Deep Space Industries vs Planetary Resources

[KelvinZero]

Bringing 1 kilogram from an asteroid back to earth (landing incl.) must already cost more than 50k, right? (50k is the price of 1kg of platinum)

Then you still haven't done any mining, which probably would require tons of equipment for extracting precious metals, unless there are asteroids made of pure platinum floating around in space.

Regarding water mining. Well that certainly sounds easier, however it would require a significant human presence in space in the forseeable future. That could be due to some utterly expensive government program (like the ISS), which would be good for business, however if a massive reduction in launch costs is behind it, bringing in water from asteroids could become less attractive.

(Im no expert but..)

I think the first obvious use (of water) would be a propellant depot to refuel the upper stage (Eg a Centaur stage) of an existing launcher after it gets to orbit. This would let us throw much larger unmanned robots at mars with our existing rockets for example.

Cheaper launch from earth would probably bring down the worth of fuel in LEO, but with cheaper launch we would probably have the ability to do something really worthwhile, and that thing would likely benefit from a depot further out than LEO, eg EML2 or even mars orbit. At some point the advantage has to move to using local resources.

[JohnFornaro]

... in well designed cubesats almost all the subsystems are completely powered down when they are not active and they often have several super low power modes beyond that.

Which is fine, and I assumed that the cubesats in question would be "well designed".

Yet another reminder that I am discussing the communication power cababilities of the illustrated cubesat, particularly when it is at opposition.

Do you think that the illustrated "fleet" of two 6U cubesats, shown within some few km of an asteroid, really have the "right stuff" to communicate their observations to the Earth at the range of expected distances, via a laser comm channel to an Earth based telescope?

'Cause I don't think so, and I would have a hard time as an investor, based on the public presentations and that illustration, in parting with substantial capital.  Just sayin'.

[Robotbeat]

Well then you don't understand the enormous difference in wavelength between visible light and radio waves.

[mlindner]

Do you think that the illustrated "fleet" of two 6U cubesats, shown within some few km of an asteroid, really have the "right stuff" to communicate their observations to the Earth at the range of expected distances, via a laser comm channel to an Earth based telescope?

Assuming your beam dispersion is low enough, what makes laser communication nice is that it doesn't fall off with distance nearly as much. The falloff rate is nearly negligible compared to radio. Radio signals fall off with 1/r^2, twice as far, four times as weak. Laser does not follow these rules and an ideal laser will have zero loss no matter the distance.

They've already done laser communication with LRO from earth's surface, news from a few weeks ago. http://www.nasa.gov/mission_pages/LRO/news/mona-lisa.html

[Robotbeat]

Laser still disperses at 1/r^2, but it just starts far more concentrated and the angle is much smaller for the same aperture size.

[joek]

Laser still disperses at 1/r^2, but it just starts far more concentrated and the angle is much smaller for the same aperture size.

Yeah, sort of, more or less, theoretically.  Which tells us nothing as to why laser's might be preferred for such applications, why free-space communications would benefit from what they have to offer, and why they appear to be the prospective choice.  Can you say why, beyond physics/optics 101, that may be?

[Robotbeat]

Laser still disperses at 1/r^2, but it just starts far more concentrated and the angle is much smaller for the same aperture size.

Yeah, sort of, more or less, theoretically.  Which tells us nothing as to why laser's might be preferred for such applications, why free-space communications would benefit from what they have to offer, and why they appear to be the prospective choice.  Can you say why, beyond physics/optics 101, that may be?

Yes. The half-angle of the beam is:

halfangle= 1.22*wavelength/diameterofaperture

So, the radius of the beam spot at a certain distance is just the half-angle times the distance to the transmitter.

Or rather:
beamdiameter=2*1.22*distance*wavelength/diameterofaperture

And of course, the area of the beam is just pi*(beamdiameter/2)^2, so that's where you get the 1/r^2 term (where "r" is "distance").

Doesn't matter if it's radio or visible, except that your wavelength is vastly different.

Postscript:
Same formula works for imaging. Except for transmitting (unlike imaging), you CAN'T get away with a thinned array or you end up losing all your power to the sidelobes... (For imaging, you just get a lower signal/noise ratio if you have a thinned array.)

(And this is the 101 version. The 201 version wouldn't use the small-angle approximation and would note that not ALL the beam would lie inside that area, and the 401 version would note that for imaging, this is the Rayleigh criterion and the formula makes some assumptions about signal/noise ratio, etc.)

EDIT:I flipped around the wavelength and diameterofaperture at first.

[joek]

Yes. The half-angle of the beam is:
halfangle= 1.22*diameterofaperture/wavelength
So, the radius of the beam spot at a certain distance is just the half-angle times the distance to the transmitter.
Or rather:
beamdiameter=2*1.22*diameterofaperture*distance/wavelength
And of course, the area of the beam is just pi*(beamdiameter/2)^2, so that's where you get the 1/r^2 term (where "r" is "distance").
Doesn't matter if it's radio or visible, except that your wavelength is vastly different.

Excuse me for being dense, but that assumes an uncollimated source?

[Robotbeat]

Yes. The half-angle of the beam is:
halfangle= 1.22*diameterofaperture/wavelength
So, the radius of the beam spot at a certain distance is just the half-angle times the distance to the transmitter.
Or rather:
beamdiameter=2*1.22*diameterofaperture*distance/wavelength
And of course, the area of the beam is just pi*(beamdiameter/2)^2, so that's where you get the 1/r^2 term (where "r" is "distance").
Doesn't matter if it's radio or visible, except that your wavelength is vastly different.

Excuse me for being dense, but that assumes an uncollimated source?

It assumes collimated, actually. Or you can say it assumes you're doing your best with your optics to get the beam spot as small as possible, within the constraint of having the aperture that size. Light is a wave (and so is matter, though the wavelength is generally FAR shorter for matter, which is why people use electrons for getting high resolution microscopic images).

And this is the far-field expression for spot size, where the distance (between the transmitter and receiver) is much, much greater than any of the other relevant lengths (such as aperture diameter, etc). I.e., the situation for a microscope may be somewhat different (though still pretty relevant, unless you get VERY close).

[joek]

It assumes collimated, actually. Or you can say it assumes you're doing your best with your optics to get the beam spot as small as possible, within the constraint of having the aperture that size. Light is a wave (and so is matter, though the wavelength is generally FAR shorter for matter, which is why people use electrons for getting high resolution microscopic images).

And this is the far-field expression for spot size, where the distance (between the transmitter and receiver) is much, much greater than any of the other relevant lengths (such as aperture diameter, etc). I.e., the situation for a microscope may be somewhat different (though still pretty relevant, unless you get VERY close).

Thanks (smack!... must have slept through it... time to go back and hit the books!).
Maybe the operative question is what size optics/power are required for DSI?

[mlindner]

Yes. The half-angle of the beam is:
halfangle= 1.22*diameterofaperture/wavelength
So, the radius of the beam spot at a certain distance is just the half-angle times the distance to the transmitter.
Or rather:
beamdiameter=2*1.22*diameterofaperture*distance/wavelength
And of course, the area of the beam is just pi*(beamdiameter/2)^2, so that's where you get the 1/r^2 term (where "r" is "distance").
Doesn't matter if it's radio or visible, except that your wavelength is vastly different.

Excuse me for being dense, but that assumes an uncollimated source?

It assumes collimated, actually. Or you can say it assumes you're doing your best with your optics to get the beam spot as small as possible, within the constraint of having the aperture that size. Light is a wave (and so is matter, though the wavelength is generally FAR shorter for matter, which is why people use electrons for getting high resolution microscopic images).

And this is the far-field expression for spot size, where the distance (between the transmitter and receiver) is much, much greater than any of the other relevant lengths (such as aperture diameter, etc). I.e., the situation for a microscope may be somewhat different (though still pretty relevant, unless you get VERY close).

beamDiameter = 2 * 1.22 * diameterOfAperture * Distance / Wavelength
beamArea = pi * (beamDiameter / 2)^2

So I've concluded Robobeat's math is wrong or this is impossible. So plugging in a few numbers. Assuming a 20cm aperture, satellite in earth's orbit at opposition 300 million km, assuming ~1000 nm infrared laser. This gives us a beam diameter of  0.1464 * 10^15 km or 146.4 million million km. I highly doubt the beam diameter is 6 orders of magnitude larger than the diameter of earth's orbit on the other side of earth's orbit as this breaks common sense of personal use of lasers.

[Robotbeat]

Ha! You're right! I switched around the wavelength and aperture diameter.

Always do a sanity-check, first...

https://www.google.com/search?q=2*1.22*2.5AU*1micron%2F%2820cm%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Google search for "2*1.22*2.5AU*1micron/(20cm)"
4562.73506 kilometers

Much better answer.

BTW, you can often do pretty well with just a collection of relatively common-sense assumptions and making sure the units work out (we call this "dimensional analysis"), like knowing that if your wavelength is smaller, so should be your spot size and if your aperture is bigger, you should get a smaller spot size, and the further away you are the bigger your spot size... Using just that:

spotdiameter is proportional to wavelength

spotdiameter is proportional to distance


spotdiameter is inversely proportional to aperture size

With just that information, you can get this formula:

spotdiameter = k*wavelength*distance/(aperturediameter)

where k is a proportionality constant. It turns out this is just the formula we get from the Rayleigh criterion, with k being 2*1.22 in our case.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp2.html#c2

[mlindner]

Ha! You're right! I switched around the wavelength and aperture diameter.

Always do a sanity-check, first...

https://www.google.com/search?q=2*1.22*2.5AU*1micron%2F%2820cm%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Google search for "2*1.22*2.5AU*1micron/(20cm)"
4562.73506 kilometers

Much better answer.

I'm completely unfamiliar with optics but isn't the half angle the angle from the center line of the beam to the edge of the beam? In that case wouldn't the radius of the beam at distance be tan(halfAngle) * distance?

[Robotbeat]

Ha! You're right! I switched around the wavelength and aperture diameter.

Always do a sanity-check, first...

https://www.google.com/search?q=2*1.22*2.5AU*1micron%2F%2820cm%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Google search for "2*1.22*2.5AU*1micron/(20cm)"
4562.73506 kilometers

Much better answer.

I'm completely unfamiliar with optics but isn't the half angle the angle from the center line of the beam to the edge of the beam? In that case wouldn't the radius of the beam at distance be tan(halfAngle) * distance?

Technically yes! But for small halfangle, tan(halfangle)= ~halfangle.

It makes very little difference, mathematically! We're talking about microradians and such, here, so they're essentially identical (beyond the limit of a hand calculator).

And I forgot to mention we were using radians... If you try this with degrees, you'll get a very wrong result.

[mlindner]

Ha! You're right! I switched around the wavelength and aperture diameter.

Always do a sanity-check, first...

https://www.google.com/search?q=2*1.22*2.5AU*1micron%2F%2820cm%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Google search for "2*1.22*2.5AU*1micron/(20cm)"
4562.73506 kilometers

Much better answer.

I'm completely unfamiliar with optics but isn't the half angle the angle from the center line of the beam to the edge of the beam? In that case wouldn't the radius of the beam at distance be tan(halfAngle) * distance?

Technically yes! But for small halfangle, tan(halfangle)= ~halfangle.

It makes very little difference, mathematically! We're talking about microradians and such, here, so they're essentially identical (beyond the limit of a hand calculator).

And I forgot to mention we were using radians... If you try this with degrees, you'll get a very wrong result.

So ~4km is roughly 1/3 the diameter of earth.
Now the next two questions are:
1) How much power density from light is reliably detectable so that if you were to modulate the beam you could detect the modulation?
2) For a benchmark, how good of pointing accuracy is possible with the best modern spacecraft? The half angle is roughly 10 micro radians. Cubesats that I've worked with have trouble getting pointing accuracy to within 1 to 0.1 degree (or 20 to 2 milli radians respectively). The best cubesats I know of (made by aerospace corp) brag about having < 1 degree pointing accuracy. We need to have 2-3 orders of magnitude better pointing for this to be possible.

[Robotbeat]

Ha! You're right! I switched around the wavelength and aperture diameter.

Always do a sanity-check, first...

https://www.google.com/search?q=2*1.22*2.5AU*1micron%2F%2820cm%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Google search for "2*1.22*2.5AU*1micron/(20cm)"
4562.73506 kilometers

Much better answer.

I'm completely unfamiliar with optics but isn't the half angle the angle from the center line of the beam to the edge of the beam? In that case wouldn't the radius of the beam at distance be tan(halfAngle) * distance?

Technically yes! But for small halfangle, tan(halfangle)= ~halfangle.

It makes very little difference, mathematically! We're talking about microradians and such, here, so they're essentially identical (beyond the limit of a hand calculator).

And I forgot to mention we were using radians... If you try this with degrees, you'll get a very wrong result.

So ~4km is roughly 1/3 the diameter of earth.
Now the next two questions are:
1) How much power density from light is reliably detectable so that if you were to modulate the beam you could detect the modulation?
2) For a benchmark, how good of pointing accuracy is possible with the best modern spacecraft? The half angle is roughly 10 micro radians. Cubesats that I've worked with have trouble getting pointing accuracy to within 1 to 0.1 degree (or 20 to 2 milli radians respectively). The best cubesats I know of (made by aerospace corp) brag about having < 1 degree pointing accuracy. We need to have 2-3 orders of magnitude better pointing for this to be possible.

Totally depends... You can technically detect a signal of just a handful of photons. This is what people do when measuring the distance to the Moon using the retroreflectors left there by Apollo. I actually don't know a handy formula relating power intensity to data rate... partially because you need to know what your noise level is going to be. If anyone does know a handy formula, it'd be appreciated!

But I think the data rate is proportional to the natural log of the signal/noise ratio, perhaps also proportional to the total power level (perhaps the natural log of it as well), and directly proportional to the "bandwidth", i.e. the amount of spectrum devoted to the communication. With light, this last one can be pretty high. Any radio engineers? Or, I guess we could google. Of course, with low light levels, you're talking about individual photons, so you have that limit, too. You can encode information in the time-of-flight of a photon (or group of photons), and the resolution of that time-of-flight will determine how much bandwidth (or rather, how much frequency spread) you are using.

I haven't taken an information theory class, otherwise I'd probably have a really good answer for you.

[Robotbeat]

Ah-HA!

http://en.wikipedia.org/wiki/Shannon-Hartley_theorem

The Shannon-Hartley Theorem says (maximum possible data rate, whatever modulation schema):

Datarate(in bits per second) = analogbandwidth(in Hertz)*logbase2(1+signalpower/noisepower)

But it's sort of difficult to use this back-of-the-napkin style, since we don't have anything good for the "noise" level... Except that it's probably pretty low... Comparable to the brightness of the asteroid we're visiting. Of course, a good way to lower the noise level is to apply a band-pass filter that will filter all light but a very narrow range of wavelengths, so that the laser light (which is very monochromatic) will be bright compared to the asteroid.

Don't forget the photon shot noise.

[Robotbeat]

Notice, much of this equation depends at least as much on your receiver as it does the transmitter.

[happyflower]

Beyond the nitty gritty details that you guys love so much to discuss (snore lol), I must admit that when DSI came out and said Asteroid 2021 DA14 is worth $200 Billion or whatever they lost most of their credibility with me anyway. I mean come on who are they talking to, children?

I mean what is the point of saying something like that? What about your costs of extraction and transportation? What about your customers? What a pie in the sky dreamers these guys turned out to be.

Oh well Still have a little bit of credibility with PR left. Lets see if they blow it or add to it.

[QuantumG]

Remember: you're not as smart as you think you are.

Assuming other people are just dumb is a good way to write off ever learning anything new.