Description and Use of a Diagram of Navigation: By which All Problems in Plane, Traverse, Parallel, Middle Latitude and Mercator's Sailing May be Instantly and Accurately Resolved. Adapted to the Capacity of All who Know the Use of Figures. Designed as an Easy Introduction, by Sensible Demonstration, to the Principles, and the Practice of Navigation. With a Diagram EngravedFellowes & Simpson, 1822 - 4 σελίδες |
Αναζήτηση στο βιβλίο
Αποτελέσματα 1 - 5 από τα 15.
Σελίδα 5
... solutions are derived . The projection of the tri- angle given by the theorem , being readily formed by the Diagram , if it be well examined in connection with the reading of the solution , the learner will soon posses , the idea of the ...
... solutions are derived . The projection of the tri- angle given by the theorem , being readily formed by the Diagram , if it be well examined in connection with the reading of the solution , the learner will soon posses , the idea of the ...
Σελίδα 11
... solution of all problems in navigation relating to plane , traverse , parallel , middle latitude , and Merca- tor's sailing . 20. For this purpose the arc represents a quarter section of the HORIZON , one side the MERIDIAN , and the ...
... solution of all problems in navigation relating to plane , traverse , parallel , middle latitude , and Merca- tor's sailing . 20. For this purpose the arc represents a quarter section of the HORIZON , one side the MERIDIAN , and the ...
Σελίδα 39
... solutions for calculating the usual prob- lems of navigation by logarithms , which will be illus- trated by examples ; and the usual method of making up the reckoning by inspection from the calculated ta- bles will be shewn ...
... solutions for calculating the usual prob- lems of navigation by logarithms , which will be illus- trated by examples ; and the usual method of making up the reckoning by inspection from the calculated ta- bles will be shewn ...
Σελίδα 50
... solutions for calculating the problems in navigation , by logarithms . 222. Theorem . - In a right - angled plane triangle , if one side as radius be equal to the distance sailed , and a contiguous angle be equal to the course , the ...
... solutions for calculating the problems in navigation , by logarithms . 222. Theorem . - In a right - angled plane triangle , if one side as radius be equal to the distance sailed , and a contiguous angle be equal to the course , the ...
Σελίδα 51
... solutions in the 2d case of plane sailing : As rad . is to diff . lat . so is the secant of the course to the distance ; and as rad . is to diff . lat . so is the tan- gent of course to the departure . Example . Given the diff . lat ...
... solutions in the 2d case of plane sailing : As rad . is to diff . lat . so is the secant of the course to the distance ; and as rad . is to diff . lat . so is the tan- gent of course to the departure . Example . Given the diff . lat ...
Άλλες εκδόσεις - Προβολή όλων
Συχνά εμφανιζόμενοι όροι και φράσεις
arc of latitude arc or angle chord circle circumference co-sine co-tangent contiguous angle course 3 points course 3 pts course 38 deg COURSE and DEPARTURE Course and Diff course and distance course sailed curve line departure 84 Diagram of Navigation difference of lati difference of latitude difference of longitude direct course dist DISTANCE and DEPARTURE distance sailed ence of longitude Example Example.-A ship sails find the COURSE find the DISTANCE give the course give the diff give the distance Given both LATITUDES Given-one LATITUDE index remaining index set index to meet meet the difference Mercator's sailing merid meridian distance meridional difference middle lat parallel sailing plane sailing proper diff radius be equal required the course right line drawn right-angled plane triangle Rule Rule.-Set the index secant Set the index ship in lat side as radius side marked equator tance tangent traverse table tude