{"id":373,"date":"2020-10-22T14:39:26","date_gmt":"2020-10-22T14:39:26","guid":{"rendered":"https:\/\/maritimesa.org\/nautical-science-grade-11\/?p=373"},"modified":"2020-10-22T14:39:26","modified_gmt":"2020-10-22T14:39:26","slug":"finding-the-course-to-steer-in-order-to-counteract-the-current","status":"publish","type":"post","link":"https:\/\/maritimesa.org\/nautical-science-grade-11\/2020\/10\/22\/finding-the-course-to-steer-in-order-to-counteract-the-current\/","title":{"rendered":"Finding the course to steer in order to counteract the current"},"content":{"rendered":"<p>The following example is used to explain the procedure to follow:<\/p>\n<p><strong>Problem:<\/strong><br \/>\nA vessel steaming at 10 knots wishes to make good a course of 110\u00b0 (T) from its 1200 position. Find the course to steer allowing for a current setting 170\u00b0 (T) at 2 knots.<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<ol>\n<li>Plot the 1200 position on the chart.<\/li>\n<li>Lay off the course to make good from this position.<\/li>\n<li>Plot the current vector, 170\u00b0 (T) 2 miles, from the 1200 position.<\/li>\n<li>Using the end of the current vector as the centre (A), find out where a distance of 10 miles will cut the course made good (B).<\/li>\n<li>AB will be the course to steer from the 1200 fix to the B position.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-111\" src=\"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-content\/uploads\/sites\/6\/2020\/10\/11.1.8.7_fig_1.jpg\" alt=\"\" width=\"900\" height=\"501\" srcset=\"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-content\/uploads\/sites\/6\/2020\/10\/11.1.8.7_fig_1.jpg 900w, https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-content\/uploads\/sites\/6\/2020\/10\/11.1.8.7_fig_1-300x167.jpg 300w, https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-content\/uploads\/sites\/6\/2020\/10\/11.1.8.7_fig_1-768x428.jpg 768w, https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-content\/uploads\/sites\/6\/2020\/10\/11.1.8.7_fig_1-260x146.jpg 260w, https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-content\/uploads\/sites\/6\/2020\/10\/11.1.8.7_fig_1-50x28.jpg 50w, https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-content\/uploads\/sites\/6\/2020\/10\/11.1.8.7_fig_1-135x75.jpg 135w\" sizes=\"(max-width: 900px) 100vw, 900px\" \/><br \/>\n<strong>Summary:<\/strong><\/p>\n<ul>\n<li>The effective speed or speed made good (line BC) must be used to calculate ETA\u2019s.<\/li>\n<li>The course steered must be used to convert abeam and relative bearings to true bearings.<\/li>\n<li>A more accurate result will be obtained if the ship\u2019s speed and current had both been laid off for a longer period eg. 2 hours.  If it were done, the length BC would have to be divided by two to give the speed made good.  For most purposes, a one hour triangle is sufficient and less likely to lead to errors, unless the triangle is very small.<\/li>\n<li>When finding the course made good, when the course steered is given, the current is laid off at the end (the DR POSITION).<\/li>\n<li>When finding the course to steer, when the course made good is given, the current is laid off at the beginning (the OBSERVED POSITION).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>The following example is used to explain the procedure to follow: Problem: A vessel steaming at 10 knots wishes to make good a course of 110\u00b0 (T) from its 1200 position. Find the course to steer allowing for a current setting 170\u00b0 (T) at 2 knots. Solution: Plot the 1200 [&hellip;]<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4,64,71],"tags":[],"class_list":["post-373","post","type-post","status-publish","format-standard","hentry","category-11-1-navigation","category-11-1-8-currents-and-tidal-streams","category-11-1-8-7-finding-the-course-to-steer-to-counteract-a-current"],"_links":{"self":[{"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/posts\/373"}],"collection":[{"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/comments?post=373"}],"version-history":[{"count":1,"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/posts\/373\/revisions"}],"predecessor-version":[{"id":374,"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/posts\/373\/revisions\/374"}],"wp:attachment":[{"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/media?parent=373"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/categories?post=373"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/maritimesa.org\/nautical-science-grade-11\/wp-json\/wp\/v2\/tags?post=373"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}