NavList:

   

Reply

Joel asked this-


| Thomas Sumner in his "New and Accurate Method of Finding a Ship's
| Position at Sea" (1843) states repeatedly that there is but one proper
| moment in the day when a single observation can determine the Latitude
| of the ship, unless the apparent time at the ship is accurately
| known.   And when the Latitude is uncertain, ther are only two proper
| instants per day (when the sun bears exactly East or exactly West)
| when the solar altitude can be used to find the Longitude by
| Chronometer with accuracy.  If one makes an observation other than at
| those times, he stresses, the unavoidable errors in measurement may
| cause frequent errors which are very great.  He also mentions more
| than once that in his latitudes, the sun in not observable in the
| proper E and W points for seven months of the year (taking into
| account that the altitude must exceed 6 or 7 degrees to be reliable.
|
| My question is :   what are the crucial issues around the sun being on
| the prime vertical (that is, either due E or due W) when his (her?)
| altitude is taken?

Because at or near that moment, from measuring altitude, you can get
longitude without any precise knowledge of latitude.

   Am I correct in thinking that the hour angle
| between the observer's meridian and the sun's meridian would be
| exactly 90 degrees when the sun bears true E or W ?

No. At midsummer, the Sun is due East much later than 6 am, and due West
much earlier than 6pm, if that answer corresponds to the question that Joel
is asking here. Only at the equinoxes would Joel's conjecture be true, and
then the Sun would be at zero altitude.

Are we dealing
| with a Pole-Sun-Zenith triangle with a right angle at the Zenith?

Yes, that's the case, when the Sun is on the "prime vertical", bearing
exactly East or West
|
| If so, could we not calculate not only the hour angle (at P) but also
| the Latitude?

No. See the following argument.

{  sin(D) = sin(L) sin(h)    would give us L and
| cos(D) sin(HA) = cos(h)  would give us HA , where D is the
| declination, L the latitude, h the altitude, and HA the hour angle .}

Both of those equations apply, when the azimuth angle Az, the bearing of the
Sun, is exactly 90 (or 270) degrees. But they behave very differently, near
that angle, in a way that we need to think about.

Start with the second expression, copying Joel's notation-
cos(D) sin (HA) = cos(h)  or, same thing, sin (HA) = cos(h) / cos(D)

Actually, that's a special case, valid at Az = 90, of the more general
equation, familiar to us as the equation we often use to get azimuth,

sin(Az) = cos(D) sin (HA) / cos(h)   which we can rewrite as-

sin (HA) = cos(h) sin(Az) / cos(D), which applies at all azimuths.

But note here that because sin(Az) departs very little from 1, over a wide
range of azimuths centred around 90, we can readily get a precise result for
HA from Joel's simpler expression, which takes sin(Az) to be exactly 1, even
if we haven't tried too hard to find the exact moment when Az = 90. And it's
not easy to predict that exact moment, because to do so requires a precise
knowledge of our position, which is what we're trying to discover.

Go back now to Joel's sin(D) = sin(L) sin(h) from which he proposes to
derive L

Again, that's a special case of the more general expression-

sin (D) = sin(L) sin(h) + cos(L) cos(h) cos (Az) which, when Az happens to
be 90, we can simplify to

sin(D) = sin(L) sin (h), just as Joel gave it.

However, if Az differs at all from 90, that simplification doesn't apply,
because the term we have taken as zero varies in the same way that a cosine
does, changing rapidly on either side of 90. And how can we tell, for that
purpose, that we are EXACTLY on the prime-vertical, with Az = 90 degrees
precisely? We can't, because we don't yet know our latitude! It's one of
those circular problems.

So the end result is that from an observation on, or near, the prime
vertical, we can get a good value for hour-angle, and knowing Sun GHA, we
can get our longitude, independent of any guess at our latitude. Latitude
isn't available from that observation, however, and it has to wait until we
can make a noon observation.

| Or is it more a question of sensitivity to errors in Latitude when the
| zenith angle is close to 90 degrees?

Yes, that's exactly the point at issue. It's actually the INsensitivity of
the time-sight to any assumption about latitude..

Did people calculate longitude
| when the sun was on the prime vertical without using any asumptions as
| to latitude,

Yes, that's the point of it. When the Sun is on or near the prime vertical,
its altitude depends only on its hour angle and declination, and is exactly
the same whatever the observer's latitude. So, knowing declination, and
measuring altitude, gives hour angle, and then from the chronometer,
longitude. The latitude didn't need to be known.

or did they always (never say "always" or "never") use an
| oblique triangle together with their best estimate or guess for
| latitude?

Well, outside of Summer, for those 7 months or so there was no way you could
observe the Sun on the prime meridian. You could for stars, those that
happened to bear East or West around dawn or dusk. For the Sun, you would
measure a Sun altitude in the morning, then, if possible, at apparent noon,
for latitude, then in the evening. Dead reckoning had to be used to obtain
morning and evening latitude from the travel since the noon observation, in
order to put it into the morning or evening calculations for longitude. If
only a few hours intervened, that was acceptable. But if it was cloudy at
that noon, you would have to estimate latitude based on a noon altitude a
day or perhaps two, before or after.

The resulting DR errors were what Sumner was pointing to, and trying to
avoid. Navigators, if they failed to get that noon sight, would frequently
reject that day's morning or evening observations as unworkable. Sumner
showed how they could be used to derive a position line, though he made
rather heavy weather of doing so.

Although it's only when the Sun is exactly on the prime vertical that the
calculated long is entirely independent of the lat., as long as you were
somewhere near that moment, the assumed lat. makes very little difference.
So in that case, even a rough DR for lat would do. There's no call to take
great care to choose the moment precisely, then, though it makes for a
slightly more complex calculation. Indeed, it's difficult to predict that
moment in advance, in that doing so calls for a knowledge of your position,
the very thing you are trying to find.

What Sumner was pointing out was that there was no need to deduce either
latitude or longitude, but it would suffice to draw a single oblique
position line from each observation, just as we do today.

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.


--~--~---------~--~----~------------~-------~--~----~
To post to this group, send email to NavList@fer3.com
To , send email to NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---

   

Reply