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.
For Friday (8/1), the reverse bearing is . 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.
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 :
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°:
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 (Or solving for 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:
Even if we move 100 km toward the Moon the arc diameter would still only be:
…a difference of 0.0002°
Meanwhile, a person 1.8 m tall at a distance of 1 km would have an arc height of:
However, if we move closer just 0.5 km the person now has an arc height of:
…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: , where x is the distance between the primary subject (the bicyclist) and the camera. Solving for x:
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?
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.
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.
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 . 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.
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:
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).
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.
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:
Pros don’t buy these lenses. Nikon and Canon’s pro support programs loan them out for free at sporting events hoping TV viewers see more black or white to influence consumers. Therefore, don’t take any of the prices that seriously. Nikon and Canon probably take a loss on the sale of each of these lenses, considering the small quantities sold. They are created mostly for bragging rights, like the unbeaten Nikon 13mm f/5.6.
It was fun to rent, but shooting anything with a lens like this is hard. Ideally I would just move closer to the airplanes, but until the FAA decides to let drones fly with planes (or the Blue Angels invite me to ride along) I’m stuck with have to use the power of optics.
The issue which shooting with such a large lense is mostly the haze and smoke trails. The heat shimmer kicks in eventually if you get enough air between the you and the object you’re trying to shoot:
You may need to embiggen to get a good look, but the planes basically look like a mosaic because of the heat shimmer — and there’s no amount of Photoshop that can fix that.
Convection causes the temperature of the air to vary, and the variation between the hot air […] and the denser cool air […] creates a gradient in the refractive index of the air. This produces a blurred shimmering effect, which affects the ability to resolve objects, the effect being increased when the image is magnified through a telescope or telephoto lens.
Your best bet is to get a polarizer (which I unfortunately didn’t have) and shoot at 90° to the sun, as shown by the Rayleigh sky model (See also: Polarizing filter). There’s actually a pretty cool tool called SunCalc that will show you where the sun will be at a given date and time — very useful for things like this.
Still, with just over 1500 photos I was bound to get some good ones. Interestingly enough, it’s not that hard to actually track the planes once you get a bead on them. Here are the 4% that made the cut:
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