Monday, December 28, 2015

Building a 12-1/2 inch Newtonian - Calculating Worm and Worm Gear Factors for a Sidereal Rate Drive


In thinking about the questions that have been asked about building the 12-1/2" Newtonian, on left, the one I felt was not adequately addressed was the design and specs for the RA drive, seen here.

Now, no doubt this will be of more interest to an "ATMer" (amateur telescope maker), but along the way of designing the drive, I came up with some general equations to describe the relationship of main gear diameter, worm TPI (threads per inch) and input rpm in a worm driven gear system.

In revisiting my notes of yore (from 1989-1990), I found the below note tucked inside my 1990 edition of the RASC (Royal Astronomical Society of Canada) Observer's Handbook, in the chapter describing time.



So, I had to ask myself, "What was I thinking?".

At first, I could not recollect what this meant - Time? Output RPM? After some thought, I guessed "Time" meant the time it takes to complete one full revolution (360°).

I then found notes in my journal dated March 4, 1990. And I recalled that what was needed was to complete a revolution in 1436.068 minutes (a sidereal day).
As you know, a mean solar day is 1440 minutes (60 x 24). Sidereal rate is 0.99726958 of the standard 24 hour mean solar day. (See https://en.wikipedia.org/wiki/Sidereal_time)


And as I recounted recently, my first "epiphany", if you will, was that if the driving rod (worm) is a 16tpi (threads per inch) rod, one revolution of the rod/worm would move the larger gear 1/16th of an inch. And depending on the diameter of the main gear, the resulting output RPM would vary.
At the bottom of my notes from 1990 is a circled equation (shown at bottom of this page):

2 Pi R x TPI
--------------- = Time in minutes to complete a 360° revolution. (I'm sure this is in an engineering handbook somewhere.)
     RPM

Since then, in the last few days (Dec 24th), I realized that that general equation needs to be expressed in terms for either the RPM or TPI or the diameter of the main gear since those are the only variables you have at your disposal to work with, and assuming sidereal rate.

2 Pi R is of, course, circumference, C, for which you can substitute Pi x D.

So, the general equation to achieve sidereal rate is:

C x TPI
----------  = 1436.068 (i.e., the desired time to complete one revolution)
RPM

And from simple algebraic transposition, solving for each variable ...

(i) C = 1436.068 x RPM
            ---------------------
                    TPI

(ii) RPM = C x TPI
                  ----------
                 1436.068

(ii) TPI = RPM x 1436.068
                --------------------
                            C

In my particular case, I used a Hurst Model T 1/2 rpm motor rated at 250 oz-in.; a 3/8 inch diameter 16tpi threaded rod and a 14.28inch diameter main gear.

And that brought me to the end of my "saga" to regurgitate how I came up with the specs for the 12.5" Newtonian's RA drive (and, later, the Declination drive) after 25 years.

William Shaheen
Superstition Mountain Astronomical League
December, 2015

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Sunday, December 13, 2015

How to Choose an Eyepiece for a Desired Field of View (FOV)


Choosing an Eyepiece for a Desired Field of View (FOV)

With so many choices these days in wide-field eyepieces it can become confusing what with the interplay between magnification and field of view. (See Reviewing the Fundamentals below.)

For example, let's say you have your eye on one of the newer 82° or 100° apparent field of view eyepieces (or, you'd like to).  But, your question is, "What size eyepiece should I get (in one line of eyepieces) to achieve a particular field of view in another line?".  But, do you want to go through the trial-and-error process of picking an EP then calculating it's magnification then the resulting field-of-view?
 
Doing a little math, it works out that for a desired, or equivalent, field of view in another line of EP's, you simply multiply the targeted TFOV by the focal length of the chosen telescope and divide by the new EP's apparent FOV.
 
Now, as we know, the magnification achieved by a given eyepiece on a particular telescope is equal to the focal length of the telescope divided by that of the eyepiece ....

I.e.,    Mag (X) = FLt/FLep     (i)

And, from that, the true field-of-view (TFOV) is equal to the apparent field of view (AFOV) of an eyepiece divided by the system's magnification, from above ...

So,    TFOV = AFOV        (ii)
                        MAG

Substituting for MAG in (i) above ...
         TFOV = AFOV        (iii)  
                      FLt/FLep

Or, stated another way ...     
         TFOV = AFOV x FLep   (iv)
                              FLt

Therefore, the focal length of the eyepiece I would need, solving for FLep, is ...
         FLep = TFOV x FLt
                          AFOV


I.e., to determine what EP focal length is needed for a desired True Field of View (TFOV), multiply by the focal length of the telescope and divide by the new EP's AFOV.

Example: Given an 8 inch f/10 SCT of focal length 2032.
You would like to have that 82°, 30mm EP, but it's expensive.

So, you're looking at a line of EPs with 72° (or, 68°).

Now, the TFOV of that would-be-nice-to-have 82°/30mm in the 8" SCT is 1.21°.

To achieve the equivalent FOV in the 72° line of EPs,
you would need .... 1.21 x 2032 / 72.   Or, a ~34mm ep. (You would probably opt for the 36mm  that is available.

In a 68° line of EP's, you would need a ~36mm eyepiece.

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Reviewing the Fundamentals


AFOV is Apparent Field of View; TFOV is True Field of View.
 
AFOV is the field of view that you see when, without a telescope, you simply hold the eyepiece up to your eye and look through it, from the front, as you would if connected to a telescope.
 
It is limited by the field stop at the bottom of the barrel. If an EP is rated (sometimes stated on the barrel) to be 68° or 82°, or whatever, that is the angle you will see just looking through the EP.
 
TFOV (True Field of View) is what you see when you attach the EP to a telescope and is hence much narrower, since it is arrived at by dividing the native Apparent Field of View of the eyepiece by the combined system's magnification.
 
Both measurements are fairly precise since the manufacturer states the AFOV and the telescope has a pretty precise focal length.
 
Here is a real world example:
Telescope is an 8 inch f/10 SCT with a focal length (fl) 2032mm; the eyepiece is a 68°, 25mm focal length.
 
Magnification is equal to the focal length of the telescope divided by that of the eyepiece.
Or, in this example, 2032/25 = 81.28x.
 
True Field of View (TFOV) is equal to the Apparent Field of View (AFOV) of the eyepiece divided by the magnification.
Or, in this example, 68° / 81.28 = .8366°.
 
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