# NavList:

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

**Re: Sextant "tilt" error**

**From:**Frank Reed

**Date:**2011 Oct 15, 10:09 -0700

Of the error that results from not getting the sextant exactly vertical, I wrote previously: "it depends on the details."

Bill Morris, you replied:

"Tilt error depends on the observed angle as well as the amount of tilt"

Yes, it does. It also depends on how we define "tilt". You mentioned a correction table on a "ball-recording octant" but since those are rather exotic, let's consider the case of a normal reflecting instrument (a standard nautical sextant).

First, what is tilt? This depends on the method used for swinging the arc. Let's stick with the "better" method: you keep the Sun or star centered in the field of view and rotate the instrument so that the horizon drops away beneath it on either side as seen through the instrument when you rock it left and right. This angle results from rotation about the axis along the line of sight to the celestial body. Call that angle A. The actual altitude of the body is h. The incorrect altitude measured to a point not directly underneath the body is the correct altitude plus some error, call it h+dh. We can draw a right-angle spherical triangle. The Sun or star is at the peak of the triangle. The triangle has two long sides of lengths h and h+dh extending down to the horizon. The angle where the side of length h intersects the horizon is a right angle. Using a standard formula for a right spherical triangle, we have:

cos(A) = tan(h)/tan(h+dh)

If you assume A has some value, let's say, two degrees, you can then solve for dh for a sequence of heights h. Better yet, we can do a small angle approximation and derive a nice handy equation that covers all practical cases. If you assume that A and dh are both small, then you will find:

dh = (0.26')*A^2*sin(2h)

where A is in degrees (and so there's no confusion, the notation "A^2" means "A squared"). The result, dh, is in minutes of arc. The variability with altitude, h, is simple enough: it starts out near zero close to the horizon, climbs to a maximum at 45 degrees, then falls back to zero again at the zenith. If A happens to be two degrees, then the maximum error, dh, is just over one minute of arc.

So does this math mean that we have to work extra hard to get the sextant vertical when observing altitudes near 45 degrees?? Not necessarily. The process of swinging the arc tells you to rock the sextant until you can see no visible gap or overlap between the celestial object and the horizon. That visibility is the same at all altitudes. If you take enough care to spot a quarter of a minute gap (and if you have the magnification for it), then you will see it and eliminate it no matter how it varies with altitude. In addition, since the required motion varies from rolling the sextant about an axis nearly aligned with the horizon, when the object is low in the sky, to yawing it about a nearly vertical axis, when the object is high in the sky, the physical process is quite different so the math only tells part of the story. In other words, there are human issues --for a sextant-wielding robot, the math would be the whole story :).

-FER

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