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> On Wed, 15 Dec 2004, Ken Muldrew wrote:
>
> > If the star is rising in the East or setting in the West,
> > then the time
> > sight and altitude could be done at the same time
>
> I don't understand several things in this sentence.
> 1) What is the meaning of "If".

It means that one should NOT use the altitude of a star near the meridian for
a time sight.

> 2) Altitude of ANY star gives you local time at the
> moment it is measured, is not it?

Yes, but if the star's altitude is not changing very much with time (near the
meridian) you will not get a very accurate measure of local time.

> What is special about "time sight"? Why not to use the star altitude
> that is needed for the lunar anyway, as a time sight?

Because you can choose a better star for time (one that is changing altitude
rapidly with time) and just calculate the altitudes necessary for the lunar.

> I tried to repeat Thompson altitude computation.
> I use the formula
> sin h=sin L sin Dec +cos L cos Dec cos LHA,
> where L is the latitude (I use Thompson's value 50d47'24")
> Dec is the star's declination (I use Thompson's value 8d22'25" N
> for Altair)
> LHA=(RAstar-RAsun-LocTime)
> I use Thompson's values RAstar=19h41m3sec for Altair
>                         RAsun =16h11m13sec
>                        LocTime=9h3m45sec.
>
> I obtain h=10d35'40".
> How did Thompson obtain a different value of 10d54'19"?

It was difficult to read but I believe he got 10d34'19". The small difference
comes from using tables.

> What does his DR long have to do with this at all?

Nothing. The thought that I expressed in a previous message was in error.

> Or he used some different method to compute altitudes?
>
> By the way, what exactly was his watch supposed to show?
> I mean what is 9h3min45sec, exactly?
> The time elapsed from his Local Noon (=sun culmination)?

Yes.

> Or the time corrected for the equation of time?

No.

> (In any case my altitude calculation does not agree with his one).

It does; I just misread his entry initially.

> > Even when Thompson uses the sun for a lunar,
> > I think he still calculates
> > the altitude even though he uses the sun for his time sight
>
> The reason of this totally escapes me.
> What is a typical interval between his Sun-Lunar and
> time sight?

A few minutes.

> > For the star altitudes this is true but for the lunar
> > altitude the right
> > ascension and declination have to come out of the almanac,
> > so DR longitude
> > is needed to get those values.
>
> But did not you con jecture that his mistake of 2 degree
> in STAR altitude is due to an error in his DR longitude?

I did but I was wrong. Sorry.

> It seems this cannot be so: the star altitude depends on
> the local time only, you don't need to know GMT and your Lat.
>
> Because this 2 degree mistake in the star altitude easily explains his
> error in the final longitude, I am trying to understand why did it
> happen. I tried to repeat his alt calculation (see above) and I don't
> understand where the difference comes from.

I get the same altitudes as well (calculating from log tables). The 2 degree
difference came from the USNO online almanac after adding 12.4 minutes
for mean time. For some reason, another 12.4 minutes has to be added to
get the same altitudes that Thompson calculates. I am still confused by this.

Ken Muldrew.

   

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