.

Yeah, I think water pipes might be my next try at the problem.

### Re: Resistance

Forgot to mention.. not only was I first to hand in the answer.. mine was the only correct answer submitted.

Meaning I beat my Math Professor in his own field. That was rather key to his negative treatment of me. He was the first Academic Bully I had ever met. I've been bullied all my life. Physical in School, Academic in College, and Rank in Military and Business. I felt compelled to prove myself to everyone since that Professor.

So, perhaps he did me a favor.

But I changed when my wife passed away in 2003. It broke me spiritually. Now I am just trying to coast to the end. I am done fighting. Although this place has awakened some of that competitive spirit again.. so I am kinda liking it here.

Best wishes ol man,
Dave :^)

BTW, never make a professor feel foolish. When I went to college, they had an art core requirement. Of course I put that off till my last semester. I was on the GI bill and out of time to repeat any semesters, so I needed to get everything done on time. The course went sorta OK. We looked at some Reynolds, and I liked those. I was enthusiastic about Matisse, and raved about Monet.

Then the prof put up a slide of a Jackson Pollock and asked me (out of a room of maybe a hundred) what i thought. I may have confused my tobacco pouches that morning as I answered honestly that it looked like the guy had gone to the hardware store, bought a can of red spray paint and then done a bad job of painting the canvas. Getting a full head of steam, I allowed as how I wouldn't pay \$2 for a piece of art that ugly.

Guess who owned that ugly art? Guess who had paid thousands of dollars for that? Guess who did tons of extra credit work, and kissed a lot of backside in order to get a "D-"?

### Re: Resistance

A salutary lesson for all contrarians, Canady. Speaking your mind does not come without a cost, but if you can back it up with a sound argument you must do it nevertheless and be willing to accept the consequences. However your pragmatic contrarian has a quick look around to see who's listening, just in case.

By the way I love Jackson Pollock and also Monet. Not so much Matisse, so to each his own I guess.

Regards Leo
Obvious Leo

### Re: Resistance

hi Dave
I feel obliged to submit my guess, though I do not know what R.T. stays for:
imagining you are asking what is the overall resistance between the two points at the left of the figure, my result is 7/6 Ohm.
total current is V.6/7
from the left to the right, you have
V/2 A across the first 2 Ohm resistor
V/7 A across the second
V/14 A across the third (V/28 coming from each 1+1 Ohm path)
and V/7 A across the fourth
for a total of V.12/14 = V.6/7

more in detal:
V.5/28 A across each 1 Ohm and 3 Ohm directly connected
of which V/7 goes across the 2 Ohm
and V/28 goes trough the more distant 1(3) Ohm
and, together with the mirroring path, carries
V/14 total current across the third 2-Ohm

========================

Is having a half a centimeter dameter aperture at the end of a 10 feet long 4" pipe the same as having a 10 feet long, half a centimeter diameter pipe?

If you quite convincedly respond YES, then I feel content, and imagine that viscosity and blood vessel elasticity account for additive seres resistance and parallel conductance in the circulatory system.

However, if you say NO, then something is wrong in your reasoning.

neuro
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Location: italy

### Re: Resistance

neuro » Fri May 02, 2014 5:05 pm wrote:hi Dave
I feel obliged to submit my guess, though I do not know what R.T. stays for:
imagining you are asking what is the overall resistance between the two points at the left of the figure, my result is 7/6 Ohm.
total current is V.6/7
from the left to the right, you have
V/2 A across the first 2 Ohm resistor
V/7 A across the second
V/14 A across the third (V/28 coming from each 1+1 Ohm path)
and V/7 A across the fourth
for a total of V.12/14 = V.6/7

more in detal:
V.5/28 A across each 1 Ohm and 3 Ohm directly connected
of which V/28 goes trough the more distant 1(3) Ohm
and, together with the mirroring path, carries
V/14 total current across the third 2-Ohm
and V/7 across the 2 Ohm

========================

Is having a half a centimeter dameter aperture at the end of a 10 feet long 4" pipe the same as having a 10 feet long, half a centimeter diameter pipe?

If you quite convincedly respond YES, then I feel content, and imagine that viscosity and blood vessel elasticity account for additive seres resistance and parallel conductance in the circulatory system.

However, if you say NO, then something is wrong in your reasoning.

You got 7/6, I got 6/5. I think I must have counted wrong. Next time I'll take off my shoes and use my toes. :>)

### Re: Resistance

Hi Neuro,

Neuro wrote:Is having a half a centimeter dameter aperture at the end of a 10 feet long 4" pipe the same as having a 10 feet long, half a centimeter diameter pipe?

Yes.. if the pipes are made of Teflon.. in other words.. we disregard surface contact resistance inside the pipe.

Also, your answer to the Resistor Cube problem is also correct. RT stands for Resistance Total. Congrats!!!

The final calculation is 5.6 / 4.8 ohms = 1.16666666666 Ohms.

When I get a few minutes... I'll post the short-cut. I have to draw it to communicate it.

Best wishes,
Dave :^)

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Posts: 3219
Joined: 08 Sep 2010
Location: Tucson, Arizona
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### Re: Resistance

This is how I figure it

neuro
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Posts: 2621
Joined: 25 Jun 2010
Location: italy

### Re: Resistance

Neuro,

I'm no expert, but I know the answer to this:

Is having a half a centimeter dameter aperture at the end of a 10 feet long 4" pipe the same as having a 10 feet long, half a centimeter diameter pipe?"

That's a definite "No."

If you need an expert, I'll try to remember the name of someone who understands apertures; it's buried somewhere back in that tangled mess of neurons I fondly call "the mind I lost."

### Re: Resistance

Hi everyone,

I promised a breakdown of how I solved the problem presented above. Let's start with this:

Simple bridge network

When I test job applicants, this is always on my test. Their version doesn't have labels RX, A or B on it to draw attention to this aspect. Looking at the symmetry of the 2 ohm Resistors, one must ask themselves.. what affect does the 4 ohm resistor have on the RT. Strangely enough, Hobbyists knock out the answer in less than 5 seconds and move on. College educated people struggle with it and sometimes ask for a Calculator. WTF? I have no idea why, but I seen it time and time again. Anyway, I use this as a litmus test to see how intuitive an applicant is.

So.. If you replace RT with a battery, you will see the voltage potential at point A and B are identical. Thus no current will flow through RX.. Thus RX could be open or a short circuit and have no effect on RT.

So, when I saw this same symmetry in the Resistance Puzzle presented.. I did the same thing as shown below:

Problem simplification steps

Upper left is the original with shorting bars on Bridge A and Bridge B. This creates the equivalent circuit to its right. That one breaks down to the circuit in lower left.. then lower center.. then to lower right.. where the final equation is:
(2*2.8)/(2+2.8) = 5.6/4.8 = 1.166666... ohms.

Anyway, I am not by any means a Math Whiz. But I do have a talent for seeing into puzzles and solving them. So I have done alright in my chosen profession.

I found my Expert by searching the net for websites examining/explaining Hydrostatic/Hydrodynamic problems and contacted him via email. Your last answer "NO' is still wrong. A 1/2 inch aperture will still drain a tank just as fast, regardless of its distance from the tank.. assuming the aperture is the same elevation in all tests, no smaller restrictions are applied anywhere inline and we ignore Friction inside the pipe of the pipe/water contact area involved. (Hence my suggestion of using Teflon piping) Also, no bends in the pipe as that would introduce Hydrodynamic issues. And.. we aren't talking about a long hose where we need to accelerate the water mass in the pipe. Have I left anything out?

Ok.. that about wraps this up for me. It's been fun.

Best to all,
Dave :^)

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Posts: 3219
Joined: 08 Sep 2010
Location: Tucson, Arizona
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### Re: Resistance

Ah, now you want to ignore friction in the pipes. I need you in my neighborhood where we are paying a pot of money to get people to line the mains to lower the friction. Then, I'm gonna tell you about Reynolds Number.

But, in spite of the fact that I went to college (I did flunk high school, if that helps), I'm an experimenter at heart. I need to think of a good physical test I can set up.

b

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