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I find a lot of this electronic stuff interesting, but one big concern of mine is finding a material on which distances can be measured as a function of electrical change - resistance, phase, or something else.
As far as I know you were asking three questions - some which may be even more useful than the one posed in your last post - if I have identified the posts correctly. These three questions were about the resistance range, the distance, and the usefulness of a three phase power supply. Did I miss anything?
Resistance range: I'd like to experiment with two at this time: about 0-2,000 ohms, and about 0-10,000 ohms.
Distance: Say, two feet - in a sheet, not a wire.
Usefulness of three phase power supply: I can not give you an informative response at this time.
Again, my questions were about what type of material would be useful for measuring those distances as a function of those resistance ranges. And, how could the use of AC be useful for measuring distances as a function of electricity in a sheet.
I could give you a sheet 2 foot square of carbon loaded plastic. If you measure the distance from the middle of opposite edges, using a multimeter probe, it might measure 1M. If you repeated the measurement by fixing an aluminium strip along the full length of opposite edges then it might read 1k.
Whay because when you measured it with the meter probe there was only a small surface area in contact with it, and when you measured it with the foil there was a large surface area in contact with it.
You haven't even said what thickness it needs to be.
First you need to know what resistivity you require, if you don't know you can calculate it.
About the paths that electrons might take from one point electric contact to another on a sheet, has anyone seen a model of how these paths - describing direction and rate of electron movement, change as the contacts are made and then removed?
Would someone tell me if I am accurately inferring from the formula R = pL/(tW) - at Sheet resistance - Wikipedia, the free encyclopedia, that all of the lower orbitals in molecules in a sheet are filled before the conduction band? Does L mean the length of the sheet or the distance between electrical connections? Can this formula be used to describe resistance when the contacts are not on the edges of the sheet? Are the connections point connections or connections that resemble a line or an edge? For what dimensions of sheets are the formulas at https://en.wikipedia.org/wiki/Sheet_resistance good for?
I'm also interested in a database that includes sheets that are 2 feet by 2 feet and can be used to measure distances between two point electrical contacts when one of the electrical contacts is moved from between 0 to 2 feet anywhere on the board in terms of resistance between about 0 and 25 ohms.
Also, I may have made at least one mistake in previous posts. First, I think that electons do not travel as fast as charge does in DC. Second, perhaps I should have said that I am presently interested in a sheet in which resistivity is uniform - not resistance - if this even makes a difference. Third, perhaps I should have said that at the same time the direction of current in AC was approximately uniform - and not that the current was uniform throughout the sheet - if this is in fact not the case.
I could give you a sheet 2 foot square of carbon loaded plastic. If you measure the distance from the middle of opposite edges, using a multimeter probe, it might measure 1M. If you repeated the measurement by fixing an aluminium strip along the full length of opposite edges then it might read 1k.
Whay because when you measured it with the meter probe there was only a small surface area in contact with it, and when you measured it with the foil there was a large surface area in contact with it.
You haven't even said what thickness it needs to be.
First you need to know what resistivity you require, if you don't know you can calculate it.
I'm still working on a lot of this stuff. Have you got types of sheets memorized - or do you have a source for identifying sheet types that you can share with me? I don't even know how to use some of these formulas - and I asked some questions that you might be able to help me with in my last post.
I'm also interested in a database that includes sheets that are 2 feet by 2 feet and can be used to measure distances between two point electrical contacts when one of the electrical contacts is moved from between 0 to 2 feet anywhere on the board in terms of resistance between about 0 and 25 ohms.
I can sympathize wth ya, bud...I myself am trying to design a pair of shoes that I can use to walk on the sun without burning my feet and, I'm taking your approach. I'm trying to do it while making sure that I learn nothing about either the sun or shoes.
People can be so stupid, can't they. I mean these "geniuses" here can't even come up with a simple database of materials that are suitable for sun-walking.
I can sympathize wth ya, bud...I myself am trying to design a pair of shoes that I can use to walk on the sun without burning my feet and, I'm taking your approach. I'm trying to do it while making sure that I learn nothing about either the sun or shoes.
People can be so stupid, can't they. I mean these "geniuses" here can't even come up with a simple database of materials that are suitable for sun-walking.
Have you become involved in your project up to a point that you have considered one or more attributes of the material or materials that may be involved in your project?
Have you become involved in your project up to a point that you have considered one or more attributes of the material or materials that may be involved in your project?
Have you become involved in your project up to a point that you have considered one or more attributes of the material or materials that may be involved in your project?
I suggest you read the Wikipedia article and make sure you fully understand the concept of resistivity before going any further.
I'll try to explain it to you briefly.
Think of a piece of wire, the thicker the wire, the lower the resistance, the sorter the wire the lower the resistance and the the longer the wire the higher the resistance, the thinner the wire the higher the resistance. The thickness and length of the wire are both directly proportional to the resistance.
To work out the resistance we need the resistivity of the material the wire is made of.
For the purpose of resistivity calculations we measure the wire thickness in cross-sectional area. A wire with a cross-sectional area of 1m² will have an area of 1m² when viewed from the end. You can calculate the thickness by transposing a = πr² formula everyone should have learnt at school.
Resistivity is measured in Ω·m, suppose a piece of wire 1m long with a cross-sectional area of 1m² and a resistance of 1Ω, in this case it can said to have a resistivity of 1Ω·m. If the length were doubled (2m) or the cross-sectional area halved (0.5m²) then the resistance would also double (2Ω) yet the resistivity of the material would be still 1Ω·m.
If you have understood what I've just explained, you should be able to calculate the resistivity of the material you require, if you know the dimensions of the sheet and the required resistance. All the formulae you need is in the Wikipedia article.
I can sympathize wth ya, bud...I myself am trying to design a pair of shoes that I can use to walk on the sun without burning my feet and, I'm taking your approach. I'm trying to do it while making sure that I learn nothing about either the sun or shoes.
People can be so stupid, can't they. I mean these "geniuses" here can't even come up with a simple database of materials that are suitable for sun-walking.
I suggest you read the Wikipedia article and make sure you fully understand the concept of resistivity before going any further.
I'll try to explain it to you briefly.
Think of a piece of wire, the thicker the wire, the lower the resistance, the sorter the wire the lower the resistance and the the longer the wire the higher the resistance, the thinner the wire the higher the resistance. The thickness and length of the wire are both directly proportional to the resistance.
To work out the resistance we need the resistivity of the material the wire is made of.
For the purpose of resistivity calculations we measure the wire thickness in cross-sectional area. A wire with a cross-sectional area of 1m² will have an area of 1m² when viewed from the end. You can calculate the thickness by transposing a = πr² formula everyone should have learnt at school.
Resistivity is measured in Ω·m, suppose a piece of wire 1m long with a cross-sectional area of 1m² and a resistance of 1Ω, in this case it can said to have a resistivity of 1Ω·m. If the length were doubled (2m) or the cross-sectional area halved (0.5m²) then the resistance would also double (2Ω) yet the resistivity of the material would be still 1Ω·m.
If you have understood what I've just explained, you should be able to calculate the resistivity of the material you require, if you know the dimensions of the sheet and the required resistance. All the formulae you need is in the Wikipedia article.
Its still taking me a while to figure this one out. At Sheet resistance - Wikipedia, the free encyclopedia, if I understand the forumula, I can substitute Wt for A. However, doing so is not intuitively clear to me. How can the resistance offered by a material have nothing to do with where the electrical contacts are made on it? I might be totally misunderstanding the formula. I am not clear specifically what L means in the formulas R=pl/A=pL/(Wt).
Where could L - in R= pL/A = pL/(Wt) - at https://en.wikipedia.org/wiki/Sheet_resistance - if I can figure out how to do the conversion, be defined on a sheet having dimensions of pie feet squared by 1/8 of an inch?
I could define trying in this context in at least two different ways. First, considering the factors involved. Second, doing so within a lifetime. And maybe there is a third possibility - comparing scenarios.
Its still taking me a while to figure this one out. At Sheet resistance - Wikipedia, the free encyclopedia, if I understand the forumula, I can substitute Wt for A. However, doing so is not intuitively clear to me. How can the resistance offered by a material have nothing to do with where the electrical contacts are made on it? I might be totally misunderstanding the formula. I am not clear specifically what L means in the formulas R=pl/A=pL/(Wt).
Sheet resistance is no different to the resistivity of a cable, just replace the cross-sectional area with W×t, where W is the width and t is the thickness.
Please attempt the following problems:
I have a sheet 2mm thick, 50mm wide and 2m long and is made of a material with a resistivity of 2.5mΩ·m, what's the resistance of the sheet?
I need a sheet 5mm thick, 100mm wide and 0.5m long with a resistance of 1k, what resistivity material do I need?
Note you'll need to convert all the prefixes to base units: i.e. replace mΩ with 0.001Ω, and mm with 0.001m before doing any calculations otherwise you'll go wrong.
I could define trying in this context in at least two different ways. First, considering the factors involved. Second, doing so within a lifetime. And maybe there is a third possibility - comparing scenarios.
Note you'll need to convert all the prefixes to base units: i.e. replace mΩ with 0.001Ω, and mm with 0.001m before doing any calculations otherwise you'll go wrong.
Actually, that wont help too much. What he needs to convert is his mind away from his childish, pointless game. There's probably not much point in suggesting that he try to convert it to anything involving electronics or technology since it's obvious that the only interest he has in any of that is in how to use it to continue his ploy.
I admit it. I was wrong. I thought he'd pretty well exhausted the range of piddly, pointless ploys to keep his game alive. But, this one was his next move. We should learn from it.
Of course, he's not going to let on that he's trying to actually understand anything but, he's now going out and doing searches on the internet to come up with some semi_related, esoteric crap that he can pretend that he needs assistance with. Ohm's Law is too "useful" so, he throws out some formulas he found on Wiki as his way to cry out for his next his next round of attention.
Okay, you gave hime a simple problem and some prompts. Let's see if he manages to actually come up with the correct answer (BTW: that grinding noise you are hearing is the rusty wheels turning in his head as he figures out how to make sure that the correct answer is the last thing he will come up with).
He has vectored around my predictions in the past but, I don't think he can vector around this one (at least not by doing what he knows is the right thing) without tipping his hand.
Sheet resistance is no different to the resistivity of a cable, just replace the cross-sectional area with W×t, where W is the width and t is the thickness.
Please attempt the following problems:
I have a sheet 2mm thick, 50mm wide and 2m long and is made of a material with a resistivity of 2.5mΩ·m, what's the resistance of the sheet?
I need a sheet 5mm thick, 100mm wide and 0.5m long with a resistance of 1k, what resistivity material do I need?
Note you'll need to convert all the prefixes to base units: i.e. replace mΩ with 0.001Ω, and mm with 0.001m before doing any calculations otherwise you'll go wrong.
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