BBC - 'Brain decline' begins at age 27

[Old Baldy]

Yes, the branching is current. No resistance in my simple example.

How does the current loop back in a perfect, infinite flat grid example? That's why I gave the "round ball" shape example. I can see a loop back in a round ball (i.e., 3 dimensional shape) but in a perfect flat shape with no depth there wouldn't be any loop back would there?

Maybe that is why some of us feel it is zero resistance. We envision the system as two dimensional?

 

[follicle84]

Well, the general idea is that if you start with a finite grid of resistors and then add more and more resistors to the outer edges of the grid so that the grid gets larger and larger, the total effective resistance across the one in the center will steadily drop. It will gradually go down and down and down, without any limit. So with an infinitude of resistors, it will reach zero (according to that point of view).

That was mind boggling stuff. The way im understanding it is by infinitelly increasing the number of resistors from the original resistor your speading the original resistors resistance across through the other connected resistors and there resistors infinitelly. Despite each resistor having its own resistance its still spred out across them all infinitelly, which would eventually bring the original resistance of the center resistor lower and lower until it reaches close to nothing (zero). Seen as all resistors are connected to each other from every point infinitelly the effect that occurs to the center resistor is the same effect for the other resistors connected to it and each other due to the infinite loop of resitors that keep getting added. Even then i find its difficult to just calculate the resistance as zero as it must have some sort of value despite the divide that is carried out infinitelly. Is this right?

 

[ali777]

what's the maximum number of years hair loss can reverse? Is it true frontal hair follicles die faster?

That was mind boggling stuff. The way im understanding it is by infinitelly increasing the number of resistors from the original resistor your speading the original resistors resistance across through the other connected resistors and there resistors infinitelly. Despite each resistor having its own resistance its still spred out across them all infinitelly, which would eventually bring the original resistance of the center resistor lower and lower until it reaches close to nothing (zero). Seen as all resistors are connected to each other from every point infinitelly the effect that occurs to the center resistor is the same effect for the other resistors connected to it and each other due to the infinite loop of resitors that keep getting added. Even then i find its difficult to just calculate the resistance as zero as it must have some sort of value despite the divide that is carried out infinitelly. Is this right?

yes and no...

the resistance of the branches has an increasing value... although infinite number of resistors in parallel would lead to zero, the quickly increasing resistance of the branches stops the value from reaching zero... so the value of the resistance is not zero...

the solution requires complex mathematical analysis. I know of Fourier transforms, but I didn't know they are used in circuit analysis.

The max number of years hairloss can reverse, in my opinion, is one hair cycle, ie at most 5 years.

I'm experiencing first hand that frontal hair follicles die faster. So, my answer is YES.

 

[Hammy070]

I think Bryan who is relatively senior is in effect showing us the way to prevent dementia or marble deficiency by way of problem solving. This should be one way among many, all faculties should be engaged. I like numerical challenges, and I also making art, cooking, playing the guitar and piano but also try and socialize often. I think the key is to renew often: learning a new instrument or language etc.

I have so many projects in the pipeline! From origami to martial arts. Few things bore me, I am very happy generally when I'm in the middle of something or planning it, I get restless and quite irritable when there's nothing to think about. It's probably ADHD related but I really pity those who live identical days for years on end. Anyway right now I'm having a coffee which I grinder myself and listening to Beethoven, two things I didn't do a few weeks ago.

I encourage everyone to look for things they wouldn't normally look for, it's just fun! YouTube Beethoven moonlight sonata, and lay back, it's heaven.

 

[barcafan]

I think Bryan who is relatively senior is in effect showing us the way to prevent dementia or marble deficiency by way of problem solving. This should be one way among many, all faculties should be engaged. I like numerical challenges, and I also making art, cooking, playing the guitar and piano but also try and socialize often. I think the key is to renew often: learning a new instrument or language etc.

I have so many projects in the pipeline! From origami to martial arts. Few things bore me, I am very happy generally when I'm in the middle of something or planning it, I get restless and quite irritable when there's nothing to think about. It's probably ADHD related but I really pity those who live identical days for years on end. Anyway right now I'm having a coffee which I grinder myself and listening to Beethoven, two things I didn't do a few weeks ago.

I encourage everyone to look for things they wouldn't normally look for, it's just fun! YouTube Beethoven moonlight sonata, and lay back, it's heaven.

Are you really worrying about the right things? at your age.

 

[Bryan]

That was mind boggling stuff. The way im understanding it is by infinitelly increasing the number of resistors from the original resistor your speading the original resistors resistance across through the other connected resistors and there resistors infinitelly. Despite each resistor having its own resistance its still spred out across them all infinitelly, which would eventually bring the original resistance of the center resistor lower and lower until it reaches close to nothing (zero). Seen as all resistors are connected to each other from every point infinitelly the effect that occurs to the center resistor is the same effect for the other resistors connected to it and each other due to the infinite loop of resitors that keep getting added. Even then i find its difficult to just calculate the resistance as zero as it must have some sort of value despite the divide that is carried out infinitelly. Is this right?

yes and no...

the resistance of the branches has an increasing value... although infinite number of resistors in parallel would lead to zero, the quickly increasing resistance of the branches stops the value from reaching zero... so the value of the resistance is not zero...

Exactly. Adding more and more branches to a finite grid of resistors will definitely lower the effective resistance of the one in the middle, but we can't just ASSUME that as we do that, the effective resistance approaches zero.

To the poster "follicle84": It's easy to think of examples where an infinite series "approaches" a non-zero (or non-infinite, for that matter) value. Here's a trivial example. Tell me what the following series approaches:

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 .....

Even though the series gets larger and larger with each additional step, clearly it "approaches" the value of exactly 2.0, not infinity. The same goes for the "infinite grid of resistors" puzzle. If you start with a finite grid and start adding extra resistors around the periphery, it's not immediately obvious whether the value across the resistor in the middle approaches zero, or some non-zero value.

 

[Old Baldy]

Bryan wrote in part: If you start with a finite grid and start adding extra resistors around the periphery, it's not immediately obvious whether the value across the resistor in the middle approaches zero, or some non-zero value

.

Yes, I see what you mean. Interesting!

Ali wrote in part: the resistance of the branches has an increasing value... although infinite number of resistors in parallel would lead to zero, the quickly increasing resistance of the branches stops the value from reaching zero... so the value of the resistance is not zero...

Yup, thanks for the explanation!

So, now what's the answer?

 

[Old Baldy]

I think Bryan who is relatively senior is in effect showing us the way to prevent dementia or marble deficiency by way of problem solving. This should be one way among many, all faculties should be engaged. I like numerical challenges, and I also making art, cooking, playing the guitar and piano but also try and socialize often. I think the key is to renew often: learning a new instrument or language etc.

I have so many projects in the pipeline! From origami to martial arts. Few things bore me, I am very happy generally when I'm in the middle of something or planning it, I get restless and quite irritable when there's nothing to think about. It's probably ADHD related but I really pity those who live identical days for years on end. Anyway right now I'm having a coffee which I grinder myself and listening to Beethoven, two things I didn't do a few weeks ago.

I encourage everyone to look for things they wouldn't normally look for, it's just fun! YouTube Beethoven moonlight sonata, and lay back, it's heaven.

I agree about a guzillion percent!

Just keep the mind active and try to learn new things.

I don't have an answer for Bryan's question. But it was interesting to try. I'm sure my old brain got used a little.

FWIW, I think the simple act of reading alot and thinking about what you've just read exercises the brain. We don't have to be "marathon" runners to hopefully prevent those dreaded brain diseases IMHO.

What's that old saying....."Idle hands are the devil's playground"? I think you could change that to "Idle brains are Alzheimer's playground".

 

[Bryan]

So, now what's the answer?

The answer (according to those sites ali777 found) is one-half the value of the individual resistors. For example, I originally stated the problem as an infinite grid of 100 ohm resistors. If you measured across any single one of them, you'd measure an effective resistance of 50 ohms.

 

[Old Baldy]

So, now what's the answer?

The answer (according to those sites ali777 found) is one-half the value of the individual resistors. For example, I originally stated the problem as an infinite grid of 100 ohm resistors. If you measured across any single one of them, you'd measure an effective resistance of 50 ohms.

I understand that, but is it correct in your opinion?

I ask this because I figured, hastily after reading your's and Ali's recent posts, that it might be 25 ohms of resistance due to the four "branches of "escape".

Am I missing the loopback part of this question? I'm having a hard time understanding that part of the problem. Or if it even exists?

By that I mean, I can envision a 25 percent loopback but can't get to a 50 percent loopback.

 

[Bryan]

It seems quite plausible to me. I'm still going to be thinking about those explanations that one guy gave for the derivation of the "one-half" answer.

 

[Old Baldy]

It seems quite plausible to me. I'm still going to be thinking about those explanations that one guy gave for the derivation of the "one-half" answer.

Bryan, I just edited my post asking another question on the loopback problem I'm having. Am I out in leftfield totally on my confusion?

 

[Bryan]

Bryan, I just edited my post asking another question on the loopback problem I'm having. Am I out in leftfield totally on my confusion?

I don't think YOU are the one who is confused! I think the guy who wrote that material is the confused one!

As I said earlier, I couldn't make much sense out of what he said, and ali777 said the same thing. But I'm going to take another look at it...

 

[Old Baldy]

Ok, thanks Bryan. You guys are better at this stuff than I am!!

 

[ali777]

1/2 is the correct answer... The very long mathematical proofs through Fourier transforms show that it's 1/2, so I'll just take it as correct.

With the 1/4 solution, there is two of them. So, if you add them you get 1/2, that's the current flowing through the resistor in question. So, its resistance must be 1/2.

 

[follicle84]

Exactly. Adding more and more branches to a finite grid of resistors will definitely lower the effective resistance of the one in the middle, but we can't just ASSUME that as we do that, the effective resistance approaches zero.

To the poster "follicle84": It's easy to think of examples where an infinite series "approaches" a non-zero (or non-infinite, for that matter) value. Here's a trivial example. Tell me what the following series approaches:

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 .....

Even though the series gets larger and larger with each additional step, clearly it "approaches" the value of exactly 2.0, not infinity. The same goes for the "infinite grid of resistors" puzzle. If you start with a finite grid and start adding extra resistors around the periphery, it's not immediately obvious whether the value across the resistor in the middle approaches zero, or some non-zero value.

I thought my concept was too simple and there was more too it i get your point now. Eventually the answer would reach 2 not infinity.

ali777 i seemed to have crossed threads on my previous post, but thanks for the response. I've corrected that post now.

More to the point of conversation im taking it some resistance is lost and gained as resistors are infinitelly connected to each. I still find it difficult to grasp the answer to the question as half. Im with old baldy on this one with 25% resistance. The only explanation for this answer i think of is that due to added and lost resistance from the resistors the resistance cant be the same due to the growing resistors connected to each other so its thought to be divided by a quater from every connected point from the resistors which them selves would have 25% resistance from further conectivity to other resistors from four sides. Since 25% resistance would now be the level of resistance for the resistors due to connectivity from four sides. A quater of a quater of resistance would be played back to the center resistor from one side of the resistor. Seen as the center resistor has four connected sides to other resistors it would recieve a quater of a quater resistance from its four points working out something like this 1/16 (resistance) * 4 = 4/16 (resistance) =1/4 (25% resistance overall) added to resistors already 25% resistance =50% resistance total. The rule would apply for every other resistor since they are binded in the same way infinite.

Even then that sounds too simple for an explanation. Ali777 how did you come up with there being two quarter solutions that are added together to form half? This problem would make a brilliant IQ question.

 

[ali777]

Even then that sounds too simple for an explanation. Ali777 how did you come up with there being two quarter solutions that are added together to form half? This problem would make a brilliant IQ question.

A. We have an infinite field. Inject 1A of current to one of the junctions, A. Because the field is infinite, we can safely assume that the the current will be equally divided in all the branches going out of the junction. Therefore the current going out on all the branches is 1/4A.

B. The same operation, but this time the current is coming out of the field. Let's say we draw 1A from one of the junctions. The current flowing through all the 4 branches coming into the junction is 1/4...

Now, we superposition A and B, like in the figure below. Superposition means adding the currents from the two calculations. So, current injected into A goes from A to B and is 1/4A. Likewise, the current drawn from B, goes from A to B and is 1/4A.... They are the same values, but they are two different currents... so the total is 1/4+1/4=1/2


Diagram: Each line is a resistor. A and B are junctions.

 ___________
|    |     |
|A___|B____|
|    |     |
|____|_____|

Get it???

 

[follicle84]

I guess i went off on one but at least i tried. Good explanation though it makes more sense now that i can see it illustrated thanks. How long did this brain tease take you to work out?

 

[Bryan]

A. We have an infinite field. Inject 1A of current to one of the junctions, A. Because the field is infinite, we can safely assume that the the current will be equally divided in all the branches going out of the junction. Therefore the current going out on all the branches is 1/4A.

Who on earth would EVER think that was a "safe" assumption?

B. The same operation, but this time the current is coming out of the field. Let's say we draw 1A from one of the junctions. The current flowing through all the 4 branches coming into the junction is 1/4...

The same issue: who would ever assume that all four branches are passing EXACTLY the same amount of current (1/4 amp)?

Now, we superposition A and B, like in the figure below. Superposition means adding the currents from the two calculations. So, current injected into A goes from A to B and is 1/4A. Likewise, the current drawn from B, goes from A to B and is 1/4A.... They are the same values, but they are two different currents... so the total is 1/4+1/4=1/2

Huh?? If there are four different branches, each one carrying exactly the same 1/4 amp of current (which itself is impossible to believe), the total is 1/4 + 1/4 + 1/4 + 1/4 = 1.

Get it???

No I don't.

 

[Old Baldy]

FWIW, follicle84 explained how I got to 25 ohms exactly the way I looked at it (i.e., 1/4 of 1/4 times 4 branches = 25 percent, or in this case 25 ohms).

Or, .25 times .25 equals .0625 resistance for each branch,
then, 0.625 times 4 branches equals .25, or 25 ohms total resistance in our problem. (This is how I envisioned the "loopback" part of the problem, [i.e., .25 times .25]?)

(Using my simple calculation way of looking at this problem, [i.e., which is probably way off in terms of really analyzing this problem], if you had eight branches you'd get 12.5 percent resistance...... and on, and on, and on.....with answers varying based on how many more/less branches you put in the problem.)

I just don't understand why I would multiply follicle84's and my end result by two Ali. But, like Bryan points out, I'm making assumptions all over the place. Sorry, I ain't the sharpest knife in the kitchen cabinet when it comes to this kind of stuff.

Suffice it to say - I don't really know the answer!!

Edit: follicle84: I like the way you used the term "playback". That makes it clearer to me than "loopback".