<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>

[{'answer_id': 6559, 'author': 'Trish', 'author_id': 8884, 'author_profile': 'https://3dprinting.stackexchange.com/users/8884', 'pm_score': 2, 'selected': False, 'text': '<p>Let\'s go about this scientific:</p>\n\n<p>$A_r=\\pi{r}^{2}=\\pi{\\frac{d}{2}}^{2}$</p>\n\n<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>\n\n<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>\n\n<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>\n\n<p>Taking this Volume and using the $l=\\frac {V_m}{A_d}$ we get:</p>\n\n<p>$l_{d=1.75}=33333.33\\frac{\\text{cm}}{\\text{kg}}=333.33\\frac{\\text{m}}{\\text{kg}}$</p>\n\n<p>$l_{d=2.85}=12539.18\\frac{\\text{cm}}{\\text{kg}}=125.39\\frac{\\text{m}}{\\text{kg}}$</p>\n\n<p>$l_{d=3}=11315.41\\frac{\\text{cm}}{\\text{kg}}=113.15\\frac{\\text{m}}{\\text{kg}}$</p>\n\n<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>\n\n<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>\n\n<p><a href="https://i.stack.imgur.com/Jydly.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jydly.png" alt="length of filament depending on diameter and density"></a></p>\n'}, {'answer_id': 6568, 'author': 'Tom van der Zanden', 'author_id': 26, 'author_profile': 'https://3dprinting.stackexchange.com/users/26', 'pm_score': 2, 'selected': False, 'text': '<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>\n'}]
['https://3dprinting.stackexchange.com/questions/6557', 'https://3dprinting.stackexchange.com', 'https://3dprinting.stackexchange.com/users/12449/']