----- Original Message -----
From: James Ellis <ejge@...>
To: <guide-user@egroups.com>
Sent: Friday, 4 February 2000 19:27
Subject: [guide-user] Re: PHA's, MOID's, Sats etc etc


> I think the argument of perihelion is the most important factor here.
This
> is the angle between point of perihelion and the point where the object
> crosses the ecliptic (asc/descending node). Now this latter point could be
> anywhere - 0.3AU, 2AU or 300AU.
> If there was a way of working out the distance of these nodes from the sun
> and if these distances were near 1AU then we would have a PHA.
> A bit of trig may come in handy - but I haven't done trig for ages. I'll
> keep searching.
> Any maths nuts out there?

Hi all,
Apologies for the delay in reply. I had to dig into my 10 yr old math stored
over my mum's house 170kms away.
To calculate the distance of the nodes from the sun, use the following
formula.

In a parabolic case where
w=argument of perihelion
q=perihelion distance AU
(remember to set calculator to radians)

1) ascending node
w1=-w*3.1415927/180
s=sin(w1/2)/cos(w1/2)
r=q*(1+s*s)

2)descending node
replace first equation with w1=180-w*3.1415927/180

eg comet LINEAR 1999 J3
w=161.9815
q=0.976797

r at asc node = 39.83AU
r at desc node = 1.001 AU

which is why we had that potential LINEARID meteor shower
last November which didn't arise.

The elliptical mathematics are more complicated (and more interesting
with regards to PHA's) I'll get around to it when I get time.
I wrote a BASIC program years ago that I'll try to
write to microsoft basic and e-mail to whoever's interested.

Cheers,
Michael Mattiazzo,
Wallaroo, South Australia
mmatti@...