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

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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


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°.
 
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Monday, May 11, 2015

Recent Work - May, 2015

After a brief hiatus, I'm back to astrophotography.

Here is a sampling of photos taken with the Celestron 14 inch EdgeHD and the Canon 6D (full frame) camera: http://www.pbase.com/wjshaheen/apod_submissions

And, here is a gallery of cumulative work done over the years:
http://www.pbase.com/wjshaheen/astro_images_by_subject

William Shaheen
Gold Canyon, AZ
USA

Monday, June 24, 2013

The Alien Contact Probability Boundary

The Drake Equation, as we all know, purports to determine the potential number of intelligent civilizations that share our Milky Way galaxy. Obviously, it makes a number of far-reaching assumptions. But, upon hearing yet another reference to it again, one that indicated it predicts millions of other civilizations, I had to think - there has to be an upper limit, or boundary, to that number based on one sheer fact: we have not heard from any of them, at least not just yet.

Now, of course I'm not the first to proffer this question (why haven't we heard from anyone yet?). What I am suggesting is that the possible number of civilizations is bounded by our not having heard from them to date. And, as time goes by without a signal, that number declines. 

Now, by intelligent life, my only assumption is that it is one that is capable of communication and it only needs to communicate, say, the value of Pi, the ratio of a circle's circumference to its diameter, since that it a universal constant and it seems the most obvious one, to me. (I believe this was actually part of a 1950's era sci-fi movie.) Furthermore, SETI notwithstanding, let's say we've had radio receivers around for nearly a hundred years. Note that I will not be dividing this number in half since I'm not assuming a signal is in response to our own transmissions.

So, if we assume a sphere with a radius of a hundred light years, how many such spheres are contained within the Milky Way Galaxy? Perhaps we should exclude the central portion of the galaxy on the assumption it is uninhabitable. But, we could make any number of such assumptions. I think that is the question since if the population density of life (previously defined) is greater than that, we should have picked up a signal from them, intentional or not.

To make this calculation, we'll use an ellipsoid volume calculator , and eliminate the central bulge, assuming that is uninhabitable. 

For the outer ellipsoid, I very wildly assumed dimensions of 100,000 x 100,000 x 10,000 light years (major axis x minor axis x vertical axis). For the inner ellipsoid, I assumed 20k x 20k x 5k light years. The net volume of the supposed inhabitable zone is then 51,313 billion light years. Dividing that by our 100 light year radius sphere of communication yields 513 billion such spheres.

What that number then represents, fraught with the assumptions as it may be, is the number of civilizations in the galaxy there would have to be in order for us to have heard from them by now (the bottom line number of the Drake Equation).

Now, I'm certainly not saying there is no other intelligent life, however you wish to define it, in the galaxy. But whenever I hear the galaxy is "teeming with life", I have to wonder, up to what point? Oh, I suppose we will hear a signal at some point, say in a couple or three hundred years or so. But, I'm betting the message won't be, "We'll get right back to you".

William Shaheen
Superstition Mountain Astronomical League

Thursday, February 16, 2012

Rusty Mountain Observatory development


The development of Rusty Mountain Observatory is well underway and at this point I'm just waiting for the new mount (Software Bisque's Paramount MX) - hopefully early April. (Hopefully sooner.)

The photo to the left shows the permanent pier, sold by the same company, bolted to a concrete foundation.


See the continuing storyboard for ongoing developements and details: http://www.pbase.com/wjshaheen/rusty_mountain_observatory

Wednesday, February 8, 2012

Rusty Mountain Observatory development

Construction is underway with the pouring of a pier foundation for Rusty Mountain Observatory in Gold Canyon, AZ.  Visit http://www.pbase.com/wjshaheen/rusty_mountain_observatory for monitoring progress.