stretch
• Member for 2 years, 8 months
• Last seen this week
• Columbia, MD, United States

You mentioned "Equation of Time," so you should be able to answer many of your questions yourself. You understand the problem. The Wikipedia article will tell you more than you want to know. ...

Answering your main question: if you were to correct the text in the book you would word it differently, deleting all the mentions of parallax except for a single sentence saying there is such a thing ...

You don't choose a coordinate system. If you're navigating you take sights to get horizontal coordinates, then look up the equatorial coordinates, and knowing the time, calculate your position. If you'...

Neither projection has anything to do with Earth's motion. They just show different ways of defining locations on the celestial sphere. The locations of stars are more or less fixed on that sphere. ...

Your original thought, that you could just specify the Sun's altitude was correct. The shadow will reach the end of your patio at the same altitude of the Sun every day. Considering where I think you ...

Made me look. On October 15 Portugal was still observing daylight saving time. The local time for continental Portugal was one hour later than UTC and the country is about 8 degrees west longitude, ...

Your star map shows the right ascension and declination of the stars. The right and left edges of the map are the 0 (and 360) degrees right ascension (or sidereal hour angle). The Earth rotates, so ...

You can do it all with arithmetic: no algebra or trig. The Navy publications you saw are designed that way. Starting with the Nautical Almanac: there are different treatments for the stars and for the ...

https://www.giss.nasa.gov/tools/mars24/ has a link to an application that may be what you're looking for.

The commonly used first approximation for the latitude of the Sun's GP is $23.5 sin(\frac{2d \pi}{365.25})$,where d is the number of days since the vernal equinox. If your sine lookup is for degrees ...

It never is exactly, directly overhead on the vernal equinox at the Greenwich meridian. You know that it happens on different days in different years so it must be happening at different times on ...

You won't find a reference that tells you the altitude of an object when it passes your meridian. That altitude depends on your latitude. Saturn for all this year is near 18 degrees south declination, ...

Actually the gravitational tug from a nearby planet is stronger than that from a nearby person. Venus is the planet that comes closest to us - it's 42 billion meters away when it's between us and the ...

The Earth's axial tilt is expressed as a single number because it's just an angle. It can be considered to be the angle between the plane of the equator and the plane of Earth's orbit, or the the ...

Using user38308's approach: First, find the distance between you and your friend, knowing lat/long of your positions. You need to solve a spherical triangle with vertices at the north pole and the ...

You shouldn't add 2010 because the date of interest is before 2010. None of the days in 2010 were in the time between 27 July 2003 and 0000 1 January 2010. Similarly, the time after 27 July 2003 until ...

Short answers: It's not just angular velocity of Earth's rotation times elapsed time. You do need to factor in declination and latitude. Using your representative date and observer location: The ...

The declination won't change. The right ascension will increase by about 15 degrees per hour or 0.25 degrees per minute. The actual number is closer to 15.04 degrees per hour of time: (366.25 * 360)/(...

On the vernal equinox, first day of spring, the zero reference point for right ascension was directly overhead at 9:37 UTC this year. It has been advancing westward at a rate of 360/365.25 degrees per ...

There's no difference in the calculations, but the Moon, Sun, planets don't have more or less fixed right ascensions and declinations like all the stars. The site you linked seems to only work for ...

You can do it in one or two days. You'll know compass directions roughly, which is good enough, by observing the Sun's east-west path. You measure the Sun's altitude at meridian passage on two ...

You could try successive approximation using your Python library. Put in your best guess date/time and keep adjusting it to get 88 degrees.

Transmitters don't slow down as they age. Red shift isn't involved, at all. As a transmitter gets further away the signal at the receiver gets weaker, so it's deliberately slowed down. Data channel ...

The way the star chart is drawn, the points labeled 1 and 3 are on the celestial equator, the 0 degree declination reference. The tip-off is that a line between them passes through Orion, the top-...

It is tabulated, and available online in several places. One of them is the Nautical Almanac at thenauticalalmanc.com. It has the position of the moon for every hour this year and a number of past ...

What I think you're trying to do is workable, but approximate. You first find the Sun's geographical position (GP), the point on the Earth where the Sun is in zenith at the time of the observation. ...

A telescope wouldn't have a single lens, I'm sure you know. If you were to use the setup in your diagram, focusing the image of the moon on a flat surface, you could find its diameter with the method ...

The north celestial pole doesn't have a right ascension. It's like the north terrestrial pole not having a longitude. Asking for it creates an exception. In any case, the RA of a point, other than the ...