## Friday, December 30, 2016

### Continuity

Spoilers for Rogue One below the fold

## Thursday, August 25, 2016

Abstract: The Breakthrough Starshot project proposes using radiation-pressure propulsion to send multiple spacecraft in the gram-range mass to interstellar targets. There are numerous engineering challenges to accomplishing this mission, but their project discusses speeds and accelerations previously only discussed for cannon projectiles and particle physics experiments. This article examines the problem from first principles to see if the project is even orders-of-magnitude possible. A preliminary check on their numbers seems to confirm that the project is possible in principle.
Radiation pressure has been hypothesized since the early days of special relativity and quantum mechanics. It is a simple consequence of the mass-energy equivalence and the photon nature of light. Quantum mechanics states that light is made of photons, and that each photon is a discrete bundle of energy. The energy $E_{pho}$ carried by each photon is proportional to its frequency $\nu=c/\lambda$, and the proportionality constant is Planck's constant $h$.

$E_{pho}=h\nu=\frac{hc}{\lambda}$

Everyone knows the famous mass-energy equivalence equation $E=mc^2$. However, that is only a special case of the momentum equation:

$E=\sqrt{p^2c^2+m_0^2c^4}$

We can solve this for the momentum of a photon, considering that its rest mass is zero.

$\begin{eqnarray}E_{pho} & = & \sqrt{p_{pho}^2c^2+0^2c^4}\\ & = & \sqrt{p_{pho}^2c^2} \\ & = & pc\\ \frac{hc}{\lambda}&=&p_{pho}c\\ \frac{h}{\lambda}&=&p_{pho}\end{eqnarray}$

Note that momentum is a vector quantity, but this just deals with the magnitude. The direction of the momentum vector is the direction the photon is traveling.

Now, we know the energy and momentum of each photon, so we know that if we throw so many joules of photons at a target, it will transfer so much momentum. The irradiance is defined as the power $P$ (energy $E$ per unit time $t$) of the light hitting each unit area of the target, so the units are W/m^2 or J s^-1 m^-2. We can calculate it from the amount of energy striking the target of known area $A$ per unit time:

$I=\frac{P}{A}=\frac{E}{tA}$

So given a certain amount of irradiance with a known photon wavelength, how much momentum does that light carry? First, figure out how many photons per time per area $I_{pho}$ that irradiance represents:

$\begin{eqnarray}I_{pho}&=&\frac{I}{E_{pho}}\\ &=& \frac{I\lambda}{hc}\end{eqnarray}$

The dimension of photon irradiance is photons per time unit per area unit (photon s^-1 m^-2 in SI units).

Now a bit about momentum. Again, everyone knows Newton's second law $\vec{F}=m\vec{a}$, but again this is a special case. Force is defined as the change in momentum per unit time:

$\vec{F}=\frac{d\vec{p}}{dt}$

For massive objects at sub-relativistic speeds, momentum is defined as the mass of the object times the velocity of the object. For relativistic conditions, we use the energy relation above, where energy still has the customary units J=kg m^2 s^-2. Working through the units of the energy equation in the special case of a photon, we find that momentum still has the same units as it does in sub-relativistic conditions.

So from this, a given amount of energy of photons carries a certain amount of momentum. A certain power of photons carries a flow of momentum per unit time, or in other words Force. A certain amount of momentum flow impinging on a certain area is the same as a force exerted on that area, or a Pressure $\rho$. So, we can calculate the pressure exerted by a given irradiance of light from the intensity in photons/time/area, multiplied by the momentum of each photon, which gives momentum/time/area which equals force/area which equals pressure:

$\begin{eqnarray}\rho&=&I_{pho}p_{pho}\\ &=&\left(\frac{I\lambda}{hc}\right)\left(\frac{h}{\lambda}\right)\\ &=&\frac{I}{c} \end{eqnarray}$

So the pressure exerted by the light is the irradiance of the light divided by the speed of light. That is amazingly simple, and notice that both Planck's constant and the wavelength cancel out. Now we can start plugging numbers.

First, let's do one of Clarke's space yachts. It weighs on the order of 1000kg and has on the order of 1km^2=1,000,000m^2 of sail, and it uses natural sunlight. How much force does it collect and what is its acceleration? Note that since the sails are reflective, the light is reversed in direction and has a momentum change of twice its original momentum, so we collect twice as much force as the momentum suggests.

$\begin{eqnarray}I& & &\approx&1400\mbox{W/m}^2\\ c& & &=&299,792,458\mbox{m/s}\approx 3\times 10^8 \mbox{m/s}\\ \rho&=&2\frac{I}{c}&=&2\frac{1400}{3\times 10^8}=9.4\mu \mbox{N/m}^2\\ A& & &=&1,000,000 \mbox{m}^2\\ F&=&P_{pho}A&=&9.4\mbox{N}\\ m& & &=&1000 \mbox{kg}\\ a&=&\frac{F}{m}&=&\frac{9.4}{1000}=9.4\mbox{mm/s}^2\approx1.0\mbox{mg} \end{eqnarray}$

So the whole sail exerts a measly couple of newtons. Pulling on a ton of mass, we get almost a full milli-g of acceleration. Not a whole lot, but in order of magnitude of that described in the story.

Now we get into the big numbers. The game Adventure Capitalist teaches us to not be afraid of big numbers, so lets go get some. The Starshot project describes using gigawatts of power on spacecraft weighing grams. Among the enormous engineering challenges are:

• Focusing gigawatts of power onto square-meters sized targets over distances of billions of meters
• Having a reflective enough sail that the GW/m^2 incident power doesn't vaporize the sail
• Having a tough enough sail that the MW/m^2 absorbed power doesn't vaporize the sail
• Having a thin enough sail to stay within the mass budget
• Building a spacecraft with a useful payload, power source, and communications system that will work over interstellar distances and stay within the mass budget
• By the way, the mass budget is about 10g total.

Putting all of these aside, they discuss a spacecraft with a mass of about 10g, a sail area of about 16m^2, and multiple gigawatts of power focused on it. Let's go with 1GW/m^2, or 16GW total, to see what we get:

$\begin{eqnarray}I& & &\approx&1\mbox{GW/m}^2\\ P_{pho}&=&2\frac{I}{c}&=&2\frac{1\times 10^9}{3\times 10^8}=6.66 \mbox{N/m}^2\\ A& & &=&16 \mbox{m}^2\\ F&=&P_{pho}A&=&16\times 6.66=106\mbox{N}\\ m& & &=&0.01 \mbox{kg}\\ a&=&\frac{F}{m}&=&\frac{106}{0.01}=10.6\mbox{km/s}^2\approx1080\mbox{g} \end{eqnarray}$

Well, that's some git-up-and-go, alright. How long does it take this to get to a target speed of a good chunk of the speed of light, say 60,000km/s (0.2$c$)?

$v=at$
$\begin{eqnarray}t&=&\frac{v}{a} \\ &=&\frac{60,000,000}{10600}&=&5660 \mbox{s} \end{eqnarray}$

That's not too bad, a little over 1.5 hours, and real close to 1 LEO period, but not what the paper is discussing, which is on the order of 10 minutes. What acceleration is needed for that?

$v=at$
$\begin{eqnarray}a&=&\frac{v}{t} \\ &=&\frac{60,000,000}{600}&=&100 \mbox{km/s}^2\\ F&=&ma&=&0.01(100,000)=1000\mbox{N}\\ P&=&\frac{F}{A}&=&\frac{1000}{16}=62.5\mbox{N/m}^2\\ I&=&Pc&=62.5(300,000,000)=18.75\mbox{GW} \end{eqnarray}$

This is about 10000g. Now we are talking cannon-type acceleration. Since the acceleration is about 10 times greater, it will require 10 times the power, or a couple hundred gigawatts. This is a good fraction of the electricity usage of the United States, so it is a large but doable amount of power. We only need it for 10 minutes. The paper discusses putting the laser in space, which would require a gigawatt-scale power source in space. I suppose a couple of square kilometers of solar cells could do that.

If all of the engineering problems are solved, this could work. There isn't anything physically impossible about it. The engineering challenges are large, but the Starshot project claims that each can be solved incrementally. We don't have to shoot at stars first -- imagine getting back to Neptune in a couple of days, and then being able to orbit once you got there?

References

## Saturday, August 6, 2016

### August 20 progress passed!

Yesterday Yukari 4 passed the August 20 progress verification point. A whole 15 days early, too. Here is video proof:

## Thursday, July 28, 2016

### St Kwan's Is Expanding

We would like to welcome Kirktopode to the St Kwan's Family! He has the goal to enter the embedded electronics field, and I have the goal of completing the robot without going crazy. We are perfect for each other!

Also, all of the robot code and design stuff is now on Github. Kirktopode is holding the master version, and I have a fork of it. I am using another one of my projects to hold the wiki in which I am documenting things in English.

## Sunday, July 24, 2016

### Robodometer

Last year I tried to implement a wheel encoder in between heats and failed. The ambient light was just too bright. So this time I am building the wheel encoder into the gearbox. I took the spare gearbox apart to see what the best way to do it is. Inside the casing is a differential gear in the form of a cylinder with gear teeth around the outside and the differential gears inside. The cylinder is sealed, held together with 4 screws. But, the outside of the cylinder is perfect for my design. The gearbox cover plate in that part of the gearbox is relatively easy to remove, and in just the right spot. If a hole is drilled through that plateau optical sensor can be set up to look through the hole and read the half of the differential cylinder face which is painted white. Two such sensors can be used to mak single-track quadrature encoder, which can tell whether the wheel is turning forward or reverse.

I have a pair of QRF1114 sensors specifically designed and obtained for this purpose. This part is basically just an infrared LED and an infrared phototransistor in the same case. It isn't directly useful as-is, but I combined it with a few resistors and made a three-terminal analog device.

## Friday, May 13, 2016

### Hooray for distributed backups!

I found a backup of my database from April 21 of this year, which was only a couple of days before the failure. That should have all of my wikis and gallery data, and therefore represents the second-most important data I have. The most important is the code, and that is backed up by means of git. I know for sure that there is a valid git repository on one of my portable USB disks

## Thursday, May 5, 2016

### Yet Another Episode in the Annals of Data Stewardship

Having learned my lesson from before, I did not set up my filesystem as one big raid0. I did a btrfs raid5 instead. When one of the disks finally did give out, it wasn't with the click of death I heard before, but with read errors. The btrfs degraded, and by mounting read-only in recovery mode, I was able to use the two good disks in order to get my data.

Or so I thought.

A word on the issue I was having. I was seeing "stale file handle" warnings, of the type you see when you are in a folder that is NFS mounted, after you lose connection. But, this wasn't an NFS system. I rebooted the system and it wouldn't come up, because the btrfs refused to mount. After manually mounting in degraded mode, many of the disk accesses reported errors in dmesg, about the generation of certain metadata being off of the expected value, often by hundreds or thousands of generations.

First, I decided that I had lost confidence in btrfs -- if it wasn't going to keep working in the presence of a disk failure, what was the point? I spent the next several days scraping data off of the btrfs and putting it wherever I could find a place for it - on the USB disks I have, on other computers, on the system disk, etc. I then replaced the bad disk and formatted them all as zfs - now possible since Ubuntu 16.04 includes a native zfs driver.

Finally, I started copying data back onto the zfs. All appeared to go well, until I tried to bring up the wiki. The LocalSettings.php file was completely blank - it had the expected value, but all bytes in the file were 0x01 . Hrm.

Turns out a lot of files were like this. Files I care about, like the database, the git repositories, etc. It seems like the newer the file is, the more likely it is to be damaged like this.

No problem, I've got backups. A raid5 is not a backup, so I had the most important data copied off onto several other systems.

Or so I thought.

My backup script runs on a cron every night, and had backed up the bad data and spread it all around over the good data.

Oops.

It isn't a total loss. I have all my code in a git repository on the big USB disk. I have an old backup (from December, I think) of all the data I considered important. I did lose a lot of video :( but I don't think I lost anything from Florida 5.

So I think.