The galaxy imaging I did this spring was so much fun that I'm considering a big project: A mosaic of the Virgo Cluster. One of the club's members suggested that such a project could result in a poster that would be suitable for fund raising at the 2018 Astronomical League Convention (which the club is hosting). I don't know anything about poster printing and marketing, so I'll leave that to others.

The mosaic, on the other hand, interests me.

My images were all taken using my TV-102; with the 0.8X FF/FR it has a focal length of 700mm and a field of view using an APS-C DSLR of about 1.3° by 1.1°. That's too small a field to make this practical--The Cluster has a size of about 15° by 15°, which is an area about 140 times the TV-102 field. When overlapping is considered it gets even more impractical. I would like to finish this within my lifetime!

Another option is my AT65 (422mm FL) with a field that's 3° by 2°. This means I'd need 38 images not counting the area lost to overlap needed for proper alignment. This is better, but still quite a task, and nearly impossible given the fickle weather around here.

How about a 200mm lens? Its field is 5.1° by 3.9° and gives about 12 images needed before overlap is figured in. That's not bad. A big galaxy like M84 manages to be about 85x74 pixels, also acceptable. It's time to pin down the cost of overlapping.

Consider a single line made of identical overlapping images. We take

*W *as the image width and

*A *as the overlap between one image and the next.

The width of sky covered by the first image in the line is

*W. *Each additional image in the line adds (

*W* -

*A*) to the line length, making it easy to write an expression for the total line width

**W**:

**W **=

*W* + (

*N*-1)(

*W*-

*A*)

We can solve this for

*N*:

*N* = (

**W - ***W*)/(

*W*-

*A*) + 1

Pretty simple, huh? We can do the same for height, letting

*M *equal the number of images of height

*H *in a column, B be the overlap distance, and

**H** the total height of the column.

*M* = (

**H** -

*H)*/(

*H*-

*B*) + 1

We can fiddle with this a little more to express the overlap as

*f*, the fraction it is of the width or height. This is useful because we'll almost certainly try to use the same fractional overlap in both directions.

*f *=

*A* /

*W* =

*B* /

*H*
*N* = (

**W - ***W*)/[

*W*(1-

*f*)] + 1

*M* = (

**H - ***H*)/[

*H*(1-

*f*)] + 1

One more change, let's define the constant

**F** equal to

* *1/(1-

*f*), giving

*N* =

**F**(

**W/***W* - 1) + 1

*M* =

**F**(

**H/***H* - 1) + 1

Notice that the right hand side of these expressions doesn't necessarily guarantee that

*N *and

*M *are whole numbers. It's up to us to round them up or down to an integer value depending on how well the resulting grid of subframes covers the target area.

One more alteration improves the convenience:

*N* =

**F**(

**W/***W* - 1) + 1

*M* =

**F**(

**H/***H* - 1) + 1

We should check this pair of equations for correct behavior in the case of no overlap. In that case *f* = 0 is zero and *F *= 1. This gives

*N* = (**W/***W* - 1 + 1 = **W/***W*
*M* = (**H/***H* - 1 + 1 = **H/***H*

The number of subframes, *N *times *M*, is equal to (**WH**) / (*WH*), the area of the mosaic divided by the area of a subframe. This is exactly right in the limit that the subframe is much smaller than the mosaic and we can ignore the need for *N *and *M *to be whole numbers.
For the 200mm lens, we have (expressing everything in degrees)

*W* = 5.1 and

*H* = 3.9. For the Virgo Cluster both

**H **and

**W **are 15, so

*N* =

**F**(1.94) + 1

*M* =

**F**(2.84) + 1

Let's assume a 1/3 overlap rule, so

**F** = 1 / (1 - 0.333) = 1.5. This gives us

*N* = 1.5(1.94) + 1 = 3.91 (round up to 4)

*M* = 1.5(2.84) + 1 = 5.26 (round up to 6)

So--as a first guess--24 images are needed to make a rectangular mosaic of the Virgo Cluster.This could be as few as 15 (3x5), 18 (3x6), or 20 (4x5) depending on the fit and composition considerations. I'll leave that for Part II.

How about a 135mm lens that gives a 9.5° by 6.3° field?

*N* = 1.5(0.58) + 1 = 1.87 (round up to 2)

*M* = 1.5(1.38) + 1 = 4.47 (round up to 5)

Only 10 images, but the galaxies would be awfully small.

And for my AT65, with a a 3.0° by 2.0° field?

*N* = 1.5(4.0) + 1 = 7

*M* = 1.5(6.5) + 1 = 10.75 (round up to 11)

77 images, a bit much for my taste, and really difficult to get done by ALCON 2018.

Just for fun, let's do the case for the TV-102.

*N* = 1.5(0.58) + 1 = 16.8 (round up to 17)

*M* = 1.5(2.38) + 1 = 19.95 (round up to 20)

That's a mosaic with 340 subimages. That's not going to happen.