It seems that when filament throughput is increased (by increasing movement speed or extrusion width/height), printing temperature also has to be increased to compensate, because the filament will have less time to spend in the melting zone. That much seems clear from practical experience. But I have two questions (or to be more precise, one question on two levels):

  1. Is there a good rule of thumb for this, to help people calibrate their settings?

  2. How much do we know about the formula governing this behavior? Can we calculate the required hotend temperature precisely based on the increased throughput?

For anyone who has studied physics / thermodynamics, this is probably simple stuff. But has the work been done for 3D printing specifically, in a way that is practically applicable?

I share the following train of thought to start off with. Let me know if I make any errors in reasoning.

  • Presumably, every material has an optimal printing temperature just above its melting point.

  • But the thermistor doesn't read filament temperature. It reads the heat block temperature.

  • Below a certain throughput, the temperature of the filament will have time to equalize with the temperature of the heat block before it leaves the nozzle.

    • For those slow speeds, heat block temperature should be set exactly to the material's optimal printing temperature.
  • For greater speeds, however, heat block temperature will always have to be higher than the mark, because the filament doesn't have time to equalize.

    • At that point, it becomes a balancing act. Find the best heat block temperature (°C) given a rate of throughput (mm³/s), the optimal printing temperature for a given material (°C), the volume of the melting zone (mm³) and < some other property of the material >, which determines how fast it heats up. I don't know what that last property is, nor can I come up with the proper unit. The material probably approaches the temperature of the environment asymptotically. This is where thermodynamics comes in, I guess.
  • Theoretically, running filament also cools down the heat block, but we can ignore this. If this effect is significant at all (is it?), this is already compensated for by the PID controller.

I'm almost certainly missing some key insights. I'm curious to know what work has been done.

  • $\begingroup$ Did you ever find the answer to this? I have the exact same question. $\endgroup$ – Andrew Ebling Mar 29 '20 at 17:09

I think I see what you're asking, but I think you may be thinking about it incorrectly. It's really all about heat being added to the system at the same rate that it's leaving. The heat block is there as a heat reservoir from which the filament draws heat for the glass transition. The heat in that reservoir is maintained by cycling the heating coil to add energy (more heat) to the systems as it's lost.

In the very local vicinity of the nozzle, the temperature will decrease slightly as it's being transferred to the filament, but because the heat block is massive in comparison to that drain, and because the heat block is a good thermal conductor that temperature reduction is very small.

I do not know what tolerance and hysteresis are built into the temp controller, but think the variation is likely small. The difference in additional heat required (more energy into the system) for any practical difference in feed rates (40 instead of 60) is thus likely to be very small compared to the filament cooling experienced immediately after it leaves the nozzle.

Bottom line: the adjustment you would want to make is not to increase the temp, but increase the duty cycle of the heating element to maintain the desired temperature.

  • $\begingroup$ It's a bit more complicated than that. The problem is that it takes a certain amount of time Δt₁ for plastic to reach the temperature of its environment (the melting zone). That heat transfer is not instantaneous. Furthermore, there is a certain amount of time Δt₂ that the plastic can spend in the melting zone before it is extruded. As you increase throughput, you decrease Δt₂, to the point where Δt₂ < Δt₁. In those cases you have to either increase temperature or use a larger melting zone. That's why E3D has its volcano upgrade. $\endgroup$ – mhelvens Jan 22 '17 at 19:12
  • $\begingroup$ Increasing temperature would help because the plastic would heat up faster, reducing Δt₁. The problem there is that it also becomes possible to overshoot your target temperature, causing the plastic to come out too hot, resulting in droopy, saggy prints. --- Using a larger melting zone does not have this problem. But if you set your temperature higher than the ideal, it becomes a delicate balancing act. That's why I'd like to see the formula that describes this dynamic. $\endgroup$ – mhelvens Jan 23 '17 at 0:17
  • $\begingroup$ I think an equation would be very difficult, but doing an FE simulation might be tractable with the parameters you have. Much depends on the thermal conductivity of the heater block, flow rates and actual melting point configuration. Because it is hard to move temperature at singular point quickly, most systems naturally rely on heat reservoirs, which is usually an "overdamped system" in control parlance. What I think you're considering is an "underdamped system" of temp control which is faster acting, although usually less stable. Interesting problem, though. $\endgroup$ – Randy Steck Jan 23 '17 at 20:40
  • $\begingroup$ Yes, you set the temperature of the heater block higher than you want the plastic to be with the hope that the plastic exits the nozzle at just the right time. Faster acting and less stable. But I'm not considering this, so much as observing that this is what everyone does if they want to 3D print faster and/or with larger diameter nozzles. It's simply the only way to do it short of buying better components. I'm just trying to understand the dynamics better so that I can calculate the optimal set temperature for a given throughput. $\endgroup$ – mhelvens Jan 24 '17 at 10:56

I see an answer not a question. It's a balancing act and there is no predefined formula. Trial and error. Keep a spread sheet. I'll dwell on this a bit.. but as someone who did speed sprinting there's really nothing else to be said other than buy an e3d and the volcano upgrade.

You will calibrate one at a time. Thin wall. Then find solid infill is the true thermal barrier. Then sparse. You will tweak with layerheights. Get thicker nozzles 0.8+. It's a game of spinning plates. Each change will wack out another.

Last you will get to where I did. You move so fast 5 layers down your print is still molten and moving. Especially on small parts.

  • 1
    $\begingroup$ It feels like you're answering a question I did not ask. The situation is: I have decided on a material, an extrusion width/height and a print speed. The hardware is also fixed for the purpose of this question, and not available for tweaking. Question: How to find the optimal temperature setting (given that I've already found it for a lower speed)? A rule of thumb would be useful, but I'm especially curious about a formula to help us understand the dynamics involved. If such a formula hasn't yet been written down, then surely it could be written down. $\endgroup$ – mhelvens Jan 8 '17 at 1:11
  • $\begingroup$ Hi mhelvens I am sorry the answer did not work for you. I see now what you are asking. Might I politely suggest considering rewriting your question with the simplified and direct question you posted as a comment? Good questions get good answers. That said I am afraid there is no universal formula. Every printer is considerably different. My answer actually still works with some rewording. Examples like my melting 4 layers down is an example of the issues that appear the faster you go and cannot formulate. Same with jerk acceleration whiplash. $\endgroup$ – StarWind0 Jan 8 '17 at 3:21
  • $\begingroup$ It really comes down to too many unique factors with your printer. You can create a formula but it would have to be for one printer. Well I shouldn't say that. I know for a fact these formulas exist but only as IP of major companies and so buried not even their employees can easily find them. Or in the major 3d printing companies and they aren't going to share. $\endgroup$ – StarWind0 Jan 8 '17 at 3:25
  • 1
    $\begingroup$ This is all based on relatively simple physics, though. Sure, such a formula might have some additional variables in there that depend on the printer, but not that many. Mostly it'll be about the hotend and the material used, and such values could be found experimentally (and gathered in a publicly available table). Or could you name some significant variables that I'm forgetting? --- And you're probably right about my question being unclear. I'll try to rewrite it a bit today. $\endgroup$ – mhelvens Jan 8 '17 at 12:37
  • $\begingroup$ Spoken like a true PHD student. I encourage you to be the one to create such formula. If I have not seen it then it is not known to most. This is an open source movement! $\endgroup$ – StarWind0 Jan 8 '17 at 18:50

No there is no rule of thumb as far as I know and no, it's not obvious for someone with skills in thermodynamics.

I know that if you want extreme speeds you need extreme overtemperatures, for example Annex Engineering uses ABS at 290 °C (about 50-60 °C overtemperature) with a Mosquito Magnum (rated at about 40 mm^3/s) to reach 60 mm^3/s:

Still, you need to find the optimal value based on your hotend, your extruder, your nozzle size and target flow rate. You may do static tests as explained here:

There cannot be a rule for such a complex behaviour.

Also, be aware that Annex Engineering uses 23 000 mm/s^2 (not a typo, really 23 thousand) as acceleration, therefore their printing head moves at basically constant speed. Also, they set the slicer with the same speed for inner/outer perimeters, infill, and so on. If you have more common accelerations like 1500-2000 mm/s^2 and different printing speeds for different features, somewhere the filament will be pushed super fast through the nozzle and everything is good, somewhere else it will slow down and it will overcook.


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