Any astronomers around?

[HP67]

I know there are some heavy-duty mathematicians on the forums. Maybe you can explain this also.

I do not understand how to calculate the time of x degrees after some celestial event. For example, let's say sundown occurs at 19:30. How do I figure 5 degrees later? I read that every degree is equal to 4 minutes regardless of latitude. So later should be a simple matter of sundown + (5 * 4) minutes (20 minutes after sundown). But my results don't match up.

Can anybody fill in the blanks? Thanks.

[Dale Reed]

360 degrees / 24 hours = 15 degrees per hour (by division).

One hour = 60 minutes.

60 minutes / 15 degrees = 4 minutes/degree (by division).

But that only works for the sun. In one day, the earth has gone around in its orbit about the sun by 1 day motion / 365.2425-ish days per year. You need to account for this, at least for the "fixed" background of distant stars.

360 degrees / 365.2425 days = 0.9856465... degrees per day = 0.0410686... degrees per hour = 0.002737907... degrees per 4 minutes.

Now if I could only remember which way to apply this offset -- add or subtract. Sorry, I'm not an astronomer. If I were, I would know which way the earth rotates with respect to which way it orbits the sun.

Perhaps a google search of "sidereal day" would help...
Maybe see Sidereal Time at Wikipedia.
Dale

(edited to fix an oops in copying a number from calculator to post)

[everettr]

(05-29-2014 05:17 PM)HP67 Wrote:  I do not understand how to calculate the time of x degrees after some celestial event. For example, let's say sundown occurs at 19:30. How do I figure 5 degrees later? I read that every degree is equal to 4 minutes regardless of latitude. So later should be a simple matter of sundown + (5 * 4) minutes (20 minutes after sundown). But my results don't match up.

It sounds like you are working with equatorial coordinates. These are in reference to the orientation of the Earth. Declination (DEC) is like latitude and measured in degrees, and Right Ascension (RA) is like longitude, but measured in hours, minutes, and seconds.
If you are observing on the Equator, on an equinox, the Sun moves perpendicularly to the horizon at sunset (DEC=0), and would be 5 degrees below the horizon 20 minutes after sunset.
One correction that I know about is for declination. Right now (May 30) the Sun has a DEC of about 20 degrees. This means that it appears 20 degrees up from the celestial equator, and travels in a circle that is smaller than the celestial equator by COS(20°). In order to move 5 actual degrees in the sky would take the Sun 20 minutes / COS(20 degrees DEC) = about 21 minutes 17 seconds.
I also am not an astronomer, but I once was a member of an astronomy club. :-)

[HP67]

Thanks guys.

@Everettr Yes, given a lat/long my sunrise/sunset values using the USNO algorithm match up with nearly every published source I can find. But other ephemeris I calculate based on those matching values can sometimes be off by several minutes and I haven't understood why.

I will try to see if I can get this correction you explained to line up with what I am seeing elsewhere. I appreciate the help. I don't like this solution though because it is predicated on having updates to the declination based on seasons or some period which although makes perfect sense and is probably absolutely correct, does complicate the issue for a program running on a calculator. To apply it I would need to have a table and I would have to see if it needs updating from year to year.

Thanks for the comments. Keep 'em coming if anybody has more ideas on this.

[HP67]

This correction definitely helps but I am still not hitting the target. I think I may be able to use a formula to calculate the declination based on the JD for a given location. If so then it would be ok for my application.

I need to do some more research on all this. Will update with any earth shattering findings.

[NetMage]

Do you have a copy of Meeus's Astronomical Algorithms published by Willmann-Bell?

[HP67]

Yes got it recently and it's on the stack of things to look at. Did you have a pointer to a specific section or just to mention it's a good book?

[Bill (Smithville NJ)]

(05-29-2014 05:17 PM)HP67 Wrote:  Any astronomers around?

I'm not an Astronomer nor am I really into it. But I did run across an amateur Astronomer that was a HP calculator enthusiast back in 2003.

I had sent Robert Burnham a HP-35 calculator manual and as a thank you he sent be a signed copy of his latest book: "Great Comets". It's a beautiful color illustrated book that covers Hyakutake in 1996 and Hale-Bobe in 1997. The forward of the book is by David Levy. It's a beautiful book which I treasure for both its content as well as the inscription: "Best wishes to a fellow HP calculator fan".

I don't remember if I ran across Mr. Burnham in the forum here or somewhere else.

Bill

[HP67]

Neat!

[htom trites]

Meeus's Astronomical Algorithms slowly works through all* of the many (sometimes subtle) complications of stellar (and planetary) computation in the sky. For some uses, some of them can be ignored. If you're at all interested in this topic, it's an essential work. There are five collections of updates and expansions. You don't need to study all of it, from front to back, each topic is independently treated.

*Well, almost all; the version I borrowed from the library did not have the USN time + position + velocity reduction from seven independent separated celestial observations. Check your GPS with your sextant. It may be that this isn't published, or is in one of the subsequent collections.

[HP67]

I'm not interested in astronomy generally, I just need a few formulas.

If anybody can help figuring x degrees after sunset or point me to a section of the book, etc. that would be great. Thank you.

Update: apparently changing the zenith value to reflect the degrees before or after the event works. There must be a way to calculate this after the fact, I'll look into that.

[CosmicTruth]

(06-05-2014 01:09 PM)HP67 Wrote:  I'm not interested in astronomy generally, I just need a few formulas.

If anybody can help figuring x degrees after sunset or point me to a section of the book, etc. that would be great. Thank you.

Update: apparently changing the zenith value to reflect the degrees before or after the event works. There must be a way to calculate this after the fact, I'll look into that.

Back in the old days (before GPS) we used to accurately set our 41C clock, scale our LAT/LONG from a USGS topo map, then we would use a solar filter to measure the angle from our backsight point to the sun direct and reverse. The calculator would be used as a stopwatch and take the horizontal angle and the exact time and the LAT / LONG and calculate the azimuth of our backsight line.

an old ephemeris is here: http://www.rollanet.org/~eksi/2008EPHEM.pdf

the ephemeris has HP programs! I know I may be over thinking your problem but also thought readers here might be interested in this.