Here's a picture of the Lick-a-chick eatery: Flickr. At the Tim Horton's across the street, there was a claim that an old light bulb was projecting the virgin Mary on a wall. Fun area.
I was expecting a lot more from this website. It looks nice you first load it up, but the information content is quite sparce. Not even pretty pictures or informative diagrams to explain concepts. Section 6.2 on aging stars as an example is dismal.
Soyuz is a great well tested, great success rate rocket, and I'm not debating the launch of stuff into near earth orbits, but living quarters for a few humans to spend a relatively long time in space. When was the last time a human went around the moon? - the Apollo years?
Canada has 1/10 the population of the US, so at most 1/10 the resources, so agencies such as CSA must concentrate focus on a few areas such as Canadarm or Microsatellites (for example check out MOST). While CSA does provide astronauts for NASA, it has no involvement in private space travel.
I read in "Dose", a free publication in Canada that, "one small snag in the plan might be that cosmonauts don't even have the kind of technlogy that would take them to the moon and back. The Russian news agency Novosti reports that even cosmonaut Alexei Leonov, the first man to walk in space, doubts that Russia's space program has the equipment to take a civilian to the dark side of the moon."
If you want a magnification (although the concept is misleading for these type of images) the HST ACS field of view is 202x202 arcsec (where 3600 arcsec = 1 degree). Now measure the *angular size* of the pic on your monitor (for easy comparision your fist at arms length is about 10 degrees). Comparing the two numbers gives you a "magnification".
A CCD is capable of detecting single photons. In fact, the Quantum Efficiency (QE) of CCDs is usually better than 90% for red colours. The human eye is usually quoted to have a QE around 1%. (so if the QE is 1% only 1/100 photons are detected). The key point is to be able to distinguish a photon from a star from "sky" photons (ie. scattered light). If a dim object is placed on a bright background, it is very difficult to detect the dim object. Just compare the night sky with and without a full moon! So the number of photons detected is meaningless without comparison to the noise involved.
To put everything into perspective, the full moon has a magnitude of -12.7 (so you can compare that to a light in your room) which gives roughly 1x10^12 photons over 3.5 days. Compare that to only 5x10^4 photons (in proper sig figs from our guestimation) from a 31th magnitude star!
Think of how many photons are directed upwards from city lights and then scattered back down towards us, ruining dark skies, not to mention the waste of energy!
The "streaks" centering on stars are diffration spikes from the secondary mirror support. The colour alternates as different wavelenghts cause different diffration spacings.
The big bright cluster is actually a member of Andromedae (M31). Very impressive! The appearance of fuzziness is because the CCD oversamples the resolution of the telescope - which is necessary for good photometry - if you want it "sharp" then just bin the pixels by 2x2 or 3x3 or whatever looks best!
The best way is to download the processed HST images and see what the count rate is for a faint star. Then multiply by the gain (in the header of the image) which will give you the number of photons detected.
A way to guestimate the number of photons is to compare the flux of the faintest star with the Sun. At the Earth's distance the Sun has a flux of 1.36x10^6 erg s-1 cm-2 and the apparent mag of the sun is V=-26.8. If we assume that we have a star with V=31 mag (the 50% completeness level is V=30.7 mag) then the flux recieved from the star is given by:
F2/F1 = 100^((m1-m2)/5) where F1 and m1 are the flux and magnitude of the sun and F2 and m1 refer to the star.
This gives 1.03x10^-17 erg s-1 cm-2. Convert the ergs into photons by the de Broglie frequency (E=hv) where we assume that a V-band photon has a wavelength of 550nm or a frequency of 5*10^14 s-1. Thus, each photon carries 3.61x10^-12 ergs which gives a rate of 2.85x10^-6 photons s-1 cm-2. So a 3.5 day exposure is 302400 secs and HST has an aperature of 240 cm so we get about 50000 photons at the entrance of the telescope. Remember.. detection of these sources means having a low background so that these photons are not lost in noise! I should also point out that HST does not leave the shutter open continuously for 3.5Hs, instead it takes a series of short exposures that are co-added.
I hope this helps (and doesn't freak people out!)
CCDs do not suffer from Reciprocity failure like film does. However there are other problems that will turn long exposures into junk (such as cosmic rays as HST is not sheltered by the earths atmosphere!). So many shorter exposures are taken and then coadded to make a 3.5 day exposure.
Slashdot Mirror
Here's a picture of the Lick-a-chick eatery: Flickr. At the Tim Horton's across the street, there was a claim that an old light bulb was projecting the virgin Mary on a wall. Fun area.
I was expecting a lot more from this website. It looks nice you first load it up, but the information content is quite sparce. Not even pretty pictures or informative diagrams to explain concepts. Section 6.2 on aging stars as an example is dismal.
The CASCA Education website is much, much better:
http://www.cascaeducation.ca/files/index.html
Check it out.
Soyuz is a great well tested, great success rate rocket, and I'm not debating the launch of stuff into near earth orbits, but living quarters for a few humans to spend a relatively long time in space. When was the last time a human went around the moon? - the Apollo years?
Canada has 1/10 the population of the US, so at most 1/10 the resources, so agencies such as CSA must concentrate focus on a few areas such as Canadarm or Microsatellites (for example check out MOST). While CSA does provide astronauts for NASA, it has no involvement in private space travel.
I think you're missing the point. Equipment and resources that were available in the 60s and 70s is no longer around.
I read in "Dose", a free publication in Canada that, "one small snag in the plan might be that cosmonauts don't even have the kind of technlogy that would take them to the moon and back. The Russian news agency Novosti reports that even cosmonaut Alexei Leonov, the first man to walk in space, doubts that Russia's space program has the equipment to take a civilian to the dark side of the moon."
Will you ever put a leech on your nuts in another movie?
Well, it seems that this protection does not apply
to Canadian currency! (okay.. here comes the Canadian-monopoly money jokes).
That's the great part about Canada! No one cares what you believe in. We should start writing "Lamp Posts" next year.
-You have seen the light!
What part of Canada do you live in? The Census has always come in an envelope.
If you want a magnification (although the concept is misleading for these type of images) the HST ACS field of view is 202x202 arcsec (where 3600 arcsec = 1 degree). Now measure the *angular size* of the pic on your monitor (for easy comparision your fist at arms length is about 10 degrees). Comparing the two numbers gives you a "magnification".
A CCD is capable of detecting single photons. In fact, the Quantum Efficiency (QE) of CCDs is usually better than 90% for red colours. The human eye is usually quoted to have a QE around 1%. (so if the QE is 1% only 1/100 photons are detected). The key point is to be able to distinguish a photon from a star from "sky" photons (ie. scattered light). If a dim object is placed on a bright background, it is very difficult to detect the dim object. Just compare the night sky with and without a full moon! So the number of photons detected is meaningless without comparison to the noise involved.
Ha! That's too funny. Can you make zoomed out jpeg of the region to find it?
Yeah.. that would help to know!
To put everything into perspective, the full moon has a magnitude of -12.7 (so you can compare that to a light in your room) which gives roughly 1x10^12 photons over 3.5 days. Compare that to only 5x10^4 photons (in proper sig figs from our guestimation) from a 31th magnitude star!
Think of how many photons are directed upwards from city lights and then scattered back down towards us, ruining dark skies, not to mention the waste of energy!
Cats: You are on the way to destruction.
The "streaks" centering on stars are diffration spikes from the secondary mirror support. The colour alternates as different wavelenghts cause different diffration spacings.
The big bright cluster is actually a member of Andromedae (M31). Very impressive! The appearance of fuzziness is because the CCD oversamples the resolution of the telescope - which is necessary for good photometry - if you want it "sharp" then just bin the pixels by 2x2 or 3x3 or whatever looks best!
The best way is to download the processed HST images and see what the count rate is for a faint star. Then multiply by the gain (in the header of the image) which will give you the number of photons detected. A way to guestimate the number of photons is to compare the flux of the faintest star with the Sun. At the Earth's distance the Sun has a flux of 1.36x10^6 erg s-1 cm-2 and the apparent mag of the sun is V=-26.8. If we assume that we have a star with V=31 mag (the 50% completeness level is V=30.7 mag) then the flux recieved from the star is given by: F2/F1 = 100^((m1-m2)/5) where F1 and m1 are the flux and magnitude of the sun and F2 and m1 refer to the star. This gives 1.03x10^-17 erg s-1 cm-2. Convert the ergs into photons by the de Broglie frequency (E=hv) where we assume that a V-band photon has a wavelength of 550nm or a frequency of 5*10^14 s-1. Thus, each photon carries 3.61x10^-12 ergs which gives a rate of 2.85x10^-6 photons s-1 cm-2. So a 3.5 day exposure is 302400 secs and HST has an aperature of 240 cm so we get about 50000 photons at the entrance of the telescope. Remember.. detection of these sources means having a low background so that these photons are not lost in noise! I should also point out that HST does not leave the shutter open continuously for 3.5Hs, instead it takes a series of short exposures that are co-added. I hope this helps (and doesn't freak people out!)
CCDs do not suffer from Reciprocity failure like film does. However there are other problems that will turn long exposures into junk (such as cosmic rays as HST is not sheltered by the earths atmosphere!). So many shorter exposures are taken and then coadded to make a 3.5 day exposure.