Space Moon


Space Moon
800.0 mm || 1/13 || f/5.6 || ISO3200 || NIKON D7000
Seattle, Washington, United States

If you just want to look at a pretty picture, you can stop now. If you’re curious about how this picture came about and are not afraid of some math, keep reading.

The Idea

Cool picture, right? Here’s what went into making it:

It all started with an email to my friend Jacob, who moved back to Alabama1 last year:

Remember when I rented that 600mm lens for Seafair a couple years ago?
I’m renting the 800mm this year…wish you were here.

On my DX body, it’s effectively a 1200mm and it also comes with a 1.25 teleconverter for 1500mm of awesomeness.

While the primary intent was to take pictures at Seafair, Jake and I went back and forth for a bit on other cool photo ideas, originally talking about astrophotography but then moving to something a bit more Earth-bound. The idea was to do an ET-esque moonshoot similar to what Switzerland-based photographer Philipp Schmidli did:

The Silhouette of a Bike Rider on Thursday, April 25, 2013 before the rising full moon. Photo by  Philipp Schmidli.

The Silhouette of a Bike Rider on Thursday, April 25, 2013 before the rising full moon. Photo by Philipp Schmidli.

Another friend, Nate, was having a bachelor party that would involve lots of biking on Friday night. I was hoping to get everyone biking on top of the hill at Gas Works park, but the technical logistics don’t really work. Let’s go through the data.

The Data

Here’s the US Naval Observatory (USNO) Calculated Altitude/Azimuth Data for the Moonset times (when the Moon is at an altitude2 of 0°):

Date Time Azimuth3 Fraction Illuminated
8/1 23:00 258.1° 0.32
8/2 23:29 252.6° 0.42
8/4 00:02 247.7° 0.53

For Friday (8/1), the reverse bearing is 258.1^{\circ} - 180^{\circ} = 78.1^{\circ}. Moving out 1000 meters4 from the Gas Works Park hill (“S”) on a bearing of 78.1° puts us right over by I-5 (“D”) — so at least we’re on land.


Gas Works Park hill (“S”) on a bearing of 78.1° puts us right over by I-5 (“D”)

Unfortunately, it also puts us at an elevation that’s ~30 meters above the top of the Gas Works Park hill — we’d be looking down at them by \textup{tan}^{-1}\left ( \frac{-30\textup{m}}{1000\textup{m}} \right ) = -1.718^{\circ}:

Note: the elevation profile (top) is the opposite direction as the map (bottom).

Moving down to the water would reduce our elevation, but now we have another problem: Queen Anne Hill is blocking the moon. We’d have to shoot looking up 2.5°: \textup{tan}^{-1}\left ( \frac{65\textup{m}}{1500\textup{m}} \right ) = 2.481^{\circ}

Note: the elevation profile (top) is the opposite direction as the map (bottom).


Going back to the USNO, we could look up and find that at 22:43 the Moon will be at an elevation of 2.5° and an azimuth of 255.0°, or about 3.1° to left of where it would be at 23:00. We could keep playing this game until we found the perfect spot, but that still doesn’t solve our issue of being too close to the subject (i.e. the bikers at Gas Works Park).


The Gas Works Park location does not work because we must be far enough away from our primary subjects (the bikers in this instance) for them to appear small enough in comparison to our secondary subject (the Moon). Just how far is a function of \textup{tan}\ \theta = \frac{opposite}{adjacent} (Or solving for theta: \textup{tan}^{-1}\left (\frac{opposite}{adjacent} \right ) = \theta, which is the equation used several times above)


The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side.

In order to make the primary subject look smaller in comparison to the Moon, we are taking advantage of the fact the arc length (θ) of an object does not change significantly when your distance from the object is very large in comparison.

Arc diameter (Y-axis) vs ratio of distance/diameter (X-axis, log plot)

For example, the diameter of the Moon is about 3,475 km and the distance from the Earth to the Moon is about 385,000 km. This means the Moon has an arc diameter of:  \textup{tan}^{-1}\left ( \frac{3475 \textup{km}}{385000 \textup{km}} \right ) = 0.5171^{\circ}

Even if we move 100 km toward the Moon the arc diameter would still only be:  \textup{tan}^{-1}\left ( \frac{3475 \textup{km}}{385000 \textup{km} - 100 \textup{km}} \right )= \textup{tan}^{-1}\left ( \frac{3475 \textup{km}}{384900 \textup{km}} \right ) = 0.5173^{\circ}

…a difference of 0.0002°

Meanwhile, a person 1.8 m tall at a distance of 1 km would have an arc height of: \textup{tan}^{-1}\left ( \frac{1.8 \textup{m}}{1000 \textup{m}} \right ) = 0.1031^{\circ}

However, if we move closer just 0.5 km the person now has an arc height of: \textup{tan}^{-1}\left ( \frac{1.8 \textup{m}}{1000 \textup{m} - 500 \textup{m}} \right ) = \textup{tan}^{-1}\left ( \frac{1.8 \textup{m}}{500 \textup{m}} \right ) = 0.2063^{\circ}

…a difference of 0.1°!

Math with Bicycles

In Schmidli’s Moon photo, the Moon is about 7 bicycles in diameter:

If the bicycle is 1.8 m long, the camera must be at a distance of: \textup{tan}^{-1}\left ( \frac{3475 \textup{km}}{385000\textup{km}} \right ) = 7*\textup{tan}^{-1}\left ( \frac{1.8\textup{m}}{x}\right ), where x is the distance between the primary subject (the bicyclist) and the camera. Solving for x\Rightarrow x = \frac{9}{5} \textup{cot}\left (\frac{1}{7} \textup{tan}^{-1} \left ( \frac{139}{15400} \right) \right) \approx 1396.0

…1.396 kilometers.

If Schmidli had only been 500 meters away, the photo would have probably looked something like this:

It still looks cool, but not as cool. So, what else could I take a picture of?

Another Idea

How about the Space Needle? Plotting the azimuth of the Moon set for all three nights to see if a good location is feasible:


Bearing of the Moon set on 8/1, 8/2, and 8/4

The southwest corner of Volunteer Park on Sunday, August 4th looks like it could work and the elevation profile looks good too:

Note: the elevation profile (top) is the opposite direction as the map (bottom).

The “Black Sun” sculpture at Volunteer park is at ~137 m above sea level and the base of the Space Needle is ~2.77 km away at ~40 m above sea level. The top floor of the Space Needle is ~158 m above ground level which makes the top floor 198 m above sea level. Google Street View confirms that we have good view. I stopped by during the day to confirm the view and find a good spot to setup.


Framing the Picture

The Space Needle “Halo” has a diameter of ~42 m which at a distance of 2.77 km gives it an arc diameter of \textup{tan}^{-1}\left ( \frac{42 \textup{m}}{2770 \textup{m}} \right ) = 0.8687^{\circ} — or about 1.7x larger than the Moon. The maximum angle of view of the Nikkor 800mm (on a DX body) is just 2° — thus the Space Needle should take up about 43% of the frame width.

I relayed my findings back to Jake:

I’ve been doing some math, and on Sunday night I’ll head to up to the SW corner of Volunteer Park. That puts me on a 248 deg bearing with the space needle and the moon will be right behind it. At 2500m away, the halo of the space needle will still appear to be about twice the diameter of the moon and if I put the teleconverter on I think the moon + spaceneedle will fill the frame just about perfectly.

So something like this…but in high def:

So that was that. I showed up a bit past 11pm on the night of August 3rd, set up my equipment, and sat around listening to passersby play the piano as they filtered out of the park. I took test photos as the Moon grew closer and closer, and then just like clockwork…almost.

Lessons Learned

I forgot to take into account the height of Space Needle relative to the elevation of the Moon, so the Moon was too far to the left for the picture I was hoping for — but pretty much dead on the calculations I made, even if I calculated the wrong thing.

The halo of the Space Needle was at an elevation of \textup{tan}^{-1}\left ( \frac{198 \textup{m} -137 \textup{m}}{2770 \textup{m}} \right ) = 1.262^{\circ}. Add 0.3° for the arc radius of the Moon to get ~1.6° of elevation. Using the USNO table, we find the Moon at an azimuth of 245.6° at 23:50. The reverse bearing is 65.6° which puts me about 100m too far south-ish for the picture I wanted. Unfortunately, the right spot was blocked by trees…so it’s kind of a moot.

Making Lemoonade

Put I still have a bunch of really cool photos and after looking at them all of a bit I think to myself, “Hmmm…this could be really cool as a multiple exposure / panorama shot.”

I went through all the photos to find the ones that had both a complete view of the Moon and at least part of the Space Needle. Three interesting things occurred as the Moon started to set:

  1. The Moon became more red as it set. This is due in part to particulates in the air (e.g. pollution) as well as atmospheric scattering (i.e. Rayleigh scattering).
  2. The luminance of the Moon decreased significantly during its setting, such that its exposure was almost the same as the Space Needle. The Moon is typically several stops above the ambient landscape, even when the landscape is lit up. If you’ve ever tried to take a photo at night, you’ll probably notice that the Moon looks blown out — this is why.
  3. The shape of the Moon gets a bit wonky as comes in line with the ground and picks up some heat shimmer.

I played around with different sets of pictures and finally settled on a set that I think flows well. I did some basic exposure and color correction so that the photos were more-or-less “correct” and then imported them into Photoshop. Once in Photoshop, it was just a matter of aligning the images to build the panorama and then masking out sections to get the multiexposure.  The Space Needle itself is a composite of several different shots because I couldn’t fit the entire Space Needle into a single shot (yea, it’s a BIG zoom lens). I also needed a properly exposed shot of the interior of the top floor and a shot of the elevator in motion. Here’s what it looks like decomposed:


  1. Roll Tide 

  2. For the record: I prefer “elevation”, but I’ll try to remain consistent with the USNO wording 

  3. E of N 

  4. at this point, 1000 m is just an arbitrary “best guess” for how far away I’d have to be 

Vaporizing Lake Washington

Sometimes I wonder about interesting things, such as how much energy would it take to boil all the water of Lake Washington:

  • The volume of Lake Washington is 2.89 km31
  • The average lake temperature is 9.71°C2
  • It takes 4.19 Joules to raise the temperature of 1g of water by 1°C: \frac{4.19 \mathrm{J}}{ 1 \mathrm{g^{\circ}C \: _{H_{2}O}}}3
  • The density of water is \frac{1 \mathrm{g}}{1\mathrm{cm^{3}}}

Putting all that together, we get:

2.89 \mathrm{km^{3}} \times \left ( \frac{1000\mathrm{m}}{1\mathrm{km}} \right )^{3} \times \left( \frac{100\mathrm{cm}}{1\mathrm{m}} \right )^{3} \times \frac{1\mathrm{g_{_{H_{2}0}}}}{1\mathrm{cm^{3}_{H_{2}0}}} \times \frac{4.19 \mathrm{J\:} }{\mathrm{g^{\circ} C _{H_{2}O}}} \times  \left ( 100^{\circ} \mathrm{C} - 9.71^{\circ} \mathrm{C}\right )= 1.093\times10^{18}\mathrm{J}

For comparison, the energy that hits Earth from the Sun in one second: 1.74 \times 10^{17} \mathrm{J}4

Basically, if we could focus all the energy from the sun that hits the earth, it would take…\frac{1.093\times10^{18}\mathrm{J}}{1.74 \times 10^{17} \mathrm{\frac{J}{s}}} = 6.281 \: \mathrm{seconds} …to vaporize Lake Washington5.

This is a vast oversimplification of the forces and energies involved, but I think it’s still a pretty good estimate.

Update: Apparently I missed one critical element, enthalpy/heat of vaporization \Delta{}H_{\mathrm{vap}}6. “This energy breaks down the intermolecular attractive forces, and also must provide the energy necessary to expand the gas (the PΔV work). For an ideal gas , there is no longer any potential energy associated with intermolecular forces. So the internal energy is entirely in the molecular kinetic energy.”7

What we have above is the energy required to bring it up to 100°C, but not to vaporize it. To actually vaporize water that’s already at 100°C, we need to add an additional \Delta{}H_{\mathrm{vap}} = 2260\mathrm{\frac{J}{g}}8

Running this number back through our calculations, we now get:
2.89 \mathrm{km^{3}} \times \left ( \frac{1000\mathrm{m}}{1\mathrm{km}} \right )^{3} \times \left( \frac{100\mathrm{cm}}{1\mathrm{m}} \right )^{3} \times \left (2260\mathrm{\frac{J}{g}} + \frac{1\mathrm{g_{_{H_{2}0}}}}{1\mathrm{cm^{3}_{H_{2}0}}} \times \frac{4.19 \mathrm{J\:} }{\mathrm{g^{\circ} C _{H_{2}O}}} \times \left ( 100^{\circ} \mathrm{C} - 9.71^{\circ} \mathrm{C}\right ) \right ) = 7.625\times10^{18}\mathrm{J}

This is still within one order of magnitude from my original answer and really only takes slightly longer for the sun to actually vaporize Lake Washington \frac{.625\times10^{18}\mathrm{J}}{1.74 \times 10^{17} \mathrm{\frac{J}{s}}} = 43.82 \: \mathrm{seconds} 9.


  2. Average of all temperature data for 2011 for the Lake Washington buoy: 


  4. According to Wolfram Alpha 

  5. ROM estimate 

  6. this is why I’m not a chemist 



  9. still a ROM estimate 

Napkin Analysis of the Sand Flea Jumping Robot

I shared this video1 with Peter, who then asked:

I saw that a couple days ago. Awesome! And has some cool practical applications. I [couldn’t] quite tell if the pitch of the robot was adjustable by the user, or of it always jumped in the same direction. Did you get a sense for that?

It was a good question and one I didn’t have an immediate answer to.

I would actually guess that I don’t have immediate answers2 to at least 50% of questions people ask me3. I have to do some amount of thinking, and sometimes even some research. I think people tend to think I know the answer off the top of my head, I assure you: I am not that smart.

I do have an inquisitive mind, I do know where to look, and I do know how to ask the right question.

I decided to remedy this question though by talking it through, instead of just giving an answer. This is basically my thought processes as it occurred. Except that I got Sin and Cos mixed up and didn’t realize it until I had finished my conclusion. So I had to redo my entire analysis, and that’s what you see here. Please note this is still really just a paper napkin answer:

As far as angle, I’m not sure. I suspect there would be some angle change.

Elevation angle can affect two things, how high it goes and how far it goes forward, and these two things are intrinsically linked through SohCahToa. Height and forward distance can also be affected by the force applied (ceteris paribus4). This gives a problem with two independent input variables (angle and power) and two dependent output variables (height and forward distance/range).

Since my primary goal is to jump, I’m going to put most of my energy into that. If I want to jump higher, I can either apply more force or make my elevation angle higher (as long as it’s < 90°). As the elevation angle nears 90° [latex]\left (\frac{\pi}{2} \right )[/latex], more of my energy goes into going up than going forward. The proportion of energy applied to going up is defined by Sin and the proportion of energy applied to going forward is defined by Cos. Also worth remembering is that the Sin[x] + Cos[x] is not a straight line, it's another parabola that peaks at 45 degrees. The biggest bang for your average buck is to angle yourself at 45 degrees and shoot. Additionally, Cos (forward) angles that are near 90° have a high rate of change (i.e. going from 80° to 81° has more of a difference than going from 10° to 11°), thus little changes in elevation angles near 90° have relatively larger impacts on how far forward I go. Conversely, Sin (height) angles that are near 90° have very low rates of change. The cross over point for rates of change between Sin and Cos is - you guessed it - at 45° . Since the goal of the robot is to jump high (not far), it would make sense to only use high angles (above 45° ). To vary height significantly though, you are going to have vary power. Going from 46° to 90° only increases height by ~93% if the force remains the same. In comparison, going from 1 degree to 45 degrees increases height by 164,000%. Math is great, but if you can't implement it, it doesn't matter so let's turn to what's practical: One of the underlying assumptions is if the robot can vary the force it uses and if it could accurately set it's elevation angle. Setting the angle is pretty easy using encoders, and accelerometers to determine which way is down (if you were jumping from an angled surface, for instance). We've also already seen that the jumping leg can move, so adding functionality for precision angle measurements (within a degree, let's say) is pretty trivial. The real question, I think, is how does it jump? Delivering energy quickly has always been a problem. Delivering a measured amount of energy quickly even more so. Based on jumping from the ground to the loading dock (1.5 meters in height at most) and then from the loading dock to the roof (probably at least 4 meters), that's about a 166% increase in height, which is not quite enough as could be accomplished by just varying the angle from 46° to 90°. Since you can't gain that height just by altering the angle alone, it makes sense to assume that the jump force setting can be altered. However, if you change the jump force setting, what does that do to the forward movement (we know it will make the robot jump higher)? It will, of course, move the robot forward even more. How much more? I don't know exactly, but probably enough to make some minor angle tweakage worth it. We would have to sit down and work on the math to verify the exact amount. I think it involves something with squaring the derivative of the force divided by the mass. Squaring always make numbers bigger, so I tend to think it would be significant. Suffice it to say, if you don't want to proportionally more forward when you jump significantly higher, you would have to adjust your jump elevation angle. Thus I would assume there may be small changes in angle elevation, but that's hard to estimate given the view-point the videos were shot at. It's also pretty easy to solve for power required and angle needed to reach a particular height while moving forward only a certain amount (once you figure out what the maths are), so at least the implementation factor is pretty easy from a computing standpoint. And I've spent way to much time on that answer.5 As always, please check my work.

  1. found via the always interesting Kottk

  2. answers that only involving recalling a specific outcome 

  3. I just made that number up, really 

  4. all other things being equal or held constant 

  5. One of the reasons I decided to blog about it, the work was pretty much a sunk cost 


I was home sick today and in between drifting in and out of consciousness1 in my bed, I came up with these sequences of numbers:

  1. 88642483261224_
  2. 7535155251_
  3. 91998188_

Can you determine which numbers comes next?

  1. I think it was some sort of sinus infection or something…massive headache, congestion, general stuffiness, sore throat, etc…pretty much felt like death 

Turning Spheres Inside Out

Have you ever wanted to turn a sphere inside out? How about without creases or sharp corners? Warning: Math content to follow!

The following 20 minute video will show you how:

via kottke


θmg = ωtF

Unit checks are a great thing, as evidenced by this little goodie brought to my attention on Facebook:

\theta m g = \omega t f

And the unit check:

\mathbf{rad} \times \mathbf{kg} \times \frac{\mathbf{m}}{\mathbf{sec^2}} = \frac{\mathbf{rad}}{\mathbf{sec}} \times \mathbf{sec} \times \frac{\mathbf{kg} \times \mathbf{m}}{\mathbf{sec^2}}

Yup. It works.