20 Jul 15
Originally posted by SoothfastNow see, was that really necessary?
To say you are a one-dimensional character would be an insult to that one dimension. Maybe you're fractional dimensional, like the Cantor set.
You kids have no sense of humor.
So when do we send up the next billion dollar probe to go investigate more rocks and gas emissions on planets.....er.......um......dwarf planets?
Damn those scientists. I bet Plutonians are still pissed about taking their planet status away. They must be conservatives or something really, really, bad.
22 Jul 15
1 edit
Now, as I understand it, it will take up to 16 months for the New Horizons probe to send back all the data it collected from the Pluto system flyby. And then, sometime around 2018, it is tentatively scheduled to make a flyby of a small Kuiper belt object:
http://graphics.latimes.com/new-horizons-future/
Then, provided the fuel holds out and all goes well, there's a chance for another flyby of some other as yet undiscovered object.
22 Jul 15
1 edit
Originally posted by sonhouseGive us the answer, please? Would it be something like "larger than the diameter of the Earth"?
Here is a problem for you: calculate the size a telescope would need to be somewhere in the vicinity of Earth to have images of Pluto even as good as they are now, never mind when they get to closest approach. That is to say, the diameter of the reflector.
(Just a WAG... Wild-assed guess.)
22 Jul 15
3 edits
Originally posted by SuzianneWell, it's a good question, which I think can be roughly translated thus: how big must a telescope in space in the neighborhood of Earth be in order to capture the same amount of light from Pluto as the LORRI telescopic apparatus on the probe was getting at the time of Sonhouse's post?
Give us the answer, please? Would it be something like "larger than the diameter of the Earth"?
(Just a WAG... Wild-assed guess.)
At the time of Sonhouse's post, the New Horizons probe was roughly 1/1000 the distance between Earth and Pluto. Light intensity is inversely proportional to the square of the distance from a light source, so that means Pluto's light was one-millionth as intense in the vicinity of Earth as where the probe was located. So to capture the same light from Pluto, a space telescope in our vicinity would have to have a million times the area of LORRI, which means 1000 times the diameter. The aperture (diameter) of the LORRI instrument is about 21 centimeters, so we'd need a telescope with an aperture of 210 meters. That's big, but maybe not impossibly big. The mirror would have to be an array of (likely hexagonal) mirrors such as the large ground-based telescopes in Hawaii and Chile have. The total area of such a monster would probably exceed the area of all the world's big research telescopes combined.
My guess is 210 meters is not big enough. Big "light bucket" telescopes are better for viewing large, dim, deep-space objects like nebulae and galaxies. For a tiny thing like Pluto, to get better resolution, you have to increase the focal ratio of the telescope quite a bit (it's been a very long time since I've dealt with this kind of stuff). But you still need the raw light-collecting capacity since Pluto is quite dim from Earth.
There are other tricky things about the optics of telescopes to consider. For instance, Pluto's disc is exceedingly minuscule from where Earth is (around 1/10 of an arcsecond), so a telescope in Earth orbit would not only need a huge aperture to get enough light, but the image would have to be hugely magnified. You need really thick lenses to do that, but thick lenses eat light. My guess is you'd really need a telescope at least 300 meters in diameter.
To get a space telescope in Earth orbit to deliver images of Pluto to match the New Horizons flyby images of July 14, now, you'd need an aperature over 80 kilometers!
Your proposed telescope with a diameter on a par with the diameter of the Earth might have the capacity to read billboards on a planet circling Alpha Centauri, but I haven't done the math.
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22 Jul 15
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Originally posted by SoothfastThere are two problems here.
Well, it's a good question, which I think can be roughly translated thus: how big must a telescope in space in the neighborhood of Earth be in order to capture the same amount of light from Pluto as the LORRI telescopic apparatus on the probe was getting at the time of Sonhouse's post?
At the time of Sonhouse's post, the New Horizons probe was roughl ...[text shortened]... he capacity to read billboards on a planet circling Alpha Centauri, but I haven't done the math.
One is light capture.
The other is diffraction of the light entering the telescope setting an upper limit on the angular resolution.
Light capture can be solved by both increasing telescope size, and in increasing exposure times and is
not the main limiting factor.
As for the other problem...
Pluto has a diameter of 2370km and is currently ~31.93 AU from the Earth. [~4,776,660,000 km]
A circle with a radius of 31.93 AU will have a circumference of 30,012,640,061 km
Divide that by the diameter of Pluto and you find the fraction of that circle occupied by Pluto...
1/12,663,561.2
The full circle has 360 degrees, 21,600 arc minutes, and 1,296,000 arc seconds. Pluto thus does indeed
occupy ~ 1/10th of an arc second.
1/10th of an arc second is 1/(60*60*10) of a degree, or 1/36,000 of a degree.
The maximum angular resolution T of any imaging device [in radians] is given by 1.220*(Y/D) where Y is the
wavelength and D is the diameter of the primary lens/mirror. [And is what we want to find].
The angular resolution for a given diameter decreases with wavelength so I will use red light 700nm as my baseline.
I also want this in degrees so I will add in a conversion factor of 180/Pi
So T = [1.220*(Y/D)]*[180/Pi]
Rearranging for D we get D = [1.220*180*Y]/[T*Pi]
So to have Pluto occupy 1 pixel we have this equation we have T=1/36000, and Y=700E-9.
Which gives a Diameter of ~1.762m ...
However we Want Pluto to be bigger than 1 Pixel. For sonhouse's challenge we want Pluto to look like this
http://pluto.jhuapl.edu/soc/Pluto-Encounter/data/pluto/level2/lor/jpeg/029861/lor_0298615084_0x630_sci_1.jpg
Which was how Pluto looked at about the time he set the challenge.
Pluto is about ~80 pixels across in that image.
So we multiply T by 1/80 [T=1/(36000*80)] to find the minimum Diameter needed...
Which gives a Diameter of ~141m
How about if we want an image like this
http://pluto.jhuapl.edu/soc/Pluto-Encounter/data/pluto/level2/lor/jpeg/029912/lor_0299123689_0x632_sci_3.jpg
Where Pluto is by my estimation ~633 pixels across...
We now need a Diameter of ~1,115 meters across.
But lets reverse things and ask, what would our angular resolution be if we had an optical telescope the size of the
Earth?
That makes D = ~12,742,000 m
Which gives us a T of ~3.84E-12
If we divide 1/36,000 by 3.84E-12 we can see how many pixels across Pluto would be...
The answer, 7,233,602 Pixels across.
Or 3052 Pixels per km, Or ~3 pixels per meter, ~9 pixels per square meter. [not accounting for spherical distortions as
you move away from the image centre].
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22 Jul 15
2 edits
Originally posted by SoothfastHmmm, If we assume the text on the billboard to be 3 meters high then a telescope would need to be the size of the Earth
Your proposed telescope with a diameter on a par with the diameter of the Earth might have the capacity to read billboards on a planet circling Alpha Centauri, but I haven't done the math.
to read it on Pluto.
Let's look at how big it would need to be to view it on a planet around Alpha Centauri...
First, let's simplify the math.
Keeping everything else constant, Our equation for the Diameter is some constant C1 divided by T And thus
the Diameter is proportional to 1 over T.
T is the fraction of a circumference [with a radius of the object distance] the object occupies at that distance.
The size of the object is a constant [~0.33m for our 3 pixel a meter resolution] and the circumference changes
linearly with increase in radius. [with a factor of 2*Pi]
This means that [the required] T will reduce by a factor of 2*Pi*"change in radius" as the radius increases.
Consequently the diameter of the telescope needs to increase by the same factor.
Alpha Centauri is ~~4.367 LY away.
There are ~~63,241.077 AU in a light year.
Pluto is ~31.93 AU or 0.000505 LY away.
This means that Alpha Centauri is ~8,650 times farther away than Pluto.
This means that the Diameter of an optical telescope needed to read a billboard with 3 meter high letters around Alpha Centauri
needs to be 8,650*2*Pi times larger than the Radius of the Earth ~12,742,000 m
This makes the new Diameter ~~692,470,060,000 m [~430,373,000 Miles or ~4.63 AU Or approximately the size of the inner
edge of The Asteroid Belt between Mars and Jupiter.]
22 Jul 15
2 edits
Originally posted by googlefudgeYou might need it to be that big for light gathering power but for resolution, you can now combine images from two telescopes spaced far apart and get the same resolving power of a single mirror the size of the separation. That technique is already in use on Earth, Keck 1 and 2 do that already.
Hmmm, If we assume the text on the billboard to be 3 meters high then a telescope would need to be the size of the Earth
to read it on Pluto.
Let's look at how big it would need to be to view it on a planet around Alpha Centauri...
First, let's simplify the math.
Keeping everything else constant, Our equation for the Diameter is some constant ...[text shortened]... AU Or approximately the size of the inner
edge of The Asteroid Belt between Mars and Jupiter.]
http://www.pnas.org/content/early/2015/03/26/1424409112.full.pdf
Scroll down a bit and you can see an image of K1 and 2 from the air, you can see how that would work, the interferometer resolution is about the same as the distance between the scopes as if it were one very large mirror.
I don't know how much separation you can get and do good work from two scopes in space, maybe laser beams to co-ordinate the scopes, 2 or more. Maybe it could work out on the back side of the moon.
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22 Jul 15
2 edits
Originally posted by sonhouseYes, I know about Interferometry. It's used extensively with radio telescopes.
You might need it to be that big for light gathering power but for resolution, you can now combine images from two telescopes spaced far apart and get the same resolving power of a single mirror the size of the separation. That technique is already in use on Earth, Keck 1 and 2 do that already.
http://www.pnas.org/content/early/2015/03/26/1424409112.ful ...[text shortened]... eams to co-ordinate the scopes, 2 or more. Maybe it could work out on the back side of the moon.
The diameter that I calculated would be the required diameter of the Interferometry array.
I do not know if it is possible [let alone practical] to create an array the size of the inner solar system.
However as you correctly pointed out the light collecting ability will massively decrease with the
decrease in collecting area.
Thus to collect enough light the observing time would have to massively increase.
This poses a problem because the target is moving [rotating] which means that for our purposes
of reading text at that distance [or just resolving features at the 0.33m scale] this wont work as the
image will simply blur over the time scales required to collect sufficient light.
EDIT: Basically the requirement here is to build a Dyson Sphere with a 3 AU radius and cover it's outer surface with
optical telescopes in a giant interferometry array.
EDIT2:
I don't know how much separation you can get and do good work from two scopes in space, maybe laser beams to co-ordinate the scopes, 2 or more. Maybe it could work out on the back side of the moon.
Thousands of km separation for the telescopes is relatively trivial [in space].
What makes it hard is not so much the separation but the NUMBER of scopes used in the array.
For scopes a few thousand km across there is actually a better way, using parabolic mirror fragments
to focus the light you can create very large telescopes.
https://www.newscientist.com/article/mg18925401-700-the-hypertelescope-a-zoom-with-a-view/
22 Jul 15
Originally posted by googlefudgeInterestingly our estimates are not that far off. You come up with a smaller aperture of 141 m, compared to my 210 m (initial) estimate. You're assuming longer exposure times to get more light. It occurred to me that that might help bring down my aperture estimate, but then I figured I'd stick with the 100 - 150 millisecond exposure times of the LORRI instrument. Of course I failed to mention that...
There are two problems here.
One is light capture.
The other is diffraction of the light entering the telescope setting an upper limit on the angular resolution.
Light capture can be solved by both increasing telescope size, and in increasing exposure times and is
not the main limiting factor.
As for the other problem...
Pluto has a diam ...[text shortened]... uare meter. [not accounting for spherical distortions as
you move away from the image centre].
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23 Jul 15
Originally posted by SoothfastWell 210^2 is only about ~2.2 times larger than 141^2 so a ~250ms exposure time on the 141m
Interestingly our estimates are not that far off. You come up with a smaller aperture of 141 m, compared to my 210 m (initial) estimate. You're assuming longer exposure times to get more light. It occurred to me that that might help bring down my aperture estimate, but then I figured I'd stick with the 100 - 150 millisecond exposure times of the LORRI instrument. Of course I failed to mention that...
telescope would get you comparable light levels. And blurring wouldn't be a factor on those time scales
at that level of resolution.
23 Jul 15
1 edit
Originally posted by googlefudgeI usually compare the resolving power of Hubble, it can resolve 0.05 arc seconds or about 1 part in 25 million, of a 360 degree circle. So at Pluto, call it 6 billion kilometers from Earth, close enough for government work🙂. So 6X2XPI is about 35 billion kilometers circumference. So divide 35E9 by 25 E6 and you get about 1,300 km or so resolution at Pluto, maybe a bit better, say 500 Km just looking at the Hubble images of Pluto. So you would need something about 500 times bigger to get 1 km res at Pluto, which would be about 1,000 meters. if you want 1 meter res, you need 1 million Km either separation of two mirrors or one mirror 1 million Km across, about the size of the sun. That would get you 1 meter. If you wanted one millimeter res, you would need a mirror of 1 billion Km diameter or something humans can make but several of them separated by 1 billion Km.
Well 210^2 is only about ~2.2 times larger than 141^2 so a ~250ms exposure time on the 141m
telescope would get you comparable light levels. And blurring wouldn't be a factor on those time scales
at that level of resolution.
So lets suppose you want some kind of useful res at Alpha Centauri and you had a 1 billion km lens effectively. Now AC is 4.3 ish light years from Earth so 8.6 X PI, that is a circle about 27 light years in circumference. 27 X 5.8E 12X1.6 makes that 2.5 E14 km. Now, Hubble at that distance cranks in at about 10 million Km or 10 billion meters resolution, give or take. Which is why you don't use Hubble to find planets around other stars. So an effective diameter of 2 billion meters, 2 million Km would give a res of about 1 million times Hubble which would be about 10 km at AC. 100 Km res at 40 LY out. 1000 Km res at 400 LY away. THAT would allow you to see planets directly at least those within say 50 ly of Earth. You would still need significantly large mirrors at each end of the interferometer leg if you wanted to collect enough light to do real science, like spectroscopy and such. Or you wait for a couple of weeks of photon collecting to get the job done.
One thing I always wondered, why don't they use these widely spaced telescopes, say the camera at Pluto now, to take images of the stars in our galaxy and compare the parallax with ones from say Hubble which as a parallax of the Earth's orbit or about 300 million Km. Well do the same with images from New Horizons ( of course it won't have the same light gathering power of Hubble but you chart at least what you can see and see the differences compared to Hubble or other Earth based scopes, now you have a parallax of about 12 billion km or about 40 times the parallax of any Earth based telescope, including Hubble. So you would be able to push the direct parallax measurements of star distances by that amount. Seems to me a worthwhile endeavor.
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23 Jul 15
Originally posted by sonhouseYour math is off. You have too many approximations and assumptions and simplifications built into whatever you are doing.
I usually compare the resolving power of Hubble, it can resolve 0.05 arc seconds or about 1 part in 25 million, of a 360 degree circle. So at Pluto, call it 6 billion kilometers from Earth, close enough for government work🙂. So 6X2XPI is about 35 billion kilometers circumference. So divide 35E9 by 25 E6 and you get about 1,300 km or so resolution at Pluto ...[text shortened]... irect parallax measurements of star distances by that amount. Seems to me a worthwhile endeavor.
I just calculated that you can get ~0.3m resolution at Pluto [current distance] with a ~12,700km diameter telescope.
And ~1000m telescope doesn't get you 1km resolution, for that you need a diameter of ~4,174m
From before...
D=(1.220*180*(700E-9))/((1/(36000*2370))*(PI)) = ~4,174m
Number in bold is the number of pixels across you want Pluto to be. 2370 is the diameter in km, which makes this
the diameter needed for 1km resolution. The number in italics is the wavelength of light you are using, optical red light
here.
Which is why you don't use Hubble to find planets around other stars. So an effective diameter of 2 billion meters, 2 million Km would give a res of about 1 million times Hubble which would be about 10 km at AC. 100 Km res at 40 LY out. 1000 Km res at 400 LY away. THAT would allow you to see planets directly at least those within say 50 ly of Earth.
Actually you don't need anything near that large to see planets that far out, which is why we are already imaging planets
with current telescope's.
To pick out a planet from it's host star all you need is enough resolution to have the planet and the star not resolve on
the same pixels on your ccd. Which means you are looking at having X-many pixels per AU as opposed to X-many pixels
per thousand km.
In fact having too narrow a field of view is actually going to be a hindrance as you then need to manage to find the planet
around the host star to image it. whereas if you had a resolution of [say] 100 pixels per AU you can image the whole
solar system in one image centred on the host start which we can find and aim at easily.
25 Jul 15
Originally posted by googlefudgeI know Hubble resolves about one part in 25 million, 0.05 arc seconds and it is about 2 meters across, so 1 meter scope should be good for 0.1 arc second, about one part in 12 million around the circle. So if the circle is say 6 billion km radius, or 6E12meters, 12E12 meters diameter times Pi, about 37 E 12 meters circumference / 25E6 = 1,480,000 meter resolution at approximate distance of pluto, or 1,480 Km. Maybe some better since I think Hubble resolved Pluto with more pixels than one. Do you know how many pixels Hubble resolved for Pluto? I may be wrong on initial estimate of Hubble resolution, perhaps it is 0.02 arc seconds, that would cut the circle in 100,000,000 parts. That would indicate a resolution of about 370 Km at Pluto which seems to better match the image Hubble produced of Pluto.
Your math is off. You have too many approximations and assumptions and simplifications built into whatever you are doing.
I just calculated that you can get ~0.3m resolution at Pluto [current distance] with a ~12,700km diameter telescope.
And ~1000m telescope doesn't get you 1km resolution, for that you need a diameter of ~4,174m
From before... ...[text shortened]... whole
solar system in one image centred on the host start which we can find and aim at easily.
25 Jul 15
1 edit
Originally posted by sonhouseAt http://www.jpl.nasa.gov/spaceimages/details.php?id=PIA00825 you can see Hubble images of Pluto, and they're said to be at a resolution of "100 miles per pixel". Since Pluto is about 1400 miles in diameter, this means the images consist of a bit over 150 pixels (just counting Pluto's surface and not the blackness of space surrounding it).
I know Hubble resolves about one part in 25 million, 0.05 arc seconds and it is about 2 meters across, so 1 meter scope should be good for 0.1 arc second, about one part in 12 million around the circle. So if the circle is say 6 billion km radius, or 6E12meters, 12E12 meters diameter times Pi, about 37 E 12 meters circumference / 25E6 = 1,480,000 meter reso ...[text shortened]... olution of about 370 Km at Pluto which seems to better match the image Hubble produced of Pluto.
EDIT: I guess my problem with all this is what exactly the significance of a "pixel" is. You can always take an image and increase the number of pixels comprising it. A simple program can take one "red" pixel and break it into a grid of 16 smaller "red" pixels.