# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: A slope example**

**From:**Gary LaPook

**Date:**2010 Dec 01, 02:00 -0800

Peter Fogg wrote: "Yep, you're right. There is another way to calculate it: Delta H = 15 x cos latitude x sin azimuth Where Delta H = rate of change in arc minute per minute of time" ________________________________________________________________________ Yep, that's the formula I posted on July 24, 2008, see: http://www.fer3.com/arc/m2.aspx?i=105929&y=200807 But, if you really wanted to be persnickety, when doing stars you should start with 15.042 as that is the rate for Aries, 15 is the rate for the Sun and planets but for practical navigation and short time periods no significant error is introduced by using 15 for both stars and objects in the solar system. gl On 12/1/2010 12:58 AM, Peter Fogg wrote: > Gary LaPook wrote: >> I get the calculated slope for Vega for the movement of the body (change in LHA) for the five minute period from 1628 to 1633 to be -58.6' doing the sight reduction and -58.35' by using the MOB table from H.O. 249. This table show -11.7' per minute of time at latitude 35� and azimuth 288� and -11.5' at latitude 35� and azimuth 290�. For latitude 30� the values are -12.4' and -12.2' respectively. Doing some simple interpolation makes the slope for the movement of the body -11.67' per minute of time and -58.35' for the 5 minute time period. > Yep, you're right. There is another way to calculate it: > Delta H = 15 x cos latitude x sin azimuth > Where Delta H = rate of change in arc minute per minute of time > > Using that formula (x5) gives - 58.3 > >> The adjustment for the movement of the ship is plus 1.1' minutes. If you use my table you can take the value for 7 knots and relative bearing of 20� (ZN-TR, 289-270 = 19, ~ 20) which is .5' and double it to find the correction to be + 1.0'. You can also do the same thing using the MOO table from H.O. 249 which shows + 11.05' (interpolating) for 700 knots and relative bearing of 19�. Divide by 100 to get the adjustment for 7 knots, multiply by 2 to get it for 14 knots and then multiply by 5 minutes to get the adjustment of + 1.1'. (You get the same value using a calculator, 14 knots divided by 60 minutes gives a ship movement of .23 NM per minute, times 5 minutes makes 1.17 NM times the cosine of the relative bearing, 19�, makes the correction of + 1.1' also.) > Thanks for this explanation. So applying this correction results in a > corrected slope of - 57.2 ? > > > >