NavList:

   

Reply

Fred,

In your original post on this subject, you spoke of using linear regression
for choosing the "best" single observation from a set. I assume the "best"
would be the one with the smallest distance from the regression line. Yes?

You also said, "The data look very nice; not as good as the lunar of that
Australian explorer Kieran Kelly discussed a few months ago, but very good."
You did not elaborate on what your criteria are to make data look "very
good". But you are probably referring to the correlation between distance
and time. Yes?

Of course, you can use linear regression in this way only if there is a
linear dependence between distance and time. Given that Frank's seven, or
so, observations were taken within twenty minutes and Jupiter was well
positioned, that's reasonable. We trust Frank that he checked this before he
shot, but you can't really take it always for granted. For small distances
or long time intervals, you might consider using a second order polynom for
your regression.

In your most recent post, you suggest to use regression for establishing
chronometer rate. That's an interesting idea. Assuming that chronometer rate
is constant over a long time, but fluctuates daily with temperature, the
time shown by chronometer would in essence be linearly dependent on GMT.
This dependence could be found via a linear regression of observations for
time of various kinds spread out over a long interval.

Used in the above way, and used correctly, linear regression could be
considered a refined way of interpolation of observational data (hence my
remark about guns and sparrows). However, when I picked up on the term
"least squares fit" that you used, which is of course at the base of the
linear regression method, I was not speaking of linear regression at all. I
was thinking of using a least square fit for the _reduction_ of the sight,
not for data preparation. I would not know how to reduce a single set of
distance observations by linear (or any other type of)  regression.

I conceded that it is possible to use a least square sum method to reduce
such a single set of distance observations. I am unable to provide at this
moment the numerical example that you requested, but I will outline in a
little more detail what I meant by saying in my previous message "solve for
the watch error for which the sum of the square of the distance residuals
becomes a minimum". The method I have in mind is nothing special; it is just
a common technique applied to the particular problem.

Say, you have a function that gives you the apparent topocentric distance
between the moon and an other given object as a function of GMT. Such a
function can be constructed by applying the inverse of any standard distance
clearing method to the geocentric distances, which in turn may be computed,
or directly obtained, from an almanac or from an ephemeris program.

        d  =  D(t_gmt, lat, lon)

You also have a chronometer (watch), showing watch time t with unknown, but
constant watch error -w.
A number of n observations (t_k, d_k), k = 1..n, are all taken with this
watch.

        t_gmt_k = t_k + w

Now, if you have exactly one distance observation (d_1, t_1) you can solve
heuristically the equation

        d_1  =  D(t_1 + w, lat, lon)

for w, and you are done. (You told us that you have a good computer. Any
crude method such as bisection will do.) But if you have two or more
observations (d_k, t_k) you get a system of equations which is
overdetermined:

        d_k  =  D(t_k + w, lat, lon)          k = 1..n

Therefore, introduce a random observation error into your distance
measurements (I am deliberately skipping a discussion of systematic error
here. If you know that you have one, eliminate it! If you don't know about
it, it's not systematic.)

        d_k + e_k  =  D(t_k + w, lat, lon)          k = 1..n


This system is underdetermined, because you have n equations and n + 1
variables. But hopeful wishing makes us search for that value of w that
introduces the minimum observation errors:

        E(w) = Sum(Sqr(e_k), k = 1..n)  .....  min

where

        e_k  =  D(t_k + w, lat, lon) - d_k          k = 1..n

In words, E(w) is the sum of the squares of the individual observation
errors (or distance residuals, as I called it in the earlier message).
Obviously, E(w) is a function of w. Look for the minimum of this function
(again heuristically, by iteration) and you have a rigorous solution for the
most probable watch error belonging to 2 or more observations.

As you can see, the individual observations need not have anything to do
with each other. They can be taken at arbitrary times, from arbitrary
objects. Their only connection is that the observation error is likely to be
the same for all. This assumption may not always apply, but it is one that
is often made, for the lack of any better knowledge of it.

If you wish, you can relax the restriction of constant watch error and feed
the known watch rate into the formula. As long as you need to iterate
anyway, you can also in the first formula replace the longitude parameter by
local apparent time. Or you can even replace the latter together with the
latitude by two altitude observations. These need not be of the distance
bodies, any two will do. They also need not be measured at the same time as
the distances, as long as they are timed with the same watch. But these
minor improvements and others I leave to your imagination.

Herbert Prinz

   

Reply