A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Haversine formulae for Great Circles
From: Brian Whatcott
Date: 2001 Nov 19, 5:51 AM
From: Brian Whatcott
Date: 2001 Nov 19, 5:51 AM
It's rare to find such throw away insightfulness (Ah! So that's what haversines are all about...) I suspect that the code in a few GPSs could use a little tweak in the way George mentioned (taken with his errata...) Thanks a lot Brian Whatcott At 09:04 AM 11/18/01, you wrote: >What an interesting contribution from Lu Abel! (copied below). Perhaps I >can add something to it. > >When I first scanned it through, my heart sank slightly. Not haversines >again! I said to myself. Surely, those went out with logarithms, as soon as >we started to use calculators and computers. > >Navigators used logs to get round the difficulties they always had doing >long-multiplication and long-division. The problem was that logarithms >couldn't be used with negative numbers, the log of a negative number being >a meaningless concept. As ordinary trig functions ranged over positive and >negative values, something had to be done. > >So a new trig function, the "versine" was invented, which never went >negative, but varied between zero and +2. The haversine, as you might >guess, is simply half the versine, so it varies, more conveniently, between >zero and +1. And then all the trig formulae that navigators used were bent >and twisted into forms that used one of these new functions, and nothing >that required a log ever took a negative value. > >Since calculators took over, the need for logs vanished, and haversines >vanished with them. > >Lu has pointed to a difficulty in the use of the standard great-circle >distance formula. Using his own notation- > >"L1, Lo1 are L/Lo of the starting point, L2/Lo2 are the L/Lo of the >destination." > >the usual expression for the distance in miles is- > >60 * acs ((sin L2 * sin L1) + (cos L2 * cos L1 * cos (Lo2 - Lo1))), > >where the angles are in degrees and acs means "the angle whose cos is..." >or arc.cos. > >As Lu has said, this presents difficulties in situations where the >destination is close to the starting point. His explanation omits an >important matter. The real problem arises in that the expression above is >trying to derive an angle from its cosine, when that angle is small. In >that case, the cosine is very near to +1, and if you plot out a cosine, you >will see that it changes very little from +1 as the angle increases. For >small angles, then, the cosine is a very "flat" function, so it just isn't >possible to derive an angle from its cosine when that angle is very small. >It's not just a problem with a computer: you would meet it if you tried to >use cosine tables instead. > >I have tried the above formula with a Casio programmable calculator, which >has an internal precision of 12 decimal digits. Even with this high >accuracy, distances of 0.1 nautical miles have started to go a bit >inaccurate, and distances which should be .01 miles are calculated as zero. >With a computer using single-precision, things would be a lot worse. So the >problem Lu has raised is a real one, in these special circumstances of >close approach. > >Now for the alternative formula that Lu quotes, to which he gives the name >"Haversine" formula. I regret that choice of name, though I can see how it >has arisen. It's distinctly off-putting, to a navigator who has never been >exposed to the concept of haversines, as it invokes the magic and mystery >of the occult craft of navigation. And it's quite unnecessary, as to use >the formula that he provides, you need no knowledge about haversines at >all. Haversines are not mentioned in the formula. So let's keep things as >simple as possible, and leave haversines out of it altogether, shall we? > >For anyone that would like to use the procedure that Lu quotes as a single >line in a program, we can recast his formula slightly. > >The distance in miles, when angles are in degrees, is- > >120*asn(sqr(sin((L2-L1)/2)^2 + cosL2*cosL1*(sin((Lo2-Lo1)/2)^2))) > >where asn means "the angle whose sine is..." or arc.sin. > >This formula works beautifully for small angles, preserving full accuracy. > >Does it have any snags? Well just as the cos function becomes flat for >angles near zero degrees, the sin function becomes flat near 90 degrees, >and this happens when the distance nears 10800 miles, on the other side of >the world. It must be rare for an accurate distance to be required to a >destination near the observer's antipode, though perhaps an accurate >azimuth may occasionally be of interest (see footnote below). So the >formula that Lu quotes remains useful over any distance that is of real >interest. He has highlighted a significant problem and provided a useful >answer. > >================================= > >At the end, Lu asks- > >"Are there equivalent "haversine" formulae for initial direction and the >L/Lo of intermediate points?" > >Consider the problem of calculating the observer's azimuth of a heavenly >body, which is almost the same as calculating the initial direction to a >destination. > >Equivalent quantities in these two problems, using Lu's notation, are- > >L1=lat, L2=dec, (Lo2-Lo1)=hour angle, distance = zenith angle= 90-altitude. >The distance here should be expressed in degrees (of 60 miles) rather than >in miles. > >Latitudes and declinations should be given signs + or - if they are N or S. >Hour angles are measured westwards. > > >Having calculated the altitude of a body, many navigators seem to obtain >the azimuth from- > >asn ((cos (90 - hour angle)*cos dec) / cos altitude) > >This is a terrible choice of formula. It gives a completely ambiguous >result, for any azimuths that are near 90 degrees (and also near 270). For >example, an angle of 80 degrees has exactly the same sine as an angle of >100, so the formula is quite unable to distinguish between these two >solutions. And I know of no way of distinguishing between these solutions >"by inspection". If anyone else knows how to do this, I would be interested >to learn. Avoid this method > >Another option is to obtain azimuth from- > >acos((sin L2 - (sin L1 cos distance)) / (sin distance cos L1)) > >This has a similar problem that we considered above, that for some azimuths >(i.e. directions very near due North and South) this can become rather >inaccurate because cos of a small angle varies so little from 1. > >========================= > >But why not obtain azimuth from an arc.tan formula, that has no such >problems? Obtain azimuth, in degrees, for an observed body, from- > >atn( sin hour angle / (cos ha sin lat - cos lat tan dec)) > >or, for a great-circle, initial azimuth is- > >atn( sin (Lo2 - Lo1) / (cos (Lo2 - Lo1) sin L1 - cos L1 tan L2) > >Hour angles and longitudes are 0 to 360 measured westwards (not a >convention that is accepted by all), so any E longitudes are negative. You >may note that as there is no mention of altitude (or distance) in this arc >tan formula, you can obtain the azimuth directly, without having to >previously calculate those quantities beforehand. > >When making the above calculation, observe the sign of tan az, and if it's >negative, add 180 degrees to the result. If the hour angle was negative, >add another 180 degrees. The result should be the true azimuth measured 0 >to 360 clockwise from North, with no ambiguities. > >It's even simpler if your calculator or computer has rectangular-polar >conversion (sometimes labelled the atn2 function). In that case you put >into the y-coordinate the value (- sin hour angle) or (sin (L1 - L2)), and >into x goes cos hour angle*sin lat - cos lat*tan dec (or its equivalent). >The result emerges as a radius (which you ignore) and an angle, which is >the angle you require. All you have to do (it you're bothered to) is to add >360 if it is negative. > >=================== > >A footnote about calculations involving positions close to the antipode. > >These are required rarely, but I can think of an example that came up on >another mailing list. > >Muslims wish to face Mecca when they pray, but this must present problems >if they live on the other side of the world. There must be a >jump-discontinuity in the great-circle azimuth of Mecca at some spot at >Mecca's antipode, which one could consider as anti-Mecca. Near there, the >faithful would need to face away from that spot, at which one might erect >some sort of marker, if it was on land. Fortunately it isn't; it's >somewhere between the Tuamotus and the Gambier islands in the Pacific. But >imagine the problems that must face a pious Muslim crew member of an >inter-island vessel plying those waters, if he wishes to pray in the right >direction. > >================== > >Note: everything in this posting assumes a spherical Earth. If its >ellipticity is taken into account, everything will get much more complex, >including the direction of Mecca. > >George Huxtable. > >======================== >Lu Abel said- > > >Almost every celestial text provides the formulae for calculating great > >circle distance and direction between two points. This is not surprising > >since the trigonometry is identical to that of sight reduction. If one > >takes any standard celestial reduction method and substitutes the starting > >point for the AP and the ending point for the body GP, Zn gives the > >direction and 60*(90-Hc) gives the distance. > > > >A couple of years ago this august group was of great help in answering for > >me an obvious question I've never seen answered in any standard navigation > >text, namely how to calculate the L/Lo of intermediate points. > > > >Recently I've done a bit of further digging on great circles. I learned > >that while the law of cosines formula typically used for Hc calculation is > >trigonometrically correct, it can produce incorrect answers on a computer > >or calculator when the starting and ending points are close because > >computers and calculators express numbers with a limited number of > >significant digits. An alternate is the "Haversine" formula, called by > >that name because of its haversine-like terms [recall that hav(x) = > >(1-cos(x))/2 = sin^2(x/2) where "^2" means squared ] > > > >The haversine distance formula goes as follows: > > > >L1, Lo1 are L/Lo of the starting point, L2/Lo2 are the L/Lo of the > destination > > > >DLat = L2 - L1 > >DLo = Lo2 - Lo1 > > > >A = sin^2(DLat/2) + cos(L2) * cos(L1) * sin^2(DLo/2) > > > >Distance = 120 * arcsin (sqrt (A)) > > > > > >For a great writeup on the haversine distance formula, see > >http://mathforum.org/dr.math/problems/neff.04.21.99.html (note no "www" at > >beginning) > > > >Having learned about this formula, I'm going to guess it's used in the > >majority of GPS sets, since they're often calculating small distances (like > >distance-to-go when approaching a waypoint) > > > >The standard reference on calculating great circles via the haversine > >formula seems to be R.W. Sinnott, "Virtues of the Haversine", Sky and > >Telescope, vol. 68, no. 2, 1984, p. 159. I've seen it mentioned in several > >writeups of the haversine distance formula. > > > >I can't access a copy to see if Sinnott provided formulae beyond the > >distance formula, so here's my question for this group: > > > >Are there equivalent "haversine" formulae for initial direction and the > >L/Lo of intermediate points? > > > >Thanks > > > >Lu Abel > >------------------------------ > >george---.u-net.com >George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. >Tel. 01865 820222 or (int.) +44 1865 820222. >------------------------------ Brian Whatcott Altus OK Eureka!