<p>Let's go about this scientific:</p>

<p>$A_r=\pi{r}^{2}=\pi{\frac{d}{2}}^{2}$</p>

<p>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 $A_{1.75}=2.40528\text{mm}^2=0.024\text{cm}^2$.</p>

<p>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?</p>

<p>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: $V_m=\frac{m}{\rho}=\frac{1000}{1.25}\text{cm}^3=800\text{cm}^3$.</p>

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

<p>$l_{d=1.75}=33333.33\frac{\text{cm}}{\text{kg}}=333.33\frac{\text{m}}{\text{kg}}$</p>

<p>$l_{d=2.85}=12539.18\frac{\text{cm}}{\text{kg}}=125.39\frac{\text{m}}{\text{kg}}$</p>

<p>$l_{d=3}=11315.41\frac{\text{cm}}{\text{kg}}=113.15\frac{\text{m}}{\text{kg}}$</p>

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

<p>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.</p>

<p><a href="https://i.sstatic.net/Jydly.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Jydly.png" alt="length of filament depending on diameter and density"></a></p>

CC BY-SA 4.0
2018-08-05T16:33:21.267Z
This is calculated based on 3.00mm. My point is that such filament is actually 2.85mm and therefore longer than usually cited.
nebogipfel
CC BY-SA 4.0
2018-08-05T16:47:42.600Z
@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
CC BY-SA 4.0
2018-08-06T09:19:27.940Z
@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
CC BY-SA 4.0
2018-08-05T15:37:19.760Z
2018-09-14T10:17:44.147Z
2018-09-14T10:17:44.147Z
Trish
<p>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. <a href="https://3dprinting.stackexchange.com/a/1452">my answer here</a>). For ABS, being significantly less dense, a 1 kg spool would be around 155 meters.</p>

CC BY-SA 4.0
2018-08-06T08:46:16.590Z
2018-08-06T08:46:16.590Z
Tom van der Zanden
<p>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.</p>

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

<p>For 1.75 it's 2.41 mm²</p>

<p>For 3.00 it's 7.07 mm²</p>

<p>Then taking the ratio and multiplying:</p>

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

<p>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:</p>

<p>For 2.85 it's 6.38 mm²</p>

<p>and:</p>

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

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

<p>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.</p>

CC BY-SA 4.0
2018-08-06T15:35:11.340Z
Does it really matter?
Carl Witthoft
CC BY-SA 4.0
2018-08-06T19:16:30.900Z
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
CC BY-SA 4.0
2018-09-18T15:34:32.463Z
If you feel one answer solves your question, please accept one.
Trish
<filament>
Is the 110 m length accurate for 3.0 mm 1 kg spool?
nebogipfel
# Is the 110 m length accurate for 3.0 mm 1 kg spool?

(Asked by: username_0 on 2018-08-05)

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.

**Tags:** filament

> **Comments:**
> **username_1**: Does it really matter?  
> **username_0**: 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.  
> **username_2**: If you feel one answer solves your question, please accept one.  

---

## 2 answers

### Answer by username_2 on 2018-08-05. Score: 3

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.

[![length of filament depending on diameter and density](https://i.sstatic.net/Jydly.png)](https://i.sstatic.net/Jydly.png)

> **Comments:**
> **username_0**: This is calculated based on 3.00mm. My point is that such filament is actually 2.85mm and therefore longer than usually cited.  
> **username_2**: @username_0 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  
> **username_2**: @username_0 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  

---

### Answer by username_3 on 2018-08-06. Score: 2

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](https://3dprinting.stackexchange.com/a/1452)). For ABS, being significantly less dense, a 1 kg spool would be around 155 meters.


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