Sextant TheoremDate: 05/13/2001 at 21:55:23 From: Yolanda Subject: Mathematical theorem behind using a sextant. The sextant is in essence a machine used to measure the distance between a celestial body and the horizon, and can be used for varying the angle between two parallel mirrors by using the Principle of Double Reflectivity and the phenomenon that the angle of the departing light ray will have been changed by double the angle between the mirrors. For example, if a body is 70 degrees above the horizon, the parallel mirrors of the sextant must be displaced by 35 degrees in order to send its reflected light rays straight through the optics. There is a mathematical theorem that proves this. What is it, and what is its proof? Thank you. Date: 05/15/2001 at 12:58:34 From: Doctor Rick Subject: Re: Mathematical theorem behind using a sextant. Hi, Yolanda. The first thing you need is a figure. I made one using the Geometer's Sketchpad program: The path of light from the celestial body to the eye is shown in red: ABCD. The two mirrors are shown in green. The celestial body is on the continuation of BA upward. The horizon is on the continuation of DC to the left. The bottom mirror is half-silvered, or silvered on one half, so the horizon can be seen through the mirror. The upper mirror is rotated until BA is in line with the star. The normals (perpendicular lines) to the mirrors are BE and CF. If the two mirrors were parallel, then the normals would be parallel. The deviation from parallel is measured by the angle BGC between the normals. It is this angle that is measured by the sextant (though not at point G). When light is reflected in a mirror, the "angle of incidence" equals the "angle of reflection." These angles are the angles that the entering and reflected rays make with the normal to the mirror. The theorem you seek can now be stated: If angle ABE equals angle EBC, and angle BCF equals angle FCD, then angle BHC (between the line of sight to the celestial body and the line of sight to the horizon) is twice angle BGC (between the normals to the mirrors). Once all this preparatory work has been done, the geometrical proof is simple. Angle EBC is an exterior angle of triangle BGC, and thus its measure is equal to the sum of the measures of angle BGC and angle BCG. Therefore (1) m<BGC = m<EBC - m<BCG Turning our attention to triangle BCH, angle ABC is an exterior angle, and thus its measure is equal to the sum of the measures of angle BCH and angle BHC. Therefore (2) m<BHC = m<ABC - m<BCH But (3) m<ABC = m<ABE + m<EBC = m<EBC + m<EBC = 2*m<EBC and (4) m<BCH = m<BCF + m<FCH = m<BCF + m<BCF = 2*m<BCF = 2*m<BCG Therefore, substituting (3) and (4) in equation (2), we have (5) m<BHC = 2*m<EBC - 2*m<BCG = 2*(m<EBC - m<BCG) = 2*m<BGC using equation (1) for the last step. Quod erat demonstrandum (Latin for "which is what we intended to prove"). - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/