Astrophotography. believe it or not… It’s not rocket science. No Physics degree required No...

Astrophotography.

believe it or not…

It’s not rocket science.

• No Physics degree required• No Computer Science degree required• No Engineering degree required• Some computer skills can be useful• Some mechanical ability is useful• A basic understanding of electronics is useful• Some understanding of the underlying

science is a great plus• While equipment continues to improve, the

core technology itself is reasonably mature – it is highly unlikely that you will need to invent anything in order to be successful

• Patience and perseverance are absolutely mandatory!

using practical and measurable considerations when making equipment choices will ultimately simplify the job

a little simple math can greatly assist in understanding how things work as well as what you might expect as a result

taking the time to develop a logical and tested workflow will improve your odds of repeatable results

It’s not black magic, either.

understanding the critical elements involved leads to better results

What’s Tonight’s Topic?

How some easy math provides answers to questions you need to know:

• How much sky does my setup cover?

• How big will an object be on my setup?

• How long was my effective focal length?

• What was the focal ratio of the system?

• How long can I expose a planet before its rotation causes smearing?

Plate Scale

206265

What does 206265 represent?

• It is the number of astronomical units in a Parsec– An Astronomical Unit (A.U.) is the average distance between the

Earth and the Sun– One A.U. = 149,597,870,691 m, so 1 parsec ≈ 3.085 678x1016

meters ≈ 3.261 564 light-years

• It is the number of arc seconds in a radian– The radian is the standard unit of angular measurement in all

areas of mathematics beyond the elementary level– It is about 57.2958° (degrees)

First Question:

How much sky does my setup cover?


Plate Scale = 206265 ÷ telescope focal length

For the 14in Meade LX200 GPS:

– Convert its 14in aperture to mm (14 x 25.4 = 355.6mm aperture)

– Multiply its aperture by its f/10 focal ratio to get its focal length (355.6 x 10 = 3556mm focal length)

– Divide 206265 by the focal length to determine Plate Scale (206265 ÷ 3556 = 58.00478065 arc seconds per millimeter)

So, our Plate Scale is roughly 58 arc seconds per millimeter


Expressing Plate Scale in arc seconds per pixel

For the 14in Meade and a typical webcam (5.6µm pixels),

Multiply the arc seconds per millimeter by the pixel size:– 58.00478065 is the Plate Scale in arc seconds per millimeter

– .0056 is the 5.6µm pixel size expressed in millimeters (5.6 ÷ 1000 = .0056)

– 58.00478065 x .0056 = .324826771

So, our Plate Scale is roughly .32 arc seconds per pixel


Determining Field of View

14in Meade and a webcam (5.6µm pixels in a 640 x 480 pixel array):• Multiply the arc seconds per pixel figure by the size of the imaging chip’s

array (horizontal and vertical)– Horizontal: 640 x .324826771 = 207.8891334 arc seconds (horizontal)– Vertical: 480 x .324826771 = 155.9168501 arc seconds (vertical)

• Convert to arc minutes (arc minutes = arc seconds ÷ 60)– Horizontal: 207.8891334 ÷ 60 = 3.46481889 arc minutes (horizontal)– Vertical: 155.9168501 ÷ 60 = 2.598614168 arc minutes (vertical)

So, our field of view is roughly 3.5 arc minutes (H) by 2.6 arc minutes (V)

Next Question:

How big will an object be on my setup?

How big is “roughly 3.5 arc minutes (H) by 2.6 arc minutes (V)?”





• How big will Jupiter be?– Divide Jupiter’s size (38 arc

seconds) by the system’s “arc seconds per pixel” figure (.32 arc seconds per pixel)

– 38 ÷ .32 = 188.75 pixels

• This image is a simulation demonstrating how much of the frame would be taken up with the Jupiter image. It is to scale but not actual size.


14in SCT, typical webcam, prime focus (to scale)





Some things to remember…

• Telescopes with longer focal lengths and larger apertures are definitely better for planetary imaging.

• Image scale can be significantly improved with the use of barlows and extension tubes.

• Scale increase is dependent on the distance from the barlow’s lens to the chip.

Image Scale – Prime Focus

Image Scale – 2X Barlow

2.4x Image Scale increase

Image Scale – Barlow w/extension

3.2 x Image Scale increase




8in SCT, typical webcam, Barlow w/ extension (to scale)

Next Question:

How long was my effective focal length?

What was the focal ratio of the system?

Calculating Focal Length (and focal ratio) from plate scale


Effective focal length can be determined from an image of known dimensions taken using equipment of known

specifications using the following formula:

F = 206.265 x P x U ÷ O

Where:– F will be the focal length as calculated from this formula, in mm.– P is the size of the object in pixels as captured– U is the pixel size of the camera in microns– O is the true size of the object in arc-seconds


In this case:

P = 357 (pixels measured across the equator)U = 5.6 (size of the pixels in the camera)O = 49 (apparent angular diameter at capture)

So, the effective focal length of the system was:

F = ((206.265 x 357) x 5.6) ÷ 49Break this down:206.265 x 357 = 73,636.60573,636.605 x 5.6 = 412364.988412364.988 ÷ 49 = 8415.612

F = 8415.612 (That is actually ~2.4x the native focal length of the 14" Meade (3556mm) - my barlow is an over achiever!)


Now that you know the effective focal length of the system that took the shot, you can also know the focal ratio of the system by dividing it by the size of the primary mirror (355.6mm)

8415.612 ÷ 355.6 = 23.66595051 (f/24)

Using this formula I was able to calculate the effective focal length and focal ratio of the following systems…

3777mm efl @ f/18

5846mm efl @ f/25

6342mm efl @ f/23

8416mm efl @ f/24

13,149mm efl @f/40

Last Question:

How long can I expose a planet before its rotation causes smearing?



• For Jupiter:– It was 38 arc seconds in size on Memorial Day

– rotational period (at the equator) is approximately 9 hours, 50 minutes

– it will take half the rotational period for a feature to transit the 44 arc seconds (4 hours, 25 minutes)

• How long can you expose?– Convert the transit time (4hours, 25 minutes) to seconds:

– 4 x 60 = 240 + 25 = 265 x 60 = 15900 seconds

– Multiply this by the prime focus resolution of the 14" LX200:

– 15900 x .32 = 5088

– Divide the result by the size of the planet (in arc seconds):

– 5088 ÷ 38 = 133.89... seconds, or approximately two minutes, 14 seconds. After that the features would have smeared.


I used the “about two minutes” true-ism when capturing the data for this image. Did I expose for too long?

Now that we know how to work this out, lets figure out how long my captures needed to be in order to avoid smearing the image when stacking:

206265 ÷ 8415.612 = 24.50980392 (giving us the system resolution per millimeter)

24.50980392 x .0056 = .137254902 (giving us arc seconds per pixel)

15900 x .137254902 = 2182.352941

2182.352941 ÷ 49 = 49.59893048 seconds! My image might have been even sharper if I had reduced the length of the .avi recording!

Disclaimer: this all assumes a perfect scenario:

• The seeing must good enough for you to capture enough undistorted frames to stack a low noise image in the exposure time allotted

• Your collimation must be very good• Your focus must be very good• Any of these other factors will affect the

sharpness of the image.

Date post:	17-Jan-2016
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Astrophotography. believe it or not… It’s not rocket science. No Physics degree required No...

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