NavList:

   

Reply

Red:

If you look at Chapter 24 of Bowditch as mentioned in my previous post,
the most accurate sailing is a Mercator sailing.

Mercator Sailings rely on Table 6 of Bowditch, Meridional Parts.  This
massive table lists, for each minute of latitude all the way from the
equator to the poles how far north of the equator each line of latitude
should be drawn on a Mercator map/chart.  Intermediate charts (say, one
spanning from 41 d to 42 d) can be drawn by assuming the bottom of the
chart is at whatever distance is given by Meridional Parts for 41 d and
then by taking the differences between the MP for 41 d and successive
latitudes to get the separation between parallels going up from there.

These tables are interesting:  One would think the Meridional Part
between the equator and one degree north or south be exactly 60.0 or
maybe even a bit more (remember, Meridional Parts give the distance from
the equator for any given latitude on a Mercator chart in terms of the
size of a minute of longitude).  Surprise, the Meridional Part for a
latitude of exactly 1 degree is 59.6!!   This says near the equator the
distance between parallels is actually *less* than the distance between
meridians!

The explanation for this is easy:  the Meridional Parts table takes into
account the slightly squashed shape of the earth.

Having said that, I'll remind you (and everyone else on the list) that
for the typical accuracy involved in navigational calculations, assuming
the earth is a perfect sphere is perfectly acceptable.   Remember that
the out-of-sphericalness of the earth is on the order of 1/4%.   In
practical navigation there are sources of error way larger than 1/4%

Lu Abel

Red wrote:
> Pierre, is that ignoring the fact that the diameter of the earth will be reduced
> as we move from the equator to the poles. so that "degrees east" and "miles
> east" will not have the same value as latitude changes?
>
> Is that "good enough" for DR, or is there a formula that would also compensate
> (for both lat/lon) as latitude changed?
>
> ----- Original Message -----
> From: "Pierre Boucher" 
> To: 
> Sent: Saturday, June 24, 2006 10:47 AM
> Subject: Re: I knew where we were, but where are we now?
>
>
>
>>40 deg 07 min N -> 40.12 (rounded)
>>44 deg 37 min W -> 44.62 (rounded) not needed
>>Course 090 -> Course N 90 E
>>distance traveled 56.88 Nm
>>
>>To find DLat (difference of latitude in minutes of latitude)
>>distance traveled x cos(course) = Dlat
>>56.88 x cos 90 = 0
>>
>>To find DLo (difference of longitude in minutes of longitude)
>>distance travel / cos(mid latitude)
>>56.88 / cos 40.12 = 74.4 = 1 deg 14.4 min (Eastward)
>>
>>New DR?
>>New latitude
>>40 deg 07 min N + 0 = 40 deg 07 min N
>>New longitude
>>44 deg 37 min W - 1 deg 14.4 min = 43 deg 22.6 min W
>>
>>Pierre Boucher
>>
>>Guy Schwartz a ?crit :
>>
>>>This has probably been discussed here before, but hopefully someone will
>
> help me out. I'm trying to find the formula for caculating your DR after
> traveling from a know position.
>
>>>For instance:
>>>Last Fix
>>>40 deg 07 min N
>>>44 deg 37 min W
>>>Course 090
>>>distance traveled 56.88 Nm
>>>New DR?
>>>It would be a big help to me if someone that knows the formula could
>
> incorparate the above example so I could follow along.
>
>>>Thank you,
>>>Guy
>>
>>
>>--
>>No virus found in this incoming message.
>>Checked by AVG Free Edition.
>>Version: 7.1.394 / Virus Database: 268.9.3/374 - Release Date: 6/23/2006
>>
>>
>
>
>

   

Reply