We did that for you in the “Calculating the Angles” diagram above. 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”. The number “114.6” is from this calculation: 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. Just how far away do we need to be in order to get the moon the same size as an object of interest: 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“. 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. 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.
We determined that angle using Stellarium, by the way. 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. On the West Coast the last moment of totality occurred at 4:24 AM PDT. 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. Very Ordinary Photo of the Lunar Eclipse with the planet Uranus in the lower left.
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