Chasing the Moon Chasing the MoonOne of my desires has been to capture the full (or slender crescent) moon when it is exactly in line with a well known landmark. One such landmark is the Pigeon Point Lighthouse near Pescadero, California. Ideally, the moon will be nearly as large as the lighthouse and directly behind it. Another possibility is to get the moon to be the same apparent size as the lens portion of the lighthouse so that at some point the moon appears to shine through the windows at the top of the lighthouse. Here is a shot of the lighthouse using a 24mm lens. Effective focal length for this is 38.5mm as expressed in 35mm terms. The lighthouse is at a distance of about 2000 feet. The angular diameter of the lighthouse in this picture is 3.3 degrees - but more on that later. This picture was taken at moonrise.
To capture the moon exactly behind or adjacent to the landmark, we must determine each of the following:
We also need to understand what scenarios may allow us to capture this phenomenon. This may not be obvious to a non-astronomer, but the moon is full when it is 12 hours AWAY from the sun - usually on the opposite side of the sky or nearly so. This means that if the day is 12 hours long, the full moon will rise when the sun sets and vice versa. Likewise a slender crescent moon rises and sets very close to the sun. To see the crescent moon there have to be about 40 minutes or more difference in the rising or setting times of each. This translates as follows:
To get full moon near an object in the west, you have to shoot near sunrise. To get a crescent moon in the west, you must shoot close to and usually just after sunset. The net of this is that a 1/2 moon is never going to be near the horizon at sunrise or sunset - except at extreme latitudes like Alaska or Antarctica. And there is one more constraint: through the seasons, the sun and moon have a maximum and minimum azimuth (direction) where they set or rise. If your vantage point has you looking too far north or too far south, the sun or moon will never arrive behind your target object.
Consider one important principle. The moon has a small angular diameter - it is only one half of a degree - that is a bit smaller than the size of your pinkynail viewed at arm-length. If you want to fill the frame with a shot of the moon you need a lens that has about a 0.75 degree field of view. I will not bore you with the math, but that means an effective focal length of 2500mm. Wow. That's not a lens, that's a telescope! Do not despair, however, because with a 200 mm lens, a 2.0 extender and 1.6 crop factor camera, you get effectively 640mm - 1/4 of the way there at the expense of making your moon 4 times smaller in the frame.
The easiest way to figure out your target coordinates is by using Google maps. Find the object. Switch to the Terrain map and get an idea of the elevation (distance above sea level) of the base of the structure. In the case of the lighthouse, it is clearly very close to sea level and the surrounding area is pretty flat.
Figure - 1
Next switch to the Satellite view and zoom in as far as possible. Right click the map where the lighthouse proper is located (a recognizable round object on the satellite photo) and choose "center map here".
Figure - 2
Centering is important, because the next step will give you the coordinates of the map CENTER. You can see that I've already placed a map point here using the "My Maps" feature of Google Maps. Notice that my placement is different from the square marker that Google Maps provides... this represents a difference of about 50 feet and may matter quite a lot!
Figure - 3
Now click the the 
item above and to
the right of your satellite photo. You get something like is shown if Figure 4.
Figure - 4
The GPS coordinates follow the &ll= in the long link. You'll have to go to the end of the line and back up to see this. You might find it easier to copy the entire thing, paste it into a text editor (like notepad), and then cut out the portion following the &ll up to then next &. You only need what is in blue above. This corresponds to a latitude of 37.181702 degrees (north) and a longitude of -122.394071 (west). There are quite a lot of tools for capturing the latitude and longitude information... feel free to find another method if the above is too complicated.
Google search works great here. Searching for "pigeon point lighthouse height" gets the following hit: Pigeon Point Lighthouse — Height: 115 FT (35 M)
That's convenient! It also helps to know the terrain a bit by having scouted first. I know, for example, that the base of the lighthouse is actually about 45 feet above sea level, and the road and ground to the East is about 20 feet below the base of the lighthouse. So the net is that the top of the lighthouse is 135 feet above most of the vantage points. You may also have noticed in Figure 1 that I marked 3 possible locations based on some scouting using my GPS and my eye. My goal was to find a vantage point with a view of the lighthouse and some interesting foreground for a wide angle shot.
Now that you know the height of the object above the horizon, and the height of the object itself you have some considerations to work through.
If you want to create a shot where the moon appears to be as large or larger than the entire lighthouse you need to move as far away as necessary so that the moon's apparent diameter (0.5 degrees) and the lighthouse apparent angular size are nearly the same. To accomplish this you must calculate how far away you need to be from an object that is 115 feet tall so that its angular size is 0.5 degrees. You can do this with any one of a number of online Trigonometry solvers. This one is nice looking, but is restricted and has some quirks, another source is this one.
Essentially the equation is this: distance = objectHeight / sin(0.5°)
After all of the trigonometry, the equation becomes a simple: multiply the height of the object by 114.6 - and that's how far away you need to be (this works no matter what units you use)!
Anyway, after all of that, we calculate a distance of 13,179 feet - 2.5 miles! At that distance, the moon would be as large as the lighthouse and as described earlier you will need a 2500 mm lens (35 mm equivalent) to nearly fill the frame with the lighthouse. A lens like that is hard to come by! Unfortunately at that distance in this example you would have to be up in the foothills at least 500 feet above sea level where there is a lot of private land, and where the net result is that you will see the ocean behind the lighthouse - no chance of seeing the moon. So that doesn't work.
Instead we will use the "ideal lens" scenario - 640 mm effective focal length. This is about 1/4 the ideal focal length, so we can get 4 times closer and just accept that the moon will appear 1/4 as big in. Moreover, we will accept cropping our photo 50% so we can now get 5 times closer - about 2800 feet. Remember, though, that the closer we get to the lighthouse, the smaller the moon's apparent size will be. On the other hand, the closer we get to the lighthouse, the higher above the horizon the moon can be to get the shot lined up. This helps because the lower in the sky the moon (or sun) is, the more atmosphere you have to shoot through.
I already scouted several locations, including one that is 2880 feet away - it's the one farthest to the right in Figure - 1. How did I know it is 2880 feet away? I used the Google Distance Measurement tool. (Well, actually I used another free online tool that allowed me to enter two coordinates and told me the distance and the bearing - azimuth). You may also be able to make a similar determination by using your GPS and grabbing coordinates at the target and the vantage points. Precision is not that important. But do consider that you want to scout an area with room to move around. If your location is a small balcony on a building... you may find the moon never quite will get where you want - at least not in the next 100 years.
I glossed over how to do this above, in part because I found and later created a javascript based tool to do it.
I found this process the most tedious so I first created an Excel spreadsheet to do the math for me. The spreadsheet also does other things: calculates the distance, calculates the azimuth and the angular size of the object based on its height. It's not yet ready to share... check back later.
All that work, and you now have to do the next step - figure out when the moon (or sun) is going to be exactly - or nearly exactly - where you need it. First let me point out that if all you want is to have the Moon at a specific azimuth and altitude - that's not going to be that hard to do. If you want the moon to be in a specific phase (e.g. full), you will find the possible dates and times are rapidly becoming smaller. And if you want to ALSO get a photo with some dusk or twilight (recently risen or set sun) ... well now you are down in the 2-3 times in a year range of possibilities. But here is how you do it (in rough details).
Use the SunMoonCalc program. Enter your Vantage point in the coordinates. Be sure to select the correct timezone offset. Now... pay attention to the constraints we must use.
A) The object we want to photograph is WEST of all of our possible vantage
points.
B) Because of A, we need the moon to be SETTING.
C) Because we know the height of our object above the horizon, and because we
know our distance away from it, we know we need the moon to be at about that
angle (2 degrees, for example) above the horizon.
D) Since we want to get foreground illumination, we will need for the sun to
be almost or just risen. This way we can keep the contrast and the moon
details. So we check between 40 minutes before and 20
minutes after sunrise.
E) We can cheat: we can take an exposure when our subject is not yet lit, but
the moon is present and combine it with the same view after sunrise where our
target is lit - or vice versa.
Figure 5 shows the SunMoonCalc tool where we are specifying a specific latitude and longitude. Notice how -122 longitude is entered as 122 W(est), and that the time zone, and daylight time are selected.
Figure - 5
Clicking the Rise/Set Criteria produces this page, which we fill out to specify that we want to define moonset as 0 to 2 degrees altitude, an azimuth of 261 to 266 degrees (west-southwest), and that we want anything from 45 minutes before to 30 minutes after sunset. As discussed earlier, combining moonset with sunrise pretty much guarantees you are at or near a full moon.
Figure - 6
And here is the result from hitting the "Display" button:
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Location: Pigeon Point, CA Latitude: 37:11 N, Longitude: 122:23:10 W Time Zone: UTC−8 (Pacific) Dates: 7 Jun 2009 to 8 Jun 2011 Moonset Altitude (center): 0° to 2° Moonset Azimuth: 261° to 266° Moonset 45 min before sunrise to 30 min after sunrise 4 dates match criteria:
SunMoonCalc © 2009 Jeff Conrad |
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chart of sun and moon azimuthAdditional Reading
"Azimuth" is perhaps most easily understood as "bearing" - the true direction - in degrees - from the starting point to the destination point. 0 is due North, 90 is East, 180 South and 270 West. It is not a coincidence that you can imagine a clock and multiply the "hour" by 30 to come up with the Azimuth. Thus the heading 4 o'clock is 120 degrees and is ESE. Azimuth can also be expressed relative to "True North" (e.g. 78 degrees West of True North) or to Magnetic North (19 degrees East of Magnetic North). 78 degrees West of True North is equal to 360 - 78 = 282.
Altitude is the number of degrees an object lies above the horizon. 0 degrees means "on the horizon", 90 degrees is directly overhead (the Zenith). A negative number means the object is BELOW the horizon.
Coordinates - the latitude (degrees north or south of the equator) and longitude (degrees East or West of the Prime Meridian). Google Maps can show you the coordinates of a location - see above for how to obtain this. An exact latitude and longitude can tell you exactly where you are. GPS devices also can display this information.
Magnetic North = the direction a compass points. Since the Magnetic North Pole is not at the actual geographical North Pole (true north) the compass heading is different from True North and depends on the observer's location on the globe. For example those in North Western Alaska will find True North is 30 degrees WEST of magnetic north. In most of California, true north is about 14 degrees West of magnetic north. To make matters worse, the Magnetic North Pole moves constantly. If this confuses you, don't use a compass, use a GPS. The difference between True North and Magnetic North is the "Magnetic Declination".
True North = the direction of the geographical North Pole - this is unchanging.
Horizon: 90 degrees perpendicular to straight down is the Horizon. The horizon is always at eye level. If you are up very high, your horizon may include things that would ordinarily not be visible (i.e. below the horizon for an observer at a lower elevation). In addition, objects like the Sun and Moon are actually visible when they are below the horizon because the Earth's atmosphere bends the light a fractional amount.
Zenith - the point directly overhead of the observer. This is at altitude 90 (degrees).
UPDATE: I used the tools and process described above to determine when the moon would be exactly aligned along the deck of this structure.
![The (Un)Conventional View [5_004224]](http://farm4.static.flickr.com/3625/3607926815_c68c769954_m.jpg)
The predictions said 12:42 am, June 7, 2009. Here is the result:
![Moon Prison [5_004269]](http://farm4.static.flickr.com/3562/3607930807_f064801f90_m.jpg)
