PDA

View Full Version : Astronomy by the way of trigonometry is blowing my mind

geedeck
2011-Jan-18, 10:16 PM
Howdy folks, I'm new around here but I hope you don't mind me asking some questions.

I'm taking an astronomy statistics class, and as someone who is mostly an art and computers guy, it's kind of 0_o for me. I've worked with my TA some but well, he has a bad opinion of anyone who would study astronomy who isn't actually trying to get a degree in it. The teacher is much better, but there isn't the same amount of time available to work with him.

So I'm working on some problems, and I can kind of punch the numbers in, but I don't really understand why. So I don't want someone to answer the problem for me, but if they could point me at good resource that explains why things are done, or can explain why I'm doing what I'm doing, man I would be grateful.

So for example, I'm trying to find the zenith distance of a star, declination delta of +20°, observed at hour angles 0, 2, 4 and 6, done from a latitude of ~40° N.

We've got this equation:

cos Dec * sin HA = cos a * sin A
sin Dec = sin B * sin a + cos B * cos a * cos A
cos Dec * cos HA = cos B * sin a + sin B * cos a * cos A

But I don't really get what I'm doing with this equation. I mean I can plug numbers in woo but... why? And I admit, trig kind of blows my mind, but well I'm tired of that and just want to understand that.

The other one I don't get is where we're supposed to find the LST for Jan 17 2011 @ 9pm for 83° W, 40° N. I've got the GST table from the 2011 Astronomical Almanac for the time. This one makes a bit more sense, I've calculated the sidereal interval (2:02:19.7) and I know the sidereal time adjustment for EST, ie 5:32:01.1 and I know that GST for Jan 18th is 7:44:17.7... Okay but I don't really get what that last number *means*. The subtext for that last one is GHA of the equinox... But I don't get what that's quantifying in real people speak. And I don't get how to pull it all together, but the TA did admit that knowing the sidereal interval gets me 1/3 the way there.

It's like someone dumped a bucket of numbers confetti on me! And it's awesome! Except it'd be really awesome if I could actually build a picture out of the confetti.

tl;dr help a spunky college student defeat his smarmy TA that looks down on an artist trying to learn more about astronomy.

antoniseb
2011-Jan-18, 11:04 PM
It is cool that you are willing to tool through this.

I'm sure your book has some things that will help with the spherical trig, but what might help you visualize it is to get a globe and try to imagine that there is a spot on it, and you are trying to determine how high or low that spot is as you rotate the globe. It is a little more complicated than just being the sign of the angle, but thankfully, it is only one or two layers deeper than that. What is making it tough, is that this is the first time you're facing the extra steps, and at first glance all the trig functions look alike.

Hopefully others will come by and give you a more thorough pointer, or hopefully even a web-site with a visualizing applet.

StupendousMan
2011-Jan-18, 11:41 PM
I'm an astronomy professor, and I think that the following method to solve your problems would be A-OK.

Find yourself a good free planetarium program, like Stellarium. Install it on your computer. Use it to set up the dates and times and situations in your homework problems. Then, look at the screen and measure the quantities you are asked to find.

I don't care if my students use a calculator to do long division. Nor do I mind if they use planetarium programs to answer astronomical questions.

Oh, and as for the second question: the "LST" is simply one of the two coordinates of a star which is currently overhead in the sky. Each star as "Right Ascension" and "Declination", right? They're like longitude and latitude. The "Right Ascension" value of a star which is overhead right now is exactly what the Local Sidereal Time is. It's like using stars to tell time.

plphy
2011-Jan-19, 12:38 AM
We've got this equation:

cos Dec * sin HA = cos a * sin A
sin Dec = sin B * sin a + cos B * cos a * cos A
cos Dec * cos HA = cos B * sin a + sin B * cos a * cos A

But I don't really get what I'm doing with this equation. I mean I can plug numbers in woo but... why? And I admit, trig kind of blows my mind, but well I'm tired of that and just want to understand that.

Hi geedeck,

those are the standard formulas for converting between terrestrial coordinates (azimuth and altitude) and celestial coordinates (hour angle and declination), although I suspect there's at least one minus sign missing somewhere. What's your precise question? How to work with these equations? Or how they come about?

... I know that GST for Jan 18th is 7:44:17.7... Okay but I don't really get what that last number *means*. The subtext for that last one is GHA of the equinox... But I don't get what that's quantifying in real people speak.

The sky is rotating with respect to the Earth (assumed fixed in this context). Hour angles are used to measure how far the sky has rotated. The local hour angle of some celestial object is the angle between the hour circle on which the body is located (that is, the great circle going from the north celestial pole through the object and on to the south celestial pole) and the observer's meridian (that is, the great circle going from the north celestial pole through the local observer's zenith and on to the south celestial pole). See this diagram for how the local hour angle is measured:

When the object's hour angle is zero, it is on the observer's meridian. The sky's rotation keeps carrying the object westward, its hour angle thus keeps increasing. For objects which are fixed on the sky dome (e.g. the fixed stars), the hour angle keeps increasing at a rate of about 15.041 degrees per hour (360 degrees in 23h 56m 04s).

One such object is the equinox. By definition, your local sidereal time is the hour angle of the equinox. When the equinox passes through your meridian, your local sidereal time is zero; it then keeps increasing at a rate of about 1.00274 sidereal hours per clock hour (24 sidereal hours in 23h 56m 04s of clock time).

Local sidereal time is different for different observers. When the equinox passes through one observer's meridian, it has already passed through the meridians of observers farther east and has yet to pass through the meridians of observers farther west. Their respective local sidereal times are therefore different, depending on their geographical longitudes.
Instead of tabulating local sidereal times for all possible observers, almanacs give the local sidereal time for Greenwich (i.e. the hour angle of the equinox as observed in Greenwich). Knowing the Greenwich hour angle of the equinox, GHA, and thus Greenwich sidereal time GST, you can easily work out your local sidereal time, taking the difference in longitude into account. Since the almanac will only give GST for Greenwich midnight, you'll also have to correct for the time elapsed since then.

Bye,
Thomas

ngc3314
2011-Jan-19, 01:04 AM
Another way to deal with this is that you re converting between two sets of spherical coordinates, analogous to latitude and longitude. One is relative to your local horizon (zenith distance and azimuth), the other set by the direction of the Earth's axis and nonrotating (RA/dec). The trig equations have the effect of changing the angles in a way that corresponds to tipping a globe from having the pole to some other point on top, and finding what location on the globe is exactly underneath (i.e. between it and the globe's center) of a distant object. Hour angle tells what phase of the Earth's rotation one is in relative to an agreed zero-point direction in the distant Universe.

geedeck
2011-Jan-19, 07:12 AM
Hi geedeck,

those are the standard formulas for converting between terrestrial coordinates (azimuth and altitude) and celestial coordinates (hour angle and declination), although I suspect there's at least one minus sign missing somewhere. What's your precise question? How to work with these equations? Or how they come about?

For now, (pragmatically) my question how to work with them. So we were also given this equation, and if I'm starting to understand it this is the one I need to use first? (per http://spider.seds.org/spider/ScholarX/coords.html#trans_he )

cos a * sin A = cos Dec * sin HA
sin a = sin B * sin Dec + cos B * cos Dec * cos HA
cos a * cos A = - cos B * sin Dec + sin B * cos Dec * cos HA

I've got Dec, HA and latitude... But I don't see where latitude fits in on the equation. Being a length and not a degree, does that mean it's 'a'?

Local sidereal time is different for different observers. When the equinox passes through one observer's meridian, it has already passed through the meridians of observers farther east and has yet to pass through the meridians of observers farther west. Their respective local sidereal times are therefore different, depending on their geographical longitudes.
Instead of tabulating local sidereal times for all possible observers, almanacs give the local sidereal time for Greenwich (i.e. the hour angle of the equinox as observed in Greenwich). Knowing the Greenwich hour angle of the equinox, GHA, and thus Greenwich sidereal time GST, you can easily work out your local sidereal time, taking the difference in longitude into account. Since the almanac will only give GST for Greenwich midnight, you'll also have to correct for the time elapsed since then.

Okay, so 9pm EStanT converts to 2am GMT
and 2am GMT converts to 2:02:19.7 GST
do I then just subtract 5:32:01.1 of ESidT from GST to get LST?

... I know there's something wrong w/ the last part. I know GSidT (GHA of the Equinox) for Jan 18th is supposed to fit in, but I don't see how. It's 7:48:14.3. But I'm unsure of how to wrap it up.

geedeck
2011-Jan-19, 07:17 AM
Oh, and thank you for the help folks. I really appreciate it.

plphy
2011-Jan-19, 10:20 PM
Okay, so 9pm EStanT converts to 2am GMT
Yes, EST is five hours behind GMT, so at 9 pm EST it's already 2 am on the 18th in Greenwich.

and 2am GMT converts to 2:02:19.7 GST
No, "2am" means that the clock has advanced by two hours since last midnight in Greenwich. As sidereal time is faster by a factor of about 1.0027, 2.0055 hours (= 2h 0m 20s) of sidereal time have elapsed. Therefore sidereal time is now 2h 0m 20s later than the sidereal time which coincided with midnight GMT.

Your almanac tells you that sidereal time at midnight GMT of the 18th in Greenwich was 7h 48m 14s. Adding the elapsed 2h 0m 20s of sidereal time gives 9h 48m 34s as the local sidereal time in Greenwich at 2am GMT = 9pm EST.

do I then just subtract 5:32:01.1 of ESidT from GST to get LST?
You are 83° west of Greenwich, and the equinox is drifting in a westward direction. When it is crossing Greenwich's meridian (i.e. local sidereal time at Greenwich is zero), it has yet to cross yours. Your local sidereal time at the same moment is therefore less than zero; it will be zero when the equinox arrives at your meridian a few hours later.

The equinox drifts westward with a speed of 15.041° per clock hour or exactly 15° per sidereal hour (360° in 23h 56m 04s of clock time, and in 24h of sidereal time). It will reach you 83/15 = 5.533 sidereal hours after it has passed the Greenwich meridian.

Your local sidereal time at any instant is therefore 5.533 sidereal hours (5h 32m 00s) less than the sidereal time occurring simultaneously in Greenwich. So you have to subtract 5h 32m 00s of sidereal time from Greenwich sidereal time to find your simultaneous local sidereal time.

9h 48m 34s - 5h 32m 00s = 4h 16m 34s local sidereal time. My GUIDE 8 software says it's 04h 16m 32s, so we agree within two seconds.

Bye,
Thomas

plphy
2011-Jan-20, 01:02 AM
For now, (pragmatically) my question how to work with them. So we were also given this equation, and if I'm starting to understand it this is the one I need to use first? (per http://spider.seds.org/spider/ScholarX/coords.html#trans_he )

cos a * sin A = cos Dec * sin HA
sin a = sin B * sin Dec + cos B * cos Dec * cos HA
cos a * cos A = - cos B * sin Dec + sin B * cos Dec * cos HA

I've got Dec, HA and latitude... But I don't see where latitude fits in on the equation. Being a length and not a degree, does that mean it's 'a'?

Yes, that would be the set of equations you use to find an object's azimuth A and altitude a when its declination Dec, hour angle HA and the observer's geographical latitude B are given. Latitude B is an angle (it's the angle the line from the Earth's center to the observer makes with the Earth's equatorial plane), so it's fine to take the sine or cosine of it.

The second equation has only one single unknown quantity, and that happens to be the altitude a in which you are interested. All the other quantities are known. What a stroke of luck. You even have all the knowns on the right side and the single unknown on the left side, so no rearranging is necessary. Inserting the knowns (Dec = 20°, HA = 4h = 60°, B = 40°, to take one of your examples) and evaluating the formula (with your calculator set to expect degrees, not radians!), we obtain:

sin a = 0.5798.

Now you know that the sine of the altitude is 0.5798. Applying the inverse of the sine function, the arcsine, to both sides of the equation yields the altitude angle:

arcsin(sin(a)) = arcsin(0.5798)
a = 35.43°

If the zenith distance z is required instead of the altitude angle, subtract a from 90° and find z = 54.57°

There is one possible problem involved which is worth thinking about for a moment. The sine function takes an angle and returns a number between -1 and 1. The angle may come from the entire interval between -infinity and +infinity. Many angles, when plugged into the sine function, yield the same function value. For example:

...
sin(-570°) = 0.5
sin(-330°) = 0.5
sin(-210°) = 0.5
sin(30°) = 0.5
sin(150°) = 0.5
sin(390°) = 0.5
sin(510°) = 0.5
...

So if you have an angle and want to know its sine, then there is a unique answer. But if you have the sine of an angle and want to know what the angle is, there's an infinite number of angles which could have produced that sine value. Which one is the right one?

By convention, the arcsine function returns that angle (out of the infinitely many possible angles with the same sine value) which lies in the interval -90° .. +90°. Fortunately, you are looking for an angle which measures the altitude of a celestial object, and altitudes must always be within the interval -90° .. +90°. So the arcsine function happens to automatically select that angle which makes sense when physically interpreted as an altitude, and you don't have to worry about which angle to select.

What if the azimuth of the celestial object was requested? Another task occurring often in astronomy.

You take equations 1 and 3,

cos a * sin A = cos Dec * sin HA
cos a * cos A = - cos B * sin Dec + sin B * cos Dec * cos HA

and divide equation 1 by equation 3. The "cos a" on the left-hand side cancel, and we have

tan A = cos Dec * sin HA / (- cos B * sin Dec + sin B * cos Dec * cos HA)

Again, all the quantities on the right-hand side are known, and the values of our example produce

tan A = 20.340

Taking the arctangent of 20.340 should yield the azimuth A.

BUT...

Here we do have a problem: again, there is an infinite number of angles whose tangent is 20.340. By convention, the arctan function returns that angle which lies between -90° .. +90°. But azimuth angles measure the whole horizon and thus cover the interval -180° .. +180°. In that interval, there are usually two angles for which the tangent function produces the value in question. The arctan function will return the angle which is in -90° .. +90°, but that may be the wrong one. In our example, we have

tan(87.19°) = 20.340 and
tan(-92.81°) = 20.340

Both 87.19° and -92.81° are legitimate azimuth angles (lying in the interval -180° .. +180°), but which one is the correct one? Do we have enough information to find out?

We do. I won't go into much detail, but note that our angle A is given as the arctangent of a fraction

A = arctan( cos Dec * sin HA / (- cos B * sin Dec + sin B * cos Dec * cos HA) )

with the numerator "cos Dec * sin HA" and the denominator "- cos B * sin Dec + sin B * cos Dec * cos HA". You can easily convince yourself that if we know the signs of the numerator and the denominator, then we can determine which quadrant the angle A comes from:

numerator > 0, denominator > 0: the angle arctan(numerator/denominator) lies in 0° .. 90°
numerator > 0, denominator < 0: the angle arctan(numerator/denominator) lies in 90° .. 180°
numerator < 0, denominator < 0: the angle arctan(numerator/denominator) lies in -180° ..-90°
numerator < 0, denominator > 0: the angle arctan(numerator/denominator) lies in -90° .. 0°

So if we know the signs of numerator and denominator separately, we also know the correct quadrant, and we pick that angle which lies in that quadrant.
(If we perform the division without taking note of the individual signs, then we only know the sign of the result, and this always leaves two possible quadrants. In our example, we know that we are looking for the arctangent of the number 20.340 which has a positive sign, but we don' t know if 20.340 is the result of a division where numerator and denominator had both positive sign or both negative sign. In the former case, the quadrant 0°..90° would be the correct one, and we'd have to pick the angle 87.19°. In the latter case, the quadrant -180°..-90° would be the correct one, and we'd have to pick the angle -92.81°).

Here is a simple recipe for dealing with all this quadrant business:

compute the numerator "cos Dec * sin HA"
compute the denominator "- cos B * sin Dec + sin B * cos Dec * cos HA"
note the sign of the denominator
compute arctan(numerator / denominator)
if the denominator was negative, add 180° to the result.

This will always result in the correct angle. In our example, we have numerator = 0.8138 and denominator = 0.040009, with the denominator being positive. Then we compute arctan(0.8138/0.040009) = arctan(20.340) = 87.19°. Since the denominator was positive, there is nothing further to do, and 87.19° is the final result. (Had the denominator been negative, we would have computed 87.19° + 180° = 267.19°, which is the same as -92.81°, the other one of our two possible results found above.)

A final note: the equations we used assume that azimuth=0° corresponds to South, which is the usual convention in astronomy. If you prefer to work with a convention assuming that azimuth=0° corresponds to North, simply add 180° to the result obtained from the equations.

Bye,
Thomas

forrest noble
2011-Jan-20, 01:48 AM
Howdy folks, I'm new around here but I hope you don't mind me asking some questions.

I'm taking an astronomy statistics class, and as someone who is mostly an art and computers guy, it's kind of 0_o for me. I've worked with my TA some but well, he has a bad opinion of anyone who would study astronomy who isn't actually trying to get a degree in it. The teacher is much better, but there isn't the same amount of time available to work with him.

So I'm working on some problems, and I can kind of punch the numbers in, but I don't really understand why. So I don't want someone to answer the problem for me, but if they could point me at good resource that explains why things are done, or can explain why I'm doing what I'm doing, man I would be grateful.

So for example, I'm trying to find the zenith distance of a star, declination delta of +20°, observed at hour angles 0, 2, 4 and 6, done from a latitude of ~40° N.

We've got this equation:

cos Dec * sin HA = cos a * sin A
sin Dec = sin B * sin a + cos B * cos a * cos A
cos Dec * cos HA = cos B * sin a + sin B * cos a * cos A

But I don't really get what I'm doing with this equation. I mean I can plug numbers in woo but... why? And I admit, trig kind of blows my mind, but well I'm tired of that and just want to understand that.

The other one I don't get is where we're supposed to find the LST for Jan 17 2011 @ 9pm for 83° W, 40° N. I've got the GST table from the 2011 Astronomical Almanac for the time. This one makes a bit more sense, I've calculated the sidereal interval (2:02:19.7) and I know the sidereal time adjustment for EST, ie 5:32:01.1 and I know that GST for Jan 18th is 7:44:17.7... Okay but I don't really get what that last number *means*. The subtext for that last one is GHA of the equinox... But I don't get what that's quantifying in real people speak. And I don't get how to pull it all together, but the TA did admit that knowing the sidereal interval gets me 1/3 the way there.

It's like someone dumped a bucket of numbers confetti on me! And it's awesome! Except it'd be really awesome if I could actually build a picture out of the confetti.

tl;dr help a spunky college student defeat his smarmy TA that looks down on an artist trying to learn more about astronomy.

I've also taught for awhile and would first ask, do you really understand where the equations come from in the first place that you are using and why they were set up that way and why you are using this or that equation for a particular application? If you cannot derive the equations in the first place then I think something is missing from your full understanding. Once you've derived these equations yourself from nothing then you have a pretty good understanding of why you are using them. After you can derive them once, then you do not have to do it again, you can simply remember them. My experience concerning math and physics has to do with students not understanding where the equations come from. There are a number of exceptions to this general idea but I don't know of any exceptions relating to an improved understanding of the subject via an understanding of where the equations came from in the first place. The equations of Quantum Chromodynamics for instance, according to what I have read, came from a long history of experimental results that enabled their equational derivation. GR came from many recorded observations that enabled a mathematical analog to be developed and tested against those observational results. What you are looking at in your example equations are relatively simply trigonometric relationships, whereby you could read the derivation of these equations from a textbook on trigonometry. Once reading of it and understanding it, you should be able to repeat the derivation without referring back to the text.