[AstroDave]

With the upcoming total solar eclipse of August 21, 2017, the Forum has seen questions about what lens(es) might be good for taking pictures of the sun. The derivation below will allow you to calculate the image scale (in arcseconds per pixel) in your camera for any lens that you might think of for eclipse pictures. Note that the angular size of the sun is very close to half a degree, or about 1800 seconds of arc. The moon is essentially the same angular size - which is why we get total eclipses in the first place! If you want your image of the sun or moon to come even close to filling the image frame, you will want image scales of a few arcseconds per pixel. i.e. suppose you want the solar image to be 1800 pixels across - a modest fraction of a modern DSLR frame - you want an image scale of around 1 arcsecond per pixel.


Here we go: suppose you are taking a picture of two stars that are separated by an angle theta (in radians; there are 2pi radians in a circle). As imaged on the sensor, the physical separation D in millimeters between the two star images will be

D = F*theta (where * means multiply)

where F is the focal length of the lens in millimeters. This is basically the small-angle formula relating arc length, angle, and radius - you might remember it as s = r*Theta from a years-ago math class.

If we rearrange this a bit, we have the image scale as

Scale = (theta / D) = (1/F) (equation 1) with units of radians/mm


Since we generally think of plate scale in units of arcseconds per pixel, we need to apply a couple of factors to equation 1.

First, multiply by arcseconds/radian = 206265 to get arcseconds per millimeter.

Then, we need to multiply by the number of millimeters per pixel. This is just the sensor width in mm divided by the number of pixels across the sensor. This will give us the desired scale value in arcseconds per pixel. Pentax lists the sensor sizes for most of their cameras in the specification section of the camera manual.


So, to get a pretty good estimate of the image scale you can expect from a lens attached to your camera, the final equation is

Scale = (206265/F) * (sensor width (mm)) / (# pixels across sensor)

For wide angle lenses, thanks to curvature across the field (the sky is not “flat” but should be considered a sphere centered on us), this result applies only to the center of the field.


Let’s look at an example, compared to an actual measurement.

I used my K-1 to measure the scale for a number of lenses by taking pictures of star fields. In a second note, I will describe an easy way to make such measurements.

For my old SMC Pentax-M 50mm F1.7 (40 years old and still working nicely!), I measure a scale of 19.4 arcseconds/mm .

A calculation using the formula above yields

Scale = (206265/50 mm) * (35.9 mm/7360 pixels)

= 20.1 arcseconds/mm


This is within a few percent of the measured value, which I suspect is well within the tolerance of just how exact is the focal length of a lens, as well as such things as exactly where in the lens the focal length is referenced to.

See my next note for a number of actual measurements of various lenses with the K-1.

(Please excuse the somewhat kludgy equations - I can not figure out how to post something here with extra spaces within the text.)

[Bob 256]

As a side note, at totality, the main event will be the corona which extends from the sun some distance. For that portion of the eclipse, a tight shot might not be the best. Take a look at some previous totality shots to get an idea of what to shoot for and use AstroDave's formulas accordingly. For the other portions of the eclipse (partially obscured disk - ho hum), get as tight as you like to reveal disk details like sun spots (if there are any - just about in a spot minimum right now). Be sure to use a blocking filter for all but totality or you might be buying a new sensor (cheaper than a new set of eyes but same applies there as well).