24 Apr 16
1 edit
Originally posted by DeepThoughtI was comparing what I thought was the concentration of Sirius light past the sun to Sirius light at 1 AU from Sirius. At 25X sol and 1AU at Sol is about 1355 watts then that times 25 would be 33Kilowatts per meter squared at 1AU from Sirius, isn't that right? And what goes past Sol therefore would be 1.1E-7 watts/meter ^2, right?
Yes, the power ratio at the collector - focused radiance from Sirius divided by unfocused direct radiance from the Sun.
You seem to be out by a factor of two for the power from Sirius at the point of first focus - the annulus is 1/2 a metre wide for the first focus which I'm guessing is what the problem there is (for a focus point more than 117 km fur ...[text shortened]... nd" - are you talking about the equivalent effect of light from the Sun being focused by Sirius?
24 Apr 16
Originally posted by sonhouseThe figure I was using was 1,368 W/m^2 for the radiance of the Sun at 1 AU - taken from Wikipedia. Based on that the figure for Sirius at 1AU from Sirius A would be 34.2KW/m^2 and the radiance by the time its light gets here 1.16E-7 W/m^2, so yes, those figures are about right.
I was comparing what I thought was the concentration of Sirius light past the sun to Sirius light at 1 AU from Sirius. At 25X sol and 1AU at Sol is about 1355 watts then that times 25 would be 33Kilowatts per meter squared at 1AU from Sirius, isn't that right? And what goes past Sol therefore would be 1.1E-7 watts/meter ^2, right?
The problem with trying to calculate this for Sirius is that Sirius is a binary system and Sirius B is a white dwarf 20 AU's from the primary with a mass just less than the Sun's and a volume about the same as Earth's(!). This would complicate the corresponding calculation at the Sirius end as you potentially have two lenses.
24 Apr 16
Originally posted by DeepThoughtNot potentially, actual 2 lenses. I only saw 1355 as the figure for Sol but I am not going to quibble over third digit changes🙂
The figure I was using was 1,368 W/m^2 for the radiance of the Sun at 1 AU - taken from Wikipedia. Based on that the figure for Sirius at 1AU from Sirius A would be 34.2KW/m^2 and the radiance by the time its light gets here 1.16E-7 W/m^2, so yes, those figures are about right.
The problem with trying to calculate this for Sirius is that Sirius is a ...[text shortened]... d complicate the corresponding calculation at the Sirius end as you potentially have two lenses.
24 Apr 16
I think the important thing to note at this stage, is that the maximum radiation which is near the first focus is less than (about half) the radiation that earth receives from the sun.
I must also add that the solar wind is an important source of energy that is not involved in the whole focus thing.
What causes comets to have tails, the electomagnetic radiation or the solar wind? How much radiation is required to get comets to start evaporating?
I do plan to finish working out the equations for how the energy tails off to infinity along the axis of focus as well as the overall pattern for a large detector ie how significantly it drops off as you go out of the axis of focus. But I have been a little busy so it might take another week or so.
24 Apr 16
3 edits
Originally posted by twhiteheadI simplified the constant, to just 4G/c^2 which is 2.973E-27. Then all you have to do to figure the deflection for any body is to do M/r and multiply that by my new constant, I call Z. Any star, Z*M/r, M in Kg, r in meters = deflection angle in radians. Assuming you know M and r that is.
I think the important thing to note at this stage, is that the maximum radiation which is near the first focus is less than (about half) the radiation that earth receives from the sun.
I must also add that the solar wind is an important source of energy that is not involved in the whole focus thing.
What causes comets to have tails, the electomagnetic ...[text shortened]... go out of the axis of focus. But I have been a little busy so it might take another week or so.
Your K for Sirius is 11951 btw.
So using my new constant, Z (M/r) is 4.02 E30/1.191 E9 times 2.973E-27 is 1.003E-5 or 0.00001003 or 10.32 arc seconds of deflection, just to compare Sol at 1.75. Inverting 0.00001003 and the multiplier for first focus is 99653 times 1.191E9 in meters (r) = 1.187 E14 meters or 113.7 E9 km, 114 billion kilometers or 74 billion miles.
When you get time could you verify my figures?
Later I will do my version of what I think Sol focuses past Sirius. 529,000 squared = factor of 2.8 E11, invert to 3.57 E -12 times 1368 W/M^2 (sol at 1 AU) = 4.8 E -9 W/M^2 at Sirius.
I figured about 5 nanowatts per meter squared for sol energy passing Sirius, 4.8 is closer, then first circle above Sirius is 7.4E9 meters so +9 and -9 cancel leaving only 4.8 times 7.4 or 35.5 watts focused 113 billion km past Sirius.
What do you think? Numbers ok? What do you think of my new constant Z=2.973 E-27 ?
24 Apr 16
Originally posted by sonhouseIf I have time I will work out the general formulas and either do a spreadsheet or a webpage where you can put in the figures of your choice and get out the results you want.
When you get time could you verify my figures?
For now, just change the constants in my spreadsheet. Swap mass of sol for sirius, and radiation figures at 1AU and it should spit out all the figures you want.
24 Apr 16
1 edit
Originally posted by sonhouseThat statement hangs on whether the two stars line up with the sun - in other words on the orientation of the axis of rotation of the system. If the same ray of light doesn't pass close to both stars it will only be deflected measurably by the one it gets close to.
Not potentially, actual 2 lenses. I only saw 1355 as the figure for Sol but I am not going to quibble over third digit changes🙂
24 Apr 16
3 edits
Originally posted by sonhouseNote that 2GM/c^2 is the Schwartzschild radius, so you could just use the Schwartzschild radius of the sun (call it r_sol) and have:
I simplified the constant, to just 4G/c^2 which is 2.973E-27. Then all you have to do to figure the deflection for any body is to do M/r and multiply that by my new constant, I call Z. Any star, Z*M/r, M in Kg, r in meters = deflection angle in radians. Assuming you know M and r that is.
Your K for Sirius is 11951 btw.
So using my new constant, Z ( ...[text shortened]... t Sirius.
What do you think? Numbers ok? What do you think of my new constant Z=2.973 E-27 ?
K = 2r_sol*(M/M_sol) = 2*(2GM_sol/c^2) *(M/M_sol) = 4GM/c^2
or, if you also want to get rid of the leading factor of 2 you could use the Schwartzschild radius of an imaginary reference star with exactly twice the Suns mass:
K = r_2sol * (M/M_sol) = 2G(2*M_sol)/c^2 * (M/M_sol) = 4GM/c^2
The radiance of the sun at Sirius will be 25 times less than the radiance of Sirius at the Sun, so I get 4.6E-9 W/m^2 ~ 5 nanoWatts/m^2. Sirius has about twice the mass of the Sun, so your K looks about right. I'll check the point of first focus later and come back.
24 Apr 16
1 edit
Originally posted by DeepThoughtI just did first focus for Sirius B, same mass as sol ~=2 E30 kg/ r=5,570,400=~ 3.6E23 times Z, 2.973E-27 = 0.001007 radian, in arc seconds 220! inverse radian= 9370 times 5.57e6 meters ~= 5.22E11 meters or 5.22E8 km (520 million Km or 326 million miles)
Note that 2GM/c^2 is the Schwartzschild radius, so you could just use the Schwartzschild radius of the sun (call it r_sol) and have:
K = 2r_sol*(M/M_sol) = 2*(2GM_sol/c^2) *(M/M_sol) = 4GM/c^2
or, if you also want to get rid of the leading factor of 2 you could use the Schwartzschild radius of an imaginary reference star with exactly twice the Suns ...[text shortened]... the Sun, so your K looks about right. I'll check the point of first focus later and come back.
220 arc seconds! That's what you get with such a concentrated mass.
So that represents first focus. But the gravity field doesn't care about the density of mass so at one Sol r the focus would still be the same for sol, about 80 billion km or 50 billion miles.
And at that r #, 1 sol, Sol energy concentrated at 80 billion km would be about 21 watts/meter^2. Still a respectable amount of energy considering the size of Sb.
24 Apr 16
Originally posted by sonhouseI am not sure what you are saying here. Higher density does mean greater radiation at the focal point because r is smaller.
So that represents first focus. But the gravity field doesn't care about the density of mass so at one Sol r the focus would still be the same for sol, about 80 billion km or 50 billion miles.
24 Apr 16
2 edits
Originally posted by twhiteheadI am just saying the gravity field at 1 r of sol is the same regardless of the size of the mass if it is smaller than r.
I am not sure what you are saying here. Higher density does mean greater radiation at the focal point because r is smaller.
So r for Sirius B is 0.008 times sol so the size of a planet with the mass of a star.
The gravity field is unchanged at 1 sol r. It wouldn't matter if the mass was a black hole the mass of the sun, at 1 sol r the gravitational attraction would be exactly the same at 1 r as it is for Sol at 1 r.
BTW, I think I worked out a formula for the focal point of any star:
Using my new Z constant, 2.973 E-27, the new formula is, **DRUM ROLL**
f= r^2/ZM.
So putting in r squared for the sun, about 4.85 E 17/ 2.973E-27*2E30 = 8.156 E13 meters or 81 billion km or 50.9 billion miles!!!!
Eufrigging RICA!
F (First focal point)= r^2/ZM
I have been trying to figure that one out for years. Thank you guys for collaborating and correcting my misunderstandings of some of the points here. The simplification of K led me to make Z and that turned out to be the key to making a general formula.
It turns out 3 minds are better than one! We all made mistakes and we corrected one another!
24 Apr 16
Originally posted by twhiteheadHe means one solar radius distance from the centre of the white dwarf, at that distance from the surface of the white dwarf the behaviour would be just the same as if it were a main sequence star with the same mass as it has now - when it was on the main sequence Sirius B is believed to have had a mass of 5 solar masses. Sonhouse's 220 arc second figure is for light skimming the surface of the white dwarf and that is much further in as it is believed to have a radius about the same as Earth's.
I am not sure what you are saying here. Higher density does mean greater radiation at the focal point because r is smaller.
I don't think your statement that there's much greater radiation at the focal point is correct. My reason for thinking this is that the radius of the annulus is vastly smaller so less radiation will reach the collector. I'll check all the figures and confirm that.
24 Apr 16
Originally posted by sonhouseThe 1368 W/m^2 figure is quoted here [1]. I double checked and you can get it from the total power output (3.85E+026 Watts) divided by the area of a sphere of radius 149597870700 metres (this is the definition of 1 AU).
Not potentially, actual 2 lenses. I only saw 1355 as the figure for Sol but I am not going to quibble over third digit changes🙂
Note during my checking I found that what I've been calling radiance I should have been calling irradiance - this is incident power per unit area. Radiance is emitted power per unit area of the emitter (in this case the surface area of the Sun) per unit solid angle. Wikipedia gives that as 2.01E+007 Watts/m^2/sr (sr = steradian).
Also these figures are for total emitted radiation, not just for visible light, but that's ok., because we are finding a ratio.
[1] https://en.wikipedia.org/wiki/Sun#Sunlight
24 Apr 16
Originally posted by DeepThoughtI don't yet have a formula for the energy curve along the focal axis, but it is my belief it starts high at the focal point and reduces as it goes to infinity. I don't know how significant the slope is. A closer focal point means less 'stretch' (delta f) which means more radiation and I would expect the effect to be more pronounced at the first focal point end. But I will await actual figures.
I don't think your statement that there's much greater radiation at the focal point is correct. My reason for thinking this is that the radius of the annulus is vastly smaller so less radiation will reach the collector. I'll check all the figures and confirm that.
24 Apr 16
Originally posted by twhiteheadDon't forget to factor in the fact that energy from further up in r numbers will add up to a larger geometric space, more square meters collected together so I think as you go out the beam will get stronger up to the pooping out point at about 8 light years for Sirius and the sun going past Sirius also. Did you read my new general formula for figuring out the first focus? f= r^2/ZM r, radius in meters, Z my derived constant of 2.973E -27 and M, mass of star in km, gives the first focal point in meters!
I don't yet have a formula for the energy curve along the focal axis, but it is my belief it starts high at the focal point and reduces as it goes to infinity. I don't know how significant the slope is. A closer focal point means less 'stretch' (delta f) which means more radiation and I would expect the effect to be more pronounced at the first focal point end. But I will await actual figures.