Enclosure math- why they're so important and how to make one. Hi guys, I've been think a lot about print quality and exactly what sets repraps back from the pros (I'm not talking about all of these little startups, most of you have them beat) but companies like Stratasys. I think there is a very good reason that commercial printers are all enclosed and actively heated. I think one of the main RepRap setbacks is trying to account for an absurd amount of thermal contraction. Most are limited to ABS and PLA, while other plastics like Nylon and Delrin (acetal) just shrink too damn much for repraps. Currently ABS and PLA are the only plastics that are really sold everywhere, but if everyone in the community could suddenly print acetal easily I'm sure it would be everywhere in weeks. I think we're limiting our community's potential by not giving heated chambers more attention. I've seen people with them, but as I realized the other day most enclosures usually aren't enough. Hear me out: [hr] Let's assume the total surface area of an enclosure is about a square meter. That's a five-faced cube with ~450mm sides, and we're going to assume the table or whatever it's on isn't giving us any heat loss. I should probably account for a few more things, and my system is a little too ideal but this is just to illustrate a concept. So now let's use 6mm polycarbonate. It seems to be a pretty simple, clear solution to a bed. at 1 m^2 and .19 W/mK for polycarbonate, and a temp difference of 60C (Room temp of about 20, goal is about 80) That's 1900 watts. This exact number doesn't mean a whole lot because of the crude assumptions, but consider that a 12V heated bed pushing 10 amps is only giving your system 120 watts. Not even one-tenth. In reality you are helping to stall the air around your print which helps, but with a moving print head there are still lots of air currents going on, and at that point all your chamber is doing is preventing against a draft or two and raising the ambient temperature by maybe 10C. Let's try another example- 1" insulation foam. It comes a lot thicker, but 1" is still very lightweight and easy to work with. I know it's hard to get in some places but here in MN it's cheap as hell and you can get it anywhere (it's pretty necessary) Their site lists the thermal conductivity at only .029 W/mK, and rounding to 25mm thick puts us at like 70 watts. That's it! Move up to the 2" stuff and you're at like 35 watts which is crazy low. Your heated bed would provide all the power you needed to keep the enclosure nice and warm. But this foam is ugly, thick, hard to find for some, not clear, not healthy, etc. etc. I had another idea I would like to do the math for: What about two very thin sheets of plastic with an air pocket in between them? The air in the middle is so stagnant that it can mostly be modeled as a solid as far as conductivity goes. This is how windows are made- I have a 6' square bay window, and if it was solid glass there's no way I could heat my house when it gets down to -20 every year. So let's try this: 5 mil polycarbonate sheets with 1" of air between them: at .024W/mK that's only 60 watts. I'm not even going to add in the polycarbonate. Obviously it's not going to function exactly like that, but it's definitely in the right range. [hr] [b][u]In conclusion:[/u][/b] Your heated bed puts about 110W into your enclosure. Here's very roughly how much heat is lost to a 20C room through a 1m[sup]2[/sup] area enclosure: 6mm polycarbonate: [b]2000 W[/b] 1" Insulation Foam: [b]70 W[/b] Thin plastic, 1" Air Gap: [b]60 W[/b] To achieve the same insulation as the air or foam, you would need to make a solid polycarbonate enclosure [b][i]20 CM THICK.[/i][/b] To do the same out of cork (common bed insulator) it would still need to be 60 mm thick! I'm about to do a series of test prints with both kinds of enclosures. I figure the foam one can be done for under $10 and the plastic one even cheaper. I'm guessing in reality I might have to do 3 plastic layers or a few inches of spacing for the air pocket idea. We'll see. Does anyone have any other ideas or comments? Let me know what you thinkThanks. https://reprap.org/forum/read.php?1,446755,446755#msg-446755 Mon, 09 Mar 2026 01:16:04 -0400 Phorum 5.2.23 https://reprap.org/forum/read.php?1,446755,494826#msg-494826 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494826#msg-494826
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cdru

I know it's an older post, but your numbers are incorrect for what you're trying to compute as you are confusing thermal conductivity with heat transfer coefficient.

Thermal conductivity is how well heat flows within a material. For our heater blocks, we want heat to flow very well so that it's even and no hot spots right next to the heater but not on the far side of the melt chamber. For this, we choose aluminum or brass because they have a "k value" in the hundreds of W/(m K). But because the thermal conductivity is so good, this is why all metal hot ends heat breaks aren't made of aluminum...it would wick all that heat away up into the cooling fins defeating the purpose of a heat break. So instead they are made of stainless steel which has a k value of around 16 W/(m K). Polycarbonate has a k value of .19 W/(m K). It would make a excellent material for a heat break as it resists the transfer of heat...if only it could withstand the heat (it can't). In terms of your enclosure, 1900 Watts is the rate of heat transfer WITHIN THE POLYCARBONATE if you had the inside surface at 80 degrees Celsius and the outside at 20 degrees Celsius. Slowly over time though the outside surface would raise in temperature as more heat from the inside was generated. That is then where the heat transfer coefficient comes into play...

The heat transfer coefficient is basically how well the material gives off heat, usually in the form of convection between solids and a fluid or gas. This multiwall polycarbonate sheet like was shown above has a coefficient of 3.7W/(m² K). I found a few solid 6mm sheets that had a coefficient of 3.5W/(m² K). A box with a surface area of 1 meter made of solid polycarbonate 6mm thick would lose 3.5 W/(m² K) * 1m² * 60 Kelvin = 210 Watts, NOT 1900 Watts.

Yeah I guess I didn't explain that well enough.

You are correct in that in a perfectly stagnate room it would be much less than 1900 Watts, but due to convective transfer and the (essentially) infinite mass of the air in the room, this is not the case. The heat transfer coefficient of air is between 10 and 100 W/m^2/K so I'm just assuming that convective air transfer >> conductive transfer through enclosure. My printer is in a large room that people are constantly coming and going from and there is often a fan on or window open or something so this is a pretty safe assumption. I'm sure if you had it in a closet or small, unoccupied room you would get a much smaller heat transfer coefficient for the air and therefore less energy lost, but you want to plan for the worst case scenario to avoid variability. If I calibrate my enclosure so that 99% of the time the air is stagnate around it and insulation is sufficient, I'm going to get fluctuation that one time that I open the door fast and cool it down a few degrees C. By assuming the "completeness" of the heat transfer to room temp I'm eliminating all external variables. I know my calculations are rough, but I'd bet they're within 50% for a setup like mine.

What you're seeing is the U value, which is called thermal transmittance. This is used often in buildings and is used to measure entire walls, or in instances like windows where there is a gas between panes. It is similar to k in that it measures heat flux through the material rather than being a heat transfer coefficient. They give it in different units because it is usually experimentally measured for each system of materials, especially in one like the poly you linked to where the heat flux isn't uniform due to the air pockets. It looks like fantastic stuff and you are very correct in your math for it's heat loss, but for a solid piece of polycarbonate there is no way the U value would get that low. If it has a k value of .19 W/mK, then for a 6m piece the U value would be the k value divided by thickness, or 32 W/m^2K. In that case, (32 W/m^2K)(60K)(1m^2) = 1920 Watts.

But in the same way that my values are exaggerated because of my assumption, so would the 210 watts you calculated which is really cool because that's a really low heat loss!]]> SamS General Thu, 09 Apr 2015 12:10:20 -0400 https://reprap.org/forum/read.php?1,446755,494622#msg-494622 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494622#msg-494622 tmorris9 General Thu, 09 Apr 2015 02:03:58 -0400 https://reprap.org/forum/read.php?1,446755,494592#msg-494592 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494592#msg-494592 skynetprinter General Wed, 08 Apr 2015 22:31:25 -0400 https://reprap.org/forum/read.php?1,446755,494321#msg-494321 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494321#msg-494321


I already had the carboard draught shield in place, so I just put 2 plastic bags over the top of the machine and taped the inner one to the cardboard. The air space between the plastic bags stops the heat escaping too quickly.
]]> dc42 General Wed, 08 Apr 2015 11:41:17 -0400 https://reprap.org/forum/read.php?1,446755,494301#msg-494301 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494301#msg-494301
Here is the build blog on the subject. -> [engineerd3d.ddns.net]

Once I get a proper door and insulate the remaining parts, I am confident I can reach the desired 60c with relative ease. Should help with the bed heat up times as well. One more thing, having a blower fan ontop with a charcoal filter has reduced the ABS fumes quite a bit once the print is done, I use the blower to evacuate the printer chamber and filter the air.

-Bruno]]> jaguarking11 General Wed, 08 Apr 2015 11:12:59 -0400 https://reprap.org/forum/read.php?1,446755,494110#msg-494110 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494110#msg-494110
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Djkingsley
60 degrees Kelvin is -213.15 degrees Celsius. I believe you meant to use 333.15 degrees Kelvin which would give you 3.5 W/m^2K * 1m * 333.15 K = 1166 W
No. The formula is for the Δt...I just happened to put what the Δt was and not (80 degrees C - 20 degrees C). The original example was 20 degree C ambient air temperature and a theoretical 80 degree C build chamber. That's 60 degrees C. If it was in K, it would 293.15 K ambient and 353.15 K build chamber. And look, it's still a difference of 60 just in Kelvin. Kevin...degrees C...when talking about a Δt it doesn't matter between the two.]]> cdru General Tue, 07 Apr 2015 23:59:00 -0400 https://reprap.org/forum/read.php?1,446755,494097#msg-494097 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494097#msg-494097
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SamS

Is getting insulation foam easy in Europe? How about Australia/NZ? I know there are some strange restrictions on it in California so I guess I just assumed it was scarce.

yes you can get it from ebay that withstands very high temps

you can also get it recycled or from standard sources. I had some that came in a box padding that seemed to be an unusual dark foam, i tested it, it shrivels slightly at 240 and withstands 100 degrees just fine.

it's the same foam as velux use in it's side pads to stop rain splases going from window edge to under tiles. made to be under hot red tiles that can get to 100 degrees.

[www.metrotile.co.uk] it's a common foam of some kind i got it in a box from something.

For an enclosure, i am tempted to find a cardboard box of the right size and fit a thermometer to the side of teh print surface and a plexiglass to a window at the front of the cardboard box, insulative and you can lift it on and off so easily from the table.]]> skynetprinter General Tue, 07 Apr 2015 21:39:40 -0400 https://reprap.org/forum/read.php?1,446755,494059#msg-494059 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494059#msg-494059 Djkingsley General Tue, 07 Apr 2015 18:26:02 -0400 https://reprap.org/forum/read.php?1,446755,494036#msg-494036 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,494036#msg-494036
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Djkingsley
I think that there is a Celsius, Fahrenheit, and Kelvin conversion issue in your equation.
Feel free to point them out. Or if I need to provide sources to back up my math I can do that as well. All my units should be metric (Watts, meters, Kelvin) and the temperature component is actually a delta temperature change so degrees Celsius or Kelvin doesn't matter...they are the same just offset 273.15 from each other. I originally had just degrees and implied Celsius so I did update my post so that I removed any ambiguity about what "degrees" I was talking about. I also double checked my units to make sure they used the proper abbreviations.

If you still don't believe me, punch in the numbers into google.]]> cdru General Tue, 07 Apr 2015 17:27:34 -0400 https://reprap.org/forum/read.php?1,446755,493986#msg-493986 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493986#msg-493986
Quote
cdru
Quote
SamS
Let's assume the total surface area of an enclosure is about a square meter. That's a five-faced cube with ~450mm sides, and we're going to assume the table or whatever it's on isn't giving us any heat loss. I should probably account for a few more things, and my system is a little too ideal but this is just to illustrate a concept.

So now let's use 6mm polycarbonate. It seems to be a pretty simple, clear solution to a bed. at 1 m^2 and .19 W/mK for polycarbonate, and a temp difference of 60C (Room temp of about 20, goal is about 80) That's 1900 watts. This exact number doesn't mean a whole lot because of the crude assumptions, but consider that a 12V heated bed pushing 10 amps is only giving your system 120 watts. Not even one-tenth.
I know it's an older post, but your numbers are incorrect for what you're trying to compute as you are confusing thermal conductivity with heat transfer coefficient.

Thermal conductivity is how well heat flows within a material. For our heater blocks, we want heat to flow very well so that it's even and no hot spots right next to the heater but not on the far side of the melt chamber. For this, we choose aluminum or brass because they have a "k value" in the hundreds of W/(m K). But because the thermal conductivity is so good, this is why all metal hot ends heat breaks aren't made of aluminum...it would wick all that heat away up into the cooling fins defeating the purpose of a heat break. So instead they are made of stainless steel which has a k value of around 16 W/(m K). Polycarbonate has a k value of .19 W/(m K). It would make a excellent material for a heat break as it resists the transfer of heat...if only it could withstand the heat (it can't). In terms of your enclosure, 1900 watts is the rate of heat transfer WITHIN THE POLYCARBONATE if you had the inside surface at 80 degrees and the outside at 20. Slowly over time though the outside surface would raise in temperature as more heat from the inside was generated. That is then where the heat transfer coefficient comes into play...

The heat transfer coefficient is basically how well the material gives off heat, usually in the form of convection between solids and a fluid or gas. This multiwall polycarbonate sheet like was shown above has a coefficient of 3.7W/m²K. I found a few solid 6mm sheets that had a coefficient of 3.5W/m²K. A box with a surface area of 1 meter made of solid polycarbonate 6mm thick would lose 3.5 W/m²K * 1m * 60 degrees = 210 watts, NOT 1900 watts.


I think that there is a Celsius, Fahrenheit, and Kelvin conversion issue in your equation.]]> Djkingsley General Tue, 07 Apr 2015 15:38:29 -0400 https://reprap.org/forum/read.php?1,446755,493961#msg-493961 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493961#msg-493961
Quote
SamS
Let's assume the total surface area of an enclosure is about a square meter. That's a five-faced cube with ~450mm sides, and we're going to assume the table or whatever it's on isn't giving us any heat loss. I should probably account for a few more things, and my system is a little too ideal but this is just to illustrate a concept.

So now let's use 6mm polycarbonate. It seems to be a pretty simple, clear solution to a bed. at 1 m^2 and .19 W/mK for polycarbonate, and a temp difference of 60C (Room temp of about 20, goal is about 80) That's 1900 watts. This exact number doesn't mean a whole lot because of the crude assumptions, but consider that a 12V heated bed pushing 10 amps is only giving your system 120 watts. Not even one-tenth.
I know it's an older post, but your numbers are incorrect for what you're trying to compute as you are confusing thermal conductivity with heat transfer coefficient.

Thermal conductivity is how well heat flows within a material. For our heater blocks, we want heat to flow very well so that it's even and no hot spots right next to the heater but not on the far side of the melt chamber. For this, we choose aluminum or brass because they have a "k value" in the hundreds of W/(m K). But because the thermal conductivity is so good, this is why all metal hot ends heat breaks aren't made of aluminum...it would wick all that heat away up into the cooling fins defeating the purpose of a heat break. So instead they are made of stainless steel which has a k value of around 16 W/(m K). Polycarbonate has a k value of .19 W/(m K). It would make a excellent material for a heat break as it resists the transfer of heat...if only it could withstand the heat (it can't). In terms of your enclosure, 1900 Watts is the rate of heat transfer WITHIN THE POLYCARBONATE if you had the inside surface at 80 degrees Celsius and the outside at 20 degrees Celsius. Slowly over time though the outside surface would raise in temperature as more heat from the inside was generated. That is then where the heat transfer coefficient comes into play...

The heat transfer coefficient is basically how well the material gives off heat, usually in the form of convection between solids and a fluid or gas. This multiwall polycarbonate sheet like was shown above has a coefficient of 3.7W/(m² K). I found a few solid 6mm sheets that had a coefficient of 3.5W/(m² K). A box with a surface area of 1 meter made of solid polycarbonate 6mm thick would lose 3.5 W/(m² K) * 1m² * 60 Kelvin = 210 Watts, NOT 1900 Watts.]]> cdru General Tue, 07 Apr 2015 14:11:27 -0400 https://reprap.org/forum/read.php?1,446755,493871#msg-493871 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493871#msg-493871 the_digital_dentist General Tue, 07 Apr 2015 10:51:41 -0400 https://reprap.org/forum/read.php?1,446755,493849#msg-493849 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493849#msg-493849 www.alibaba.com]]]> sungod3k General Tue, 07 Apr 2015 09:41:02 -0400 https://reprap.org/forum/read.php?1,446755,493833#msg-493833 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493833#msg-493833
Quote
sungod3k
I like the idea and even calming down the airflow would be positive, however my main concern would be hotend cooling. Which would require a either a big hose getting outside air onto the hotend and model cooling fans or a pressurized fan that sits outside and blows in air though a smaller tube.

I have been thinking about pressurized fan already independent of the enclosure. It would make aiming the model cooling airflow much easier if you had 2-4 little tubes blowing on you model instead of mounting a fan (or two) and funnels that cover only half of the model anyway,
]]> the_digital_dentist General Tue, 07 Apr 2015 08:57:36 -0400 https://reprap.org/forum/read.php?1,446755,493813#msg-493813 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493813#msg-493813 sungod3k General Tue, 07 Apr 2015 07:56:55 -0400 https://reprap.org/forum/read.php?1,446755,493803#msg-493803 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493803#msg-493803 rsilvers General Tue, 07 Apr 2015 07:40:58 -0400 https://reprap.org/forum/read.php?1,446755,493774#msg-493774 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493774#msg-493774
I have been thinking about pressurized fan already independent of the enclosure. It would make aiming the model cooling airflow much easier if you had 2-4 little tubes blowing on you model instead of mounting a fan (or two) and funnels that cover only half of the model anyway,]]> sungod3k General Tue, 07 Apr 2015 06:14:10 -0400 https://reprap.org/forum/read.php?1,446755,493683#msg-493683 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,493683#msg-493683
The heated bed heats it up. I am running it now with PC filament at 60C inside using a bed set to 120C.


]]> rsilvers General Mon, 06 Apr 2015 22:18:45 -0400 https://reprap.org/forum/read.php?1,446755,471129#msg-471129 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,471129#msg-471129 www.thingiverse.com]
If you use 1/2" sheet, just scale the parts by 50% and print. If you don't need to take the enclosure apart and reassemble it just use tape to hold the pieces of foam or whatever you use for the walls and top together.

My printer has a 450 watt heater and keeps the enclosure very warm without any additional heat source. I keep the door propped open a little to keep it from getting too hot inside the box. You don't have to go nuts with insulation to keep things warm. If your bed heater is insufficient to keep the box warm enough it is very cheap to put in a few incandescent light bulbs that will heat and light the enclosure at the same time.]]> the_digital_dentist General Sun, 15 Feb 2015 17:47:02 -0500 https://reprap.org/forum/read.php?1,446755,470911#msg-470911 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,470911#msg-470911 i don't have 3d printer yet, but i will buy one soon, and i'm already looking for an enclosure, and i found your thread. So, to answer one of your questions : yes we can easily find isolation foam here in France if this is what you meant
also i was wondering if using some reflective sheet (like this) could help maintening the warm inside (or maybe just alu sheet like the one we use for cooking)
also, i think it would be "safer" to include some charcoal filter somewhere in the enclosure to avoid bad smell when opening
and finally, what about some dessicant (like silicagel) inside to avoid 'vapor' due to temperature ?]]> loneplayer General Sun, 15 Feb 2015 05:29:32 -0500 https://reprap.org/forum/read.php?1,446755,447226#msg-447226 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447226#msg-447226
Anyway, I am getting my enclosure done from lexan and we'll se how it goes. Especially when I install a hexagon hotend and try some high temp prints]]> t0bb General Wed, 24 Dec 2014 05:33:26 -0500 https://reprap.org/forum/read.php?1,446755,447166#msg-447166 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447166#msg-447166 The vacuum idea is very new and not reliable yet.]]> SamS General Tue, 23 Dec 2014 21:32:44 -0500 https://reprap.org/forum/read.php?1,446755,447143#msg-447143 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447143#msg-447143 Windows protect from heat loss because there is a vacum in between the panels (or it should be if they are made correctly)
..so no air current possible there.]]> t0bb General Tue, 23 Dec 2014 19:17:42 -0500 https://reprap.org/forum/read.php?1,446755,447135#msg-447135 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447135#msg-447135
Most of the heat is lost from the belt holes on the top. That's the only place I can feel heat at all, even touching the lexan. I will be printing some covers to block most of those openings on top.

My brother just got an IR camera, so we will be looking at it with that after Christmas.]]> Paul Wanamaker General Tue, 23 Dec 2014 18:46:06 -0500 https://reprap.org/forum/read.php?1,446755,447131#msg-447131 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447131#msg-447131
Paul, yeah that's exactly what I had in mind. Here in Minnesota they regularly stock up to 2" (maybe not that R value but still) Have you gotten the chance to get an IR profile of your heat loss or anything like that? How'd you do the joints? I'd be interested to know where/when the most heat loss is coming from.]]> SamS General Tue, 23 Dec 2014 18:32:11 -0500 https://reprap.org/forum/read.php?1,446755,447122#msg-447122 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447122#msg-447122
Lexan panels are fairly pricey so you may want to reserve those for the door. I used double panels which I bent.

You may want to consider foam panels like these: R-matte Plus 3 has R3.2
Only about $13.00 from Home Depot here in the US for a 1/2" x 4' x 8' sheet, and can take reasonable heat. That's what I used.

They also have 3/4" available and possibly thicker (probably not needed).

I'm using a 200w bed and 100W in incandescent lights. The interior is about 60C and I have little warping. Testing is ongoing tho as my printer is new.]]> Paul Wanamaker General Tue, 23 Dec 2014 18:05:04 -0500 https://reprap.org/forum/read.php?1,446755,447099#msg-447099 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447099#msg-447099 tmorris9 General Tue, 23 Dec 2014 17:05:51 -0500 https://reprap.org/forum/read.php?1,446755,447096#msg-447096 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447096#msg-447096
If you're looking to maintain a decent temp though, I don't think that greenhouse stuff (at 5mm anyway) will work. If I were you I would find out the equivalent heat coefficient of that material, because even if it's like .05 W/mK or something like that you'll be wasting your money on a 40W heater...40W is really small. Remember, my math gave the IDEAL scenario so although it probably helps a lot, to hold 80C you'd need a hell of a low more. I think you can get regular sized heated bulbs in the 200-300 rage and if I were you I'd go for that with only a 5mm wall.

As far as the patent goes, you're right ryan_m, Stratasys has like 5 patents on filament spools so I don't know exactly where the enforcement line is. Check out all of their patents- it's absurd. [www.google.com]

What I'm proposing though is that we come up with something so easy that you don't really have to buy anything special for. No one would sell anything besides maybe the brackets because it would be absurd to- seran wrap, garbage bags, foam, etc. The reason I think this is possible is that I'd bet 99% of reprap owners who can fit a build chamber don't care if it's 5mm thick or 50 mm thick because space isn't usually that crucial.

Is getting insulation foam easy in Europe? How about Australia/NZ? I know there are some strange restrictions on it in California so I guess I just assumed it was scarce.]]> SamS General Tue, 23 Dec 2014 16:55:31 -0500 https://reprap.org/forum/read.php?1,446755,447072#msg-447072 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447072#msg-447072 tmorris9 General Tue, 23 Dec 2014 15:45:53 -0500 https://reprap.org/forum/read.php?1,446755,447070#msg-447070 Re: Enclosure math- why they're so important and how to make one. https://reprap.org/forum/read.php?1,446755,447070#msg-447070 tmorris9 General Tue, 23 Dec 2014 15:38:19 -0500