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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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) = FL_{t/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°.

 

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

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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®

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Skies Over Picketpost Mountain

Live video for viewing the conditions over Picketpost Mountain

Click here: Skies Over Picketpost Mountain

(Video streamed by BroadCam Streaming Video Server)

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Rusty Mountain Observatory development

The development of Rusty Mountain Observatory - Phase II 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

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Rusty Mountain Observatory II 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_ii for monitoring progress.

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f-ratio myth - indeed

After reading of Stan Moore's "f-ratio myth"*, I decided to test the concept. And, I agree with him (that it is a myth). Afterall, when you adjust the f-ratio on a camera, all you're really doing is stopping down the lens - i.e., reducing the aperture. Makes sense - and I think this test supports his contention. 

Both images below were taken with an SBIG ST-8300M (monochrome CCD camera) on an 11 inch Celestron EdgeHD SCT using 10 minute exposure times. (The images were taken as part of testing the Optec Lepus focal reducer and the f/10 was a baseline.)

See individual images at: http://www.pbase.com/wjshaheen/fratio_myth__indeed

If f-ratio determined "brightness", then the f/7 image should be twice as bright as the f/10. As it is, the difference is simply in the scale (and resolution) of the two images.

*( See http://www.stanmooreastro.com/f_ratio_myth.htm )

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Horizontal blooming in the Kodak KAF8300 sensor, when binned

Referring to the image here: http://www.pbase.com/wjshaheen/image/134819312/large
it turns out that what was suspected to be tracking errors, resulting in the "protrusion", or bulging to the right in star images, is actually caused by a characteristic inherent in the Kodak KAF8300 sensor known as "horizontal blooming", when binned. See discussion in the QSI support group here: http://tech.groups.yahoo.com/group/QSI-ccd/message/8682

In the process of centering an object the other evening, I was taking 0.5 sec binned images and noticed that even that short of an exposure exhibited the smearing I've been wrestling with for several months now. See the examples in this album:
http://www.pbase.com/wjshaheen/horizontal_blooming_in_the_kaf8300_ccd

Note the 2x, 3x, and 4x images exhibit progressively larger "bleeding" to the right, whereas the 1x1 example show no anomolies, other than a misshapen star due to atmospheric turbulence.

To attempt to isolate the cause and rule out tracking errors, I devised a simple test:
I rotated the camera approximately 45 degrees. (I normally image with the camera x-axis aligned to the RA axis.) The resulting binned image still showed the blooming directly and solely in the horizontal axis.

I then took similar images using Nebulosity2 instead of ImagesPlus - same result. The next morning, I reviewed past images taken with the SBIG ST-8300M - same result. Here is an example showing both tracking error as well as horizontal blooming:
http://www.pbase.com/wjshaheen/image/134819488/original
Note the distinct oval shape more characteristic of tracking errors (in both axes) versus the bleeding into adjacent pixels caused by horizontal blooming.

After trying a different USB port and eventually a different computer, and still seeing the same, consistent results, I browsed the QSI support group and found this is a known and common problem.

The good news, however, is that the fine detail strucure in an image would not be affectd by this issue (I had been wondering why there was no noticeable "smearing" in my M42 and M16 images, when tracking went well). And, some have suggested this could be addressed in post-processing.

But, as the saying goes, "the proof is in the pudding". So, here's a sample unbinned 10 minute image, dust mote notwithstanding:
http://www.pbase.com/wjshaheen/image/134819756/large

The area is in the neighborhood of PGC54526, RA 15h 17m, DEC 07d 00m, just north of M5.

With the Optc Lepus 0.62x reducer due to arrive any day, I will be able to image unbinned, albeit at a less comfortable 0.58 arc-sec scale. But, I'm very pleased that I've finally gotten to the bottom of a long-standing issue.

Regards,

Bill

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Testing the Optec Lepus 0.62x Focal Reducer

Testing the Optec Lepus 0.62x Focal Reducer with the 11" Celestron EdgeHD telescope and the SBIG ST-8300M (monochrome) CCD camera. The telescope's stock rear adapter (with SCT threads) was replaced with an Astro-Physics 2" Visual Back (AP part # ADASCTLC) in order to install the focal reducer as close as possible to the OTA, as recommended by Optec.

I'm pleased to report that, overall, I found the reducer to produce very satisfying images, resulting in a reduction factor of 0.69x. See http://www.pbase.com/wjshaheen/optec_lepus_062x_focal_reducer_testing for full scale test images as well as an f/10 baseline image.

Update 24-Apr-2011: I resumed testing after resolving several issues having to do with the baseline - i.e., SCT collimation, guide-camera flexure and focuser flexure (yes, the Feather Touch focuser resulted in a slight drift, which was not mirror shift).

Visit http://www.pbase.com/wjshaheen/optec_lepus_062x_focal_reducer_testing and see the 4 images of M57 (Ring Nebula) taken this date - on the bottom row.

Some coma is noticeable in the lower left corner only and may be due to a tilted plane, which itself may be the result of, again, the leading edge of the reducer stopping at the edge of the internal field flattener. This is being researched with the vendor.

On another note, I have a professional source whose contact at Celestron indicates they will be releasing a reducer this summer. It will have a respectable back-focus (90 to 110 mm) but may not have the coverage originally desired. It will however cover an APS size chip and certainly the 8300.

Update 11-May-2011:
Update on the question of the reducer possibly not seating squarely (the leading edge of the reducer stopping at the edge of the internal field flattener), I just received word from Optec that they "have developed a different mounting configuration for the Edge HD scopes. For the C11 HD and C9.25 HD scopes a spacer is required to keep the lens housing from touching the Edge HD retaining ring." . (At the time I ordered mine from OPT, they had the former version in stock.)

Optec is sending me the new lens and housing gratis and I will be testing it in about a week.
(Thanks, Jeff!)

As an aside, I've replaced the ST-8300M with a QSI-583wsg - solves a host of flexure issues up and down the imaging train as well as with any mirror flop or guidescope flexure. The Starlight Xpress Lodestar connects easily to the QSI's guideport, given the standard C-mount adapter) and so far I've had no problem locating a guidestar.

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It pays to collimate your optics

I have to admit I've frequently given collimation (aligning the optics of a telescope) the short-shrift in my haste to produce an image. But, after examining the NGC3628 photos taken lately and some visual viewing the other night, I took another set of images after doing a "proper" collimation.*

Here's the image from 3/28/2011 (prior to collimating):
http://www.pbase.com/wjshaheen/image/133518647/large

This is from last night (after collimating):
http://www.pbase.com/wjshaheen/image/133590305/large

(Feel free to explore the Original sizes.)

And, here is the side-by-side comparison:
http://www.pbase.com/wjshaheen/image/133590304/original

Further, here's an excellent paper on how to collimate:
http://legault.perso.sfr.fr/collim.html

I for one will never sell collimation short again.

Bill

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