NavList:

Two days ago I had my first opportunity this season to do a "noon-like" meridian sight of Venus in the middle of the day. Venus crosses the meridian around 3pm local mean time everywhere on Earth. It is just now barely bright enough to be seen in a sextant, and it's also barely bright enough to be seen naked-eye.

The "Venus +ONE" scenario has become one of my favorites. I've named it this way to keep it parallel with my "Polaris +ONE" scenario. For Polaris the other body, the "plus one", could be any bright star in twilight. For Venus in daylight, the "plus one" is usually the Sun. The beauty of either of those scenarios is that we get a high-quality fix with quick math, no plotting, and usually no dead reckoning run. The first object, Venus at meridian in daylight or Polaris in twilight, gives us a fast latitude, and then a short calculation from the second body's altitude yields longitude. And that's it.

Step one: find Venus. This is difficult this week, but as Venus brightens over the next few months, it will be much easier. To find Venus, you need an estimate of time of meridian passage and the approximate meridian altitude. To find the time, look at today's almanac data (any source you like) and examine the planet's GHA. GHA is the longitude of the subStar point for Venus. So when the planet's GHA matches your estimated longitude, Venus will necessarily be due South or due North of your position -- on the meridian. This can be done in your head. Looking at the almanac data two days ago, the GHA of Venus was 69.1° at 1900 UT/GMT. My estimated longitude was 71.5°. That's a change of 2.4°. I know the Earth turns 1° in four minutes of time, so the GHA of Venus will match my longitude in a little less than ten minutes, at 1910 UT. The approximate altitude at that time is given by the co-latitude plus the Dec. My co-lat was a little less than 49°. This is the altitude of the celestial equator at the meridian. The Dec of Venus was 17.6° so the expected meridian altitude would be 66.5° roughly. Given the several degrees in the field of view of a sextant, that's plenty close enough. Note that this is really a "true altitude". The sextant altitude would be slightly higher because of dip and refraction, but at this point that's inconsequential. So I preset my sextant to 66.5°, face as nearly 180° T as possible --due South, and I see nothing. 

Finding Venus is tricky. I started 15 minutes before predicted meridian passage. Even when it's brighter than this week, you need something in the middle foreground, like tree branches on land or like rigging on a vessel, to guide your visual search. After finding nothing on first try above, I moved to a spot with a craggy tree in front of me. Next I checked to see just which little branch at the top of the tree was nicely aligned with where I expected to find Venus. Now I went for the heavy guns and grabbed a pair of 7x35 binoculars. I aimed at that spot above the little branch, and there it was. Easy as could be! I moved about half a meter back and about a meter left and from that spot Venus would be sitting right close to end of an easily-identified twig. So I put the binoculars down, grabbed the sextant again, aimed at that twig, and... nothing! I tried again... Still nothing. Next I flipped the sextant upside-down and held it in my left hand. Now I aimed directly at that twig through the horizon glass (the index mirror is pointing roughly at the horizon at this point. And there it was. It jumped out easily. There was some reduction in contrast in the optics, but having seen it once through the sextant I could confidently turn it over and manage it in the usual fashion: sea horizon in the horizon glass and Venus reflected from the index mirror. 

From here it's mostly math. I shot a couple of altitudes of Venus near meridian passage, averaged those. Then a minute later I turned to the southwest and shot four altitudes of the Sun, recording the UT along with the altitudes. Here's the work:

appox UT:   19:10
avg Hs        66°16
decimalize: 66.27°
corr:            -0.09°
corr'd:         66.18°
ZD:             23.82°
Venus Dec 17.65°
Lat=ZD+Dec     41.47°  =41°28.2'  (should be 41°28.6' so: CLOSE  0.4n.m. err)

Next the Sun. Visually estimated azimuth 225°.

avg UT:   19:11:40
avg Hs     40°40.2'
decimalize: 40.67°
corr:           +0.17°
corr'd:         40.84°
ZD:             49.16°

need Sun GHA, Dec at 19:11:40
GHA: 106.80°, Dec: +3.89°

ABC method (*) requires ZD: 49.16°, Lat: 41.47°, Dec: 3.89°

A: 0.87476
B: 0.06010
C: 0.81466
HA: 35.45°

Sun is falling, so subtract HA from GHA
lon: 71.35°   =71°21.0'   (should be 71°22.0' so: NOT BAD  0.7nm err)

That's all there is to it. That's a "Venus +ONE" fix. By the way, I can also easily draw a rough error ellipse around my fix since I took note of the 45° difference in azimuth between the Sun and Venus.

Frank Reed
Clockwork Mapping / ReedNavigation.com
Conanicut Island USA

* The ABC method is what I teach in my modern celestial workshops. You calculate the HA of the Sun from the standard spherical triangle:
cos HA = (sin alt - sin Lat sin Dec)/(cos Lat · cos Dec) or equivalently,
HA = acos(cos ZD / cos Lat / cos Dec - tan Lat · tan Dec), or (as above):
HA = acos(A - B) where A and B should be obvious. And then
Lon = GHA +/- HA. You add HA when the body is rising (in the east) or subtract when it's falling (in the west).