Topic Review (Newest First) |
05-02-2010 11:47 PM | |
AdamLein |
Quote:
Originally Posted by RandyBC
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What I think now after more consideration is that the error of 1 degree 8' does not necessarily translate into 68nm between assumed and actual position.
Actually, it occurs to me you could be plotting the intercept in the wrong direction. You should be plotting the azimuth line towards the the sun and through the assumed position. Sounds like you are indeed doing this. If the observed altitude is greater than the computed altitude, then you mark the intercept in the direction of the azimuth. If the computed altitude is greater, mark the intercept in the reciprocal direction. For example, if the azimuth is SW and computed altitude is 20 degrees, then: - observed altitude 18°52' --> intercept is plotted 68 miles SW of AP. - observed altitude 21°08' --> intercept is plotted 68 miles NE of AP. If you plot it in the right direction, you should be just right. If you plot it in the wrong direction, then you will be approximately 136 miles wrong. In general if your error is double one of the numbers involved in the computation, sounds like a positive/negative or direction problem. |
05-02-2010 10:45 PM | |
Faster |
Randy... I just tried plotting exactly 1 NM on GE at the equator both along the equator E-W (technically not guaranteed) and N-S (technically for sure) I did get exactly 1.0 NM both directions for 1 minute of long. or lat. Are you sure your distance setting was NM? - on my GE it defaults to statute miles each time you open it. |
05-02-2010 09:51 PM | |
RandyBC |
Hi Adam....I checked distances at the equator to make sure that Google Earth was measuring 1' as 1nm. That was done to validate how Google Earth measures distance. I was using the sun, and no I did not put corrections in for semidiameter or dip. The assumed position was a quite far away so it is no surprise that the error was on the large side. What I think now after more consideration is that the error of 1 degree 8' does not necessarily translate into 68nm between assumed and actual position. Rather, if I draw an azimuth through the assumed position and advance a perpendicular through the azimuth 68nm toward the sun, then that perpendicular should more or less cross through my actual position on my back porch. So using Google Earth to double check that my porch was 68nm from my assumed posistion was incorrect. I still can't figure out why Google Earth doesn't measure 1nm between 0N,0W and 0N 1.0W at the equator. |
05-02-2010 08:57 PM | |
AdamLein |
Randy: still not clear on why you checked any distances at the equator. For now I'll assume you used Google Earth to measure the distance from your Assumed Position and your Backyard Position and got 136 nm. That sounds like a very large error to me. Some things to consider: - Have you checked that your sextant is properly calibrated? - Are you using the sun? Are you correcting for semidiameter (you should not if you are using an artificial horizon and bringing the sun down to match his own reflection). - If you are using a planet or the moon, are you correcting for parallax? - Are you using a star? Are you sure it's the one you think it is? I have not been able to take sights from stars, and have not tried planets. The sun is difficult enough for me right now. The other likely candidate for error is in the calculations somewhere. Are you doing it by hand, or by some computer program, or what? |
05-01-2010 02:29 PM | |
chris_gee |
This can be confusing. Latitude lines run EW but are spaced in NS distances and on the right of the chart. Longtitude lines run NS but give EW angles. The distance between them at the equator is the biggest and at the poles zero. So stand on the pole and turn around and you do a 360 in longtitude, but your latitude stays the same. What you are doing is trying to change your longtitude at the pole by 1 minute 1/360 of arc true but not a mile - only a mile at the equator. There is a formula that gives changes in longtitude as you near the poles. Someone once used the table to test my knowledge or rather more to establish superiority. I had shelved it as irrelevant as I had no intention of sailing near the poles and we were in the tropics at the time. |
05-01-2010 12:08 PM | |
RandyBC |
I do understand what you were saying Flybyknight, which is why I used the equator. Thanks Adam for the correction in terminology. I see where what I said may cause confusion. What I was measuring on Google Earth was the the distance between my assumed position and known position (from my back porch) and then comparing that known difference with the result I got after reducing the sight. The difference between Ho and Hc was 1 degree 8' which should put me approx 68nm from my assumed position. What I really was interested in was whether the calculated distance off would jive with the known measurement. I went to double check the result on google earth the measured difference was about twice that. That led me to input the coordinates at the equator where 1' of longitude should be 1 nm. I wasn't getting that result which made me question if Google Earth was measuring distance correctly. Thanks again Adam for you input, you obviously have a special interest celestial navigation. |
05-01-2010 10:12 AM | |
AdamLein |
Flybyknight is only partially correct. You are measuring a difference of one minute of longitude, that's true, and that is not generally equal to one minute of arc. However you are doing it at the equator, and as a simple approximation, one minute of longitude at the equator is indeed one nautical mile. Without knowing more about what exactly you're doing, it's hard to say. One thing I can say is that "calculated distance" and "assumed distance" are not terms I know of in celestial navigation. You have an "assumed position", and from this position, you calculate an altitude and azimuth of the body you're observing. I don't know of any easy way to check these values with Google Earth. Furthermore, if you assume a position somewhere on the equator and then actually measure altitude in Hinton, Alberta, of course your results will be way off. Finally, computing distance between two points is done by a simple formula, and is not really something you frequently do in celestial navigation. If that's all you're trying to do, Wikipedia has a few formulas, some quite simple, some quite complex. For me, the Spherical Law of Cosines suffices. Google Earth probably uses something much more complex, so don't expect the values to match exactly, though on the equator the approximation should be quite good. Sounds like a lot is missing from your story. Please explain exactly what you're trying to do and what procedure you're using so we can help. |
05-01-2010 05:26 AM | |
Flybyknight |
I hope I understand your question correctly. 1 minute of arc should be measured (with dividers) on the latitude scale, not longitude scale. 1 minute of arc in nautical miles at one latitude is vastly different at another latitude. |
04-30-2010 07:16 PM | |
RandyBC |
What is wrong here?? I'm teaching myself celestial navigation and I wanted to double check my results on google earth. The calaculated and assumed distance was out by a factor of 2. I wanted to double check how Google Earth measures distances and inputed two points, 0 0.0n, 0 0.0w and 0 0.0n, 0 1.0w. That is 1' of arc and that should equal 1nm should it not? When I measure between these 2 points I do not get that answer. Why not? |
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