Hi, I'm here with a rather weird question. I'm just a beginner, and I got involved with a small plane construction project, for which we'll need 2x2 meters hot wire cutter that we are designing. I was doing some research on what gauge wire will fit this purpose the best, and do some calculations to understand what power source should be chosen. But since the size is so big, I can't find info about the appropriate diameter of the wire that needs to be selected as well as calculations for the power source. We'll need to cut 1.5x1.7 blocks of extruded polystyrene (XPS) to make the ribs od the wings.
In my opinion, we should use a 20-24 gauge wire with resistance 1-4 Ohms/ft respectively, and 24V source, which will require a ridiculous amount of current.
Could you please give any ideas on how to optimize the design/do the calculations or what components to choose. Any suggestions or ideas would be very appreciated.

 

Does it happen to be a Rutan Quickie? (https://en.wikipedia.org/wiki/Rutan_Quickie ) Since you are talking about wing ribs, I suppose not.

However, there are Quickie build sites that can help with that question.

 

What do you mean by "a ridiculous amount of current"?

It is not about how much current. It is about power.
You can calculate power by using any one of the three formulas:

P = V x V / R
P = I x I x R
P = I x V

Make sure the power supply can provide the desired amount of power.

 

What do you mean by "a ridiculous amount of current"?

It is not about how much current. It is about power.
You can calculate power by using any one of the three formulas:

P = V x V / R
P = I x I x R
P = I x V

Make sure the power supply can provide the desired amount of power.

Alright then, for example, if I want to use 26 gauge Nichrome 60 wire(2.67 Ohms/ft) I'd need P = (24x24)/(2,67) = 215 W, and I = 215/24 = 8.9 A. Do I not need to account for the length of the wire?
And if so, what would be the best gauge to choose, so it would be strong enough for 2m and easy enough to heat up.

 

What? You did not include the length of the wire?

Go back and try again. This time include the length of the wire!

 

What? You did not include the length of the wire?

Go back and try again. This time include the length of the wire!

That's literally what I asked..... P= (24x24)/(2,67) x 6.5 . I = 1495/24 = 62 A. But my question is the same, what gauge would be the best?

 

The best gauge is the one that cuts the foam without breaking.

 

The best gauge is the one that cuts the foam without breaking.

Why can't you reply normally, I've told you that I'm just learning, and I'm asking for help, and you are just being sassy?

 

You lost me at:

we need 2x2 meters hot wire cutter

2 by 2 meter wire?

 

You lost me at:

2 by 2 meter wire?

It's gonna be a CNC cutter, and the size of it will be 2x2 meters, so the wire needs to be 2 meters long in order to reach the other side.

 

The best gauge is the one that cuts the foam without breaking.

Wire sag (and drag), particularly for wing ribs, is also important. I suspect there are theoretical studies of both factors, particularly in industry where huge sheets are cut. However, Quickie and some other Rutan designs used foam and provide a resource for practical experience.

For model airplanes, anything wider than a meter or so is usually built with panels so far as i know for DIY. CNC machining has also replaced a lot of foam cutting. See: https://www.rcgroups.com/forums/showthread.php?3034755-Cnc-foam-cutting-service-available It refers to a 60" cut, but that may be in two pieces. If it were for me, I would give CNC machining instead of wire cutting serious thought.

 

OK, here's my take on a two meter length of wire with a resistance of 2.7Ω per foot (I'll just assume a meter is roughly the same as a yard since I'm a yankee). 3 feet of 2.7Ω wire is roughly 8Ω. 24V ÷ 8Ω =3A. That's 72 W.

Apparently there have been more posts while I was typing.

 

OK, here's my take on a two meter length of wire with a resistance of 2.7Ω per foot (I'll just assume a meter is roughly the same as a yard since I'm a yankee). 3 feet of 2.7Ω wire is roughly 8Ω. 24V ÷ 8Ω =3A. That's 72 W.

Apparently there have been more posts while I was typing.

Yeah, I did some calculations as well and here's what I got.
I choose 3 wire sizes 26 gauge(0.4mm, 2,67Ohms/ft), 32 gauge(0.2mm, 10,55Ohms/ft), 36 gauge(0.12mm, 26,7Ohms/ft). The power required for them is 1400 W, 354 W and 140 W respectively, as 2 meters is 6,5 ft. So the needed current is 58 A, 14 A and 5.8 A respectively.

 

I cut some foam panels to insulate a room. The panels were four feet by eight feet by two inch thick. I used welding wire, diameter I think was 0.050". I also used my welder as a power source. When the wire got hot it sagged greatly. To solve that problem I changed the way the wire was connected to a 10 foot long 2X4. I set one up rigid and the other end was on a spring tensioned pivot. I only had two inches to cut through, but in some cases the length was eight feet. Wasn't the fastest cutter around but it did quite nicely. Nice straight cuts. The FIRST piece of foam I cut I did with a table saw. What a fricken mess that was. That's why I opted for the hot wire cutter method.

While the wire was cutting through there was sag (or lag, as I cut from top down). For that reason I had to go slower. But my cuts were being made straight and not following some contour. Now, my welder is not capable of "Insane" current. And I don't know what the resistance was for my wire length of just over 10 feet.

IF you have a 2 meter block (6 feet) (I don't feel like looking up the conversion), that's double the resistance - so that's half the amperage for a wire long enough to cut through such a block. Depending on the amount of tension you hold the wire to, your form (shape) cutting will vary. I don't know if that will become an issue for you.

 

If you want to ignore the length of the wire, you need to know how much power per foot is required to cut the foam.

Suppose that you need 10W per foot.
Then the current = sqrt(10/2.67) = about 2A
Voltage required per foot = 10W / 2A = 5V
Hence 5-foot of 26 AWG wire will require a power supply delivering 2A @ 25V = 50W.

For longer wire, you need more voltage, not more current.

 

Yeah, I did some calculations as well and here's what I got.
I choose 3 wire sizes 26 gauge(0.4mm, 2,67Ohms/ft), 32 gauge(0.2mm, 10,55Ohms/ft), 36 gauge(0.12mm, 26,7Ohms/ft). The power required for them is 1400 W, 354 W and 140 W respectively, as 2 meters is 6,5 ft. So the needed current is 58 A, 14 A and 5.8 A respectively.

Not sure if you're aware of this - and I'm not trying to be snarky, your "ZERO POINT FOUR mm wire should NOT have two thousand six hundred seventy ohms per foot. You used a comma instead of a decimal point.

I (amps) = E (volts) over (divided by) R (ohms). So 24V ÷ 2.67Ω = 8.9887A (round to 8.99 (rounded to 9A)). That's PER FOOT. You have three feet. So 2.67Ω (per foot) times 3 (feet) is 8.01Ω (in three feet). Substitute the 2.67Ω with 8Ω and you get 3A.

 

If you want to ignore the length of the wire, you need to know how much power per foot is required to cut the foam.

Suppose that you need 10W per foot.
Then the current = sqrt(10/2.67) = about 2A
Voltage required per foot = 10W / 2A = 5V
Hence 5-foot of 26 AWG wire will require a power supply delivering 2A @ 25V = 50W.

For longer wire, you need more voltage, not more current.

That's right. The longer the wire the higher the resistance, which translates to lower current, which also translates to lower wattage. You can't control the resistance of the wire, so to maintain the wattage (voltage times current) then you have to change the voltage. There's no way you're going to need 1400 W. That kind of wattage will burn the house down.

 

Not sure if you're aware of this - and I'm not trying to be snarky, your "ZERO POINT FOUR mm wire should NOT have two thousand six hundred seventy ohms per foot. You used a comma instead of a decimal point.

I (amps) = E (volts) over (divided by) R (ohms). So 24V ÷ 2.67Ω = 8.9887A (round to 8.99 (rounded to 9A)). That's PER FOOT. You have three feet. So 2.67Ω (per foot) times 3 (feet) is 8.01Ω (in three feet). Substitute the 2.67Ω with 8Ω and you get 3A.

Yeah, the coma was a mistake.

Thanks for the explanation, and I don't want it to turn into an electronics class, but I'm confused. Say we have "24V ÷ 2.67Ω = 8.9887A (round to 8.99 (rounded to 9A))" per foot. If we have 6,5 feet shouldn't we multiply it, as each foot will have 2.67 Ohms of resistance and 9A to push electros thru?

 

comma is used instead of decimal point in Europe.
26 AWG nichrome wire is 2.67 ohms per foot.

 

"24V ÷ 2.67Ω = 8.9887A only works for 1 foot of wire.
Your wire is longer than 1 foot. Hence the voltage across the 1-foot wire will be lower than 24V.

See my post #15.