# Would like to use Nichrome wire to make a small heater



## Yoyizit (Jul 11, 2008)

no burn ever at 42°C(108°F)
burned in 30 sec at 54°C(129°F)
5 sec at 60°C(140°F)
1 sec at 71°C(160°F)
But these values are only for skin with normal blood flow. You may not have that.
http://en.wikipedia.org/wiki/Raynaud's_phenomenon

There was another post about using nichrome wire. It was kind of tricky to find the right wire size.

For a prototype I'd use the car's 12v electrical system and wrap several inches of the wheel circumference with a few dozen 1/4w resistors wired in series. 

Depending on the pitch of the winding 1/8w into each resistor should be warm enough unless the wheel dissipates some of the heat.

For 25 resistors @ 1/8w each the total power would be ~3w. 14v squared divided by 3 = 65 ohms, about 2.7 ohms for each resistor.

Probably you should go smaller than 2.7 ohms and add a series adjustable voltage dropping resistor to give you some latitude on the heat generated.

Try Hosfelt, Digi-key, allelectronics, Mouser or Jameco for resistors in quantity.

Once you zero in on the winding pitch and watts per inch you can pick the Nichrome to do the same thing.

A more sophisticated version would use thermistors and feedback to maintain a fixed temperature. Depending on the thermal inertia of the wheel it might take a while to get to a comfortable temp., with or without feedback.


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## vsheetz (Sep 28, 2008)

A cursory Google search show many varients of heated steering wheel covers. Perhaps using a commercially made unit, or a commercially made unit as the basis with self-modifications would be a way to go.


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## GerGa (Oct 7, 2009)

Im not really understanding your calculations, I dont know what kind of resistor you are suggesting. 
Also I thought the equation is p=I squared times R. Where did you get 14 squared divided by total power? Oh wait you did power equals volts squared divided by resistance but I have never seen resistance or resistors expressed in watts..
Sorry Im confused I have no idea what you did..


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## micromind (Mar 9, 2008)

Quite a few years ago, I made some heated mirrors using nichrome wire from a 1500 watt household heater. 

12 volts applied to about 4' of the wire resulted in burning the frost off the mirror in less than a minute. If left on for about 5 or 6 minutes on a warm day, the mirror is just about too hot to touch. The little loops of bare wire on the outside of the mirror will get too hot to touch in about 30 seconds. 

I'm not sure how long the entire wire was, I seem to remember using about 1/2 of it for both mirrors though. 

Rob


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## Gigs (Oct 26, 2008)

It's funny how no one ever admits they are building an herbal vaporizer.


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## Yoyizit (Jul 11, 2008)

GerGa said:


> Im not really understanding your calculations, I dont know what kind of resistor you are suggesting.
> Also I thought the equation is p=I squared times R. Where did you get 14 squared divided by total power? Oh wait you did power equals volts squared divided by resistance but I have never seen resistance or resistors expressed in watts..
> Sorry Im confused I have no idea what you did..


The resistors are the upper two devices. A 1/4w resistor is about 1/4" long.
http://www.willcoxonline.com/X10mods/3components.jpg

P=I²R = V²/R so 
R = P/I² = V²/P

On second thought, get a heating pad or blanket, wrap some of it around the wheel and adjust the power for comfort. 

At this setting, if the blanket is taking 80w and the blanket is one square foot (144 sq. in) and your hand grip is 12 sq. in. then you need a heater that spreads the heat and gives you 0.6w per sq. inch [80/144] and 6.7w [12(80)/144] for one hand.

P = VI, I = P/V, so 80w @ 120vac = 0.7A. You'll need an ammeter to check this current draw and you can assume you have a constant 120vac.
For a given I, your power in watts @120v is (120)I.

Finding a suitable material with the right sheet resistivity to generate this heat at 12 to 14vdc is another problem. A wire or resistor concentrates the heat in a small volume so this is kind of opposite to what's desired.

You also might need slip rings of some kind to get the power to the rotating steering wheel.

If you have a local chapter of Volunteers for Medical Engineering and they are willing to take on this project the whole thing may be free.


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## GerGa (Oct 7, 2009)

Well I am not too worried about the heat will be distributed right now since the hardest part is the electrical. But at quick thought I am thinking something along the lines of something metal wrapped over the wheel like flashing tape, etc..just something metal to distribute the heat although nichrome tape would be awesome but there are none to little choices for that that I have seen.Yes I realize I probably shouldn't have something conductive touching the nichrome/connection. But whatever that doesn't matter right now.. 
I realize now that xxwatt resistor means it's power limit and are you saying that I should probably get a resistor with a rating of 1/4 watt limit and that at (14?)v(why 14 instead of twelve?) from the car shod produce an 1/8w each for 25 resistors to get sufficient heat, which at 25 resistors is a resistor rated at under 2.7ohm. And you are saying that to produce sufficient heat at whatever volts you mean is probably not quite 1/8 watt, or a 2.7ohm resistor. How much less resistant a resistor do you reckon I might try out(of course at 25 resistors). I don't really mind wasting a little money on trial and error since resistors are pretty cheap.

I am somewhat familiar with electronics, I have seen and soldered some components such as resistors before and also have been in school for EET for only under a month now so I have only learned ohm's law, power law, coulombs kvl kcl and a couple others I think and we have done some studying with series and parallel circuits with resistors. I should have recognized the derivative of the power law but I have only seen that a little bit. 

Thank you very much for helping me out, 
Greg


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## Gigs (Oct 26, 2008)

Car batteries at 13.8 volts normally, more when they are charging.


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## Yoyizit (Jul 11, 2008)

Beg or borrow a heat pad and ammeter first to get some idea of what power is necessary.

Yes, auto voltage may go to 15.5v in the winter to maximize battery charging while not unduly shortening the life of the battery.

For predictable results we may want to stabilize the voltage fed to this at something under 12v so we get predictable results. An LM317 comes to mind.

A 1/4w resistor dissipating a 1/4w gets pretty hot, so for a WAG I figured 1/8w. That will halve the temp. rise above ambient for what 1/4w will give you.

Temp. rise above ambient is proportional to watts dissipated into a volume with some surface area. Doubling the area should cut the temp. rise in half.


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## GerGa (Oct 7, 2009)

So how does a voltage regulator circuit kit that utilizes the LM317 sound? Although I don't need that huge heat sink, I think this one looks perfectly functional:http://cgi.ebay.com/Voltage-Regulator-Kit-AC-DC-in-DC-out-Based-on-LM317_W0QQitemZ260479385900QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item3ca5c8012c 
With your guess at of a total of 3w the current would be something like 213mAh at 14v and I read up on the LM317 and apparently it's minimum current is 150mAh which seems kind of small so I should be fine with that.


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## Yoyizit (Jul 11, 2008)

GerGa said:


> I think this one looks perfectly functional:http://cgi.ebay.com/Voltage-Regulator-Kit-AC-DC-in-DC-out-Based-on-LM317_W0QQitemZ260479385900QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item3ca5c8012c


That thing has way more parts than you need and all "the heavy lifting" has been done for you.

With electronic heater controls I'd be concerned about the Load Dump Transient that vehicles have. 
National Semiconductor has app. notes online and some of their stuff that is supposed to run off car voltages have a few extra components to suppress this transient. The least you need is a series diode with a PIV of 1000v because the transient goes way negative.

I'd first do some kind of prototype with just a heater and maybe a series dropping resistor. The watts/in² required is the most uncertain part of this project.
Imagine or measure or calculate the surface area of a 25w incand. bulb. It's way too hot to touch, so it gives you some idea of how many w/in² you need. Try two in series (1/4th the power) to try to bracket what kind of power you need. Give them a few minutes for the glass temp. to stabilize.
By mixing and matching lamps in series you might get pretty close to a comfortable temp. Remember that an incand. lamp acts like a resistor whose value is current-dependent. See
"Voltage, light output, and lifetime" in
http://en.wikipedia.org/wiki/Incandescent_light_bulb

A Variac would come in handy for this experiment but they are pretty hard to borrow. You could use a lamp dimmer but then you need true RMS meters to measure the watts inputted.


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## GerGa (Oct 7, 2009)

Ok I definetely would not just do some guessing/calculations and just slap it together and hope it works. I will do as much as I can to figure out how to do this(right power for the right heat according to how it is distributed). What about these power supplies, that say they have load dump protection for cars, one ihas adjustable voltage and one is 6v only.
http://www.powerstream.com/dc6.htm 

http://www.watt-power.co.uk/PSD05%20Vehicle%20DC-DC%20Converters.PDF


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## Yoyizit (Jul 11, 2008)

http://en.wikipedia.org/wiki/Load_dump
I think a 1000v PIV diode, a 35v Zener and a resistor or fuse will probably take care of this transient.

Then the 317, 2 resistors, 2 small caps and possibly a heat sink [by mounting to metal in the engine compartment] and you're done.


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## GerGa (Oct 7, 2009)

So how could I go about finding out where and in what orientation to connect the diodes? Also how do I choose what diodes - just on allelectronics there are two 1000v diodes, 1 and 3 amp, and also I have not had much luck in finding a zener diode in 35v except here which is 5w but I am not sure what power rating to look for; is that one fine? 
Also how could I go about figuring out what kind of fuse or resistor to use and how to hook that up? 

And your mention of the 317 and a couple resistors and caps seems like the standard circuit design for a v-regulating circuit that uses the 317 - (just trying to state my understanding)


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## Yoyizit (Jul 11, 2008)

GerGa said:


> So how could I go about finding out where and in what orientation to connect the diodes? Also how do I choose what diodes - just on allelectronics there are two 1000v diodes, 1 and 3 amp, and also I have not had much luck in finding a zener diode in 35v except here which is 5w but I am not sure what power rating to look for; is that one fine?
> Also how could I go about figuring out what kind of fuse or resistor to use and how to hook that up?
> 
> And your mention of the 317 and a couple resistors and caps seems like the standard circuit design for a v-regulating circuit that uses the 317 - (just trying to state my understanding)


I'll post a schematic or a description; first let's find the watts/sq. in. and decide if you want the whole steering wheel heated and if you want it heated for the whole time you're driving.


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## GerGa (Oct 7, 2009)

Would I be better off trying to get a grasp for the power need with other things like bulbs or heating pads and resistors with a power source similar to the car? I have an ac/dc adapter adjustable 3-12v, although it does have a 300mA minimum, I could just hook up the 317 variable power supply to that to get lower mA if desired, correct?


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## Yoyizit (Jul 11, 2008)

GerGa said:


> Would I be better off trying to get a grasp for the power need with other things like bulbs or heating pads and resistors with a power source similar to the car? I have an ac/dc adapter adjustable 3-12v, although it does have a 300mA minimum, I could just hook up the 317 variable power supply to that to get lower mA if desired, correct?


The power requirement, whether at 1v or 1000v, will drive the whole design. The heater "doesn't know" what volts and amps are causing it to heat. Using 120vac from your house, and light bulbs or heating pads as loads, is more convenient than zapping your car's elec. system.

And if the power needed to heat your hands over the surface area of the steering wheel is more than 200w or so you might need to beef up your car's alternator.

I can hold onto a 13w CFL for quite a while. The surface area that I am touching is about 5 in. sq. So for my hand I can use about 2 to 3 w per sq. in. And my hand cuts off the free air supply to the bulb so the bulb gets hotter than normal.

For a 100 sq. in. steering wheel surface I might need 200 to 300w, less if I just have hand pads. 
. . .Maybe it's time to shop for a new alternator, or use an aux. battery just to run the wheel heater.
The plot thickens and the scope of the design widens to include your car's power system.


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## GerGa (Oct 7, 2009)

My heating pad: 
area=160sqr inch 
rated power= 45w 
measured resistance=90ohm 
Turned it on medium which is pretty warm, left it sitting on for more than 5 minutes to stabilize temp, measured volts:around 40-50v 

Whole steering wheel: 
approx area=140sqr inch 

The heating pad does not receive constant varying voltage, it gives intermittent bursts of voltage. my wild guess is 400ms on, 400ms off 
So I can't tell exactly the voltage with my analog meter but it was around 40-50v. 
My meter only measures to 150mA...


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## Yoyizit (Jul 11, 2008)

GerGa said:


> My heating pad:
> area=160sqr inch
> rated power= 45w
> measured resistance=90ohm
> ...


The measurement of this pad's power consumption might be difficult because of the duty cycle bimetallic temp. control. 
http://www.evenheat-kiln.com/controls/inf/inf.htm
You'd need to use a wattmeter for a time interval and then calc. amps.

Since you know the pad's resistance, knowing the amps and duty cycle will tell you the power.
If you have some power resistors you can make a shunt and then use your voltmeter to measure current.

Anyway, with the pad at med, 23w/160sq in = 140 mW/sq. in., so the wheel would need 20W. At 12v this is not even 2A and the resistance of the wheel's heater = 144/20 = ~7 ohms. 
So far so good.

At this point you should maybe try to make a small heater for one hand out of nichrome or resistors and mount it on the wheel. 
You can power it directly from the battery with the engine off. If you don't fuse this circuit and you short out the battery your wires connecting this circuit will melt very quickly.

The winding pitch will influence the mW per sq. in. This prototype can be much hotter or much colder than the calcs would indicate, so be careful.


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## GerGa (Oct 7, 2009)

Why shouldn't I just get a connector for the ashtray lighter socket? It's still on with the car off, I tested it no load at 13v.
Also how would go about fusing the circuit? 

Edit:Also, if your worried about a bad connection there, which I'm sure it does have a bit of resistance, I could just solder in the wires to where the lighter socket is soldered in, and use some high power connectors that I have laying around. 

It's not that I don't want to put in a fuse of my own, it's just that much easier to not have to worry about running wires all the way around the engine compartment and in through the dash.


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## Yoyizit (Jul 11, 2008)

GerGa said:


> Why shouldn't I just get a connector for the ashtray lighter socket? It's still on with the car off, I tested it no load at 13v.
> Also how would go about fusing the circuit?
> 
> Edit:Also, if your worried about a bad connection there, which I'm sure it does have a bit of resistance, I could just solder in the wires to where the lighter socket is soldered in, and use some high power connectors that I have laying around.
> ...


Yes, use that socket. It might be fused at 10A, or not at all.


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## GerGa (Oct 7, 2009)

So I'm not really sure what nichrome to get. 
I suppose what I might aim for for testing would would be 20w for 140inch^2. 
And we'll say a 7.5 ft length of wire is wound around the wheel. A a 20w total that is 2.66w/ft. I'll pick 8v for head room to go up and down in power. 8v/(2.66w/ft)=~3ohm/ft 
So I might look for wire that is around 3ohm/ft. 
Are my calcs correct? 
thanks


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## Yoyizit (Jul 11, 2008)

GerGa said:


> So I might look for wire that is around 3ohm/ft.
> Are my calcs correct?
> thanks


I think so, but how will you spread the heat?
Maybe you should use
http://www.google.com/search?client=safari&rls=en&q="heat+tape"&ie=UTF-8&oe=UTF-8
of the right resistance per foot or per sq. inch. The manu's rated volts are irrelevant for this application but it does have to withstand the current.
20w @ 8v would be ~4.5 kw @ 120v.


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## HandyPete (Mar 23, 2008)

Has anyone mentioned a cars steering wheel rotates? How are you going to wire all this up? Your going to pull the wheel and go through the steering column? Use the lighter outlet and have 5 feet of wire dangling around?

I have a hard time believing someone could actually do this in an "elegant" manner.


_pete


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## Yoyizit (Jul 11, 2008)

See above "slip rings" or modify the
http://www.thefind.com/cars/info-steering-wheel-clock-spring


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## GerGa (Oct 7, 2009)

If outside the steering wheel, the wire could be inside a mechanism that has excess wire wound inside on a wheel, like a measuring tape, and it would be mounted under the steering wheel column. Then the wire might have to be guided somehow to make sure it takes a clean path from the pulley to the wheel, not getting stuck inside the crevace between the wheel and column, etc. I have seen it with a usb cord light for a computer. I am trying to take this one step at a time though. 
I found this website where 120v 6watt per foot wire is the most wattage. 120v and 6w is .05ohms/foot. Even though I don't need as many watts that is still wayyy too much current for my volt range.


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## Yoyizit (Jul 11, 2008)

GerGa said:


> Even though I don't need as many watts that is still wayyy too much current for my volt range.


Yeah. We need to do some impedance matching here.
The constraints are the voltage, the power and the length of the heater. 
That leaves us with the resistance per foot or per sq. inch and not much else.

Maybe you could make a "braided" network of 1/8 w or 1/4 w resistors, some in parallel and these in series with other paralleled resistors. Hosfelt, Allelectronics, etc. have reasonable prices.
Commonly available values are above 10 ohms but you can use paralleling to get almost any value.


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## GerGa (Oct 7, 2009)

So your saying trials with resistors will give me a starting point for trials with nichrome? 
If I wanted to mock up 1 foot of resistors like a 21awg nichrome wire which is .831ohms/ft http://www.heatersplus.com/nichrome.htm , then I would connect a bunch of resistors 1 foot long to make .831ohms of resistance(using whatever combination of series and parallel is needed)? Correct? 
I am also fuzzy on how a resistor is going to produce a similar amount of heat, or might there be not too much difference and it mostly comes down to having resistors about the size of wire and similar resistance per length..? 
Edit: I tested some resistors and I got them pretty hot, they are 1/4w resistors. I did some calculations and I really see why you said that maybe I should only do part of the wheel, it takes so many ohms to get low enough wattage. With the resistors I guesstimated with an ohm value a reasonable watt value for the whole steering wheel, within my limitations of volts. I came up up with 300ohms per foot at 8ft to give the right power amount for the whole wheel. But, nichrome is much less resistive about 8ohm/ft at 31gauge and 27ohmft at 36gauge;this is way too little(therefore way too hot), and an only an unreasonable amount of wire would up the resistance and down the wattage;within my voltage limitations. 
So now to find resistance wire that has very high resistance... 
Very high gauge wire and a close wind seems the way to go; best heat distribution for the power I need, but god is the wire so expensive...I really need high resistance 
wire.. 

EDIT #2 - Oh ****. I can use resistors. doh. :thumbsup:


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## fireguy (May 3, 2007)

Cabela's ragg wool/thinsulate gloves $15.00


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## GerGa (Oct 7, 2009)

fireguy said:


> Cabela's ragg wool/thinsulate gloves $15.00


 Thanks but if you had such a problem with the cold you would understand that gloves never work. Mittens are superior. Also this is a chance in the day to warm up my hands since most building are too cold summer or winter(over airconditioning). God now ssome people probably think im a snob..


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## Yoyizit (Jul 11, 2008)

GerGa said:


> So your saying trials with resistors will give me a starting point for trials with nichrome?
> If I wanted to mock up 1 foot of resistors like a 21awg nichrome wire which is .831ohms/ft http://www.heatersplus.com/nichrome.htm , then I would connect a bunch of resistors 1 foot long to make .831ohms of resistance(using whatever combination of series and parallel is needed)? Correct?
> I am also fuzzy on how a resistor is going to produce a similar amount of heat, or might there be not too much difference and it mostly comes down to having resistors about the size of wire and similar resistance per length..?
> Edit: I tested some resistors and I got them pretty hot, they are 1/4w resistors. I did some calculations and I really see why you said that maybe I should only do part of the wheel, it takes so many ohms to get low enough wattage. With the resistors I guesstimated with an ohm value a reasonable watt value for the whole steering wheel, within my limitations of volts. I came up up with 300ohms per foot at 8ft to give the right power amount for the whole wheel. But, nichrome is much less resistive about 8ohm/ft at 31gauge and 27ohmft at 36gauge;this is way too little(therefore way too hot), and an only an unreasonable amount of wire would up the resistance and down the wattage;within my voltage limitations.
> ...


Resistors give much more flexibility in the resistance value than nichrome but they don't easily wrap around curved surfaces.

A 1/4 w 1 ohm resistor with 0.5v across it gets pretty hot. With 0.35 v the temp. rise above amb. is only half as much. If you knew the thermal resistance, θ, of your resistors in °C/w you could calculate the rise in free air. 
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics

Metal alloys are much more conductive than carbon composition resistors. 

One way to use skinny wire is to wind it around an elec. insulating flexible core so your actual wire length would be very long but the length of this "heated string" would be only a few feet.

BTW, #30 AWG copper wire wrap wire is 10' per ohm but it's delicate.

8' of wire @ 20w @ 8v = 3 ohms = 3/8 of an ohm per ft.
One thing spreadsheets are good for is error-checking, once you get them debugged.



fireguy said:


> Cabela's ragg wool/thinsulate gloves $15.00


The heat problem comes mostly from the inside.


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## GerGa (Oct 7, 2009)

I don't know, I had 7 360ohm resistors in parallel, which is about 51.5 ohms. These are on a pcb. Judging by their size and spacing and the heat I felt at my max of 12v, Ive established they are a good estimate for each 2 inch segment of resistors 8 feet around the wheel. I will most likely have a little something on the wheel to insulate from the resistors(just in-case its too hot for the wheel), some sort of heat conductive material sandwiching the resistors(perhaps aluminum tape, and yea the resistors have to be electrically insulated..), and lastly a cover over that made of acrylic or leather or what-ever will make it look like a real steering wheel(but of course not too thick). I don't thing the resistors being a weird shape over the wheel will be a problem, these 1/4inch ones are not huge, the cover over them will smooth it out, and I can deal with a little bit of bumpiness. 

Considering all that, plus warm up time; I may want to go ~slightly~ higher in power at 12v. And an adjustable supply is really needed for an adequate warm-up time(ability to turn down power once at a comfortable temperature). 

So say I want an 8foot stretch coiled around the wheel at 12v, at 51.5ohms/2 inches, which is 2.8w per 2 inches, and to get my total watts I can just multiply that by 6 inch=16.8w per foot, then at 8' 16.8x8=134watts, correct? 

The thing is I don't know how to go from there to figure out how to wire up a proper amount of resistors to get my desired watts per foot.


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## Yoyizit (Jul 11, 2008)

GerGa said:


> I don't know, I had 7 360ohm resistors in parallel, which is about 51.5 ohms. These are on a pcb. Judging by their size and spacing and the heat I felt at my max of 12v, Ive established they are a good estimate for each 2 inch segment of resistors 8 feet around the wheel.
> 
> 8foot stretch coiled around the wheel at 12v, at 51.5ohms/2 inches, which is 2.8w per 2 inches, and to get my total watts I can just multiply that by 6 inch=16.8w per foot, then at 8' 16.8x8=134watts, correct?
> 
> The thing is I don't know how to go from there to figure out how to wire up a proper amount of resistors to get my desired watts per foot.


2.8w each 2" of length for proper heating
7 ea. resistors in parallel for each 2" of length for proper heat distribution
8ft x 12"/ft = 96" of length
96/2 = 48 resistor clusters of 7 resistors each

each cluster needs 12/48 v across it to generate 2.8w.
R = V^2/P = [(12/48)^2]/2.8 = 0.022 ohms
7 ea. resistors x 0.022 ohms per resistor cluster = 0.154 ohms, max, for each resistor, for 2.8w min. for each cluster.

Each "resistor" could be 0.154 ohm x (10'/ohm) = 1.54' of #30 wire wrap wire or some shorter length of Nichrome wire larger than 31 gauge.

check
48 x 2.8w = 134w.
and
R = 144/134 = 1.07 ohms total 8' strip resistance

BTW, this network is equivalent to 8' of 7 ea. paralleled strands of resistance material, with each strand having a resistance of 48 x 0.154 ohms = 7.4 ohms.

check
7.4/7 = 1.06 ohms.

12v/1 ohm = 12A, so we are out of the LM317 class unless we use some tricks.

#40 AWG copper would be approx. 1' per ohm, so it is just about the right resistance per foot for this application. But, it would be delicate. 
The strands in lamp cord are probably larger than #40 so then you would need less than 7 parallel strands, but your heat distribution may suffer.

This problem is easy in theory but finding the parts for this and keeping the bookkeeping straight is proving difficult. 
If you program a spreadsheet with all of this, the bookkeeping problem disappears and you can plug in many different lengths and resistances until you find some optimum between price, availability, durability, etc..

And I don't think you can solder Nichrome.


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## GerGa (Oct 7, 2009)

Hmm that estimate was a little aggressive but I didnt realize it would be 12A. I'd like to limit it to less than 10A considering the lighters in cars are usually 10A. 
I can go less on the power,2.3w/ft is too much. 
I am thinking the nichrome at around 30 gauge with a 10foot or longer length sounds good. I will have to look at it more for trying to get the amps down and consider possibly not heating the whole wheel. 
Thanks


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## Yoyizit (Jul 11, 2008)

GerGa said:


> Hmm that estimate was a little aggressive but I didnt realize it would be 12A. I'd like to limit it to less than 10A considering the lighters in cars are usually 10A.
> I can go less on the power,2.3w/ft is too much.
> I am thinking the nichrome at around 30 gauge with a 10foot or longer length sounds good. I will have to look at it more for trying to get the amps down and consider possibly not heating the whole wheel.
> Thanks


12v/[<10A] = >1.2 ohms.

As you narrow down the design you may bump into other tradeoffs. There are plenty of surprises left in this project.

If you really need 12A or more you can use an aux. battery and a current-sharing arrangement.

I'd wind the wheel with something close to what you want and actually try it. The power, etc., can be worked out later. There may also be a seasonal variation in how much power is needed.


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