Geometry and The Moon

Please do not run away. We are about to use adult language here. For example we will be using the word trigonometry. Still here? Good.  Here is a very pedestrian looking lunar eclipse photo taken with a 280mm lens*, cropped.

Near and Distant Neighbors

Very Ordinary Photo of the Lunar Eclipse with the planet Uranus in the lower left.

This past lunar eclipse several of us put our heads together to try to come up with a more creative photo than the one above. We had a trigonometry problem, however. On the West Coast the last moment of totality occurred at 4:24 AM PDT. We were brave enough to be out at any time of night – even if it meant extreme sleepiness in our day jobs but our problem was that the lowest the moon would be in the sky at the last bit of totality was 32.6 degrees above the horizon. We determined that angle using Stellarium, by the way. Unfortunately there is pretty much nowhere to go to get a nice large moon near an interesting object when the moon is almost 33 degrees high.

Wait: Why do we want the moon and the object to be similarly sized? Here is why… we want the moon to be noticeable like the Fantasy version below, not merely “present” like the real photo on the right. Even bigger would be better, right!?


Notice above right (Reality) and below how tiny the moon is compared to the building in the foreground?  Indeed, if you see a photo taken from anywhere on the West Coast where the eclipsed moon is significantly lower in the sky or larger than shown against foreground, you know it has been “photoshopped“.

Plan C: San Jose City Hall Eclipse Sequence

In short, it is nigh impossible to get the large moon effect with an altitude (angle) of 32 degrees here is why:

Calculating the Angles

Calculating the Angles

Just how far away do we need to be in order to get the moon the same size as an object of interest:

114.6 x object size

In other words, an object that is one foot tall, requires us to stand 114.6 feet away to make the 1/2 a degree angular size of the moon the same angular size as that 1 foot tall object.  The number “114.6” is from this calculation:

1 / TAN (0.5 degrees)

Yeah, that is trigonometry. Using still more trigonometry it is possible to calculate how high above the horizon a 9 inch tall object has to be so that it is “moon sized”.  We did that for you in the “Calculating the Angles” diagram above. Once you calculate the distance from the camera of 85.9, you can multiply that by the sine of the angle to calculate a height of about 46 feet! Here is the trigonometry:

Height = 85.9′ * SIN (32 deg)

You can go one step farther and calculate the distance from the object with ‘distance = 85.9 * COS(32 deg)’.

Of course after all that calculating you will still need to find a location, have contingency plans for weather and so on. At StarCircleAcademy we have built some tools and put together materials to help in all these endeavors.  We teach these things in our NP111 Catching the Moon Webinar.

The Road To The Temple

Below is where we ended up. This image is from our friend and co-conspirator Andy Morris.

Lunar Eclipse over Temple by Andy Morris of PhotoshopScaresMe

Four of us plotted and schemed to get an interesting shot. Above is Andy Morris’ result.  Click the image and you can read a great article about how he created the shot using Photoshop Skills at his site: In fact, it’s a great article which we strongly encourage you to read. You’ll learn how he composited the images together in Photoshop as layers.

The Long Conversation to Pick a Location

Andy has more details including how alcohol played a part in the process. Mostly I, Steven, was the wet blanket explaining why the geometry was all wrong.

  • The Stanford (Hoover) Tower looks like it is shrouded in trees from the needed angle
  • Bank of Italy (formerly BofA) in SJC doesn’t work
  • The main problem with the wind turbines is that the angle to the top of them is something around 12 degrees above the horizon which is 40 moon diameters below the eclipse.
  • Here is why the GG Bridge doesn’t work…
  • This seems to be the best solution I could find: the Coit Tower…
  • Darn. It would appear the coast is out. Forecast calls for Fog from SF to HMB
  • This might make an interesting foreground (see below)… Somebody want to check if they will mind us being on their property in the wee hours?

*Ok, we lied, it was actually a 70-200mm lens with a 1.4 TC on a full frame camera, but the net is the same: 280 effective mm focal length.

Where did you go and what did you get in your planning efforts?  Post a comment and link below… we’d love to see what you came up with!

6 thoughts on “Geometry and The Moon

  1. Craig Stevenson

    I’ve been ham-and-egging it with pretty good results. For bristlecone pines and saguaro cacti, I’m figuring a minimum of 0.7 mi. and over a mile is better. A mile and a half helps me get a better depth of field with a 500 mm and a doubler. I do need to do the webinar, it just seems I’m out shooting or committed to something else when they come up. I’d prefer having instructional CDs, so I could go at my own pace.

  2. Marsha

    Being mathematically challenged, I like it when you do the math for us. After attending several of the StarCircleAcademy Catching the Moon webinars, I knew I would be facing the high-in-the-sky, small-moon-syndrome. I wanted to show the phases of the eclipse from beginning to end, so needed to calculate how much sky this would take. Add to that challenge, I also wanted foreground. Here is the result.


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