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Russell Sher asked an interesting question, copied below.

It rang a bell with me, because I had struggled to understand Sumner's
working a few weeks ago, and met similar problems.

It was good to see a contribution from Bill Noyce, who hasn't been
on-list
for a while, but is always good on mathematical stuff.
Nevertheless, I think there may still be something worth adding.

=====================

Russell correctly quoted the usual formula for deriving the time
interval from local apparent noon,  as-

LHA = arccos{ [ sin(Ho) - ( sin(Lat.) * sin(Dec) ) ] / [ cos(lat) *
cos(Dec) ] } (eq.1)

which we could just as well write as-

cos LHA =  [ sin(Ho) - ( sin(Lat.) * sin(Dec) ) ] / [ cos(lat) *
cos(Dec) ]  (eq.2)

where LHA is the angle from noon expressed in degrees, or convertable
to hours, at 15 degrees to the hour.

Sumner doesn't use that formula, but a more convenient one for
logarithmic use, as Bill has suggested, which is-

Vers LHA = [Cos (Lat - Dec) - Sin Ho] / [Cos Lat * Cos Dec] (eq.3)

Sumner describes this as- "Bowditch's third method" for finding the
apparent time. My more modern Bowditch of 1977 refers to that same
formula as used in  20th century methods by Martelli and by Goodwin,
and those are only trivially different in using Cos (zenith angle)
rather than Sin (altitude), which are the same thing.

 When logs are involved, the Versine is particularly useful because it
can never go
negative (and logs get meaningless for negative numbers).

The Versine of an angle is just (1-Cos of that angle). So the Vers of
0 deg is 0.0, that of 90 deg is +1.0, that of 180 deg is +2.0, and
it's always in that range, 0 to +2.  (Readers may have come across it,
slightly modified, in terms of the Haversine, which as its name
suggests, is just half of the Versine, so ranging between 0 and +1.)

Those that delve into a Bowditch of the Sumner era should find the
method described somewhere there.
Frank's recent guidance was-
"For Bowitch's Navigator from 1826, go to google books here:
http://books.google.com/ and enter "Bowditch date:1826-1826" in the
search box.  Click "Search
Books" and it should take you right to it."

If you look at Sumner's working, you will see that his time interval
from Noon, in hours and minutes, is expressed as "log. rising", and I
expect you will find somewhere in that Bowditch a table of
"log.risings" for that very purpose. This is a table of log. Versines,
but modified for the purpose, for reasons which will become clearer.
It tabulates, against hours and minutes rather than degrees, a
quantity log (10,000 Vers LHA)



=========
Here's a short trigonometrical diversion, for those that enjoy such
things, about tweaking one formula into the other.

A standard trig conversion formula for the cos of the difference
between two angles is, as Bill Noyce suggested-
Cos (Lat - Dec) = Sin Lat * Sin Dec + Cos Lat * Cos Dec  (eq.4)

so instead of Sin Lat * Sin Dec in (eq. 2) we can write
Cos (Lat - Dec) - Cos Lat * Cos Dec,  to end up with

Cos LHA = [Sin Ho - Cos (Lat - Dec) + Cos Lat * Cos Dec] / [Cos Lat *
Cos Dec]

or Cos LHA = 1 + [Sin Ho - Cos (Lat - Dec) ] / [Cos Lat * Cos Dec]

or 1- Cos LHA = [Cos (Lat - Dec) - Sin Ho] / [Cos Lat * Cos Dec]

Now (1 - Cos (angle)) is the definition of the Versine of an angle so

Vers LHA = [Cos (Lat - Dec) - Sin Ho] / [Cos Lat * Cos Dec]
and this was (eq.3) above. QED.

End of diversion.
===========================

Now we can start to tackle some of Russell's questions.

He has correctly worked out that the two log sec terms to be added in
were to deal with the denominator, dividing by Cos Lat Cos Dec.
Looking up log sec and adding, was exactly the same as looking up log
cos and subtracting, and subtractions would be avoided wherever
possible.

He asked- "Question 1. Why is value of declination added to value of
latitude at this point?" You can see that the eq.3 asks for Cos
(Lat - Dec), and as Bill points out, North being positive and South
negative, making that subtraction equates to summing those numbers.
Sumner would have done it a bit differently using a "~" symbol, which
means "take the sum if they are in different hemispheres, and the
difference if they are in the same", which amounts to the same thing
as algebraic subtraction. Navigators in those days were never trusted
to subtract a negative
quantity!

Russell's question 2 asked for the source of  "nat. cos 26920", which
is in fact
10,000 times the cosine of the summed angle of 74 deg 23', and
similarly 21076 is 10,000 times the sine of the altitude. The 10,000
is a fiddle-factor, put in  so that when using 5-figure tables, no
decimal
points need to be entered. The final lookup table is similarly fiddled
to compensate.

It was common then to refer to log trig tables such as
log cos simply as cos tables, the log part being understood and
unstated, so when a table really was just a cos as opposed to a log
cos, it would be labelled as "nat. cos", natural cosine.

Next, those two quantities are subtracted, to end up with a difference
of 5844, which would really have been .05844 if that 10,000
multiplication had been omitted. Taking its log gives 3.76671. Then
those log sec figures are added in , to handle the denominator,
Sumner's result being 4.00506.

Next step is a lookup table for the final answer. I haven't looked for
it in an early Bowditch, but presumably there's a table somewhere
labelled "log. rising". This will show, for different values of time
interval from noon (before or after) a value which is actually
log ( 100,000 Vers LHA) or log (100,000 (1 - Cos LHA)) .

So, open the "log. rising" table, search into it for the value
4.00506, and look along the edge for the corresponding time interval,
which turns out to be 1 hour 43 minutes 59 seconds (using the
interpolations provided). Sumner knows it's before noon, so his local
apparent time must be 10h 16m 01s, in the morning. Then he has to
allow for the equation of time and compare with his chronometer to
find the time difference, hence longitude. All standard stuff.

The procedure, above, is one which any navigator of the period, with a
chronometer, would go through once, or a few times, each day, so it
would be a matter of complete routine, that Sumner would whizz through
hardly even thinking about what he was doing. It had been whittled
down to make it as simple and straightforward as possible.

George.

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.


=====================
=====================

Russell Sher wrote-

For those of you who enjoy a bit of maths, perhaps you could take a
look and explain this to me - I hope I've laid out the question
clearly, but if I haven't, let me know and I'll send further info -
thanks.

I am currently reading the book 'Line of position navigation'  which
discusses Sumner's method of calculating a line of position. (I think
several list members mentioned the book previously). The book includes
actual workings of Sumner's calculations. I worked through some of the
examples but cannot quite understand Sumners application of logarithms
as used here.  I can use a calculator to get the result using the
formula, but am trying to use his method with logarithms. Perhaps some
of you can shed some light on this...

The formula to solve would seem to be the familiar  LHA = arccos{ [
sin(Ho) - ( sin(Lat.) * sin(Dec) ) ] / [ cos(lat) * cos(Dec) ] }
(Ho here is the corrected altitude of the body) (remember to enter
South declination as a negative quantity if you want to confirm the
workings)

I agreed with the result obtained by using a calculator, but I would
like to understand the use of logarithms. (Obviously since calculators
weren't around then, logs were used).

In the book in appendix A, page 15, the following figures are given:

Latitude 51 degr. N,  Decl. 23 degr. 23' South
Sun's altitude (with all corrections applied) 12 degr. 10'

The solution begins as:
Latitude 51 degr. N ---  sec.  0.20113  --I'll call this [value 1]
Dec.  23 degr. 23' S.  -- sec 0.03722  -- I'll call this [value 2]

I managed to work that the reference above to 'sec' means the natural
logarithm of the secant of each of the angles. (remember that
secant(x)  = 1/cos(x) , then take the logarithm of the secant to get
each of the two above values)

Now I understand that since the denominator of the formula for LHA
given above is [cos(lat) * cos (Dec)], the two 'sec' values when added
would appear to be equivalent of this denominator (since adding logs
is equivalent to multiplying the original numbers) and since sec is
1/cos, this forms the denominator - correct?

Now I get lost...

Then Sumner adds latitude  51 degr. +  Decl. 23 degr. 23'  - he
writes:
Sum:  74 degr. 23'        Nat. cos 26920 --I'll call this [value 3]
Alt. 12 degr. 10'            Nat.sin 21076  -- I'll call this [value
4]

Question 1. Why is value of declination added to the value of latitude
at this point?

continuing...
I calculate the cosine of 74 degr. 23' = 0.269199
and the sine of 12 degr. 10' = 0.2107561

Question 2. So where do [value 3] and [value 4] come from? - they
appear to be a scaled up (by 100000) equivalent of the the two values
just above for cos(lat. + dec) and sine(altitude). (Ok his values are
rounded off - so there are slight differences.). I assume Nat. cos and
Nat. sin means natural cosine and natural sine (as opposed to log. cos
or log. sin ??)

Then sumner subtracts: [value 3] - [value 4] = 5844 and then he takes
the logarithm of this:  log(5844) = 3.6671 - call this [value 5]

Then sumner takes [value 1] + [value 2] + [value 5] = 4.00506. and he
writes:
1 hour 43 min 59 sec from noon = log rising = 4.00506   (the value
just calculated).

(now, if I use a calculator I obtain LHA = 25 degr. 59 min which
agrees (in time) with 1 hour 43 min 59 sec) -- using 15 degr. = 1
hour)

Question 3. But how does Sumner associate 1 hour 43 min 59 sec. with
log 4.00506 ??
Question 4. Although I agree with his final result from the formula,
how does it fit together to give the correct answer from the use of
logs as in the example?


regards
Russell

===========================

and Bill Noyce responded-

Good questions!
Although the formulas we use with calculators are relatively
straightforward,
they're awkward iv you have to compute them with log tables.  For
example,
to evaluate
  LHA = arccos{ [ sin(Ho) - ( sin(Lat.) * sin(Dec) ) ] / [ cos(lat) *
cos(Dec) ] }

you would start by looking up log(sin(Lat)) & log(sin(Dec)) and adding
them.
Next you would need to find the antilog of the result, add sin(Ho),
and take
the log of the result -- accumulating roundoff error along the way
too.

There were numerous formulas devised to avoid coming "out of logs" --
or
equivalently, expressing the formula as a product.  I assume the
formula
Sumner used was the standard one in his day for reducing a time sight.
Let's see how it works.

As you point out, log(sec(Lat))+log(sec(Dec)) forms the denominator of
the
formula -- no trouble there.

From the angle-addition formulas we memorized in high school,
cos(Lat+Dec) = cos(Lat)*cos(Dec) - sin(Lat)*sin(Dec)
so your [value 3] - [value 4] comes out to
100000*{ cos(Lat)*cos(Dec) - sin(Lat)*sin(Dec) - sin(Ho) }

Aside from the extra factor of 100000, and the presense of
cos(Lat)*cos(Dec),
this looks like the numerator of our present-day formula.  (Recall
that Lat
and Dec are in opposite hemispheres, so our modern approach would
make one of them negative, so the product of their sines would also be
negative.)

Finally we use logs to multiply this by the product of the secants,
ending
up with
5 + log( [cos(Lat)cos(Dec) - sin(Lat)sin(Dec) -
sin(Ho)]/[cos(Lat)cos(Dec)] )
which equals
5 + log( 1- [sin(Lat)sin(Dec) + sin(Ho)]/[cos(Lat)cos(Dec)] )

This can now be looked up in a "log rising" table that lists
LHA against 5 + log(1+cos(LHA)).

I apologize in advance for any blunders above, but I hope this gives
the
flavor of the approach.  I'm sure the derivation of the formula is
discussed
in Bowditch, for example.

    -- Bill

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