Monday, December 28, 2015

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, 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 this tucked inside the 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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