Photographing Solar Eclipses
Author: Bill Kramer
Last update: 3 JUL 2014 BK
Mounting the Camera/Telescope
Do you really need to pack a tracking mount?
Mounting of a camera or camera/telescope combination requires some consideration. A tripod or rigid post capable of holding the equipment is essential when using any sort of long lens. Even a shorter lens will benefit from a mount because during the total solar eclipse you may be too excited to hold a camera steady.
The question is whether you can use a regular tripod or do you need something better suited to astronomical picture taking - a tracking mount.
Any experienced astrophotographer will tell you that an equatorial mount is best when taking pictures of things in the sky. An equatorial mount is one that tracks the sky during the course of time. The most common has one axis pointed at the pole (north or south) allowing it to rotate slowly matching the turn of the Earth.
The Earth rotates about an axis once a day. That means that the stars along the equatorial part of the sky move (relative to an observer on Earth) once a day or 360 degrees per day, 15 degrees per hour, 0.25 degree per minute, 0.004 degree per second. It works out that the stars move at about 15 arc seconds per second of time at the equator.
As the target star is located closer to either pole the speed slows down by a factor of the cosine of the declination angle until you are at the pole; at which point the stars in that region do not appear to move relative to the Earth bound observer (like the North Star which is pretty darn close to the 90 degree mark and thus appears in the same position of the sky every clear night).
Astrophotographers use an equatorial tracking mount when taking pictures of the stars. The mount moves very slowly and follows the star across the sky. To move the mount a small electric motor is often employed. The size and weight of the mount, motor, and counter weights will often be several times that of the telescope and camera.
Note that many commercial tracking mounts do not account for the slightly different speeds that the sun and moon are traveling relative to the back ground stars. To obtain 100% tracking of the sun you will need a variable speed drive that can be set to keep the sun centered. The sun is moving at rate slightly slower than the background stars when viewed from the surface of the Earth (1/365th of a day slower).
The issue is that equatorial mounts tend to be heavy. When packing for a solar eclipse chase weight can be very important. Heavy counter weights and mounts can mean the difference between regular luggage and paying for overweight luggage.
There are two reasons you might elect to bring along an equatorial mount.
- To assist in keeping the image centered.
- To allow for longer exposures.
Looking at the first situation - A tracking mount will keep the image in the center (when set up right) but if your focal length is not very long then you can get away with out one. For every minute of the eclipse the sun moves about a quarter of a degree in a fixed frame. To see if you can keep the image in the view calculate the field of view and size of the image. The total image will shift a quarter degree for each minute; or one and a half degrees for a long, six minute eclipse.
Determining if you really need to bring an equatorial mount is a function of the focal length of the camera/telescope you are planning to use as well as the types of images you hope to obtain. The situation can be summarized as saying that the longer the exposure and the longer the focal length, the greater the chance you need to haul a heavy equatorial tracking mount (with a small motor) along.
You can calculate the maximum exposure without star trails if you know the detailed size of the imager or film plane.
Mounting a Camera/Telescope
Do you really need to pack a tracking mount?
The maximum exposure without smearing (no star trails) for a fixed mount is a function of the focal length and the type of receptor.
The type of receptor may be film or some electronic device such as a CCD (Charged Coupled Device) or CMOS imaging chip. What we need to know about the receptor is the density of the individual light receivers (pixels for a chip) or in the case of film, the resolving power. Given that data the speed which the object appears to be moving is applied to find what the maximum exposure can be before the image smears. For nicely arrayed electronic imaging systems the math is pretty straight forward. Film resolving power power is matter of discovering the Modulation Transfer Function (MTF) for the film from the manufacturer.
(Math behing the calculator) It is important to note that this is a theoretical approach and only serves as a reference point for starting. There are other factors involved in getting "perfect pictures". Field experiments and lots of practice taking pictures of the moon and stars are highly recommended.
Yes, you can do this at home!
The easiest way to determine the maximum exposure you can use without any star trails or blurring for your camera and lens is to use the following equation:
Exposure (t) seconds = Magic_Number * cosine(Declination) / Focal Length (mm)
The Declination of the object in the sky can be obtained from tables and charts. As the declination approaches 0 degrees the cosine approaches 1, and as the declination goes to 90 degrees at the poles the cosine value becomes 0. This equation is to be applied for celestial objects between those two extremes.
The Focal Length is the effective focal length of your optical system. Often times this is simply the number printed on the zoom lens.
The harder number to find is the "Magic_Number". The magic number is a factor that is found through experimentation and will vary for different cameras.
Using a fixed mount and your camera you can set up a basic test to learn what the maximum exposure time for any camera will be, shooting at any declination. You need a good clear night and some time to accumulate the data. I originally did this experiment using film (Kodak Tri-X) and can assure you it is much easier with a digital camera!
To save you the cost of processing some film the numbers I use are 1000 for very grainy film and 300 for pretty good film.
How to get the Magic_Number for your camera:
Use a standard lens, (50mm is recommended), find a place you take longer exposures of the stars near the celestial equator. The best (easiest) target is the belt of Orion. The northern most star of the belt almost lies on the celestial equator (declination is zero). Orion is not visible all year long, so you may need to consult a star atlas for other candidates. Sometimes the planets are in the right position making for another easy target.
On a clear night with the best possible conditions (fair temperature, low humidity, no winds) set up your tripod and camera facing Orion. Focus the camera as best you can manually. Automatic focus is not going work. Using a cable release (or remote control) take a series of pictures starting at 5 seconds and incrementing by another five seconds up to about 45 seconds. The result will be a set of images at 5, 10, 15, 20, 25, 30, 35, 40, and 45 seconds.
Take the images back to the computer and zoom in on each of them. The examples above show a series of photos taken with a 50mm lens and a Canon Digital Rebel. Each exposure was progressively longer and in exposure 3 you begin to see trails.
Now look at your own pictures. At what exposure do you start to see streaks or star trails? At what exposure do you see no movement at all?
I recommend that you repeat this experiment a few times to make sure your data is consistent.
Multiple the time exposure of the longest exposure that does not show any star trails by the focal length of the lens in use. For example, let's say that you notice no trail at an exposure of 10 seconds and do notice the start of one at 15 when using a 50mm lens. Multiple 10 by 50 for a factor of 500. If the movement shows up slightly at 10 seconds but not at all at 5 seconds then use the five value to achieve the starting factor. If you want to try for a more precise factor then repeat the experiment bracketing the two values and use a smaller increment.
The factor just found by experiment can be plugged into the equation above as the "Magic_Number".
Glenn Schneider, "The Umbraphile" from Steward Observatory: "The question is: Without tracking, do I need to worry about intra-exposure blurring?" This question has a built-in assumtion that the sharpness of an image is limited (without tracking) by image motion across the field of view due to Earth rotation. That may not be the case, such as other effects such as image focus could dominate. Assuming it is, however, that then in part depends upon what the "spatial frequency" cut-off response of the detector or film is and will set what the maximum exposure duration would be.
Sampling theory basically says that you must put (at least) TWO sensing elements (pixels, or chunks of silver halide grains for film) across a "resolution element" or you will "undersample" the image at the focal plane and loose information. So, to answer this question you need to know what the sampling density of photosensors. (When one assumes a "diffraction limited" optical system, this says for imaging point sources like stars that your image sensr must lay down two sensing elemets across an optical resolution element which is essentially the Airy resolution limit. (1.22 lamda/d).
For electronic detectors that is simply the number of "picture elements" or "pixels" per linear spacing since they are laid out in neat and constaly spaced rows and colums. For film, that depends upon the (size and) density of photosensitive silver halide grains in the photographic emulsion. For film (but also for electronic detectors) the limits of its "spatial frequency" response is often expressed by its ability to sharply record alternating pairs of equally spaced and equally wide light and dark lines and is expressed ususally in "line pairs per millimeter". Some additional details about that are follow. Of course OTHER things can degrade (blur, smear) an image, but the question is for what maximum exposure time will image "smear" due to Earth rotation dominate.
The limiting spatial frequency reponse is determined by laying down a physical line-ruled mask on the film (applicable also for electronic detectors), so one doesn't have to worry about the quality of forming an image of lines by an optical system and exposing the film with the mask in contact on top to light. When the perfectly opaque (black) and transmissive (white) lines are "far apart" relative to the pixel (or silver halide grain) spacings, the recorded center of the black and white lines are 0% or 100% intensity and the "picket fense" is a sharp square-wave. As the lines get closer together and thinner and approach the spacing of the sampling elements, the recorded edges of the alternating light and dark lines "roll off" and become soft and the center of the dark lines become less dark and the center of the bright lines become less bright. The "input" square-wave (alternating dark and light lines) becomes modulated like a sine wave of decreasing amplitude as the line spacing approaches the sensor element spacing -- and when it reaches (or gets smaller than) the "Nyquist limit" (of basically two sensing elements per line spacing) what is recorded on the image will have "zero modulation" and appear uniform gray. The ratio of the high and low recorded line intensities is the image "contrast". When the lines spacing is far apart relative to the spacing of the sensing elements the recordable contrast is large -- but declines with smaller line spacings. The contrast that can be acheived with a given film as a function of the (black/white) line-pair spacing is, essentially, what is known as a "modulation transfer function". If the MTF is 100% then there is a perfect transfer, no loss of contrast, and the sharp lines remain sharp. If the MTF is zero (or very small), as when the line-pair spacing is smaller than the spacing of the sensing elements, there is no modulation and all you get is a uniform gray. When the number of line-pairs per millimeter is equal to twice the number of sensing elements per millimeter the sampling is equivalent to being at the "Nyquist limit" for resolving the line-pairs. That, actually - for uniform sampling (like pixels spaced evenly in an electronic detector) can be computed directly. For those who think more digital than analog, this is equiavlent to the high frequency cut-off point of the Fourier transform of the power spectrum of the input image.
Math behing the Calculator- The first step is to calculate the field of view (actually, the angle of view) for the camera and lens. Using the same units of measurement for the width of the receiver (W) and the focal length (F) we get an angle value as the result.
Field of View (FOV) = 2 * arctangent (W / (2 * F))
Note: A simplified version of this formula can be used to obtain the field of view because the angles tend to be small when using a telescope and camera. Field of View (radians) = W / F = FOV
Skip this step for film, applies only to digital cameras. Pixel density (pD’) and pairs density (pD) from sizes (X and Y) of the chip and total pixel count (P).
pD' = square root of (P / (X * Y))
pD = pD’ / 2
The film Modulation Transfer Function needs to be located from the manufacturer of the film. Generally this information will be supplied in the form of a curve from which you will have to interpret the number. Find the value where the response (%) drops to 20% (a cut off resolving power for most applications) and use the associated spacial frequency value (cycles/mm). This number is typically between 50 and 100.
The next step is to convert the "density value" into angular units. Note that this is really not the density in the case of the film. For film this value is the resolution limit as defined by the input MTF value.
The value to be produced is the distance between sensor elements expressed as an angle (SD). For this we need to know the field of view as well as the density or linear resolution in terms of pairs/mm (D) and the size of the sensor plane in mm (W).
SD = FOV(degrees) * 60 * 60) / (D * W)
We know the eclipse is moving at a relative speed of about 15 arc seconds per second of time (for a stationary camera system). NOTE: Because we are imaging eclipses which occur within 23 degrees of the celestial equator the declination aspect of the star trails will be ignored. This set of calculations is based on an object at or near the celestial equator.
Now we can determine the time (t) it takes to cross a pair.
SD / 15 = t
The resulting value is the exposure time where smearing or blurring across the pairs will start to appear. Set your exposures to a value less than that and you do not need to track the image using a heavier equatorial mount. Of course, the focusing is left up to you!
-Programmed by Bill Kramer March 2009 based on an idea by Glenn Schneider and a question posted at the SEML.