# Density of GEL-LAY and LAYWOO 3D print material?

I'm looking for the specific density of the GEL-LAY and LAYWOO 3D materials by manufacturer CC Products.

It isn't noted on their website or on the spool or the box the spools came in. I've looked for hours on Google and various websites, from resellers to people who tested it, without being able to find it.

• You mean the density of the material as on the spool, yes? Nov 19, 2018 at 10:56
• @Trish Yes, the density of the material on the spool.
– Sava
Nov 23, 2018 at 19:40

I can't provide the end answer, but if you already have the material, you should be able to measure this yourself quite simply.

Measure and cut a sample of filament, and weigh it. For example, a 10 meter length with a 1.75 mm diameter will have a volume of:

v = pi * r2 * l

v = pi * (0.175 cm/2)2 * 1000 cm

v = 24.05 cm3

Density is mass divided by volume. If your sample weighs 18 g, this would be

d = m / v

d = 18.0 g / 24.05 cm3

d = 0.748 g/cm3

Note that the accuracy of this measurement will depend on the accuracy and precision of your measurements. A household kitchen scale might not be good enough for such small weights. In order to get a good weight measurement, you may need to use a much longer (and heavier) sample of filament.

• Not sure that I have an accurate enough balance, although it could probably work with the small balance we used in the office to weigh letters. I'm thinking your solution could work if I would be to print a cube using the material. Volume of the cube is easily calculated, mass on the balance, then just use your equation to find density. You think it would work?
– Sava
Dec 6, 2018 at 20:04
• Your suggested method will work to measure the average density of the cube as printed. However, when printing with an FDM printer, the resulting object is rarely solid. Typically, the outer shell is only a few layers thick, and the inner volume is printed with an "infill" that is sufficient to support material printed above it and adds strength. This will be determined by various setting in your slicer software; there are options for different infill patterns and approximate density (%). For almost all prints, the inside of the object will be mostly air. Dec 7, 2018 at 17:19
• I know, but I can set it to 100% infill to print it solid, thus have a solid cube of x cm3 then weigh it and calculate.
– Sava
Dec 7, 2018 at 17:21
• Yes, that could work. Printing a solid cube with sufficient size seems likely to be problematic, and will certainly take a long time. if you have raw filament, you only need to know the length to calculate volume. If you are careful, you could still use that filament to print things with, so less wasteful as well. Dec 7, 2018 at 17:56

Indeed, the properties of this filament are kept rather secret, so to find out what the density is, you need to either contact the filament supplier or the manufacturer for accessing the data sheet or calculate this yourself. The answer below expands on the "calculate it yourself".

Density is defined as $$\rho = \frac{m}{V}=\frac{[kg]}{[m^3]}$$. The use of this formula has been show in this answer. The drawback of that answer is that it is an approximation that relies on a uniform piece of filament that requires cutting off expensive filament and relies on assumptions rather than actual calculations. Furthermore, the weighing of a small piece of filament is much less accurate of a small piece than for a larger piece or the whole spool (for the same scale, so a decent kitchen scale might be usable when more weight is concerned). You could improve the density calculation by measuring the diameter at various sections and make a better approximation based on the average diameter, but still that would need you to unroll the spool and carefully measure a piece of filament (and cut it). The advantage of that answer is that it is far easier than my proposition.

The method that is proposed here relies on a well known method to calculate the density of materials that is called hydrostatic weighing. Hydrostatic weighing uses the displacement of a fluid due to a submerged object to determine the density of the object. Any submerged object will displace the fluid surrounding it by it's own volume, as such you would need to measure the rise of the volume level to read the volume of the submerged product. This can be done accurately by using methods that include containers of known dimensions, known fluids and even an overflow method and weighing.

If the filament comes on a spool, you would require an identical spool to prevent removing it from the spool. But, I read that it is sold in bundles, not on spools. Without a spool would make it even easier to calculate the density as you do not have to subtract the spool weight and volume, the answer continues as if you have it on a spool. This is purely necessary so that you would not need to cut off filament or unroll the whole spool. The suggestion below let's you measure the whole spool, so weight is measured more easily as there is a lot more.

Theoretically, you could put the filament in a fluid which is known to not affect the filament properties (so not water for GEL-LAY!) in a bath of known dimensions. Once the spool and filament are completely submerged, you could measure the volume rise. If you do the same for the empty spool, you also know the volume of the spool alone. If you also are able to weigh the empty spool and the full spool (before you plunged them in the "bathtub"/container), you now know the volume and the weight of the filament, dividing the weight (full spool weight minus empty spool weight) by the volume (full spool submerged volume minus empty spool submerged volume) will give you the density.

$$\rho_{filament} = \frac{(m_{full\ spool}-m_{empty\ spool})}{(V_{full\ spool}-V_{empty\ spool})}$$

Now let your filament dry for a long, long time! :)

• I'm not too keen in submerging an entire roll of filament in whatever liquid that would be safe to use, wouldn't know what to use for Gel-Lay anyway, but that's still an interesting solution to my problem. Though drying time would be horrendous indeed! :p
– Sava
Dec 6, 2018 at 20:00