# NavList:

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

Message:αβγ
Message:abc
 Add Images & Files Posting Code: Name: Email:
Re: Bygrave sight reduction by slide rule
From: Paul Hirose
Date: 2018 Jan 7, 20:40 -0800

```For a very short distance (high altitude) test of the Bygrave formulas
on a slide rule, solve for the separation and position angle of two
components of a double star. To make the test more extreme, let the star
be Polaris, which is a triple system. Coordinates of the two main
components from Wikipedia:

02h 31m 49.09s  +89° 15′ 50.8″  α UMi Aa
02h 30m 41.63s  +89° 15′ 38.1″  α UMi B

If Aa (the star visible to the naked eye) is the "observer" and B is the
"celestial body", then

89.26411 lat
89.26058 dec
0.28108° LHA

W = arctan (tan 89.26058 / cos .28108)
W = 89.26058

At slide rule accuracy the cosine of such a small angle is virtually 1,
so W equals the declination.

X = 90 - 89.26411 + 89.26058
X = 89.99647

X is in the range 0 - 180, so "altitude" is positive.

Z = arc tan(tan 0.28108 * cos 89.26058 / cos 89.99647)

That formula is a good lesson on techniques to extend the range of the
ST scale.

1. Tan 0.28108 is not on ST, so set it as tan 2.81.

2. Cos 89.26058 (red 89.26058) may be set directly. Since red 89 = black
1, begin at black 1 and move left .26°. (A few rules include the red
numbers on ST. But usually they're omitted since the numbers are obvious.)

3. The method in step 2 won't work for cos 89.99647. Therefore, delete
nines to get a value you can set: 86.47. Begin at black 4 (= red 86) and
move left .47° to get cos 86.47.

Though it's not immediately obvious, the same principle is at work in
step 1 and step 3. I.e., cos 89.99647 (red 89.99647) is equivalent to
black .00353, which is outside the range of ST. By shifting the decimal
point three places you get black 3.53, alias red 86.47, which is on the
scale. This flexibility with the decimal point is possible on ST because
the sine or tangent of a small angle is practically proportional to the
angle. That's not true on the S and T scales!

If the angles were literally equal to the numbers on ST, the
characteristic of the tangent would be -2, i.e., the result is between
.01 and .1. But if you include the decimal point adjustments in steps 1
and 3, the characteristic is 0 and so the arc tan is read on T. With the
slide in its final position (index at the tangent on D), set cursor to
the D index.

Z = 134.26° on T at the cursor. Correct angle is 134.13.

Hc = arc tan (cos 134.26 * tan 89.99647)
Hc = 89.99494

The cosine can be set directly. Change the multiplication to division by
cot 89.99647, which you set as red 86.47 as in the Z computation. Read
84.94, adjust to 89.99494 to account for the deleted 9s. Separation
angle is the complement of that angle: .00506°. Correct angle is .005054
(18.19″) and error is .02″.

It wasn't easy, but a 10 inch slide rule and the Bygrave formulas
performed well in this test.
```
Browse Files

Drop Files

### Join NavList

 Name: (please, no nicknames or handles) Email:
 Do you want to receive all group messages by email? Yes No
You can also join by posting. Your first on-topic post automatically makes you a member.

### Posting Code

Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.
 Email:

### Email Settings

 Posting Code:

### Custom Index

 Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd)