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Re: (meteorobs) Zenith Attraction, some numbers




me>>Jim,
>by any chance do you have the zenithal attraction shift for the pi-puppids? 
>Of IMO's working list, that probably would have the greatest shift with a 
>geocentric velocity of 18 km/s. 
>geozay<<

jim>>Using the classical equations from Lovell (Meteor Astronomy, 1954):
*   Vg^2 = Va^2 + 124.9
where:
vG = geocentric (observed) speed
Va = apparent speed (Earth velocity + meteor heliocentric velocity)
In this case:
Vg = 18 km/sec
va = 14 km/sec
*   tan((1/2) * dZa) = ((vg -va) / (vg +va)) * tan((1/2) * Za)
where:
Za = radiant zenith angle (deg) = 90 deg - radiant altitude (deg)
dZa = zenith attraction (deg)
In this case:
tan((1/2) * dZa) = 0.121 * tan((1/2) * zA)
Giving dZa as a function of Za:
dZa(30 deg) = 3.7 deg
dZa(60 deg) = 8.0 deg
dZa(90 deg) = 14 deg
This last gives the zenith attraction with the radiant located at the
horizon.  <<

Jim, are these results for the Even days or Odd days of the month?  :o)   
Thanks!
George
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