Is the 110 m length accurate for 3.0 mm 1 kg spool?

Question

I've been seeing the 330 m and 110 m length estimates for 1.75 and 3.0 mm spools (of PLA, presumably). But a moment's thought staring at that will raise an obvious question.

Using πr², we get the area of the filament in square millimeters (rounding to two decimal points)

For 1.75 it's 2.41 mm²

For 3.00 it's 7.07 mm²

Then taking the ratio and multiplying:

(2.41/7.07) * 330 m = 112 m which is close enough to 110 m.

BUT as all makers of filament and makers of extruders know, 3.00 mm is just a rounding off of the real dimension, which is 2.85 mm. Now do that:

For 2.85 it's 6.38 mm²

and:

(2.41/6.38) * 330 m = 125 m which is at least 120 m.

So, whoever calculated the approximation of 110 m did the calculation based on the rounded rather than actual dimension. What am I missing here?

My point is not that the 330 and 110 would be inaccurate given the dimensions of 1.75 and 3.00. Rather, my point is that the 3.00 mm diameter is not what is really so; it's actually 2.85 mm and therefore the answer is longer than 110 m.

The figure does not appear in just one place, but _has been copied numerous places taking it for granted that it is correct_. I consider _that_ to matter, yes. — nebogipfel, Aug 06 '18 at 19:16
If you feel one answer solves your question, please accept one. — Trish, Sep 18 '18 at 15:34

Trish · Answer 1 · 2018-09-14T10:17:44.147

Let's go about this scientific:

$A_r=\pi{r}^{2}=\pi{\frac{d}{2}}^{2}$

As a result the crossections are $A_{3}=7.06858\text{mm}^2=0.0707\text{cm}^2$, $A_{2.85}=6.3794\text{mm}^2=0.0638\text{cm}^2$ and $A_{1.75}=2.40528\text{mm}^2=0.024\text{cm}^2$.

Volume of a cylinder is $V_{A_d,l}=\times {A_d} {l}$. Turned around to get a length from Volume and Area we get $l=\frac {V_m}{A_d}$, but what is V?

We know the density of comercial PLA is about $\rho=1.25 \frac{\text g}{\text{cm}^3}$, and we know $m=\times V \rho$. So: $V_m=\frac{m}{\rho}=\frac{1000}{1.25}\text{cm}^3=800\text{cm}^3$.

Taking this Volume and using the $l=\frac {V_m}{A_d}$ we get:

$l_{d=1.75}=33333.33\frac{\text{cm}}{\text{kg}}=333.33\frac{\text{m}}{\text{kg}}$

$l_{d=2.85}=12539.18\frac{\text{cm}}{\text{kg}}=125.39\frac{\text{m}}{\text{kg}}$

$l_{d=3}=11315.41\frac{\text{cm}}{\text{kg}}=113.15\frac{\text{m}}{\text{kg}}$

If the filament is more on the dense side $(\rho>1.25\frac{\text g}{\text{cm}^3})$, then it will have a smaller volume and thus be shorter than this estimate.

To show this better, a graph: This is the length of a filament spool in dependancy of the density. The values were calculated for the usual diameters with their closest neighbors rounded to 0.1 as absolute diameters and run over a broad range of densities commonly used in plastics - 0.7 g/cm³ to 2 g/cm³.

This is calculated based on 3.00mm. My point is that such filament is actually 2.85mm and therefore longer than usually cited. — nebogipfel, Aug 05 '18 at 16:33
@nebogipfel I added the calculateions for 2.85mm too. Remember, Most 3mm filaments are more on the thicker side than the 2.85, but if your supplier has this number down perfectly, then you should get those. Also, keep in mind that I do assume the density - PLA can start at 1.2 and go up to 1.45 g/cm³ — Trish, Aug 05 '18 at 16:47
@nebogipfel also, added a (huge) graph that should cover enough density variation to be true for almost all plastics - the lower group is the "3.0" diameter (+/- 0.05mm), the upper the "1.75" one — Trish, Aug 06 '18 at 09:19

score 2 · Answer 2 · answered Aug 06 '18 at 08:46

The 110 meter figure is for 3.0 mm filament that actually is 3.0 mm in diameter. As you (correctly) note, most 3.0 mm filament is often actually closer to 2.85 mm. In that case, the correct figure is indeed around 125 meters. Note that this figure is for PLA filament (see e.g. my answer here). For ABS, being significantly less dense, a 1 kg spool would be around 155 meters.

Is the 110 m length accurate for 3.0 mm 1 kg spool?

2 Answers2