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Need help using Nichrome wire for a foam cutter

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Ian602

Feb 4, 2016
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I having trouble using Nichrome wire to cut styrofoam. I'm running 18 volts AC. When the span of the Nichrome wire is 1 foot, it cuts perfectly.
However, when the span is 3.5 feet, it doesn't cut at all. The Nichrome wire is 26 gauge and each end is connected to the transformer using standard household Romex. I have about 4 feet of Romex on each end.
I don't understand enough about electricity to know why it works with only 1 foot but not with 3.5 feet. What should I do to solve the problem? If I need to add more power, what is a cheap and easy way to do so? I'm currently using a train transformer. Thanks.
 

Mark Cox

Feb 3, 2016
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Not an expert, but the longer the wire the more resistance so you need more power to create the same amount on heat. If you were using 18 volt @ 1000 ma (1 amp) you will likely need more like 3000 or 3 amps to generate the heat you need.
You can get a variable transformer so that you can vary the current and or voltage or amount of heat.
I'm sure someone on here can give some more accurate information with regards to what optimal would be. Same would apply if you went to 6 inches it might be too much and melt wire.
 

Bluejets

Oct 5, 2014
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Ian,
I just took a look at the transformer I used to use to cut foam cores for model aircraft.
Mark was on the money with his reply and I'll give you some of the details of the system I have here.

Transformer I use is a Ferguson which came from an ice making machine.
Windings are 2 x 20V at 5 Amp.
I run these in series to give 40V at 5 Amp maximum.
To control the heating for different size nichrome and differing lengths, I run the transformer through a standard light dimmer installed on the primary side. (mains side)
These will handle 400VA no worries as the loading above is 40V x 5 Amp = 200VA.

When I say differing lengths and size of nichrome, I refer to perhaps cutting a wing core with a fine nichrome at around 900mm to 1200mm long or at the other end of the scale, cutting small grooves in the wing cores for carbon fibre or timber spars maybe 5mm wide with a tip fashioned to the groove shape from fairly course nichrome maybe only 100mm long.

Foam must be cut at just the right temperature.
Too little heat and the cutting wire drags in the centre causing an incorrect profile with regard to the patterns.
Too much heat and the core gets under-cut as the wire passes over the pattern.
The dimmer requires one to do some trial and error but once the setting for a particular setup is established, the cuts can be repeated accurately.

I'll see if I can find the actual size of the nichrome I used and post it back here.

Actually the whole system worked so well that the people from Jabiru Ultralights had me knock one up for them many years ago.
 

duke37

Jan 9, 2011
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I assembled a cutter using a dimmer and battery charger. I also fitted a 100W bulb in the primary in case the dimmer was not balanced.
We tried several wires, Nichrome, stainless (guitar string) and suture wire. The very high temperature capability of Nichrome or Kanthal is not necessary.
The wire will expand so will need a spring bow to keep it taught.
The current necessary will not change if the length is changed but the voltage will
 

mgrass

Sep 25, 2011
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POWER DENSITY...watts/ ft, or watts/sq in., or sq meter.
power density!
Your 1' wire at 18v has a certain amount of power/ft. You are increasing the length by 42/12 = 3.5. So now you need 3.5x the power. But you've also increased the resistance by that same factor of 3.5 To get 3.5 times the power you need 3.5 times the current. Just for sake of argument lets say your wire is 18Ω at 1 foot. I=E/R=18v/18Ω=1A. P=IE=1a x 18v =18w. You need 3.5 x 18w=63w. Working backwards P=IE....63w=18v x i....
i=63/18=3.5amps will give us 63w. Our wire is 18Ω/ft so 18 x 3.5'=63Ω. So a single wire 3.5' long will only draw 18/63= .285 amps. The resistance we need for 3.5amps at 18v is 18/3.5=5.14Ω. 63Ω/5.14Ω=12.26 (remember this number 12.26) I used 18ohms just to have a number. This all still applies. So..... If you have enough nichrome wire cut 12 (12 is the closest multiple of 12.26) pieces 3.5' long and twist them together (a variable speed drill works great for this). If your power source can handle the extra current, you now have watt you need! Why 12.26? when you put two resistors is parallel, the equivalent resistance is 1/2. 10 resistors(of the same value, they all have to be the same or the math becomes different)..10 resistors would be 1/10. so we need 5.14Ω. Our resistor(our 3.5' wire)is 63Ω. How many 63Ω resistors need to be in parallel to get 5.14Ω? 63Ω/5.14Ω= 12.26 so, 12 resistors of 63Ω in parallel has an equiv. R of 63/12=Ω. If you go back to the 3.5 and the 3.5..
3.5X3.5=12.25 imagine that! Just remember that you need 3.5x the current from you 18vac. As mentioned in previous replies, you could use a higher voltage. P=IE and E=IR so I=E/R.. substituting P=(E/R)E or P=ExE/R. Filling in...63w=Esquared/63Ω or
63w X 63Ω = Esquared so 3969=Esquared.so E=63v
If you're careful you could use a triac based lamp dimmer and your 3.5' long wire. That wire will be double hot!!! and dangerous. As mentioned previously, you now can control the power to the wire. Your lamp dimmer needs to be able to handle your current and as he said an incandescent light bulb in series will help protect your dimmer. How much current=3.5 x 18v/R. So what is the resistance of your 12" piece of wire?
 

Colin Mitchell

Aug 31, 2014
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"To get 3.5 times the power you need 3.5 times the current. "

Where do you get this rubbish from????
You have absolutely no idea what you are talking about.
 

mgrass

Sep 25, 2011
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Where do you get this rubbish from????
You have absolutely no idea what you are talking about.
Really? The last time I checked P=IE So 10a x 10v=100w 3.5 x 100= 350w (3.5x10a)x10v=350w
maybe my calculator does work anymore.
So yes if you need 3.5 times more power you need 3.5 times the current.
 

mgrass

Sep 25, 2011
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Where do you get this rubbish from????
You have absolutely no idea what you are talking about.
I just checked to see if they've changed what was in all of my text books and according to http://ewiaei.org/formulas--laws.html , they still use the same formula for power. Pwatts=Voltage X Current. And I know they haven't changed the algebra or are you saying that they have changed algebra too?
 

Colin Mitchell

Aug 31, 2014
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"To get 3.5 times the power you need 3.5 times the current. "
A little knowledge is a dangerous thing.
What you are talking about does not apply in this situation.
You have never produced a hot-wire cutter. I have.
What you are saying is entirely incorrect.
Think about the actual situation and not simply apply mathematics to a hypothetical problem.
 

mgrass

Sep 25, 2011
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You sir know not of what you speak. I
"To get 3.5 times the power you need 3.5 times the current. "
A little knowledge is a dangerous thing.
What you are talking about does not apply in this situation.
You have never produced a hot-wire cutter. I have.
What you are saying is entirely incorrect.
Think about the actual situation and not simply apply mathematics to a hypothetical problem.
You sir know not of what you speak. I have designed and built and use them quite often. I have three in my shop as I type. I am not misusing anything. You said I was full of rubbish. What I said is correct. And it works very well.
 

Colin Mitchell

Aug 31, 2014
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We are talking about the requirements for a wire 3.5 ft long.
If 18v heats up 1 ft of wire, then 18v x 3.5 will heat up 3.5ft of wire.
The current delivered to 1ft of wire will be the same as for 3.5ft of wire. The current does not alter.
 

mgrass

Sep 25, 2011
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The power density needs to be the same and I stand by that. I typed the response very quickly; I'll have to go back a check it and admit anything that's wrong. However, you took exception to my statement " to get 3.5 times the power you have to have 3.5 times the current." Your comment to that was that I was full of rubbish. My statement is correct. So you make mistakes too! I'm sure we have better things to do.
Have a splendid evening!
 

Colin Mitchell

Aug 31, 2014
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I don't know where you get the idea that 3.5 times the current is needed. No-where in this example is 3.5 times the current needed.
Look at your original reply. You go on and on with completely incorrect statements and assessments. - twisting three strands together etc.
You have NEVER produced a hot wire cutter, otherwise you would have never offered these absurd results.
You clearly state that: "you've also increased the resistance by that same factor of 3.5 To get 3.5 times the power you need 3.5 times the current." This is absolutely incorrect.
 
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Colin Mitchell

Aug 31, 2014
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To start with, the wire is connected to an AC source and it is known that 18v AC is suitable. So no other calculations are necessary. The supply just has to be increased to 63v AC and use the same current capability as the 18v supply.
 

Bluejets

Oct 5, 2014
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To start with, the wire is connected to an AC source and it is known that 18v AC is suitable. So no other calculations are necessary. The supply just has to be increased to 63v AC and use the same current capability as the 18v supply.

Agreed....
 

mgrass

Sep 25, 2011
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If
To start with, the wire is connected to an AC source and it is known that 18v AC is suitable.
That is correct. You are telling him to change the voltage by a factor of 3.5 making him buy a new transformer. Whereas I'm saying to change the current by changing the resistance. I know that I have more nichrome wire than I do various transformers. And not knowing someones background and expertise, I'd prefer to stay with a lower voltage. The only factors we were given was 18v, 1', and 3.5'. Not knowing what current his 1' wire was drawing, the VA rating of his transformer, the size of any other transformer he may have, his expertise, etc., I figured he was more likely to have enough nichrome wire to make what he needs vs having the correct voltage transformer, buying a transformer (they aren't cheap) . There was nothing wrong with my logic. My solution was to vary the resistance of the nichrome wire, which is what he had already done by going from 1' to 3.5'! How many 63v transformers do you have. Still not knowing what his current draw is we can't even recommend whether a 600w triac based dimmer would work for him. I myself would use a wire from a direct feed from line voltage. I know that it will burn me; I know that's it's has the voltage to kill me and will take the necessary safety measures like a GFCI, etc.. I would not build or suggest that approach for some one else without knowing their background. So, I keep the voltage low and increase the current by reducing the resistance. If the kerf of the cut needs to be very narrow, then the solution will require a thin wire which means a higher resistance and therefore a higher voltage. But we don't have to use nichrome. We could use a piece of steel wire (which has a lower resistance /ft. I've never searched for steel wire that's awg 18 to 30, but it would work at a low voltage. So where is the rubbish?
 

duke37

Jan 9, 2011
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For the same resistance, 3.5 times the current will give 12.25 the power.
P = I*I*R

In fact extending the length does not alter the current density required so the current should stay the same but the voltage will be higher.
Putting several strands in parallel to lower the resistance will work but things will not be in proportion since the wire area will be greater. Also there may be problems with an uneven cut.
 

mgrass

Sep 25, 2011
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For the same resistance, 3.5 times the current will give 12.25 the power. P = I*I*R
Hello Duke, Glad you could join our friendly discussion.
P=IE vs P=i*i*R The later contains the variable R, which you have to change during the computation between Colin's solution and mine. Colin's will require 3.5r and i, while mine requires e, 3.5i, and r/3.5.
To calculate with the i*i*r mine p=(3.5i)(3.5i)(r/3.5).
The 3.5 in the denominator cancels one of the 3.5 in the numerator giving us p=3.5(i*r)... so 3.5P, not 12.25p.
For Colin's solution (i)(i)(3.5r) also gives 3.5P Now I do make mistakes, but I think that I have that correct.
Putting several strands in parallel to lower the resistance will work but things will not be in proportion since the wire area will be greater. Also there may be problems with an uneven cut.
I agree! You have more surface area to dissipate the heat and you therefore must melt more foam. So, yes, there is that factor! And The uneven cut has always added a nice texture to my cuts, and, sometimes those aren't desirable.
I still see what I said as being correct; yes you will have to allow for the increased surface area of the larger wire and the uneven cut (but if the wire is salvaged from a blow dryer....well, I prefer free over having to buy something...even if it's his money).
 
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