Please answer the questions in my previous post; if you do so properly, then I'll provide more detail.
What does that mean?Distance: Say, two feet - in a sheet, not a wire.
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?
What does that mean?
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.
Resistivity - Wikipedia, the free encyclopedia
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.
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?
Deja vu. Read people's replies twice before replying.
What about my reply indicated that I didn't read it thoroughly?
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.
He's taking the p1ss.
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.
The fact that it is clearly a joke.
For your convenience:
What is clearly a joke to you? How does my reply qualify?
If you seriously believe crashsite is trying to make shoes that would allow him to walk on the surface of the sun...then that would explain a lot.
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).
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.
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.
**broken link removed**
[latex]R = \rho \frac{L}{A} = \rho \frac{L}{W t}[/latex]
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