Re: For Henri Wilson.
From: Paul B. Andersen (paul.b.andersen_at_hia.no)
Date: 09/15/04
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Date: Wed, 15 Sep 2004 22:12:52 +0200
"Henri Wilson" <H@..> skrev i melding news:qe7fk0d9t1kvhqla2l0qflg9k2mbtksj0h@4ax.com...
> On Tue, 14 Sep 2004 23:17:37 +0200, "Paul B. Andersen" <paul.b.andersen@hia.no>
> wrote:
>
> >
> >"Henri Wilson" <H@..> skrev i melding news:6s8ck05ijicnrhh3mi2q7hqkvcbviujlrv@4ax.com...
> >> On Mon, 13 Sep 2004 23:28:25 +0200, "Paul B. Andersen" <paul.b.andersen@hia.no>
> >> wrote:
> >> >
> >> >You have lost, Henri.
> >> >And you know it.
> >> >
> >> >Paul, having a great time
> >>
> >> ..yes! Avoiding: sin(t/T) + sin(t/T+pi) = 0
> >
> >Claiming that 2 flashes + 2 flashes = 0 flashes again, Henry?
> >
> >2 + 2 = 4
> >Pin it up above your computer.
> >
> >Paul
> >
>
> Paul I have now given you about five good reasons why you wont see the flashes
> that YOUR version of the ballistic theory expects.
You are referring to these flashes, right?
Paul B. Andersen wrote in June 2004:
| I think we now can sum up what the ballistic theory
| predicts HD80715 should look like.
| From:
| http://arxiv.org/PS_cache/astro-ph/pdf/9912/9912158.pdf
| You can see from the spectrum that the H_alpha line split ca. 2.5A,
| which is a relative split of 3.8*10^-4.
| This means that the amplitude of each component is ca. 1.9*10^-4.
| The amplitude of the radial (from observer) component of
| the orbital speed will thus be v/c = 1.9*10^-4, or v = 57 km/s.
| The period is 3.8 days.
| From:
| http://astro.estec.esa.nl/hipparcos_scripts/HIPcatalogueSearch.pl
| we find that the distance is 24 parsecs.
|
| These number will give the ratio (2*pi*d*v/c^2)/p
| the value 9.
| So we will during one period see five stars all the time,
| and two additional stars part of the time. These two last
| stars will either merge in a brilliant flash, or suddenly
| appear in a brilliant flash and then split.
| These flashes will be brighter than 1 for 0.036 period,
| brighter than 10 for 0.0004 period,
| brighter than 100 for 0.000004 period.
| By adding the brightness of the stars, we will get
| the following light curve:
|
| "phase" is normalized, one period = 1.
| "brightness" is relative to the brightness of a stationary star
| N is number of stars seen.
|
| phase brightness N
|
| 0.0 1.22 7
| 0.1 1.21 7
| 0.17 1.97 7
| 0.18 2.45 7
| 0.19 5.90 7
| 0.1913 60.00 7
| 0.191310 infinite 7->5
| 0.2 0.66 5
| 0.3 0.64 5
| 0.4 0.63 5
| 0.5 0.62 5
| 0.6 0.63 5
| 0.7 0.64 5
| 0.8 0.67 5
| 0.808719 infinite 5->7
| 0.8089 21.6 7
| 0.809 11.80 7
| 0.81 3.90 7
| 0.9 1.34 7
| 1.0 1.22 7
|
| Note that the integral over one period is 1,
| that is the average brightness is 1.
|
| The above is for one of the stars, you can get
| the light curve for both stars by translating
| the above half a period and adding.
Henri Wilson responded:
| I can get these figures from my program.
|
| Surprisingly, they agree exactly with yours..... proves my program is
| correct.... not that I ever doubted it.
|
| So I could have saved you all that time and trouble.
| Just click your mouse a
| few times and...there is your curve.
... which MY version of the ballistic theory predicts, eh? :-)
And one of the reasons why we don't see the flashes
predicted by the ballistic theory is: sin(t/T) + sin(t/T+pi) = 0?
> You are so brainwashed you have lost your powers of reason.
Since the reason in your remark:
"Avoiding: sin(t/T) + sin(t/T+pi) = 0"
escapes me, I suppose I must be brainwashed.
But I am confident that you will explain the reason
in your remark:.Won't you?
Paul, still having a good time
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