## 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 Alabama ((Roll Tide)) 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:

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 altitude ((For the record: I prefer “elevation”, but I’ll try to remain consistent with the USNO wording)) of 0°):

Date Time Azimuth ((E of N)) 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 meters ((at this point, 1000 m is just an arbitrary “best guess” for how far away I’d have to be)) 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.

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 $\textup{tan}^{-1}\left ( \frac{-30\textup{m}}{1000\textup{m}} \right ) = -1.718^{\circ}$:

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°: $\textup{tan}^{-1}\left ( \frac{65\textup{m}}{1500\textup{m}} \right ) = 2.481^{\circ}$

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).

## Math

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)

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.

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:

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

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.