Home More Articles

Calibrating the Analemma Scene

It became immediately obvious in the project of photographing an analemma at the location I chose, that a lot of number crunching would be needed. Ths location I chose for the camera was in a sensitive area of a public-use airport. This meant that I could neither leave a camera on-site for an entire year, nor erect a fixture that would let me place the camera in the same exact location for every shot. So it was imperative to find a defendable way to take sun-shots and superimpose them accurately onto a carefully documented background scene.

In the original analemma photo, Dennis diCicco solved the problem of year-long registration by mounting the camera in his house. This gave him complete control of everything but the scene in the background of the astronomical event. Frank Zullo dealt with it in Arizona by shooting the Sun from his house and superimposing the analemma onto a beautiful background scene that he shot from a nearby, but different, location. Pál Váradi Nagy dealt with the problem by excruciating mathematical exorcism that would soon become very familiar to me.

Building the background scene, a 150-degree panorama, turned out to be the easy part of the operation. All I had to do was find a suitably scenic morning, go take a few dozen photographs, and stitch them together accurately. It's been done a thousand times before. But how could I be sure that that I could put the sun shots in the proper location against that background? Answering that question led me to confront several others:

What follows will cure just about anyone's insomnia, but it explains how I'm sure that my analemma image shows exactly where you'd see the Sun if you stood on the camera spot for a whole year, and blinked twenty-four times.

Horizontal Axis

Testing the horizontal calibration is easy. Re-stating one of the pixel-mapping equations,

(1)
x = (π / 180) (f / δ) φ + x0

which was developed in the main article:

There are eight landmarks in the scene whose exact location is known, thanks to the functionality of Google Maps. The bearing of each (φ) is a straightforward calculation, as described in the companion article How Long Is Your Lens?

Site Coordinates Bearing
from Camera
Pixel*
Camera 41.166369
-73.123600
Church spire 41.155080
-73.123439
179.38 1950 1838
VOR 41.160724
-73.124496
186.81 2396 2309
Wind Sock 41.163297
-73.125423
204.07 3465 3405
Control Tower 41.163489
-73.128290
230.79 5171 5100
WICC
South antenna
41.159218
-73.164002
256.76 6693 6749
WICC
North antenna
41.161005
-73.164659
260.15 6908 6964
RESCO stack 41.162481
-73.209440
266.53 7290 7369
Candy Stack 41.170680
-73.184041
275.39 7832 7911

* !! first numbers: dumpster4; second numbers: dumpster5

The panorama was carefully assembled to conform to Eq. 1 for a focal length of 20 mm, so it's no surprise that agreement is quite good. The size of the component panes was adjusted so that Eq. 1 would be satisfied. That provides the confidence needed to properly superimpose the Sun images to form the analemma – at least in azimuth.

!! comment about linearity and accuracy of slope

The range between the control tower and the Candy Stack got special attention. This area needs the best accuracy because that's where the Sun images were superimposed on the backgrund scene.

!! more comment about linearity and accuracy of slope

Vertical Axis

Calibrating the vertical axis is more tedious because of the trigonometry involved. We're projecting the celestial sphere onto a plane. The figure uses the projection paradigm (upright image) instead of the one for thin lenses (inverted image) because it's a bit easier to see what's going on. The math is the same either way. We're interested in the distance on the sensor between the images of the Sun and the horizon, to be compared to the results predicted by the vertical projection equation

(2)
y = y0 - (f / δ) tan(θ)

which was also developed in the main article:

The exact ratio f / δ is used here because we will be concerned with the exact focal length of the lens.

To test the accuracy of the equation, I took a series of pictures of the Sun at known times, and compared them to the elevation obtained from NOAA's Solar Position Calculator. The pictures look like this example, which has been cropped to remove a lot of image below the horizon. Ideally, the horizon would be in the center of the test image, but that's not always easy to arrance precisely. So we do a bit of math to get around that. However, it is important for the horizon to be level in the test shot. It's possible to correct that too, but it's much simpler to just hold the camera level.

Ideally, the camera should be level, with the horizon centered vertically. This is the assumption in Eq. 2, and it's the formula that was used to construct the analemma photo. This means that half of the camera frame is filled with stuff below the horizon, which is basically wasted information for our purpose. Here's the geometry for a scene where the camera was looking up a bit to get the Sun and the horizon in the same picture.

The right side of the figure is adjusted to show how things are seen from the viewpoint of the sensor. Its "horizon" is always the at center of the picture. Here the natural horizon is below center by some angle αH, and the Sun is αS above centerline. Because the lens is focussed at infinity (our subject is 93 million miles away), the sensor's distance from the lens is f. The offset angles for the Sun and the horizon therefore satisfy

sS / f = nS δ / f = tan(αS)
sH / f = nH δ / f = tan(αH)

The Sun's elevation is the sum of the two angles.

(3)
α = αS + αH = arctan(nS δ / f) + arctan(nH δ / f)

where nS and nH are the number of pixels between the center of the frame and the Sun and horizon, respectively. If the calculated value of α matches the Sun's elevation obtained from the Solar Position Calculator, the calibration is good.

The focal length of the lens is an important parameter here. With an affordable zoom lens, the only positions you can repeat reliably are the extremes. In my case, that means 18mm, which is the nominal focal length for all of these calibration shots. The actual focal length of the lens at this stop is slightly different, even though the EXIF metadata says 18.0mm. So there's a calibration shot for that too. This crop is from the center of a test shot where the horizon was placed as close to the center as care would allow, to minimize the effects of the lens's known distortion at that end of its zoom range. If you click it to pop up the full-sized version, you can count the pixels yourself. The angle these two markers subtend at the camera location is known very accurately, so I used the method discussed in How Long Is Your Lens? to calculate the actual focal length, 18.4 mm.

Also because of the barrel distortion at 18mm, it's important to locate the Sun as close as possible to the center of the test image. It was a while before I realized just how important this was – not just for measuring the Sun's elevation, but also to minimize the error if I didn't get the horizon 0.00° level. Barrel distortion puts each pixel closer to the center of the image than it should be. Measurements with this lens have been corrected for its known distortion, which is significant: -3.5%.

Perspective distortion

There is a small, but measurable effect on magnification for any part of the image that is off-center. This is not a lens aberration; it's an unavoidable effect of perspective. For distant objects on centerline, the magnification is approximately f / s, the ratio of focal length to object distance. But the sensor isn't curved, it's flat. So for off-center objects, f must be replaced by f / cos(α), where α is the object's angular separation from the center. This effect is very visible in pictures taken with fisheye lenses, and with a visitor seen through a goorway peephole. As the focal length increases, the effect quickly becomes very small.

For an object that's n pixels off-center, the offset angle is

(a)
α = arctan (n δ / f)
and the magnification ratio due to perspective is
(b)
h2 / h1 = 1 / cos(α)

Combining Eqs. b and c, and skipping a few intermediate steps that require trigonometry and the Pythagorean Theorem,

(c)
h2 / h1 = sqrt (1 + (n δ / f)²)

In the calibration photos used here, this error should be applied to the height of the Sun as measured above the center of the frame. As it turns out, the easiest way to deal with it is really to make it negligible; that is, to make n δ / f as small as possible. Even with an 18-mm lens (the worst case in my equipment), this isn't hard to do. If the Sun is within 350 pixels of center frame, the error is less than ½%. Within 150 pixels of center, it's below 0.1%. (With a 100mm lens, this error is less than 0.5% over the entire frame.) This is much smaller than even the error in measurement that would arise by misrepresenting the time of day by several seconds – a reasonable comparison to make, because the solar elevation depends on the time of day, and the best estimate we have easily available is the time stamp on the calibration picture. It's fairly easy to keep the camera synchronized to USNO time within seconds, but I didn't get as obsessive about it as diCicco did with his digital clock synchronized to WWV.

 

You can't just take a picture of the Sun, and see things on the horizon at the same time. The sensor doesn't have enough dynamic range for that. But I was able to find an exposure that allowed me to locate the center of the Sun with reasonable accuracy, while still keeping enough detail to locate the horizon. That was to meter for the scene while ignoring the Sun, and then add a 64X neutral density filter. This is an acceptable compromise at both ends. The Sun is blown out, but not so badly as to keep me from estimating its center. The horizon is deep in shade, but can be brought up in Photoshop well enough to measure where it is. The little photo on the left shows about how it started. A lot of darkness and an overexposed Sun. But if you click the thumbnail on the right, you'll see that there's enough detail in the shadow to find the horizon. Now, we just need to deal with that blown-out Sun.

Zooming in on the Sun, we can set a threshold at the very highest brightness level, and center a target there. After the threshold mask is removed, the target still locates the center of the Sun quite accurately.

The center pixel value (camera axis) is just half the height of the image. Together with the values for the Sun and horizon, we now have enough information to find nS and nH in Eq. 3.

For this particular photo, Eq. 3 yields a measured solar elevation of 29.3°, in almost perfect agreement with the Solar Position Calculator's prediction (29.4°). The method used to locate the Sun makes it very unlikely that this is off by more than one pixel, leaving the horizon as a possible source of error. Even if the horizon is mis-located by ten pixels (unlikely), Eq. 3 yields values between 29.2° and 29.5°.

It isn't necessary to wait a whole year to find out if the calibration is good, just long enough to take as many pictures as needed to catch the Sun in the appropriate range of elevation.
!! <better explanation>: I did that, and the next figure shows that the agreement was satisfactory: