A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: dip, dip short, distance off with buildings, etc.
From: Frank Reed CT
Date: 2006 Jan 15, 21:22 EST
From: Frank Reed CT
Date: 2006 Jan 15, 21:22 EST
Bill you wrote: "OK with treating ABCA as a right triangle given the ratio of R to H and h, but semi-clearly an oblique triangle in almost all cases once the horizon "dips" due to height of eye/curvature/refraction and B is raised above the horizon." The triangle isn't a right triangle (and you don't need to assume it is), although it will be nearly so in many practical cases. But let's continue with the development of the equation behind Table 15. To recap, you have point C at the center of the Earth, point A at height h above the Earth's surface, and point B some distance away at height H above the Earth's surface.Those three points make a big triangle ABC. Our GOAL is to find the angle in the triangle at point C (since, if we multiply that by the Earth's radius, we get the distance). Now what do we measure with our sextant or theodolite? Well, we're at point A, and we measure altitudes above the horizontal which is, by definition, the plane perpendicular to side AC of the triangle. Now how is that related to the angle at point A in the big triangle (the angle "CAB")? Clearly that's just 90 degrees plus the measured altitude. So we KNOW the angle CAB. This is equivalent to the measured parameter (or at minimum, it's related to the measured parameter by a simple relationship). So we know the corner angle at point A (the observer), which I named "gamma", and we are seeking the corner angle at point C (the center of the Earth), which I named "phi". But, uh-oh, we're stuck with another angle --the one at point B. Wait... no we're not. Since it's a simple plane triangle, all three angles must add up to 180 degrees. That means that the angle in the big triangle at point B is NECESSARILY equal to 180-(gamma+phi). Ok so far?? If you haven't drawn a picture of this yet, you can't possibly be ok here
so please make sure you've got a picture of this. And if you don't want to draw your own picture, see image 488 in the archive (see below). Now we are in a position where we can apply the law of sines (the ordinary plane trig law of sines for oblique triangles...) to the big triangle ABC. Set it up as sin("angle at A")/(R+H)=sin("angle at B")/(R+h). And work from there. The angle at B is 180-(gamma+phi) so the the sine of that is the same as the sine of (gamma+phi). We now have sin(gamma)/(R+H)=sin(gamma+phi)/(R+h) which is equivalent to (R+h)/(R+H)=[sin(gamma)*cos(phi)+sin(phi)*cos(gamma)]/sin(gamma) or (R+h)/(R+H)=cos(phi)+sin(phi)*cot(gamma) Next we replace gamma by using its relationship with the measured altitude (assumed to be corrected for dip) which is gamma=90+alt. For any angle x, it is always true that cot(90+x)=-tan(x), so that gives us (R+h)/(R+H)=cos(phi)-sin(phi)*tan(alt). We could stop here since this equation can be solved iteratively (glorified "trial and error") to get the value of phi for any combination of h,H, and alt. With phi in hand, you then get distance from d=R*phi. And if you want to incorporate refraction, just replace R by R'=R/(1-beta) where beta is typically somewhere between 0.13 and 0.4 depending on the lapse rate. But if we don't want to stop here we can start throwing in some assumptions about the relative "smallness" of certain angles and ratios. For starters, it is certainly true that h and H are very much smaller than the radius of the Earth in cases of practical interest. So it is an excellent approximation to replace (R+h)/(R+H) by the expression 1+h/R-H/R. That gives us 1+h/R-H/R=cos(phi)-sin(phi)*tan(alt). Next, we go to the "small angle" approximation for the trig functions of the angle phi. This is a very good approximation whenever the angle phi is less than about 0.1 (a tenth of a radian or roughly 5.7 degrees). In our case, the angle phi will rarely exceed even 1 degrees, so we can replace cos(phi) by 1-(1/2)*phi^2 and sin(phi) by phi (if you don't believe this, try calculating the sine and cosine of 1 degree or 2 degrees using these equations... don't forget to convert to radians first). Replacing cos(phi) and sin(phi) now yields 1+h/R-H/R=1-(1/2)*phi^2-phi*tan(alt). Let's cancel 1 from both sides and introduce some temporary names for tan(alt) and (H-h)/R. We'll set f=tan(alt) and g=(H-h)/R. Then we have (1/2)*phi^2 + f*phi - g = 0. Aha! (say it out loud) This is just a plain old quadratic equation which we learned how to solve in first year algebra. The solution of a quadratic equation of the form a*x^2 + b*x + c =0 is x=[-b+sqrt(b^2-4*a*c)]/(2*a) or the same thing with a negative sign in front of the square root. So we can now solve "our" quadratic equation. The solution is phi = -f+sqrt(f^2+2*g). Bringing back the original constants (replacing f and g), we have phi = -tan(alt)+sqrt[(tan(alt))^2+2*(H-h)/R] And of course, distance = R*phi, so the distance from the base of A to the base of B is d=sqrt[(R*tan(alt))^2+2*R*(H-h)] - R*tan(alt). As I mentioned above, you can incorporate refraction simply by replacing R in this equation by R'. Finally converting everything to minutes of arc and nautical miles and picking one very specific value for the terrestrial refraction reproduces the formula in Bowditch exactly. You can go beyond the Bowditch equation by experimenting with different values for the refraction. By the way, there is another way of deriving this result. You can start with refraction explicitly at the beginning of it all by working from the so-called "refraction invariant". More work, same result (which is re-assuring!). -FER 42.0N 87.7W, or 41.4N 72.1W. www.HistoricalAtlas.com/lunars