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First time poster here. First of all please excuse my english mistakes but most of all my terminology and ignorance on the subject.

Now to the question! A friend of mine asked me if I could help him convert his ceramic 3d printer extruder from compressed air to a piston system (driven by a leadscrew linear actuator). As of now, he told me, he uses 4 bar of pressure to make the clay extrude at the desired rate and that's all I know plus the nozzle and cartridge internal diameter (respectively 1 and 65 mm). What I'm looking for is a stepper motor/lead screw combination that can exert enough linear force to move the clay through the nozzle (the rate of extrusion would be given by the stepping rate of the motor).

Here is what I've done as "homework" but I'm a firmware developer so I'm hoping to get some guidance in solving this problem. Since the nature of this problem is not only practical but also not needing particular precision, what I've done are essentially ballpark, or back of the envelope, calculations.

  1. converted bar to N/mm2 (4 bar -> 0.4 N/mm2)

  2. multiplied N/mm2 by the cartridge cross sectional area ($0.4\text{ N/mm}^2 \cdot 32.5^2\pi\text{ mm}^2$) to get the total force exerted on the clay of ~1320 N (I'll round this to 1400 N to have some margin)

Knowing the force needed I picked a lead screw from a catalog to get some specs for the next calculations, what I chose is a screw of 10 mm diameter and 2 mm/rev lead

  1. using this Wikipedia article, I calculated the needed torque to raise a load of 1400 N, assuming that's the clay's resistance according to step #2. (I used the raise formula with the stainless steel coefficient of friction to purposely oversize the system a bit to have some overhead). So the terms were:

$$\begin{alignat}{2} F &= \text{load on the screw} &&= 1400\text{ N} \\ d_m &= \text{mean diameter} &&= 0.01\text{ m} \\ \mu &= \text{coefficient of friction} &&= 0.2 \\ l &= \text{lead} &&= 0.002\text{ m} \end{alignat}$$

$$ \text{Torque} = \frac{Fd_{m}}{2} \left( \frac{l+ \pi \mu d_{m}}{ \left( \pi d_{m}- \mu l \right)} \right) $$

As a side note I also used the formula with the lead angle and angle of friction to be sure of my results for which I used:

$$\begin{alignat}{2} \phi &= \text{angle of friction} &&= 11.3° \\ \lambda &= \text{lead angle} &&= 3.6433° \end{alignat}$$

$$\text{Torque} = \frac{Fd_{m}}{2} tan(\Phi + \lambda)$$

for a total result of 1.86 Nm of torque (from both formulae!)

Well that's all nice and gives an answer that I like (a torque of 1.9 Nm is quite common for steppers).

But the fact that I like it doesn't mean that it's correct!

What makes me suspicious is that 1400 N of force is more or less what's needed to hold up a 140 kg man (please bear with me on that... I'm trying to use some "everyday experience" to get to a correct answer) and I can extrude the clay from the cartridge with my bare hand... I'm sure I'm not able to put 140 kg on that!

So what I'm asking is essentially if I followed the right path in tackling this problem.

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  • $\begingroup$ Is my question really not worthy of a single comment saying "you're doing right" or "nope, you got it wrong"? even without details, just a comment to point me in the right direction $\endgroup$ – zakkos Sep 30 '16 at 10:00
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I'm unable to post a comment as I don't yet have 50 reputation!

It looks to me like you've gone about this problem in a sensible and logical manner. I have a few questions, however.

  1. Can you confirm if the pressure that your friend uses is really "4 bar" (absoute), or 4 barg (gauge, or above atmostpheric)? You have calculated for 4 barg.

  2. Your suspicion seems to stem from a mismatch between your human strength and the numbers. This is an exceptionally good way to 'sanity check' your answers! How does the rate at which you are able to push the clay through 'by hand' compare to the 4 bar(g) compressed air system? If the air can push it through at double the rate, then it will require approximately double the force. Does this clear up the discrepancy?

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  • $\begingroup$ Thank you Will for your time. In the end everything worked out, I don't know the exact force that the piston makes since I don't have a proper gauge but it's ok for the job. 1) The pressure measured is indeed in barg, so 4 barg. 2) I don't know the rate of extrusion of the compressed air system, now it's around 1mm/s. It might have been double that or more This can absolutely be why I was having trouble figuring out if my calculations where at least in the correct range. Again, thanks for your time. Since I was asking essentially for a check and you provided it I'm accepting your answer $\endgroup$ – zakkos Nov 24 '17 at 14:08
  • $\begingroup$ Thank you! I'm now over the limit and can comment in future :) $\endgroup$ – Jonathan R Swift Nov 27 '17 at 9:35

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