20 Apr 16
1 edit
I have added some extra detail:
http://whereitsat.co.za/GraviationalLensing3.jpg
Solve for x in terms of delta r, alpha1, K and d or any other combination of knowns.
x is the radius of the receiver required to capture all energy going through a ring of inner radius r and width delta r.
Hint: if delta r = 1m. x is not 1m.
20 Apr 16
Originally posted by twhiteheadA one meter area has 9.8 nanowatts in it coming from Sirius. There is room for 5+ billion boxes around the sun and the energy in that area should come together at a focal point and I am not expecting it to be like a nice lens so what else is going to happen? By delta r you mean change in the radius number I assume and delta f you mean where on the focal line do the 'boxes' focus? It's not a big deal to figure out and I already know the trig. If the radius doubles and the angle goes in half the focus is 4 times the previous integer number of the radius. I don't really care about the finer details, for Sirius, the focus like the energy coming in for any other star also, the first focus is around 550 AU and the last major energy focus line is about 500,000 AU out or 8 ly.
What you are doing appears to be mostly guesswork and very wild calculations. What an astronomer can do is highly accurate calculations that can give a very good estimate of what we would see and such calculations can be verified by observations of other gravitational lenses. 'Simply guesswork' is not a good characterization.
[b]I chose a 1 meter box b ...[text shortened]... sked you to work out the relationship between delta r and delta f. I want you to think about it.
I want to know about other effects, like how close is the original 1.75 number around the rim of the sun, like is it still 1.75 at the suns equator, is there much difference between poles? It seems like there should be but maybe such a small change as to not be worth crying about but it would be nice to know. The sun bulges at the equator so would that change the picture?
20 Apr 16
Originally posted by sonhouseNo, it doesn't come together at a focal point, that is what the diagram is supposed to help you understand.
A one meter area has 9.8 nanowatts in it coming from Sirius. There is room for 5+ billion boxes around the sun and the energy in that area should come together at a focal point ....
and I am not expecting it to be like a nice lens so what else is going to happen?
See the diagram. Lets assume for a moment that it is a nice lens ie a gravitational lens. It works a little differently from a convex glass lens.
By delta r you mean change in the radius number...
Its labelled on the diagram. In your case it is 1m ie the width of the ring around the sun through which the light from Sirius passes that we wish to measure and track.
I assume and delta f you mean where on the focal line do the 'boxes' focus?
Correct. Over what distance does the energy that passes through the 1m wide ring actually end up on the focal line.
It's not a big deal to figure out and I already know the trig.
Then stop waffling and figure it out.
I don't really care about the finer details,
The finer details are everything. You have missed a very massive 'finer detail'.
for Sirius, the focus like the energy coming in for any other star also, the first focus is around 550 AU and the last major energy focus line is about 500,000 AU out or 8 ly.
I see you are still imagining distinct focal points. Light doesn't travel in integers, get that into your head!
There are not 'energy focus lines'. There is only one single focal axis from the sun to infinity. The energy does gradually tail off, but not nearly as fast as you believe, nor is it nearly as strong close up as you believe.
I want to know about other effects, like how close is the original 1.75 number around the rim of the sun, like is it still 1.75 at the suns equator, is there much difference between poles? It seems like there should be but maybe such a small change as to not be worth crying about but it would be nice to know. The sun bulges at the equator so would that change the picture?
Get the basics right with the assumption that the sun is a perfect sphere first. Then worry about imperfections later.
20 Apr 16
1 edit
Originally posted by twhiteheadI can figure f for any number, what is the big deal about that? I don't have an overall formula, I just crank in the r 1.3344 whatever and figure the deflection angle at that height and go from there. I am missing something here?
No, it doesn't come together at a focal point, that is what the diagram is supposed to help you understand.
[b]and I am not expecting it to be like a nice lens so what else is going to happen?
See the diagram. Lets assume for a moment that it is a nice lens ie a gravitational lens. It works a little differently from a convex glass lens.
By ...[text shortened]... ith the assumption that the sun is a perfect sphere first. Then worry about imperfections later.
The deflection angle is a linear function of altitude. And I did your K, 5907. So 5907 divided by r in meters gives me the angle in radians. I already know you invert the radian number and you have the multiplier, that number times r gives f.
For instance, r=1, 5907/696300000 meters =8.48E-6 and invert that to get around 117000 times r gives f. Is there something esoteric here?
Not nearly so strong as you think? What is it then? Are you saying the intensity of light from Sirius is not 9.8 nanowatts per meter squared? Are you saying there is not room for 5 odd billion of those meter squared available in the closest ring around the sun, or that there would be room for 10 billion of them at r=2?
21 Apr 16
3 edits
Originally posted by sonhouseSuppose we are interested in the light focused from Sirius hitting a disk of radius h at a distance x from the Sun. The light hitting the centre of the disk has passed the Sun at a distance of closest approach r. The light just skimming the edge of the disk of radius h passed the Sun at a distance r + dr, and would be focused at a point x + dx had it not hit our disk. So, from earlier, we have:
A one meter area has 9.8 nanowatts in it coming from Sirius. There is room for 5+ billion boxes around the sun and the energy in that area should come together at a focal point and I am not expecting it to be like a nice lens so what else is going to happen? By delta r you mean change in the radius number I assume and delta f you mean where on the focal lin ...[text shortened]... ut but it would be nice to know. The sun bulges at the equator so would that change the picture?
x = r^2/K
x + dx = (r + dr)^2/K = r^2/K + 2rdr/K + O(dr^2) = x + 2rdr/K
Where I've dropped the term in order dr^2 as it's tiny by comparison to the rest. So we have that:
Kdx = 2rdr
Using similar triangles we also have that the radius of the disk where the light is being collected is:
h/dx = (r + dr)/(x + dx) ~ (r/x)
whence
h = rdx/x
Eliminating dx we have:
dr =(1/2) Kxh/r^2 = (1/2)(r^2/x)xh/r^2 = h/2
so the annulus has width h/2. The power output of Sirius is 25 times that of the Sun. Lets look at the inner focus (for light from Sirius just grazing the Sun's surface), this is at 542 AUs. At 1AU the radiance is 1,368 W/m^2, so at 542 AU's that's 0.0044 W/m^2 or 1.4 milliwatts on our 1m radius disk (of area pi).
Now lets look at Sirius. It's power output is 25 times that of the Sun which gives a radiance at 1 AU (from Sirius) of 34.2kW/m^2. Now, according to Wikipedia the equatorial circumference of the Sun is 4.379 billion metres so we have an area of 2.189 billion metres squared for the annulus through which light from Sirius will be projected onto the target disk. We now need the distance between the Sun and Sirius in astronomical units. 8.6 lightyears is 540,000 astronomical units. We need to use the inverse square law and (5.4E5)^2 = 2.9E11. Putting it all together I get that the light from Sirius focused on a 1 m radius collector at 542 AUs from the Sun will be 240 Watts making it 170,000x brighter than the Sun at the collector.
To my surprise you're right. This is wildly against my intuition. So much so that I've spent an hour trying to find what is wrong with the calculation (it's correct). Interesting that the background object is so much brighter than the foreground object.
21 Apr 16
Originally posted by DeepThoughtBut that is how Einstein rings work. The background focus agent is further away and the object focused is going to be amplified by some factor.
Suppose we are interested in the light focused from Sirius hitting a disk of radius h at a distance x from the Sun. The light hitting the centre of the disk has passed the Sun at a distance of closest approach r. The light just skimming the edge of the disk of radius h passed the Sun at a distance r + dr, and [i]would be focused at a point x + dx had i ...[text shortened]... ulation. Interesting that the background object is so much brighter than the foreground object.
Anyway, I found 1/10th of your number, and why do you need to go through so much to get the sun's radiance at 500 AU? It is just 500 ^2 inverted times 1400 Watts. 1 AU =1400 watts/ meter squared and 500 times away is 1/25000 of 1400 watts. I come up with 56 milliwatts per square meter.
21 Apr 16
Originally posted by sonhouseGetting the Sun's radiance at 542 AUs was one sentence in my working. The stuff I had to go through was to work out the width of the annulus around the Sun through which light from Sirius passes on it's way to hit the disk (which you seem to have just assumed was 1 m^2 for a 1 m^2 disk, although you're only out by a factor of 2). Also the focus is at 542 AU's (which is a figure from the Wikipedia page).
But that is how Einstein rings work. The background focus agent is further away and the object focused is going to be amplified by some factor.
Anyway, I found 1/10th of your number, and why do you need to go through so much to get the sun's radiance at 500 AU? It is just 500 ^2 inverted times 1400 Watts. 1 AU =1400 watts/ meter squared and 500 times away is 1/25000 of 1400 watts. I come up with 56 milliwatts per square meter.
The reason for the factor of 10 is that 500*500 = 250,000 - you have 25,000.
21 Apr 16
Originally posted by DeepThoughtI'll double check all that later, but I suspect you are getting it wrong.
Putting it all together I get that the light from Sirius focused on a 1 m radius collector at 542 AUs from the Sun will be 240 Watts making it 170,000x brighter than the Sun at the collector.
To my surprise you're right. This is wildly against my intuition. So much so that I've spent an hour trying to find what is wrong with the calculation (it's correct). Interesting that the background object is so much brighter than the foreground object.
It would help if you stuck with the symbols used in the diagram:
http://whereitsat.co.za/GraviationalLensing3.jpg
I'll try to also add more to the diagram such as known figures.
21 Apr 16
Originally posted by DeepThoughtArrggh, I hate when that happens. Shows what you get when you use the Windows calculator🙂
Getting the Sun's radiance at 542 AUs was one sentence in my working. The stuff I had to go through was to work out the width of the annulus around the Sun through which light from Sirius passes on it's way to hit the disk (which you seem to have just assumed was 1 m^2 for a 1 m^2 disk, although you're only out by a factor of 2). Also the focus is at 5 ...[text shortened]... Wikipedia page).
The reason for the factor of 10 is that 500*500 = 250,000 - you have 25,000.
21 Apr 16
Originally posted by sonhouseI was having the same problem with the calculator on Ubuntu, so I used a spreadsheet as it's a lot easier to see what's happening and I could go back and make corrections.
Arrggh, I hate when that happens. Shows what you get when you use the Windows calculator🙂
21 Apr 16
Originally posted by DeepThoughtAnd me with a frigging dozen calculators🙂 Latest one was given to me, I had wanted it for years, the HP11C. The one the guy gave me was encrusted with dirt and such, he is a retired forensic financial investigator, and rock musician🙂 So it cleaned up nicely though.
I was having the same problem with the calculator on Ubuntu, so I used a spreadsheet as it's a lot easier to see what's happening and I could go back and make corrections.
Even though the 11C was first made in 1982 or so, they are still being made by HP and go on Ebay for $70!
I like it because its not so daunting as my HP48GX which is admittedly ten times as powerful but I like the simplicity of the 11, and helps me learn RPN and programming.
The 11 was the first fully programmable from HP where it had conditional tests and goto's and such. Only a couple hundred steps available but that is plenty to learn the basics on.
At work I use the Casio 115 series, cheap, less than 20 bucks but it has a very useful feature for me, hour, minute and second arithmetic. One of the machines I work on is called a sputtering tool, sputtering most anything onto a substrate of some kind, glass, silicon, alumina, whatever and we sputter aluminum, silicon carbide, SI02 and chromium onto our substrates.
The thing is we want to know how long the process takes, the operators ask me when it will be finished and when to expect to have to load the next batch.
The 'carrier' moves the 'pallet' which contains the substrates being coated, moves slowly back and forth over the 'target' which has the material to be sputtered.
So we would wish the timing of that mechanical movement would be stable but it isn't as stable as we would like, so the first thing I have to do is measure the exact time for one 'scan', one pass of the carrier and pallet by the target. That number we would like to make exactly 60 seconds but it varies from 68 to 74 seconds depending on how many passes have happened. So I have to multiply the pass time by the number of passes, say 260 passes at 75 seconds per pass = 19500 seconds/3600 gives 5.42 hours. So enter the time of day, say 8 am in the hours button, then go plus 5.42 and it says 13;25, or 1:30 pm. That is a very handy function for my work.
But I am slowly learning the ins and outs of RPN and HP programming.
I wonder if there is a program we could write that would take a given star, known distance, power out, power received at Earth and then crank out the energy deposited in the focal area without having to do all the intermediate steps. That would allow me to add many more stars to the list, Alpha Centauri, Tau Ceti and so forth.
21 Apr 16
Originally posted by sonhouseOf course there is, and I will write it for you just as soon as you work out the equations I asked you to work out. Once you have the equation, you don't need to be relying on rule of thumb guesswork and wild estimations, you can get an exact figure in seconds form a spreadsheet, or I could even do a little JavaScript program for you. But there is no point when you are using the wrong equations. Garbage equations, garbage results.
I wonder if there is a program we could write that would take a given star, known distance, power out, power received at Earth and then crank out the energy deposited in the focal area without having to do all the intermediate steps.
You probably have a programmable calculator in your collection that could easily do it too, but again, you need the equation first.
21 Apr 16
4 edits
Originally posted by twhiteheadSorry, I am going full tilt at work with process development, maintenance of our sputtering machines, teaching the operators proper use of same, long work hours and commute times.
Of course there is, and I will write it for you just as soon as you work out the equations I asked you to work out. Once you have the equation, you don't need to be relying on rule of thumb guesswork and wild estimations, you can get an exact figure in seconds form a spreadsheet, or I could even do a little JavaScript program for you. But there is no poin ...[text shortened]... lculator in your collection that could easily do it too, but again, you need the equation first.
Could you give me a list of the equations? I have limited time to work on these projects considering all the projects I have at work. Equations plus units. I think my little 11C could handle all of those even with it's limited memory.
Right now I am involved in a big puzzle about the hardness of a sputtered coating of Silicon Carbide, why it is 1/3 the hardness it was a few years ago. I subscribe to an industry journal, Vacuum and Coating magazine, just found a piece by Dr. Robern N Casellano of "The Information Network' and I emailed them about our present dilemma. That is just one issue going on for me here. My thinking is the way SIC is manufactured, they grind down the stuff to a certain size, say 100 Nm, then press into the proper shape at some elevated temperature, maybe with some kind of binding polymer and heat it up enough to drive out the binder, that is just my guess as to how they make the stuff. Annealing would probably work but it would take a lot higher temp than the substrate circuitry can take, namely 500 degrees C. If it takes 1200 C to make the SIC crystalize it would destroy our circuits so it is a big problem.
Sorry to dump on you but just showing some of the work I am involved in right now.