HELP! Does anybody know the resisitivity of 5mm diameter 24SWG constantan wire? Thanks !!!!!!!!!!!!!!?
966971
ok resistivity formula is:\n\nresistivity = [resistance x length]/ cross sectional area..\n\nyou didnt give me enough info to work with\n\nhope it helped
755125
LNA designed with a single transistor for RF with a smith chart and power circles, or a general amplifier design with op amps?\nNot having enough data from your question I would suggest you use a low noise op amp, it has the bias taken care of. \nLet me know if thats not the case
685980
most wire is marked as 600 volt that is the max voltage the insulation can handle before breakdown.\nI think you need the current handling ability of the wire, which is determined by the wire size. 14 AWG is 15 amp., 12 AWG is 20 Amp., 10 AWG is 30 AMP. any more than 30 amps you need a lot more information as it is dangers at any voltage or current.
780548
Normally wire-size measurements are in millimeters and gauge. It will be easier for you to communicate with who you are ordering from if you do the computations to change it to millimeters and then transfer that to gauge of wire.\n\nI suggest you go to a refiner like David Fell & Co. (see link below.) The should be able to supply you with any gauge wire in gold that you need.
297024
Nichrome (Nickel Chrome) wire is one of the more common alloys used as a heating element wire.\n\nIn a way it is correct that the higher the resistance, the more heat that will be generated.  Looking at Ohm's law, V = I * R and P = I * V.   P being power, which in a resistor is exhibited as heat.  P = I^2  * R  or P = V^2 / R\n\nWhat this is saying is that while although more resistance will produce more heat, you will also need to increase the voltage (which will increase the current) to get that heat.  So, more resistance is not the simple answer, you need to match the resistance with your electrical supply capabilities to optimize the heat output.\n\nIt would probably would be better to figure that the more current flowing through the resistor, the more resulting heat you will get.\n\nAs far as wire goes, a smaller diameter wire will have a higher resistance, and will get hotter using less current.  A larger wire will have a lower resistance and require more current to get as hot as the smaller wire.  But, again referring back to Ohm's law, the overall heat exhibited will be the result of figuring the power output for the wire (voltage multiplied by current).  The larger wire will be capable of a higher power output before it surcumbs to the heat and melts because it can carry a higher current.\n\nFor the sake of comparison, let's say that we have a short piece of thin wire and a longer piece of thick wire.  Because of the difference in lengths, they will have the same resistance.  If they are both subjected to the same electrical power supply (same volts, same amps), they will both produce the SAME amount of heat.  The smaller wire will have a higher heat DENSITY because it needs to produce the same thermal energy over a smaller area.  It will appear hotter because of this.  But, if both were used to heat water, they would in theory do exactly the same thing.
1060195
You seem to looking for very precice and specific information where the slightest misunderstanding in communication could be an expensive and time consuming mistake. As this is your first antenna built, much of the experience has been played by ear i'm sure, asking advice in places like this.\n\nAbout an hour ago, my sister asked what I'm doing and I said I'm answering questions. As a scientist, she caught on to the fact that anyone could say anything, regardless of the accuracy of the advice, and be trusted. \n\nGiven that your after accurate specifics, I'd suggest you contact a group or discussion panel that specialises in antennas for radio astronomy, and ask the people that acutally know what they are talking about.
910217
There are two issues to consider, cable heating and voltage drop.  Of the two, voltage drop is by far the easiest.  You need to determine what the allowable voltage drop in you cable is.  The voltage drop is just I x R where I is the current and R is teh round trip cable resistance.  The specification of 100 kW isn't enough.  If its a high tension 100 kV line, the current is only 1 amp and a run of 16 AWG wire would deliver about 99,999 volts out of the original 100,000 volts to your load.  If its a 100 volt system then a 4 AWG wire would only deliver only 20 volts to the load.  \n\nThe formula for wire resistance (in AWG) is that 10 AWG has 0.001 ohm/ft.  Each increase of 1 in the gauge increases resistance by the cube root of 2 (about 25%).  You calculate by picking what voltage drop you can live with, for the known current figure the resistance, knowing the length figure the resistance per foot, and then pick the AWG using the above.\n\nThe second concern is wire heating.  Although you could do the heat transfer calculations, instead look up the allowable ampacity for any wire in an electrical engineers or NEC handbook.
420078
Here's a link to the standard American Wire Guage table.  It has it all listed:\nhttp://www.eng.auburn.edu/~tdenney/ENGR1110/wireawg.html
929646
Depending on the output current you desire and what current the +24 VDC can supply a simple 3-terminal regulator like a 7812 should do the trick. It should be able to handle the 24 V input. You may need to add a 0.01µF Cap from Ground to the +5 VDC output to dampen any oscillations. There are usually a number of cookbook circuits in the data books. \n\nGood Luck!
704473
The stress in the cable is m*g/A.  The Young's modulus of steel is about 200*10^9 N/m^2 (pascal).  Therefore the strain will be\n\nm*g*/(A*200*10^9).\n\nThe elongation is strain*length.\n\nFor the second part, replace g above with (g+a), where a =2.7m/s^2\n\nFor the last, compute the mass from S = m*(g+a)/A, The elastic limit is Sm, so the max mass is m = Sm*A/(g+a)\n\nNote: keep the units straight.  Use kg, m, sec, N in all calcs then convert to the asked-for units last.
967084
I would use a length of nichrome wire (see links below) stretched between a couple copper blocks. You could make some sort of pinch block out of the copper and connect a power supply to the copper. (or drill a hole through it with a set screw, tap a hole and us a pan head screw)  You will have to insulate the copper from the rest of the world so you don't short something out.  G-10 (a glass/epoxy resin) makes a good moderately high temperature insulator.  One of the contacts should be fixed and the other allowed to "float" and allow for changes in the length of the wire due to heating.  A spring could be used to provide tension or an air cylinder.  The power supply will depend on the length and diameter wire you choose.  If there is a problem with the electrical contacts heating up, you could run some cooling water through them.  Just drill a hole through it and 1/8 NPT pipe tape both ends.  You can plumb some water to it and it will keep the copper chilled.
870231
In the American Wire Gauge (AWG) scale, 0000 gauge represents a wire having a diameter of 0.46 inch and 36 gauge represents a diameter of 0.005 inch (5 mils). Diameters for the other gauges are obtained by geometric interpolation, meaning that the ratio between successive diameters is a constant, except for necessary roundoff. Thus n gauge wire has a diameter of .005·92((36-n)/39) inch. The metric wire gauge number is equal to 10 times the diameter of the wire, in millimeters; thus a metric 8 gauge wire has diameter 0.8 millimeters.
926295
Convert all this to MKS units and find the cross-section of that wire.\n\nE (stainless-steel) is given to be 18 x 10^10 N/m^2.\nCross-section is from pi(r^2).  This must be in m^2, not mm^2.\n3.14159(0.125*10^-3 m)^2 = 4.909 x 10^-8 m^2 cross-section.\n\nYoung's modulus (E) is given as,\n\nE = (delta F / cross-section) / (delta length / original length)\n\nDelta F is the tension in the wire, thats the first part of solving this problem.  So rearrange the equation for delta F, renaming delta F as "tension" and substitute the value given in place of E:\n\nTension = (18 x 10^10 N/m^2) * (8 x 10^-5 m / 0.028 m) * (4.909 x 10^-8 m^2)\n\nTension in the wire is therefore 25 N.\n\nThe magnitude of force on the tooth is 25 sin 30 deg or 12.5 N\n\nWarning!:  While the reasoning is sound, double-check all of my math.  Also, note that I rounded the intermediate answer 25 N.
104622
I am trying to figure how large a .745" round wire can be shaped into a rectangle shape wire. \n\nAssuming the wire is malleable, find the area of a circle with diameter .745 or radius = .3725\n\n(.3725)squared times pi = .4359156156 sq in.\n\nFor a square to have this area, each side would be the square root of that number which is .6602390595.\n\nFor a rectangle that is not a square, pick any 2 numbers whose product is .4359156156.  Choose one dimension and divide by it to get the other.
971687
why do you need it?  how much do you want to spend to figure out what it is, or what alloy it is. \n\nyou can just do the density measurement\nmeasure the radius of the wire and length. \ndo the pi r squared length deal and then weigh it, and run it through the density equation. \n\nyou need a ruler and a calipers or micrometer that reads in inches. \n\nso lets say you have .125 inch wire.  divide that by 2 to get the radius. =  .0625"\nthen square it.  .0039" and multiply by 3.14. \n\nand multiply it by the length in inches  \n\nthen weigh it.  in lbs. \n\ndensity of al is between .096 lb/cu in. and .100 lb/cu in\n\nyou might want to buy wire that is certified as the type of aluminum you need.  the stuff is cheap.  you can get at the electical supply store or maybe a hardware store. \n\nwiring used for electical applications is 8000 series alloy for modern applications \n\nolder wire is 1350 alloy.
225165
The first thing that you really need to pay attention to is the dynamics of the wire itself and not the physical size.  Your statement is very bias as I could say that why do fiber optic lines tranmit so quickly and the wire holding up the golden gate bridge is not quicker???\nThe quickest and best answer unless you have degrees in physics is the density of the fibers.  It all comes down to density and a mirid of other factors, but density is the key my friend.\n\nHappy conducting my fiber friend!
135604
Are you talking about the wire to supply the resistance heating wire or the resistance wire its self?\n\nThis is not something you should undertake in any way shape or form. It is something for a trained professional who has the experience and knowledge to install such a system safely.