February 2009 - Utah-Astronomy - mailman.xmission.com

Build your own "Andy"
by Chuck Hards 12 Feb '09

12 Feb '09

http://cgi.ebay.com/8-5-F-12-professional-refractor-telescope-lens-NEW_W0QQ…

1 0

SPOC 1 meter
by Patrick Wiggins 12 Feb '09

12 Feb '09

Those who have not been around SLAS a pretty long time may not have heard that early on the very first scope planned for SPOC was to have been a 1 meter (ok, some called it a 40 *inch* but I try not to use such four letter words <g>). Or that the primary mirror was to have been made of several smaller mirrors glued to a slab of granite and ground and polished as a single piece of glass. I have not been able to find any pictures from back then. But a few hours ago the son of Jerry Montgomery (Jerry supplied the SPOC dome) emailed several scans of pictures from back then. They should eventually end up on the SLAS website but in the mean time here's a sampling, one showing Jerry standing inside the mirror cell and another showing the prototype setup for grinding the big mirror. Pictures were taken probably in 1975. http://users.wirelessbeehive.com/~paw/temp/PRESPOC.JPG patrick p.s. In case you are wondering the optics never materialized and the project was scrapped.

4 9

Big Binos
by Dale Hooper 12 Feb '09

12 Feb '09

My annual observatory Improvement & Modernization (I&M) money is about to become available. This year I'm planning on using some of it to buy some "Big Binos". I've been discussing it with Ron V. but I also wondered if anyone else on this list might have some input. Here is the current list that I am considering (in rank order): Oberwerk 25x100 ($400) Zhumell Tachyon 25x100 ($345) Barska 25x100 ($330) Orion GiantView 25x100 ($350) If you have some experience with any of these - or others, I'd appreciate hearing from you. Thanks and clear skies, Dale.

2 3

Re: [Utah-astronomy] Big Binos
by Wayne Sumner 12 Feb '09

12 Feb '09

I have the Barska 25 X 100 bino's that I bought from Patrick Wiggins about a year ago. They're great! While I can see a bit of color fringing around the moon and other really bright objects, they are wonderful for viewing comets and galaxies. Wayne A. Sumner Math/Physics/Astronomy/Engineering Boy's Tennis Coach Northridge High School Davis School District (801) 402-8610 >>> Dale Hooper <Dale.Hooper(a)sdl.usu.edu> 02/12/09 12:00 PM >>> My annual observatory Improvement & Modernization (I&M) money is about to become available. This year I'm planning on using some of it to buy some "Big Binos". I've been discussing it with Ron V. but I also wondered if anyone else on this list might have some input. Here is the current list that I am considering (in rank order): Oberwerk 25x100 ($400) Zhumell Tachyon 25x100 ($345) Barska 25x100 ($330) Orion GiantView 25x100 ($350) If you have some experience with any of these - or others, I'd appreciate hearing from you. Thanks and clear skies, Dale. _______________________________________________ Utah-Astronomy mailing list Utah-Astronomy(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://gallery.utahastronomy.com Visit the Wiki: http://www.utahastronomy.com

1 0

U.S. Satellite Destroyed in Space Collision
by Patrick Wiggins 11 Feb '09

11 Feb '09

I can't wait to see what the conspiracy theory people make of this. :) http://www.space.com/news/090211-satellite-collision.html patrick

1 0

Re: [Utah-astronomy] quick question
by Canopus56 10 Feb '09

10 Feb '09

Or as a four year said at a star party last year, holding out boths hand with fingers spread, "A bunch more than this many!" - Kurt

2 1

ante crepuscular rays
by Jim Gibson 10 Feb '09

10 Feb '09

Daniel, I like your shot of ante crepuscular rays. I used to use that term at work on occasion. When co-worker would outline a complex software solution on the board to a difficult problem I would reply that, “Its ante crepuscular to me!”. The co-worker would go Huh? Then I would say,” It is so far over my head I have to turn around backwards to see light” *;o) Jim

1 0

Re: [Utah-astronomy] quick question
by Canopus56 09 Feb '09

09 Feb '09

Joe asked: > Hi, Can anybody tell me right off-hand how many arc-minutes the night sky is? Thanks, Joe What Daniel said. My calculator returns a slightly different number than Daniel's: Total celestial sphere: 148,538,682 arcmin^2 Percent of the total celestial sphere you see across 24 hours at latitude 40N: 88.3% Arcmins^2 visible at 40N latitude in 24 hours: 131,159,656 arcmin^2 - Kurt P.s. - The math: r = 1 radian = 180 degrees / pi = 57.3 degrees r = 1 radian = ( 60 arcmins / deg ) * 180 degrees / pi = 10,800 arcmins / pi r = 1 radian = 10,800 arcmins / pi A = 4*pi*r^2 pi = 3.141 A = 4*pi*(10,800/pi)^2 = 4*(pi/pi^2)*(10,800)^2 = (4*(10,800)^2)/pi =~ 148,538,682 arcmin^2 Appended find excerpts from two post that I did on sci.astro.amateur in 2005. They also discuss the related question of "how many square degrees does an observer at latitude x see over a 24 hour period?" From 40N latitude, you see a truncated sphere (truncated at about S40 degs) and not the full sphere. http://members.csolutions.net/fisherka/astronote/astromath/fracsphere.gif Clear Skies - Kurt =============================== sci.astro.amateur post 8-21-2005 canopus56 (excerpt) The observer, at whatever latitude they are standing, always see 1/2 of a sphere in the local horizon system. As you and others have noted, the area of a sphere is 4*pi*r^2 and where an infinite unitized sphere is used, r = 1 radian = 180 degrees / pi = 57.3 degrees. So, the area of sphere in steradians is: 4*pi*(180/pi)^2 = 4*(pi/pi^2)*180^2 = (4*180^2)/pi = 41253 deg^2 A local observer's view is 1/2 the local horizon system, and, in steradians is always a fractional part of angular area of a total sphere (steradians in square degrees), or 41253 deg^2. So - 1/2* 41253 = 20626.5 deg^2 If the local observer looks at the hemisphere of the night sky using another coordinate system at a rotated angle as the reference - in this case the North Pole of the celestial coordinate system from 40 deg north latitude - that changed angle isn't going to change the physical fact that there are only 20,626.5 degs^2 (steradians) in the hemisphere of the local horizon system. sci.astro.amateur post 8-24-2005 canopus56 Thanks for the further explanations. I was having problems relating the 1/2*(1+cos(lat)) formula to any geometric construction of the problem. With the additional tips you provided, it now makes sense to me. To resummarize - The surface area of a sphere is - S = 4*pi*r^2 There are 41252.96 square degrees on a sphere. At any instant during the night, you see 1/2 (50%) of the celestial sphere or 20626.5 sq. degs. The observer's local horizon can be vizualized as a layer or plane, titled at the origin of the celestial sphere at the observer's geographic latitude - 40 degs north in my case - that bisects the sphere. Across twenty-four hours of daylight and nighttime, the observer's local horizon rotates 360°. That rotation inscribes the bottom of the celestial sphere with a cone shape. Think of taking an apple and cutting a cone of the bottom of it such that the sides of the cone have angles equal to your geographic latitude with respect to the equator of the apple. The "cone" shape is the part of the earth blocked from the observer's view. That sphere with a cut-out cone shape can be abstracted as simple truncated sphere. Turned upside down, the truncated sphere looks like one of the spherical light globes used in out lighting fixtures that we amateur astronomers hate so-much. E.g. - http://www.thelightingcenter.com/products/*/*/1033 With the truncated sphere model, we can apply (per Brian), Archimedes Hat-Box-Theorem to find the area of the truncated sphere. http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html That area of the truncated sphere is given by: S_cap = 2*pi*r*h Surface area of zone or cap "h" is the length of the line vertically through the origin of the sphere, which at starts the sphere's apex and ends in the truncated layer at one end of the sphere. See - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - or for an upside down sphere - http://www.thelightingcenter.com/products/*/*/1033 The answer sought in this thread is "what is the ratio (or fractional part) of the truncated sphere to the entire sphere." or - S_trunc = Fraction * [4*pi*r^2] (See http://en.wikipedia.org/wiki/Solid_angle ) The fractional part seen by an observer across a single day is: Fraction = [2*pi*r*h] / [4*pi*r^2] - which solves to: Fraction = h / 2*r - Our celestial sphere is a unit sphere << http://en.wikipedia.org/wiki/Sphere >>, so r=1 and Fraction = h / 2 The answer to the question "what fraction of the sky can you see from any latitude" boils down to what is "h" for any latitude in Archimedes Hat-Box-Theorem: http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - except we are applying the theorem to the "big" part of the upside truncated sphere - http://www.thelightingcenter.com/products/*/*/1033 - and not the small cap shown in - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html Take a great circle through longitude 0 degrees on the sphere. From that 2-D planar construction, it can easily be seen that "h" is the sum of cosine of 90 degrees, plus the cosine of the observer's latitude. http://members.csolutions.net/fisherka/astronote/astromath/fracsphere.gif The cosine of 90 degrees = 1, so - h = 1 + cos(latitude) - and Fraction of the sky that is observable across twenty-four hours of darkness and daylight is: Fraction = [1+cos(geolatitude)] / 2 In my case at 40 north degs, that's the 88.3% you guys calculated, or about 36426 sq. degrees. Thanks again for the further explanation of the geometric construction. The extra tips helped me to piece it together. Sorry to be so dense initially. - Canopus56 P.S. - If you approach the problem by finding the surface area of the small cap cut-off the end of the truncated sphere - that is diagram at Wolfram's website - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - then the area of the small cap is - Fraction_small_cap = [1-cos(geolatitude)] / 2 - or 11.7% at 40° north latitude. P.P.S. - Amateur astronomers primarily are concerned with the fraction of the sky that we can see during the night. Even of the shortest day of the year at 40° north, daylight and nighttime are equally divided - about 12 hours each - and the Earth rotates nearly a full 180° at night. As a consequence, we might see the full 88% as "dark sky" around the longest day of the year. Otherwise, night time at 40° north latitude is shorter than 12 hours and the Earth rotates less than 180°. So, at 40° north, we see some fraction of the night sky between 88% and the 50% of the night sky that can be seen at any one instant during the night.

4 5

Re: [Utah-astronomy] New Zealand Dark Sky
by Jerry Foote 09 Feb '09

09 Feb '09

Hi Erik, I refurbished a 24" telescope on Mt John last spring. The sky is very dark but the seeing is not as good because Mt. John is situated in a valley with high mountains on each side and the down slope winds stir up the air. The telescope we were refurbishing was an old Optical Craftsman mount and Arne Henden (AAVSO director), Dirk Terrell (SWRI, Colorado) and I went down to bring the telescope up to modern control standards and fitting it with the capability to be operated remotely by the AAVSO. The mountain has a 1.8m telescope used for micro-lensing events, a 1 meter used for follow-up work the 24" AAVSO scope, a 24" Boller & Chivens and a 16" used by an organization called Earth & Sky that gives star parties on selected nights. A link to the University of Canterbury, Department of Physics is here: http://www.phys.canterbury.ac.nz/research/mt_john/index.shtml Some images that Dirk took while we were there are here: http://www.boulder.swri.edu/~terrell/images/nz2008mar/ Image #4 of this set is Arne & I looking over the Lake Tekapo and image #3 & #22 are the 24" telescope we were re-furbishing. A link to Earth & Sky is here: http://www.earthandsky.co.nz/ NZ is a beautiful place with the South Island outstanding. Cindy & I spent 7 days touring after we finished the work on the telescope. Jerry Foote ScopeCraft, Inc. 4175 E. Red Cliffs Dr. Kanab, UT 84741 435-899-1255 jfoote(a)scopecraft.com

6 9

OT: Darwin Bicentennial week
by Richard Tenney 09 Feb '09

09 Feb '09

For those interested in the topic, there are a series of lectures this week at BYU on Darwin, interestingly enough. Probably at least a few (if not all) have some LDS-themes, given the location (and some of the lecture titles). You can see the agenda here: http://sciencebysteve.net/?p=654 I'm planning on being at most (if not all) of them. There is also a lecture at U of U Wednesday, 7:30 PM in the Biology Auditorium (where the usual F.O.S. lectures are held titled: "Darwin Day Celebration: Exercise and the Evolution of Humankind"; more info here: http://unews.utah.edu/p/?r=020409-2 /R

1 1