Friday, 30 August 2013

solar - Local Standard Time Meridian - LSTM (Units)

In a time, the word "standard" normally means the civil time, recognized by the government in the place of interest. I'm not sure about other countries, but in general conversation in the USA, "standard" also means that daylight saving time is not in effect. "Daylight" time (as in "Eastern Daylight Time") means the civil time, recognized by the government, and daylight saving time is in effect.



These can be expressed in the ISO 8601 format; the standard gives the example "15:27:46+01:00" meaning the time zone is one hour ahead of (east of) Greenwich.



In astronomy, time can also be the local time of the observer, ignoring the existence of time zones. This is local mean time, and is obtained from Universal Time by adjusting for the longitude of the observer. Usually the exact form of Universal Time that would be used would be UT1.



Local apparent solar time is based on the actual position of the sun; it is the time given by a sundial. It can be about 15 minutes ahead of or behind local mean time.

visible light - Why doesn't a lighter flame cast a shadow?

Imagine that you have a titanium screen. It's thin, so it's difficult to see. You can use a torch to heat some areas to incandescence. These areas become easily visible.



Now if we take a very bright light source and shine them at these lit areas, we won't see a shadow that gives us any information about the pattern of the light. Both the cool and hot areas interact with the bright light beam in the same way. The grid itself may be visible, but the pattern won't reflect anything about the relationship between the dark and light regions.



The flame is the same thing. It contains matter (such as fuel or soot particles), but similar matter is also present outside the flame. The imaging light may be slightly attenuated by these particles, but there is no sharp division in their location that corresponds to the shape of the flame. This lack of sharp division will prevent imaging.

amateur observing - Free source of printable star charts in the format of the Millennium Star Atlas?

I've been looking for sources of printable electronic star charts - there seem to be many different ones available, either as part of planetarium software or just plain charting packages. But there's such a richness of offerings that I'm having trouble figuring out which of those, if any, can produce printable charts in the format that I'm looking for.



I like the general format of the Millennium Star Atlas - black stars on a white sky background, and with horizontal lines representing 1 degree of declination and vertical lines representing 4 minutes of RA (see sample page at http://www.skyandtelescope.com/wp-content/uploads/GD-Sept2011-Chart.pdf ).



I'd like to be able to specify a range of RA and declination and get that chart.



Can anyone suggest any free star chart printing software that can print similar output for any desired sector of the sky?



I'm looking at Cartes du Ciel (http://sourceforge.net/projects/skychart/) which might have that capability, but if so, I haven't yet figured out how to enable it.

Wednesday, 21 August 2013

optics - Magnification of a telescope

enter image description here



I took an image of Jupiter through my 8" Dobsonian Telescope, attaching a DSLR and a 1.25" Barlow Lens where the eyepiece goes, as shown in this video: https://www.youtube.com/watch?v=reFxoF3XoaU



Through numerous online sources, I learnt that to find the angle of view of my image, the magnification of my image needed to be calculated, and that this value along with the given field view for my eyepiece.



http://www.rocketmime.com/astronomy/Telescope/Magnification.html:



The above website answers the following question, the calculations of which I attempted to mimic.




My first telescope was a Meade 6600 -- they don't make it any more -- it's a 6-inch f/5 Newtonian scope. It came with a 25mm eyepiece. So... what was the magnification I was getting with this scope?




Here are the calculations I did in hopes of getting the above result:



$$textrm{Diameter} = 8'' = 203.2 textrm{mm}$$



$$f_{textrm{ratio}} = frac{textrm{focal length of objective}}{203.2 textrm{mm}}$$



$$therefore textrm{focal length of objective} = 203.2 cdot 5.9 = 1200 textrm{mm}$$



$$textrm{Magnification} = frac{textrm{focal length of objective}}{textrm{focal length of eyepiece}} = frac{1200}{x}$$



As shown in the video (the first link above), I didn't use an eyepiece to take my picture - I used a barlow lens, a couple of adapters, and a DSLR. So at this point, I am not sure what value to use for the "focal length of the eyepiece." How can I proceed to calculate the magnification?

Saturday, 17 August 2013

space telescope - How is the cost of JWST distributed on different parts of development and operations?


So most of the costs should end up as salaries to engineers.




Even in a project that is solely people-based, most of the costs do not end up as salaries to engineers. Salary is typically a bit less than a half of the cost. Those engineers get benefits. Health care in the US isn't cheap. They get holidays, vacation, and sick leave. They receive matching funds on their 401K contributions. In the US, employers and employees both contribute to the cesspool lockbox sometimes known as "Social Security". The employer share is part of the cost of having an employee (and companies charge for it). The employer also pays for unemployment insurance and other items on behalf of their employees. Those engineers sit at desks in an office or a cubicle, and have a nice computer on which they do their work. They use the employer's servers, printers, copy machines, conference rooms, break rooms, and coffee machines. All of that costs money. These overhead costs are the first of several indirect costs.



The accountants who write those engineers their paychecks aren't charging to any project. They're still paid by the company, as are the executives of the company. The executives also typically don't charge their time to a project, and those executives can be paid quite handsomely. This leads to another category that adds to quantity that companies charge the government, "general and accounting". Finally, for-profit companies need to keep their shareholders / stakeholders happy. They need to earn a profit. This leads to a third category that companies charge the government, "profit".



When you add up all of those indirect costs, they come close to and oftentimes exceed direct costs (i.e., chargeable salary). Note that this is for a project where the only costs are the people working on it.



Building a spacecraft notches things up a lot. Cleanrooms are expensive. Control rooms are expensive. The flight computers are expensive. The bolts and fasteners are expensive. The custom, one-off sensors are expensive. The spacecraft's thermal control system is expensive. The spacecraft's structure is expensive. Launch is expensive. Every piece of hardware, every fabrication facility is expensive.





Finally, cost overruns, requirements changes, and delays can represent a significant portion of the cost of a poorly managed project. JWST has experienced numerous cost overruns, requirements changes, and delays. Big contractors have learned that the only way to win big government contracts is to underbid. They would lose their shirts if the government kept to that initial contract. But it never does. The government instead changes the requirements, changes the payments, and changes the schedule. Those changes are beyond expensive.

Thursday, 15 August 2013

orbit - LLR & Orbital Motion

As I understand it, light that is emitted from a source is not imparted with the motion of the source and so always follows a "straight line". If this is correct, I am having a difficult time conceiving how the Lunar Laser Ranger experiments can detect photons.



In this experiment, a pulse of light is aimed towards a retro-reflector and it is reflected back. It takes about 2.5 seconds round trip and due to diffraction, the returning pulse covers a circle of approximately 20kms in diameter.



The Earth's orbital velocity is approximately 30km/s and in the time the pulse takes to make the round trip, the detector would be 75kms further along the Earth's orbital path. If my first paragraph is accurate, and the light pulse is not imparted with the Earth's orbital velocity, then how is detection of the returning light pulse achieved?



I hope this makes sense and thank you.

Wednesday, 14 August 2013

the sun - How does the appearance of the analemma vary with latitude

Analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. Note that it says nothing about time- it can be any fixed time. Analemmas created at different times of the day have slightly different shapes. For example, see the solar analemmas taken at the same place at different times.



700 UTC



Solar analemma at 0700 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu



1400 UTC



Solar analemma at 1400 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu



The best way to understand the effect of latitude on the shape of the analemma is to consider the effect of earth's axial tilt on the shape of the analemma.




If the Earth’s orbital path was elliptical, but its axis was not tilted, the analemma curve would be oval shaped. At the Equator, this line would be a straight line spanning from left to right or West to East.



The 23.5-degree axial tilt of the Earth affects the Sun’s apparent position in the sky – as the year progresses and the Earth continues to spin at an tilted axis and orbit around the Sun, the Sun seems to move up and down (North-South) in the sky. This has the effect of generating the two loops of the figure 8.




You're correct that the tilt of the analemma varies with the latitude. Observers in the Northern hemisphere will obtain an analemma curve with the broader loop at the bottom. This reverses for observers in the Southern hemisphere, where the broader loop is on the top of the curve. At equator, the anlaemma lies on its side. At north pole, only the top of the analemma would be visible, while it is the opposite at the south pole.