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by knowing where those are with relation to the earth

On Apr 15, 2008, at 7:50 PM, Gary J. LaPook wrote:
> Gary LaPook writes:
>
> OK, so you end up with a position defined by the lat-long of the  
> spot on the surface of the earth directly between the spacecraft  
> and the center of the earth (the spot that the spacecraft is  
> directly above and where a person on the surface would measure 90�  
> with his sextant to the spacecraft) and the radar distance. But is  
> this useful for space navigation? how do you relate this to an  
> inertial frame or sidereal frame or to determine if you are on  
> course to the moon or mars?
>
> gl
>
> Fred Hebard wrote:
>> Yes, it does. One gathers elevation information with radar  
>> ranging; it's the same problem, you're just at a different  
>> elevation, so there's a larger (!) dip correction. It was the  
>> method proposed by Weems, et al, in the delightful book, "Space  
>> Navigation Handbook," Navpers92988, US Govt Printing Office: 1962  
>> 0-628762. On Apr 15, 2008, at 7:27 PM, glapook@pacbell.net wrote:
>>>
>>> Gary LaPook writes: But that doesn't solve the problem. The only  
>>> reason that CN works on the earth is that the direction of "up"  
>>> varies with your position on the earth. The altitudes measured on  
>>> earth (and in aircraft) rely on the direction of "up" for the  
>>> measurement. The sea horizon used with a marine sextant is where  
>>> it is due to the local gravitational field which causes water to  
>>> assume a shape at right angles to "up" and gravitational "down."  
>>> A bubble sextant uses a bubble to sense "up." Because local "up"  
>>> changes at a constant rate of one nautical mile per minute of  
>>> altitude we can find our place on or above the surface of the  
>>> earth. This relationship does not hold on the way to the moon. gl  
>>> On Apr 15, 2:12 pm, Fred Hebard  wrote:
>>>>
>>>> I believe they measured altitudes from a limb of the Earth, more- 
>>>> or- less in the "normal" way. On Apr 15, 2008, at 4:00 PM, Gary  
>>>> J. LaPook wrote:
>>>>>
>>>>> Gary LaPook wrote:
>>>>> If I remember correctly, the Apollo spacecraft had a sextant on  
>>>>> board used to mesure angles of celestial bodies in order to  
>>>>> compute their position in space on the way to the moon, (maybe  
>>>>> only as a backup.)
>>>>> gl Fred Hebard wrote:
>>>>>> So it would have to be sun/moon/planet-star distances. I  
>>>>>> suppose those are limited by the low degree of parallax of the  
>>>>>> planets and sun, not to mention one has to know where one is  
>>>>>> on earth to determine the "position" of other bodies in the  
>>>>>> solar system, which I guess would be a circular argument. On  
>>>>>> Apr 15, 2008, at 12:54 PM, Lu Abel wrote:
>>>>>>> Fred: You're right about traditional surveying. But your  
>>>>>>> proposal is to use star-to-star distances to locate one (if I  
>>>>>>> understand correctly) in 3-D space relative to some very  
>>>>>>> distant stars. Imagine a couple of stars several hundreds of  
>>>>>>> light-years away (that's on the order of 10^20 cm). Suppose I  
>>>>>>> move a few cm closer to them. By how much would the angle  
>>>>>>> between them change? Not by much at all. Lu Fred Hebard wrote:
>>>>>>>> Lu, Why billionths of an arcsecond? One arcsecond gets one  
>>>>>>>> to 1/60th of 100 feet in traditional surveying, or about 50  
>>>>>>>> cm. One- thousandth of an arcsecond would drop one to 5 mm.  
>>>>>>>> I wonder if refraction is a problem here. Fred On Apr 15,  
>>>>>>>> 2008, at 12:33 PM, Lu Abel wrote:
>>>>>>>>> Fred: In theory, yes; in practice, no. To position oneself  
>>>>>>>>> using star-star distances would require require measuring  
>>>>>>>>> angles to billionths of an arc-second. Maybe something an  
>>>>>>>>> astronomer could do, but not something you or I are going  
>>>>>>>>> to do with our sextants! BTW, I remember a conversation  
>>>>>>>>> with a radio- astronomer about 20 years ago where he said  
>>>>>>>>> that his team had measured the distance between two  
>>>>>>>>> radiotelescopes on opposite sides of the US to within a cm  
>>>>>>>>> or so using a technique called long-baseline  
>>>>>>>>> interferometry. But the whole experiment took them a year  
>>>>>>>>> or so... Lu Abel Fred Hebard wrote:
>>>>>>>>>> Completely unrelated, but stemming from the same article.  
>>>>>>>>>> The author states that height can only be known to some  
>>>>>>>>>> few cm or whatever because of variations in gravity, if I  
>>>>>>>>>> remember correctly. It would seem that this is due to our  
>>>>>>>>>> tradition of assuming we are on the surface of a spheroid  
>>>>>>>>>> or ellipsoid when doing navigation. Confining ourselves to  
>>>>>>>>>> a surface makes the trig easier, but couldn't one position  
>>>>>>>>>> oneself with greater accuracy (with feet firmly planted on  
>>>>>>>>>> earth, not on a boat) using only stars or stars plus the  
>>>>>>>>>> sun, ignoring the earth's horizon, by measuring star-star  
>>>>>>>>>> distances? Make it a true 3-D problem. Or would  
>>>>>>>>>> uncertainties in the positions of stars still hamper ones  
>>>>>>>>>> efforts, especially uncertainty in their distance from us?  
>>>>>>>>>> Fred Hebard On Apr 14, 2008, at 9:50 PM,  
>>>>>>>>>> frankr...@HistoricalAtlas.net wrote:
>>>>>>>>>>> The fascinating article which Fred Hebard linked: http://  
>>>>>>>>>>> www.physicstoday.org/vol-59/iss-3/p10.htmlincludes a  
>>>>>>>>>>> detailed discussion about the problems of gravitational  
>>>>>>>>>>> time dilation and extremely accurate clocks. That's the  
>>>>>>>>>>> main topic, and it's great stuff. The article also  
>>>>>>>>>>> mentions leap seconds and navigation: "Celestial  
>>>>>>>>>>> navigators --that vanishing breed-- also like leap  
>>>>>>>>>>> seconds. The Global Positioning System, however, cannot  
>>>>>>>>>>> tolerate time jumps and employs a time scale that avoids  
>>>>>>>>>>> leap seconds." So here's my question: what's the best way  
>>>>>>>>>>> of doing celestial navigation if leap seconds are dropped  
>>>>>>>>>>> from official time-keeping? I don't think it should be  
>>>>>>>>>>> all that difficult to work around, but I'm not sure what  
>>>>>>>>>>> the best approach would be. Assume we get to a point  
>>>>>>>>>>> where the cumulative time difference is, let's say, 60  
>>>>>>>>>>> seconds (that shouldn't happen for decades, so this is  
>>>>>>>>>>> just for the sake of argument). Should we treat the  
>>>>>>>>>>> difference as a 60 second clock correction before working  
>>>>>>>>>>> the sights? Or should it be a 15 minute of arc longitude  
>>>>>>>>>>> correction after working the sights? Or something else  
>>>>>>>>>>> entirely?? -FER Celestial Navigation Weekend, June 6-8,  
>>>>>>>>>>> 2008 at Mystic Seaport Museum:www.fer3.com/Mystic2008
>
>
> >



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