23 Apr 16
Originally posted by twhiteheadMy figures for power from Sirius at first focus comes in at about 40 watts/meter ^2.
It must have been late last night (and a busy week) because I got a number of things wrong.
I am using an Excel spreadsheet by the way.
I now agree with your figure and will proceed.
I did the same thing for light from Sol passing Sirius, found K to be about 11,900 and the amount of light focused was actually about the same as Sirius focus past Sol. That would be because Sirius is so much bigger than Sol, the area close to Sirius surface is bigger so the area of the first ring is also bigger. Sol produces about 5 nanowatts/meter^2 at Sirius, about half of Sirius coming our way.
I also did Alpha Centauri A, K there is about 6500 and light from Sol passing AC A at first focus which is about 112 billion km out, coming in at about 100W/M^2 from Sol.
23 Apr 16
2 edits
Originally posted by twhiteheadWhere did you get the number for Sirius energy level at Sol? 1.1E-13? Figures I saw published were a lot stronger. 9.8E-9. 10 nanowatts per meter squared.
Please check my latest figures:
http://whereitsat.co.za/GraviationalLensing6.jpg
As usual they don't agree with anyone else's 🙂
Radiation at focus for Sirius, you have listed what looks a lot like 800 Kilowatts/M^2.Both those figures seems wildly out. I got something more like 50 watts/M^2 based on how many 1 meter squares I could put around the sun at 1r, disregarding corona blockages and such. I see the radius of the sun printed as 696300 Km or 696.3E6 meters for a circumferance of about 4.2E9 meters times 9.8E-9 gives me about 42 watts/M^2. That is all the energy there is in that much area.
If you do the sun, radiance level at Sirius, same distance, 500,000 AU roughly, it goes down by a factor of 2.5E11 and invert 4E-12 times 1355 W/M^2 gives about 5E-9 W/M^2 at Sirius. About half what Sirius gives us at Sol.
Doing Sirius backwards, at 9.8 W/M^2 times 500,000 squared I get 2450 W/M^2 at 1 AU. That sounds reasonable considering the power Sirius puts out.
23 Apr 16
Originally posted by sonhousePlease refrain from using different units all the time lets stick to a particular unit, in this case W/m^2
Where did you get the number for Sirius energy level at Sol? 1.1E-13? Figures I saw published were a lot stronger. 9.8E-9. 10 nanowatts per meter squared.
My figure is based on DeepThoughts origional post where he says that Sirius:
At 1AU the radiance is 1,368 W/m^2
I then use the inverse square law and the distance between the sun and Sirius.
I think I have the distance from Sirius to the sun in AU wrong, so I'll double check that.
The spreadsheet is here if you want to see how everything is calculated:
http://whereitsat.co.za/GraviationalLensing.xlsx
Radiation at focus for Sirius, you have listed what looks a lot like 800 Kilowatts/M^2.
One step at a time. If my first figure is wrong, the rest will be too.
23 Apr 16
1 edit
Originally posted by twhiteheadThe problem there with the 1300 watt thing, that is the radiance from OUR sun. And U have stuck with W/M^2 and meters when called for and when I used Km I said Km. Sometimes mentioned miles as an aside but I would never do calcs using miles.
Please refrain from using different units all the time lets stick to a particular unit, in this case W/m^2
My figure is based on DeepThoughts origional post where he says that Sirius:
At 1AU the radiance is 1,368 W/m^2
I then use the inverse square law and the distance between the sun and Sirius.
I think I have the distance from Sirius to the sun in ...[text shortened]... e 800 Kilowatts/M^2.
One step at a time. If my first figure is wrong, the rest will be too.[/b]
For instance, I used 696,300,000 as the radius of the sun, in meters. Times the invert of 8.48E-6 (radians of divergence angle by the sun) or 117,000 or so giving about 8.15 E13 meters or 8.15 E10 Km to first focus.
23 Apr 16
Please confirm that my figures are correct for:
1. Distance Sirius to sun in AU (I have 5.4E8 DeepThought has 5.4E5 )
2. radiation of sun at 1 AU
3. radiation of sirius at 1 AU (I got that from DeepThought who says it should be about 25 times that of the sun)
I use those figures and the inverse square law to calculate the disputed figure.
23 Apr 16
Originally posted by twhiteheadMost of them seem to. I've redone the calculation on a spreadsheet because I didn't save it last time.
Please check my latest figures:
http://whereitsat.co.za/GraviationalLensing6.jpg
As usual they don't agree with anyone else's 🙂
So going through the figures as you have them on the jpeg (but skipping the various constants and distances):
delta f = 117 km (agree)
area of annulus = 2.1875 E9 (I get 2.1856 differ on 3rd decimal place, did you use 3.14 for pi?)
area of collector = 3.14 (obviously I agree)
magnification of energy = 6.96 E8 (I did the calculation differently, but combining your own figures this looks right)
radiance of Sirius at 1AU (from Sirius) = 3.42E4 (agree)
radiance of Sirius at sun = 1.154E-13 (I have 1.156E-7 so we need to check)
radiance of sun at 542 AU = 4.543E-3 (I have 4.65 E-3 so we differ in 1st decimal place)
You present it differently but I have:
radiation of Sirus passing through annulus = 252 Watts
radiation from sun hitting disk = pi*radiance at 542 AUS = 0.014 Watts
ratio = 252/0.014 = 17000 (I seem to have lost a factor of ten relative to my earlier calculation).
23 Apr 16
1 edit
Originally posted by twhiteheadIts just about 500,000 AU. Don't forget we are working with inexact numbers here so we don't need to go 5 or 6 significant figures when we are just looking for basic ballpark stuff.
Sorry, he said 34.2kW/m^2. and that is what I have in the spreadsheet.
Where we differ is the distance between Sirius and the sun in AU.
Wiki gives distance to Sirius at 8.6 ly, where one light year is 9.28E12 Km which is about 8E13 km or 8E16 meters. Divide by 148 E6 Km (one AU) and you get 539,000 odd AU to Sirius. That is a factor of 2.9E14 less radiance here than there. So times 9.8E-9 and I come up with about 2800 W/M^2 at one AU from Sirius. That doesn't jive with the reported 25X luminosity which according to that would be 33,875 W/M^2. That doesn't jive with the 9.8 E-9 W/M^2 I saw either, that would be more like 121 E-9. 22 times what I said. if so, first focus for Sol would be 22X40 watts I first calculated or about 900W/m^2 at first focus.
Size of Sirius, 1.7X Sol. (radius) so Sirius radius clocks in at about 1.1E6 km or
1.1E9 meters.
or about 14 E6 Km around, circumference. So using that figure in meters, 14E9 meters times Sol radiance at Sirius and I come up with 70 Watts/meter^2 at first focus of Sirius of the light from Sol.
K for Sirius, about 11,900 divide radius of Sirius, 1.1E9 gives a radian number of 0.0000108 or in arc seconds, 2.22 divergence angle near surface of Sirius.
Invert the radian # gives 92400 as multiplier of radius in meters gives first focus of light from Sol past Sirius of about 1.0E14 meters or 1 E11 km or 39 billion miles, somewhat closer than Sol's first focus so for Sirius, about 420 AU for its first focus.
23 Apr 16
2 edits
Originally posted by twhiteheadJeez that has to suck🙂 One wonders where the number I got for luminosity at Sol of 9.8E-9 w/M^2 came from.
OK, I have figured it out. I have been copying my distances off Google which uses a comma as a decimal point. Excel takes it and thinks its a separator and adds three zeros! That is also why I had another figure wrong earlier in the thread.
If Sirius cranks out 25 times Sol's radiance, it should be cranking 33Kw /meter squared at 1 AU.
I wonder if the units for my number of 9.8 was per square foot? That would be 10.75 X 9.8, which comes out at 105 Nanowatts per square meter. I think that is the source of my errors for first focus.
So I come out with 900 watts/meter squared at first focus with new numbers for radiance from Sirius, coming in at about 120 Nanowatts per meter squared..
That makes more sense.
23 Apr 16
Originally posted by DeepThoughtI have updated my spreadsheet with corrected distances.
You present it differently but I have:
radiation of Sirus passing through annulus = 252 Watts
radiation from sun hitting disk = pi*radiance at 542 AUS = 0.014 Watts
ratio = 252/0.014 = 17000 (I seem to have lost a factor of ten relative to my earlier calculation).
Still disagree about the radiations experienced at the focus.
23 Apr 16
Originally posted by sonhouseThere are figures at the surface of the sun, at 1AU and at ground level (after passing through the atmosphere). We have used 3.42 as it was quoted to be the average for what strikes the top of earth's atmosphere.
One wonders where the number I got for luminosity at Sol of 9.8E-9 w/M^2 came from.