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(meteorobs) Viewing elevation for maximum rate (Update)



Thanks to Rob McNaught for pointing out a problem with my
assumption #1,
1. the number potentially visible increases linearly with airmass (X)
should be,
1. the number potentially visible increases as airmass squared (X^2)
This better accounts for the visible surface area of the meteor layer.

This changes the rate equation to:
n = Z * X^2 * r^(-0.5*(X+1)*A - 5*log((X+1)/2) )

The quadratic equation to solve for maximum n in terms of X
becomes:

(0.5*A*ln(r)) * X^2 + (ln(r)*5/ln(10) + 0.5*A*ln(r) - 2) * X - 2 = 0

This makes quite a difference, now ALL showers have a maximum
rate at lower observing elevation angles, not at the zenith. (Bob,
your 35 years of experience have not gone to waste!)

From this formula, we can make a table showing the optimum
elevation angle to see the maximum meteor rate, for different
atmospheric extinctions (A) and shower population indices (r):

                     A
 r          0.15   0.28   0.50
-------------------------------------
2.1        6°      10°    15°
2.5        12°     17°    23°
3.0        21°     26°    35°

This is for a range of showers bright (2.1) to faint (3.0), and
atmospheric transparency excellent (0.15) to poor/muggy (0.50).

NOTE: As Marco Langbroek pointed out, do NOT use these
elevation for your FOV in serious observations reported to the
IMO/NAMN! You will cut off your field by the horizon as well as be
inconsistent with the standard elevations of 60°. These results are
just to point out what elevations appear to have the greatest rate.
They can be used as a guide for pointing a camera to maximize
your chances of capturing one, and particularly useful for
telescopic meteor observations, where seeing the maximum rate
per unit area of a small piece of sky is the main consideration.

Thanks to all for your input.
Mike Linnolt

On 6 Aug 2001, at 17:36, Mike Linnolt wrote:

> On 6 Aug 2001, at 14:31, Robert Lunsford wrote:
>
> > below your field of view. By aiming your camera low it is viewing a
> > thicker column of atmosphere which in theory should produce more
> > activity. The zenith is actually the worse area to view meteor activity
> > both visually and photographically.
>
> When I read this my initial thought was it appears this way - the
> lower you look the more airmass you see therefore more meteors
> will be visible. However, on further analysis it appears not to be so
> in most cases. As you look lower, atmospheric attenuation and
> inverse-square law dimming of the meteors will reduce the
> numbers, (since on the average they are further away) but how
> much?
>
> I did a quick back of envelope derivation and got the following
> results:
>
> I assumed 3 factors governs the number of meteors seen as a
> function of elevation (or airmasses (X) where X=1/sin(elev)).
>
> 1. the number potentially visible increases linearly with airmass (X)
> 2. atmospheric extinction reduces their average magnitude
> 0.5*(X+1)*A, where A is the average attenuation in magnitudes per
> airmass (typically 0.28 mag)
> 3. inverse square dimming in magnitudes is 5*log((X+1)/2)
>
> Then, using the familiar "r" population index, we can write the
> meteor rate (n) as a function of airmass and magnitude dimming
> (M) will be n = Z * X * r^(-M) where Z is the rate at the zenith.
>
> Combining (2) and (3) to get M= 0.5*(X+1)*A + 5*log((X+1)/2) we
> get n = Z * X * r^(-0.5*(X+1)*A - 5*log((X+1)/2 )
>
> To find at what airmass (X) this rate is a maximum, we differentiate
> it with respect to X, set it to zero and solve for X.
>
> This yields a quadratic equation in X:
>
> (0.5*A*ln(r)) * X^2 + (ln(r)*5/ln(10) + 0.5*A*ln(r) - 1) * X - 1 = 0
>
> Since X must be at least 1 airmass, this tells us that r < exp(2/(A
> + 5/ln(10))) for a solution to yield a maximum rate value at some
> elevation BELOW the zenith. So for a typical value of A=0.28 we
> see that r < 2.26.
>
> So, in conclusion only the bright showers with population index <
> 2.26 would benefit from looking below the zenith to achieve a better
> rate. For a very bright shower like the Quadrantids with r=2.1, the
> maximum rate will occur by looking at airmass (X) = 1.19 or 57°
> elevation above the horizon. If you observe from a very transparent
> sky with A=0.15, for the same bright shower with r=2.1, the best
> occurs at X = 1.35 or 48° elevation.
>
> While this is based on a simple model, it shows that only for the
> brightest showers could it benefit to center your view somewhat
> below the zenith, certainly not near the horizon. For average or
> fainter showers the best results will be observed at the zenith.
>
> Sincerely,
> Mike Linnolt
>
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