Today is the scheduled launch date for the Falcon Heavy demonstration. As of this writing, the launch is on schedule for an 11:30 MST launch.

There is very little official guidance from SpaceX as to what to expect. Elon Musk has stated that minimum mission success is clearing pad 39A far enough such that any further failure doesn't destroy that pad. There has never been a catastrophic failure at pad 39A, and Musk would like to keep it that way.

However, the plan is to do a boost, then three burns of the upper stage. The first finishes launch to LEO. The second is about 30 seconds, and seems to put the booster into a GTO-like orbit. Lifting a heavier spacecraft into full GTO takes about a minute, so there is some hint that this will go into an elliptical orbit that is short of GTO. Part of the demonstration is a 6-hour coast. They are doing it on this flight because the upper stage is very similar to any normal Falcon 9 upper stage, and any demonstration on this stage would apply there. This coast is what would be needed for a 3-burn GSO direct insertion, that apparently is very interesting to the military. For one thing, it would demonstrate that the upper stage could put a GPS satellite directly into its target orbit, like the much more expensive Delta IV medium. A bit more oomph and a similar endurance would put a spacecraft directly into GSO.

In any case, the consensus on NasaSpaceflight is that the target high orbit is one with a period of around 6 hours. After this coast, the second stage would be back at perigee, ready to take maximum advantage of the Oberth effect.

SpaceX has claimed that they will put the payload (A cherry-red Tesla Roadster) into an "earth-mars heliocentric orbit". The launch window for Mars is in May, so they will be launching 3 months out of the window, but since this is such a light payload, they should have plenty of C3 and probably could target Mars if they wanted. However, I think that they will instead target an orbit with periapse at Earth and a C3 typical of launching to Mars. The payload will reach the vicinity of Mars orbit, but Mars will be far far away by then. In fact, to be responsible about Planetary Protection, they should launch into an orbit which will not actually intersect Mars orbit at all, so that there is never any possibility of the car impacting Mars.

Running the numbers based on the Trajectory Planner 1.1.1 from Orbit Hangar, I get a departure C3 of 23.9 km^2/s^2, with a departure today and an arrival on October 17, 2018. The flight time is 252 days. This C3 is high for a Mars launch, but should be doable with such a light payload.

If they are targeting Mars, then the launch vehicle must be able to adjust azimuth in order to target Mars at any time during the window. If they are just going for a given C3, they can use the same azimuth whenever they go. Since ASDS is parked somewhere definite to catch the core stage, I estimate that they are targeting a fixed azimuth independent of launch time.

There are no signs of high-gain antennas or solar panels on the payload, so it is almost certain that once the battery runs down, the payload will become inert. However, the payload is an electric car, with many many amp-hours of battery life. The car radio might run for hours or days.

# St Kwan's Home for the Terminally ADD

Nihil est tam malus ut non possis plus

## Tuesday, February 6, 2018

## Monday, August 21, 2017

### Live, from the eclipse path

The Kid Attractor - works on adults too. |

Cookie Monster |

Just after first contact |

- It did get noticeably darker as the eclipse passed 50%
- We could see the shadow approach from the west. It was hazy that direction, and I could see a band of dark start at the horizon. It looked like it was getting vertically wider rather than closer.
- Totality itself cannot be done justice in pictures. I took pictures but used my eyes too. Light level was close to that of just after sunset. The sky was a deep blue purple. We saw the diamond ring, which was MUCH brighter than the rest of the corona. The corona is white, and rather narrow with 3 long streamers (more than one solar diameter) One at 12:00, one at 1:30 and one at 7:00. The disk of the moon was the same color as the sky.
- I thought I got crowd shots, and everyone around said that it was totally worth it. Unfortunately for some reason either the video was never captured or was deleted :(
- No one bothered to stay long after totality - we just all packed up and ignored the reverse of the spectacle we had just seen.

See you all in 2024!

## Saturday, July 22, 2017

### Today's Episode of "It's my responsibility, but..."

"...The datasheet diagrams didn't have the pins labeled."

This time, it is the encoder board.

Here, it isn't obvious which pad in the footprint goes with which pad on the part.

It doesn't matter anyway, because a closer examination of the footprint and the Digikey list of optical sensors reveals that

Well, back to the fab again, for both boards. I'll use the QRE1113 parts that I have to test if the parts even fit in the hole. If not, I'll have to use the EE-SY193.

All parts of this project are my responsibility. It can't be otherwise, since there isn't anyone I report to or who reports to me. My partner is a special case.

This time, it is the encoder board.

It doesn't matter anyway, because a closer examination of the footprint and the Digikey list of optical sensors reveals that

*I bought the wrong part*. The footprint on the board is marked QRE1113, but it is actually for an Omron EE-SY193. D'oh!Well, back to the fab again, for both boards. I'll use the QRE1113 parts that I have to test if the parts even fit in the hole. If not, I'll have to use the EE-SY193.

All parts of this project are my responsibility. It can't be otherwise, since there isn't anyone I report to or who reports to me. My partner is a special case.

## Friday, July 21, 2017

### Dagnabbit!

That beautiful purple board I received yesterday won't work. If it had been stuffed and plugged in, it would have immediately shorted out anything plugged into it.

One of the features on the board is a super-wide (for a 6-mil PCB) strap between the two adjacent 5V pins coming from the Pi. When I had a close look at it, I saw a thin little gold crescent around part of the hole. It took a little bit for it to dawn on me that this was the ground plane, which was visible through the mask, plated in gold, and not protected from shorting with the other contacts:

The effect is everywhere on the board. I saw it first on the strap in the upper-left, but this magnified version shows it on the motor power section and connection to the Arduino. Note the crescents around the top of D3~ and RST, and the slivers of ground plane visible through the mask around the isolator footprint.

I don't think that there is anything that can be done to fix it. I also don't think that the encoder board is affected, so at least I can still use that. I'll try to stuff it, but I will check continuity closely. If any of the solder bridges the 6mil gap, then the short will exist.

I'm pretty sure the problem comes from settings in the ground plane. Since OSHPark can make 6mil boards, I take full advantage of the feature. Unfortunately, this interacts with a bad default in Kicad. Fortunately, that is easy to change, but it would have been nice to know fifteen dollars and two week ago. The money isn't a big deal, it's the time.

Kicad was even trying to tell me that something bad was about to happen. Not from the DRC (although that would have been nice) but in the images. Here is the old, bad design:

This option has settings in several places: The global setting I talk about above, the zone setting (which I haven't found), the footprint setting, and the individual pad setting. The later settings in the list have priority over the earlier ones, but the earlier ones are used as defaults.

In general with modern fab processes, you should set this value to zero. This will describe a hole in the solder mask the exact size of the pad. The fab can and will edit this to match their process, so don't worry about getting it too small.

Thanks to a YouTube video by My 2uF for help finding this setting.

One of the features on the board is a super-wide (for a 6-mil PCB) strap between the two adjacent 5V pins coming from the Pi. When I had a close look at it, I saw a thin little gold crescent around part of the hole. It took a little bit for it to dawn on me that this was the ground plane, which was visible through the mask, plated in gold, and not protected from shorting with the other contacts:

The effect is everywhere on the board. I saw it first on the strap in the upper-left, but this magnified version shows it on the motor power section and connection to the Arduino. Note the crescents around the top of D3~ and RST, and the slivers of ground plane visible through the mask around the isolator footprint.

I don't think that there is anything that can be done to fix it. I also don't think that the encoder board is affected, so at least I can still use that. I'll try to stuff it, but I will check continuity closely. If any of the solder bridges the 6mil gap, then the short will exist.

I'm pretty sure the problem comes from settings in the ground plane. Since OSHPark can make 6mil boards, I take full advantage of the feature. Unfortunately, this interacts with a bad default in Kicad. Fortunately, that is easy to change, but it would have been nice to know fifteen dollars and two week ago. The money isn't a big deal, it's the time.

Kicad was even trying to tell me that something bad was about to happen. Not from the DRC (although that would have been nice) but in the images. Here is the old, bad design:

The purple rings around the pads are the soldermask gap. You can see a slight red tinge around the edge where the mask gap overlaps the ground plane. |

The OpenGL preview shows it too, perhaps in an easier-to-understand form. |

To fix it, use the Dimensions/Pads Mask Clearance menu option |

Change the Solder mask clearance value from its default... (Yes, I design my boards in US units. You got a problem with that?) |

Change it to zero. |

The borders of all the pads are now black, indicating the solder mask gap doesn't span the space between the pad and the ground plane |

In general with modern fab processes, you should set this value to zero. This will describe a hole in the solder mask the exact size of the pad. The fab can and will edit this to match their process, so don't worry about getting it too small.

## Thursday, July 20, 2017

### Yukari 5 parts!

## Friday, December 30, 2016

## Thursday, August 25, 2016

### Radiation-pressure propulsion for nano-spacecraft

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 \(h\).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.

\[E_{pho}=h\nu=\frac{hc}{\lambda}\]

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}\]

&=&\frac{60,000,000}{10600}&=&5660 \mbox{s}

\end{eqnarray}\]

\[v=at\]

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.

\[\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}\]

&=&\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

Lubin, P. A Roadmap To Interstellar Flight

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