22 Apr 16
Originally posted by DeepThoughtOK, fair enough. It would help if you declared all your variables in one place rather than making me go through multiple posts to find out what you are talking about.
That is dr in my notation and comes out at half the radius of the disk. dx, in my notation is the distance of the focus of light just skimming the collector disk from the point of first focus. The point of my calculation was to get dr. You might want to read it properly before claiming that it's wrong.
So are you saying that the width of the annulus is 0.5 m?
22 Apr 16
Originally posted by DeepThoughtI have said it a number of times already:
Why do you think this?
Because the equation clearly has the deflection angle tending to infinity as r tends to zero whereas it is obvious that in reality at some point inside the sun, the trend reverses ie the deflection angle tends to zero as r tends to zero.
22 Apr 16
Originally posted by twhiteheadI cottoned on and edited my previous post, for convenience here is the edit, with a little extra material:
I have said it a number of times already:
Because the equation clearly has the deflection angle tending to infinity as r tends to zero whereas it is obvious that in reality at some point inside the sun, the trend reverses ie the deflection angle tends to zero as r tends to zero.
We don't need a more sophisticated calculation, the force of gravity drops linearly inside the Sun until it is zero at the centre. At a point a distance r from the centre of the Sun, but less than R the radius of the Sun the force of gravity is only due to the material inside the sphere of radius r.
The formula:
Deflection angle = K/r
has K = 4GM/c^2, but M goes as r^3. So if we introduce a new constant k where K = kR^3 for R = radius of Sun, we have:
Deflection angle = kr^2. For r < R.
This is the formula we need, we don't need to use the full technology of General Relativity as the weak field approximation is fine inside the Sun.
22 Apr 16
1 edit
Originally posted by DeepThoughtWouldn't that be K/R^3? our regular K is 5907 and multiplying by some cube root will always be more than 1 so the diffraction number would keep getting bigger that way.
I cottoned on and edited my previous post, for convenience here is the edit, with a little extra material:
We don't need a more sophisticated calculation, the force of gravity drops linearly inside the Sun until it is zero at the centre. At a point a distance r from the centre of the Sun, but less than R the radius of the Sun the force of gravity is ...[text shortened]... he full technology of General Relativity as the weak field approximation is fine inside the Sun.
Ah, sorry, the number does go down as R approaches zero it would be zero. So with that formula, the 0.7R depth would give you 0.343 times 8.48E-6 radians or 0.0000029 radians or almost exactly 0.6 arc seconds. So its a smooth slide down. At R=0.5 then dif angle is 1/8th arc second. This isn't like a linear function, isn't this a parabolic shape?
22 Apr 16
Originally posted by twhiteheadYes, but bear in mind this is for the first focus, for foci further out the annulus has width 1m for the reasons we discussed earlier.
OK, fair enough. It would help if you declared all your variables in one place rather than making me go through multiple posts to find out what you are talking about.
So are you saying that the width of the annulus is 0.5 m?
22 Apr 16
OK, please tell me what I am doing wrong.
See updated diagram with new formulas and figures.
http://whereitsat.co.za/GraviationalLensing4.jpg
I have:
y = r / tan ( (K / r) - alpha)
In reality alpha can be ignored.
And it gives me the first focus at 8.2e13 metres or 0.54 AU.
This clearly disagrees with your figure of 540 AU.
At first I thought I might be mixing up radians and degrees, but I don't think so.
22 Apr 16
2 edits
Originally posted by twhiteheadMight just be a number thing, you are exactly 2 orders of mag off.
OK, please tell me what I am doing wrong.
See updated diagram with new formulas and figures.
http://whereitsat.co.za/GraviationalLensing4.jpg
I have:
y = r / tan ( (K / r) - alpha)
In reality alpha can be ignored.
And it gives me the first focus at 8.2e13 metres or 0.54 AU.
This clearly disagrees with your figure of 540 AU.
At first I thought I might be mixing up radians and degrees, but I don't think so.
Your bottom number K is very close to mine, 5910, I got 5907. So divide that by r in meters, 696300000 and you get 0.00000848 and change which is 8.48E-6 Radians and convert to arc seconds, 1.75.
Nice thing about radians is just invert and that is the multiplier of the radius, in this case 117,000 times 696300 Km = 8.14E10 km or about 50 billion miles. 1/8.48E-6=117,000 and change.
22 Apr 16
Originally posted by sonhouseWhy are we converting to arc seconds? Which formula uses arc seconds?
Your bottom number K is very close to mine, 5910, I got 5907. So divide that by r in meters, 696300000 and you get 0.00000848 and change which is 8.48E-6 Radians and convert to arc seconds, 1.75.
22 Apr 16
Originally posted by twhiteheadThat is just the proof the radian number is correct since the published number is always arc seconds, thats all. Arc seconds seems to me a pain in the butt, radians are a lot more useful but it does give you a reality check. If you come up with 3E-5 or something and convert that radian number to arc seconds you will be way off. (that is 6.1 arc seconds)
Why are we converting to arc seconds? Which formula uses arc seconds?
22 Apr 16
2 edits
Originally posted by sonhouseLeave out the miles, that's an archaic american unit. 🙂
r* 1/radian = focus so 1r =696300 km and 1/radian= 117000 and multiply the two you get 80 odd billion km or 50 odd billion miles or about 540 AU. Just a simplification I found.
So, 80 billion kilometres, by that I take it you mean 80,000,000,000 ?
= 8e10 km = 8e13 metres
Google says 1 AU = 1,496e+11 metres
Therefore you should have 0.5348 AU.
Where does your 540 come from?