<p>Let's start with the obvious: this printer would need to be really big. Not because of a large print volume, but because it needs to collect a lot of sunlight or needs a really big focussing array. The linked question states that the array there, about 0.6 m² large, has roundabout 600 W of power to focus on that one point.</p>
<h1>Power draw needs</h1>
<h2>What powers are we dealing with?</h2>
<p>A typical laser cutter uses a laser tube that at least 20 W for thin material and up to 300 W for thicker material. But we need to weld steel, so we need to be roughly equivalent to a cutting laser for the same material. We are not talking mere hundreds of watts, we are talking an <a href="https://pentalaser.com/en/view/88.html" rel="nofollow noreferrer">industrial 2000 to 20000 W</a> in a laser of less than a millimeter in diameter. Why do I say the later? Well, power need scales with the area, which scales with the square of the radius.</p>
<p>Let's use a ballpark, a nice round 10000 W Laser with a somewhat large 1 mm² crossection. We're talking about the ballpark of 100 Gigawatt per square meter.</p>
<p><span class="math-container">$L=P/A=\frac{10000\ \text W}{0,001\times 0.001\ \text m^2}=10^{10}\frac{\text W}{\text m^2}$</span></p>
<h2>Solar power harvester size</h2>
<p>Luckily, we don't need to illuminate a whole square meter, so we only need some 10 Kilowatts of sunlight for our application. But we need this number to calculate how much sunlight we need to <em>harvest</em> in our smelting machine.</p>
<p>At the stratosphere, Earth gets about 1400 W/m², and on a sunny day, about 1 kW/m² makes it to the surface, we call this the solar constant <span class="math-container">$S$</span>. Now, if we compare the ballparks, we get to quickly see the ballpark size of our machine:</p>
<p><span class="math-container">$P/S=A=\frac{10 \text{ kW}}{1 \frac {\text{kW}} {\text m^2}} = 10\ \text m^2$</span></p>
<p>10 square meters of harvesting area gets us the same power. Incidentally, this scales linearly with the item discussed in the linked question, as that one already uses square meters.</p>
<h2>Sizing down</h2>
<p>But 10 square meters of mirrors into a focal point and then lenses to a focal point is huge, can we make it smaller? And to that I must say <em>somewhat</em>. First of all, we could get our focal point smaller: We need to get 10 Gigawatt per square meter for our <span class="math-container">$L$</span>, but we can turn two screws here: what if we get from 1 mm² to only 0.1 mm²?</p>
<p><span class="math-container">$P=L*A=10^{10}*(0.0001\times 0.0001)\ \text W=100 W$</span></p>
<p>100 Watts focussed on 0.1 mm² is a sixth of the power that array can deliver, so totally feasible in terms of power, as long as you can achieve such a small focal point.</p>
<h1>Conclusion</h1>
<p>Yes, with a focussing array good enough or a mirror array large enough you could achieve the powers needed to melt metals on a spot focus. Note though, that you need to have a really good focus setup that creates pretty much a solar-powered laser, which means that such a machine will be extremely expensive due to the high precision machinery needed for that - and that this focusing aperture will be most likely the largest part of your machine. We're talking building-size scale. I'd be cheaper and easier to just put lots of solar panels onto the roof of the building you run a conventional metal 3D printer in.</p>

CC BY-SA 4.0
2020-06-06T11:51:53.397Z
2020-06-07T10:42:18.360Z
2020-06-18T08:29:41.667Z
Trish
<p>Could concentrated sunlight be used to build a metal 3D printer, sintering or melting metal powder?</p>

<p>Related question on physics stack exchange <a href="https://physics.stackexchange.com/questions/143049/what-temperature-is-achieved-in-focus-point-by-5000-flat-1x1cm-mirrors-onto-a-sa">https://physics.stackexchange.com/questions/143049/what-temperature-is-achieved-in-focus-point-by-5000-flat-1x1cm-mirrors-onto-a-sa</a></p>

CC BY-SA 4.0
2020-06-04T21:43:32.027Z
Yes that could be possible, I've seen that [before](https://youtu.be/6ajzOaauYa4), but that printer used a Fresnel lens that melted sand to glass. But I doubt whether it would be practical. Are you coming from a planet builder perspective?
0scar
CC BY-SA 4.0
2020-06-05T12:46:32.903Z
I was mainly wondering whether oi could be practical to make cheaper printers without an expensive laser
stckmower
CC BY-SA 4.0
2020-06-05T13:52:11.477Z
That expensive laser will create a highly concentrated focused bundle of energy (light), it is probably undoable to positioning 5000 mirrors to achieve the same accurate focal point.
0scar
<metal-printing>
Sunlight powered metal 3D printer
stckmower
# Sunlight powered metal 3D printer

(Asked by: username_0 on 2020-06-04)

Could concentrated sunlight be used to build a metal 3D printer, sintering or melting metal powder?

Related question on physics stack exchange <https://physics.stackexchange.com/questions/143049/what-temperature-is-achieved-in-focus-point-by-5000-flat-1x1cm-mirrors-onto-a-sa>

**Tags:** metal-printing

> **Comments:**
> **username_1**: Yes that could be possible, I've seen that [before](https://youtu.be/6ajzOaauYa4), but that printer used a Fresnel lens that melted sand to glass. But I doubt whether it would be practical. Are you coming from a planet builder perspective?  
> **username_0**: I was mainly wondering whether oi could be practical to make cheaper printers without an expensive laser  
> **username_1**: That expensive laser will create a highly concentrated focused bundle of energy (light), it is probably undoable to positioning 5000 mirrors to achieve the same accurate focal point.  

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## 1 answers

### Answer by username_2 on 2020-06-06. Score: 3 | Accepted answer

Let's start with the obvious: this printer would need to be really big. Not because of a large print volume, but because it needs to collect a lot of sunlight or needs a really big focussing array. The linked question states that the array there, about 0.6 m² large, has roundabout 600 W of power to focus on that one point.

# Power draw needs

## What powers are we dealing with?

A typical laser cutter uses a laser tube that at least 20 W for thin material and up to 300 W for thicker material. But we need to weld steel, so we need to be roughly equivalent to a cutting laser for the same material. We are not talking mere hundreds of watts, we are talking an [industrial 2000 to 20000 W](https://pentalaser.com/en/view/88.html) in a laser of less than a millimeter in diameter. Why do I say the later? Well, power need scales with the area, which scales with the square of the radius.

Let's use a ballpark, a nice round 10000 W Laser with a somewhat large 1 mm² crossection. We're talking about the ballpark of 100 Gigawatt per square meter.

$L=P/A=\frac{10000\ \text W}{0,001\times 0.001\ \text m^2}=10^{10}\frac{\text W}{\text m^2}$

## Solar power harvester size

Luckily, we don't need to illuminate a whole square meter, so we only need some 10 Kilowatts of sunlight for our application. But we need this number to calculate how much sunlight we need to *harvest* in our smelting machine.

At the stratosphere, Earth gets about 1400 W/m², and on a sunny day, about 1 kW/m² makes it to the surface, we call this the solar constant $S$. Now, if we compare the ballparks, we get to quickly see the ballpark size of our machine:

$P/S=A=\frac{10 \text{ kW}}{1 \frac {\text{kW}} {\text m^2}} = 10\ \text m^2$

10 square meters of harvesting area gets us the same power. Incidentally, this scales linearly with the item discussed in the linked question, as that one already uses square meters.

## Sizing down

But 10 square meters of mirrors into a focal point and then lenses to a focal point is huge, can we make it smaller? And to that I must say *somewhat*. First of all, we could get our focal point smaller: We need to get 10 Gigawatt per square meter for our $L$, but we can turn two screws here: what if we get from 1 mm² to only 0.1 mm²?

$P=L\*A=10^{10}\*(0.0001\times 0.0001)\ \text W=100 W$

100 Watts focussed on 0.1 mm² is a sixth of the power that array can deliver, so totally feasible in terms of power, as long as you can achieve such a small focal point.

# Conclusion

Yes, with a focussing array good enough or a mirror array large enough you could achieve the powers needed to melt metals on a spot focus. Note though, that you need to have a really good focus setup that creates pretty much a solar-powered laser, which means that such a machine will be extremely expensive due to the high precision machinery needed for that - and that this focusing aperture will be most likely the largest part of your machine. We're talking building-size scale. I'd be cheaper and easier to just put lots of solar panels onto the roof of the building you run a conventional metal 3D printer in.


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