# NavList:

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

**Re: 360 degree slide rule trig**

**From:**Paul Hirose

**Date:**2017 Feb 08, 22:44 -0800

Last year I posted two messages on slide rule trig beyond the marked range of the scales. http://fer3.com/arc/m2.aspx/360-degree-slide-rule-trig-Hirose-nov-2016-g37154 http://fer3.com/arc/m2.aspx/360-degree-slide-rule-trig-Hirose-nov-2016-g37223 Finally I've gotten around to writing the final part. Previous messages explained how to get 360° of sine, cosine, and tangent on a slide rule. Now I'll go into the coordinate transformations between rectangular (easting and northing) and polar (course and distance). With these transformations a slide rule can solve plane and Mercator sailing. It helps if you remember my suggestion to associate sine with easting and cosine with northing. Example: 50 miles are made good on true course 070. Then northing = 50 * cos 70 and easting = 50 * sin 70. Set right C index to 50 miles on D. Set cursor to cos 70 (red 70) on S. Read 17.1 miles northing on D. Set cursor to sin 70 (black 70) on S. Read 47.0 miles easting on D. The corresponding latitude difference is simply the northing: 17.1 minutes. To get the difference in longitude, divide easting by the cosine of latitude. With the cursor at 47.0 on D, suppose latitude is 30°. Then set cos 30 (red 30) to the hairline and read longitude difference 54.3 minutes on D at the C index. Example: 88 miles made good on true course 265 from latitude 40. Set right C index to 88 miles on D. To find cos 265 on scale S, use the back and forth method described earlier. Zero is at the right end of S, 90 at the left, 180 at the right. Proceeding left from there, find 260 at black 10. Exit S at the left, enter ST at the right, and find 265 at black 5. Set cursor there. Sines and cosines from ST are in the range .001 to .01, so 88 miles * cos 265 = -7.7 miles northing. For easting we need sin 265. Zero is at the right end of S, 90 at the right, 180 at the left, and 265 is 5° from the right end, at black 85. Set cursor there, then set cos latitude (red 40) to the cursor. Read -114.5 minutes easting on D at the C index. The opposite problem is the determination of course and distance when easting and northing are known. For this conversion from rectangular to polar coordinates: 1. set C index to longitude difference on D 2. set cursor to cos latitude (red numbers) on S 3. set C index to northing on D 4. read course on black T 5. set course on black S to cursor 6. read distance at C index on D Explanation: steps 1 and 2 convert longitude difference to miles of easting. Step 3 divides that by northing. The quotient is on C. Step 4 takes the arc tan of the quotient to yield course. Step 5 divides easting by sin course to obtain distance. Example: from latitude 40° the destination is 50′ north and 30′ east. What are the course and distance? After step 2, easting (23 miles) is on C at the cursor. At step 4, course = 24.7. At step 6, distance = 55 miles. That algorithm fails when easting exceeds northing, since the tangent of course exceeds 1 and therefore exceeds the range of a single T scale. The solution is to exchange easting and northing. Now the quotient is a cotangent, which is readable on red T. The first two steps are unchanged, but in the third step easting becomes a divisor instead of dividend: 1. set C index to longitude difference on D 2. set cursor to cos latitude (red numbers) on S 3. set C index to cursor 4. set cursor to northing on D 5. read course on red T 6. set course on red S to cursor 7. read distance at C index on D Example: from latitude 40° the destination is 30′ north and 50′ east. What are the course and distance? After step 2, easting (38.3 miles) is at the cursor on D. Read course 51.9 at step 5. Distance is 48.5 miles. The final setting (to find course) is on S, at either a black or red number. To avoid confusion, it helps to remember my suggestion to associate sine with easting and cosine with northing. For instance, in step 4 above, the cursor is set to northing, and so in step 6 a cosine is set opposite that value. The topic is 360 degrees of trig, so for a final problem we'll compute a course outside the 0 - 90 range. From latitude is 20° the destination is 5.5° south and 3.5° west. Easting = 3.5 * cos 20 = 3.28 great circle degrees. Evidently the course is between 180 and 225, so when reading it on T, think of the 10° graduation as 190°, 20° as 200°, etc. Course = 211. Divide 3.28° easting by sin 211 (= sin 31) to get distance 6.40°, or 385 miles.