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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@huxtable.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!

   

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