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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.
```
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