<h2>My background</h2>

<p>I am not a polymer chemist.  I know some, and I've talked about UV curing resins with them.  These are my conclusions based on that informal education.  Your expertise may well be better.</p>

<h2>Minimizing curing on the printer is good</h2>

<p>For the maximum print rate and best layer adhesion, the resin should cure during the printing exposure the minimum amount that is consistent with the object supporting itself during the print, being removable from the bed, being rinsed in isopropyl alcohol, and stand on its own during curing.</p>

<p>If you are curing too completely during exposure, you won't get optimal mixing of polymer chains between layers.  The layers will be attached by the glue-like action of the new layer on the old, rather than having their polymer chains cross-linked and extending between layers.</p>

<p>If you are curing too little, the object won't be strong enough to support itself and survive post-processing.  Soft, gummy bits may dissolve, or be swept away by currents in the cleaning solution.  Small details may not be robust enough.  Supporting structures (as added supports or parts of the model) may not be strong enough to resist gravity and handling.</p>

<p>If you are curing too completely, your print will take longer than it could take.</p>

<h2>Curing is not binary</h2>

<p>Curing will take place spontaneously over a long time.  If it didn't, a bottle of resin might last a very long time.  But, curing is not a self-catalyzing process that runs quickly to completion.  If it were, then the first stray UV that came through an open bottle top would turn the contents solid.  If exothermic (which it seems like it must be), the bottle would get hot.</p>

<h2>UV curing may fail</h2>

<p>UV curing the inside of a large, opaque object probably doesn't happen.  Before I get all excited, though, I would need to place some bounds on "large" and "opaque".  </p>

<p>Absorption of UV light depends on the pigment or dye used in the resin.  This absorption is never absolute.  It is not 100% gone after the first, thinnest penetration.  The light is attenuated by an amount per distance it penetrates.  </p>

<p>Transmission is the complement of absorption, and the numbers are easier to work with, so lets work with transmission factors of T rather than absorbtion factors of A.  <span class="math-container">$T=1-A$</span></p>

<p>If a factor of T is passed for millimeter, then one centimeter into the object the light intensity is <span class="math-container">$T^{10}$</span> of what it is on the surface, which is a small, but non-zero, number.</p>

<p>Keep in mind that opaqueness depends on wavelength.  For example, Window glass is transparent (very little absorption) to visible light, but highly attenuating to UV light.  Were I to design a black resin, I would look for a black pigment that was relatively transmissive to UV.</p>

<h2>Will it cure?</h2>

<p>A low UV dose delivered to a 0.2mm layer of resin will partially cure the resin.  1 mm into an object, the dose is lower, but it still exists.  1 meter into the object, the absorption is probably too high pass a useful level of UV.  </p>

<p>If the transmission factor is 0.8 for a 0.2 mm layer, it is <span class="math-container">$0.8^5$</span> for a 1 mm layer (0.33).  It would take only three times as much UV exposure to cure a 1 mm thickness as a 0.2 mm thickness.  If the object were 1 meter thick, the transmission to the inner bit would be <span class="math-container">$0.8^{5000}$</span>, which is a very tiny number, roughly <span class="math-container">$2.82×10^{-485}$</span>.</p>

<p>Finally, consider if the object is or must be truly be solid.  UV curable resin is relatively expensive.  Many UV printable objects include drainage paths for uncured resin to flow out of the object during printing.  Perhaps your object could similarly be hollowed out.</p>

CC BY-SA 4.0
2020-05-14T16:13:56.617Z
The question was generic, but of course the implications were: attainable strength and reasonable amount of hollowing. What would be nice is to get some reference values for some common resins. I mean, getting UV only 1-1.5 mm in the object would be quite bad.
FarO
CC BY-SA 4.0
2020-05-14T14:29:26.630Z
2020-05-14T14:29:26.630Z
cmm
<p>From what I know about the various resins, once they are printed they are cured enough. Some of the transparent ones require additional curing, however the manufactures of the machines only test based on the max size that their machines can print. </p>

<p>The resin never stops curing apparently, going into diminishing returns; curing slower and slower as time goes on. Only way you will know for sure is by experimentation. You should get a poly-carbonate tube, place black tape along one side then spray paint the rest black. This will leave a window that you can use for future observation. Seal the bottom, and pour in uncured resin. Leave it under a UV source for 2 weeks then peal the tape off to see how much of it has cured. Then come back here and ask Trish to do the math on what the co-efficients of absorption are for the material :)</p>

CC BY-SA 4.0
2020-05-14T13:08:19.663Z
I thought that after lifting them from the vat they are still flexible. Maybe it applies only to transparent ones.
FarO
CC BY-SA 4.0
2020-05-14T13:25:37.963Z
Meanie to let me do the math! I point to the Penetration depth formulae and Beer-Lambert's laws: photons penetrate to some degree dependant on the energy of the light and the material structure penetrated. An article that popped up with the relevant keywords is [this](https://www.researchgate.net/publication/49966993_Energy_absorption_and_exposure_buildup_factors_for_some_polymers_and_tissue_substitute_materials_Photon_energy_penetration_depth_and_chemical_composition_dependence) but I have not yet read more than the abstract. Note that the paper used gamma ray and polymers that substitute ...
Trish
CC BY-SA 4.0
2020-05-14T13:26:36.740Z
tissue - the material closest to the resin might be the nylon and water substitute. While not a good analogue, note the general shape of the graphs, which shows the general behavior of high energy photons in matter - little interaction till some depth (mean free path, MFP), then a huge peak and then a falloff of energy deposition. This can be altered by altering the energy (=wavelength) of the photon beam and depends on the material...
Trish
CC BY-SA 4.0
2020-05-14T19:34:29.680Z
@trish at the energy levels of the UV, I think thus would behave more like light than gamma rays -- more driven by scattering and diffusion than particle interactions.  The energy is more like vibrational energy than ionization energy.  The activator is special in that it is the path between these low energy photons and the creation reactive radicals.
cmm
CC BY-SA 4.0
2020-05-15T11:59:40.580Z
@cmm Gamma radiation is light of 10^19 Hz/100+ keV. UV is 10^13-10^16 & <100 eV. Between the two lies X-ray, but all the 3 behave exactly the same but for the energy and free lengths shifting around. Gamma radiation, X-Ray, UV and visible light, just as much as microwaves and radar are all the same: ***LIGHT***. It's the amount of energy deposition that acts either ionizing (gamma) or heating (Infrared). Lifting an electron to a higher level to make the atm more reactive is technically ionizing!
Trish
CC BY-SA 4.0
2020-05-15T12:14:34.930Z
I mainly linked to the article to show the pattern: to deposit the energy in the material needed to start curing, there needs to be several factors given: The energy of the wave needs to be in a specific, material dependant band. The Mean Free Path has little to no influence on the position of the energy position, but the larger the MFP, the faster the energy deposits, telling us a density dependency of the process. The main interesting chart is on p.123: The 15 keV = 0.826^-12 m = 362*10^18 Hz chart shows us a huge dependency on the material for penetration depth.
Trish
CC BY-SA 4.0
2020-05-15T22:03:46.143Z
Trish, I may be wrong, but I think there are several mechanisms at play.  One has to do with lifting electrons to higher energy levels, one has to do with having that life be to such a high energy level that the electron is functional free from the atom, but another has to do with adding energy to vibrational modes of the atomic bonds, which is not an electron process.  You probably know the chemistry.  What exactly is changed by the UV light?  I strongly suspect it is a bond change, I am often wrong.
cmm
CC BY-SA 4.0
2020-05-16T06:44:48.797Z
Would Autoacceleration affect the polymerization rate any?
user77232
CC BY-SA 4.0
2020-05-14T12:51:21.430Z
2020-05-14T12:51:21.430Z
user77232
<p>From what I understand, UV curing of resin prints works by starting a chemical reaction that hardens the resin permanently.</p>

<p>Also, a curing step after print is needed to speed up the print and also to reduce the curing during print, which would cure resin beyond the current layer.</p>

<p>However, what is the transmissivity of UV light in partially cured prints? If I print a more massive object, or if I use an opaque resin, how deep will the object harden properly? Absorption is always exponential, meaning that it decreases quickly with depth.</p>

<p>Depending on the resin, how thick prints can be effectively cured? This information is not provided with the resin, which I actually would expect from reputable manufacturers.</p>

<resin>
Can UV cure inside opaque or more massive printed object?
FarO
# Can UV cure inside opaque or more massive printed object?

(Asked by: username_0 on 2020-05-14)

From what I understand, UV curing of resin prints works by starting a chemical reaction that hardens the resin permanently.

Also, a curing step after print is needed to speed up the print and also to reduce the curing during print, which would cure resin beyond the current layer.

However, what is the transmissivity of UV light in partially cured prints? If I print a more massive object, or if I use an opaque resin, how deep will the object harden properly? Absorption is always exponential, meaning that it decreases quickly with depth.

Depending on the resin, how thick prints can be effectively cured? This information is not provided with the resin, which I actually would expect from reputable manufacturers.

**Tags:** resin

---

## 2 answers

### Answer by username_1 on 2020-05-14. Score: 1 | Accepted answer

## My background

I am not a polymer chemist. I know some, and I've talked about UV curing resins with them. These are my conclusions based on that informal education. Your expertise may well be better.

## Minimizing curing on the printer is good

For the maximum print rate and best layer adhesion, the resin should cure during the printing exposure the minimum amount that is consistent with the object supporting itself during the print, being removable from the bed, being rinsed in isopropyl alcohol, and stand on its own during curing.

If you are curing too completely during exposure, you won't get optimal mixing of polymer chains between layers. The layers will be attached by the glue-like action of the new layer on the old, rather than having their polymer chains cross-linked and extending between layers.

If you are curing too little, the object won't be strong enough to support itself and survive post-processing. Soft, gummy bits may dissolve, or be swept away by currents in the cleaning solution. Small details may not be robust enough. Supporting structures (as added supports or parts of the model) may not be strong enough to resist gravity and handling.

If you are curing too completely, your print will take longer than it could take.

## Curing is not binary

Curing will take place spontaneously over a long time. If it didn't, a bottle of resin might last a very long time. But, curing is not a self-catalyzing process that runs quickly to completion. If it were, then the first stray UV that came through an open bottle top would turn the contents solid. If exothermic (which it seems like it must be), the bottle would get hot.

## UV curing may fail

UV curing the inside of a large, opaque object probably doesn't happen. Before I get all excited, though, I would need to place some bounds on "large" and "opaque".

Absorption of UV light depends on the pigment or dye used in the resin. This absorption is never absolute. It is not 100% gone after the first, thinnest penetration. The light is attenuated by an amount per distance it penetrates.

Transmission is the complement of absorption, and the numbers are easier to work with, so lets work with transmission factors of T rather than absorbtion factors of A. $T=1-A$

If a factor of T is passed for millimeter, then one centimeter into the object the light intensity is $T^{10}$ of what it is on the surface, which is a small, but non-zero, number.

Keep in mind that opaqueness depends on wavelength. For example, Window glass is transparent (very little absorption) to visible light, but highly attenuating to UV light. Were I to design a black resin, I would look for a black pigment that was relatively transmissive to UV.

## Will it cure?

A low UV dose delivered to a 0.2mm layer of resin will partially cure the resin. 1 mm into an object, the dose is lower, but it still exists. 1 meter into the object, the absorption is probably too high pass a useful level of UV.

If the transmission factor is 0.8 for a 0.2 mm layer, it is $0.8^5$ for a 1 mm layer (0.33). It would take only three times as much UV exposure to cure a 1 mm thickness as a 0.2 mm thickness. If the object were 1 meter thick, the transmission to the inner bit would be $0.8^{5000}$, which is a very tiny number, roughly $2.82×10^{-485}$.

Finally, consider if the object is or must be truly be solid. UV curable resin is relatively expensive. Many UV printable objects include drainage paths for uncured resin to flow out of the object during printing. Perhaps your object could similarly be hollowed out.

> **Comments:**
> **username_0**: The question was generic, but of course the implications were: attainable strength and reasonable amount of hollowing. What would be nice is to get some reference values for some common resins. I mean, getting UV only 1-1.5 mm in the object would be quite bad.  

---

### Answer by username_2 on 2020-05-14. Score: 1

From what I know about the various resins, once they are printed they are cured enough. Some of the transparent ones require additional curing, however the manufactures of the machines only test based on the max size that their machines can print.

The resin never stops curing apparently, going into diminishing returns; curing slower and slower as time goes on. Only way you will know for sure is by experimentation. You should get a poly-carbonate tube, place black tape along one side then spray paint the rest black. This will leave a window that you can use for future observation. Seal the bottom, and pour in uncured resin. Leave it under a UV source for 2 weeks then peal the tape off to see how much of it has cured. Then come back here and ask Trish to do the math on what the co-efficients of absorption are for the material :)

> **Comments:**
> **username_0**: I thought that after lifting them from the vat they are still flexible. Maybe it applies only to transparent ones.  
> **username_3**: Meanie to let me do the math! I point to the Penetration depth formulae and Beer-Lambert's laws: photons penetrate to some degree dependant on the energy of the light and the material structure penetrated. An article that popped up with the relevant keywords is [this](https://www.researchgate.net/publication/49966993_Energy_absorption_and_exposure_buildup_factors_for_some_polymers_and_tissue_substitute_materials_Photon_energy_penetration_depth_and_chemical_composition_dependence) but I have not yet read more than the abstract. Note that the paper used gamma ray and polymers that substitute ...  
> **username_3**: tissue - the material closest to the resin might be the nylon and water substitute. While not a good analogue, note the general shape of the graphs, which shows the general behavior of high energy photons in matter - little interaction till some depth (mean free path, MFP), then a huge peak and then a falloff of energy deposition. This can be altered by altering the energy (=wavelength) of the photon beam and depends on the material...  
> **username_1**: @trish at the energy levels of the UV, I think thus would behave more like light than gamma rays -- more driven by scattering and diffusion than particle interactions.  The energy is more like vibrational energy than ionization energy.  The activator is special in that it is the path between these low energy photons and the creation reactive radicals.  
> **username_3**: @cmm Gamma radiation is light of 10^19 Hz/100+ keV. UV is 10^13-10^16 & <100 eV. Between the two lies X-ray, but all the 3 behave exactly the same but for the energy and free lengths shifting around. Gamma radiation, X-Ray, UV and visible light, just as much as microwaves and radar are all the same: ***LIGHT***. It's the amount of energy deposition that acts either ionizing (gamma) or heating (Infrared). Lifting an electron to a higher level to make the atm more reactive is technically ionizing!  
> **username_3**: I mainly linked to the article to show the pattern: to deposit the energy in the material needed to start curing, there needs to be several factors given: The energy of the wave needs to be in a specific, material dependant band. The Mean Free Path has little to no influence on the position of the energy position, but the larger the MFP, the faster the energy deposits, telling us a density dependency of the process. The main interesting chart is on p.123: The 15 keV = 0.826^-12 m = 362*10^18 Hz chart shows us a huge dependency on the material for penetration depth.  
> **username_1**: Trish, I may be wrong, but I think there are several mechanisms at play.  One has to do with lifting electrons to higher energy levels, one has to do with having that life be to such a high energy level that the electron is functional free from the atom, but another has to do with adding energy to vibrational modes of the atomic bonds, which is not an electron process.  You probably know the chemistry.  What exactly is changed by the UV light?  I strongly suspect it is a bond change, I am often wrong.  
> **username_2**: Would Autoacceleration affect the polymerization rate any?  

---

