user:Tango has deleted one of you contributions on the reference desk. If you'd like, if may be discussed on the talk page. Buddy431 (talk) 17:53, 31 December 2011 (UTC)[reply]
This notification could have been better worded. Perhaps:
Hi ToE. I have given quite a bit of thought to the various ideas we posted at the Reference Desk. I can explain why specific air range (air distance travelled per unit mass of fuel) is lower for an aircraft flying towards the equator, relative to another flying east or west. (It is because, to follow a meridian, the Coriolis force must act on the aircraft and this must be a component of thrust. Less than 100% of thrust is available to balance drag so more thrust is required to maintain airspeed, or airspeed will be slower if thrust is unchanged.)
However, I have been unable to find an explanation for why SAR might be higher for an aircraft flying towards the pole. Consequently I must concede specific air range will be the same for an aircraft flying north as for one flying south. I have posted my latest thoughts below yours. See my diff. Thanks for the discussion! Dolphin(t) 11:58, 29 January 2012 (UTC)[reply]
Section is archived here. I had promised to answer a few specific questions later, but VR bumped his archiving rate from two weeks to one day, so I assume he just wanted to let the entire discussiondie. -- ToE 00:57, 16 February 2012 (UTC)[reply]
Just swap the words around: Nunquam Fidicen Semper Sarciens. Each pair of words is a clause in itself, so the endings of the nouns are not affected by the change (it's not like you're swapping the subject and the object of a verb). -- AnonMoos (talk) 23:54, 7 June 2012 (UTC)[reply]
Sarciens is a participle, and participles can have some characteristics of verbs, nouns, or adjectives in various contexts (in Latin or in English). However, in this sentence, it functions as a noun... AnonMoos (talk) 07:31, 8 June 2012 (UTC)[reply]
Not without making it a lot more complicated... AnonMoos (talk) 09:58, 8 June 2012 (UTC)[reply]
Thanks again! Don't worry, i don't feel offended when you said that. I like frank people anyway and i'm always glad to hear criticizes from other people. I'm not going to lie but criticizes are actually helping me to be a lot better. I'm also glad to admit things that made up of me. I just asked another question in reference desk with the section named "number 26". Hope you can help me again. Hope i don't bother you too much. I have been trying to study astronomy for months now. Its knowledge is like infinity (i know it's not but it seems like so to me). It comes in so many variety ways that one has to spend all of their high school years to master it and even then there are still some problems that they don't even know. I got to admit that i have a long way to go!! I know for the fact that i'm not going to master all of it this year for sure. Well i think i really need to push hard now. 2 weeks from now is when the competition holds. So i have 2 more weeks to study and that's it. I just going to just my best to get as much out of it as possible, the more = the better.
By the way, are you living in England? Nice to meet you! I'm living in the United States. If possible then we can chat through yahoo or something? It's faster that way since i can ask and talk to you directly. I think i can learn a lot faster by doing so. I'm thinking perhaps you can spend like 10 minutes everyday to teach me astronomy for the nest 2 weeks? We can arrange the time. Anyway is fine to me except my school time. It's ok if you don't want to. If you want to then we can E-mail each other to give contact info through Email this use function on Wikipedia. You know this is my dream since i was very little, i have done everything that i possibly could to try to fulfill my dream but i need a teacher like you to help since no one else can. You're going to my savior if you help me. I have nothing to pay back for you now but one thing i can guarantee is i won't forget you. When i grow up, i'm promised to help you back when you need to to show my grateful for what you have done to me. I know it's kind of a long time from now. My goal is to get to this, sometimes in my high school years! Thanks in regard!Pendragon5 (talk) 20:47, 19 February 2012 (UTC)[reply]
No, I'm not in England, but am on a sailboat currently cruising Southeast Asia. With dodgy and intermittent internet access, this is probably my best means of communicating. If you are concerned about overtaxing the ref desk, you are welcome to vet problems with me if you think they may just be a matter of applying equations. I'm not particularly well versed in astronomy, but am happy to help where I can.
Is this your first shot at the IAO, and how many more tries will you have? Is the practice test you are working available online, or something that was mailed to you? Is there a forum of students discussing the practice test and other elements of preparing for the IAO? -- ToE 01:20, 20 February 2012 (UTC)[reply]
NICE! Southeast Asia is where my home country belongs to before i moved to America! Specifically is Vietnam. Whenever you got a chance to visit Vietnam, let me know. I still have relatives there in Ho Cho Minh city. I'm sure they will welcome you. They don't know how to speak English though. But well i can work things up for you easily. You can stay at my relative's house or eat meals there. So should i just post problems on your talk page here? To be honest, i don't get as much help out of reference desk as much as i got out of you. I think you're really good at astronomy for the most part. Most people know nothing. The practice test is from my teachers. I think the national Science Olympiad passed them out for participate schools. There is a forum but well students are usually don't use it anyway. The forum is just useless in my opinion. Well this is my first year of participate in astronomy yea. I'm currently a sophomore in high school. During my freshman year, i was going to do it too but i missed it because i forgot the deadline. It's not how many more tries, it is not simply just one shot in. There are regional, state, nation competitions and last international. The one 2 weeks from now is the state one so yea. Technically i have 2 more years so i would consider that 2 more tries. On the test there are two parts, one is general knowledge about astronomy and the other one is the specific topic they want you to focus on. Every year they changed the specific topic material. Like what stars, objects in space... they want you to focus on. The specific topic part is pretty easy to me i think, i have no problem with that. It's just like normal studying for my classes but the general knowledge concepts ones are hard. All the problems i have been asking on here are fall into the general concepts parts. The general part is like "everything" about astronomy can be possible in there so the more you know about astronomy the better. I have no clue what other elements needed for astronomy. Science Olympiad is all independent. Students have to start learning from scratch. Their high school teachers won't be able to help the students either because they don't even know the material. How is each student going to prepare for it is one's own business. There is no guild line or anything, you study whatever way you want. The method i'm using to study for it is try to do past events tests.Pendragon5 (talk) 20:25, 20 February 2012 (UTC)[reply]
Can you help me on number 16? Thanks!Pendragon5 (talk) 20:27, 20 February 2012 (UTC)[reply]
By the way, thanks a lot for your effort on number 26. I have strong feeling that they must have made a mistake or they supposed to be a lot more clear on what exactly are they asking. I just gave up on that one for now. Thanks!Pendragon5 (talk) 20:48, 20 February 2012 (UTC)[reply]
Did you get an answer for Q15, as it is required for Q16? Also, what does the spectral information in Q16 tell you? Anything? -- ToE 20:53, 20 February 2012 (UTC)[reply]
Yep i got the answer for question 15. It is about 218 km/s. Star C and D have the same orbital velocity i think. Well spectral tells me about the star's temperature bases from this. What is the H are they talking about? I'm kind of clueless on this one sorry.Pendragon5 (talk) 21:09, 20 February 2012 (UTC)[reply]
Hα. What does it mean that the observed spectral line from the star has a different wavelength than that observed in the lab? -- ToE 21:18, 20 February 2012 (UTC)[reply]
I don't get your question "What does it mean that the observed spectral line from the star has a different wavelength than that observed in the lab?". Do you mean like the question states the observed spectral line from the star has a different wavelength than that observed in the lab? I think so yea. I don't have a picture in my head what angle are trying to solve? I don't know how to interpret spectral line yet. The answer for this should be from 50-70 angles (all they wanted is really rough estimate).Pendragon5 (talk) 21:35, 20 February 2012 (UTC)[reply]
I don't know what does red shift has to do with the angle in this question. Sorry if i'm acting stupid right now but i'm really clueless on even which direction to go to solve the problem. I don't even understand the question them self.Pendragon5 (talk) 22:00, 20 February 2012 (UTC)[reply]
I think that I do see what they are getting at, and if so, then this question has a lot of unwritten assumptions as well. Have you calculated the redshift z and radial velocity? -- ToE 22:13, 20 February 2012 (UTC)[reply]
I just went for a walk as a break. Just got back. Anyway i don't know how to use the formula so i didn't get to calculate the redshift z and radial velocity. I don't understand the radial velocity term either. It's basically how fast the star moving toward us or moving away from us right? Then radial velocity should be the same as the orbital velocity of Star C, which is 218km/s. Can you do it as an example? If possible draw on the piece of paper to help me visualize. I don't even know what angle are they talking about.Pendragon5 (talk) 23:09, 20 February 2012 (UTC)[reply]
(outdent) The unstated assumption here is that the binary star system has no significant overall motion with respect to sol, so the redshift is entirely due to the orbital motion of the stars. Depending on the angle of inclination of their orbit, only a fraction of their orbital velocity will be radial to us. If their orbital plane were perpendicular to us, then none of their orbital velocity would be radial to us and there would be no redshift. At the other extreme, if we were seeing their orbital plane edge-on, then at points in the orbit all of the orbital velocity is radial to us, while at other points none would be. The other unstated assumption here is that the redshift varies with time -- that it is different at different points in the orbit, and that the value given for the star's Hα is just one measurement. If what they gave you represents an extreme value, then you should be able to calculate the inclination. If it was just some random value within its range, then you should be able to determine a range of possible values for the inclination. So give me some numbers. λ is wavelength, and the assumption we are all making (well, most of us) is that the wavelength of Hα emitted by the star is the same as what is observed in the lab. Then you should be able to use the non-relativistic approximation where c is the speed of light. What is z? What radial velocity does that imply? Then think about the geometry of the inclined orbit. -- ToE 00:32, 21 February 2012 (UTC)[reply]
I don't know how to find the inclination. I can't picture the problem in my head. I have never do this type of problem before. Sorry that i still don't get the concepts. Anyway speed of light is 186,282 miles per second. 656.5386/186,282 = so z = approximately .00352? I'm so not understand anything on this, i have been trying to understand it for few hours now T_T. I deserve to be called dump now. Can you just do it for me as an example? If by then i still don't understand then i won't bother you anymore.Pendragon5 (talk) 00:57, 21 February 2012 (UTC)[reply]
Try again to calculate z. The formula you quoted for z didn't have c in it; only the two wavelengths. The formula with c is for calculating the radial velocity v. (See redshift.) And keep the units in all your work as it will help you catch a lot of errors. z is a unitless term, but your calculation above would have yielded nm/(m/s) or units of time. (Did you catch that nm was nanometers, a common unit for measuring the wavelength of light?) I've got to run now; we'll work out the geometry later. How much trigonometry do you have? (What is needed for this problem can be easily introduced from scratch.) -- ToE 01:14, 21 February 2012 (UTC)[reply]
Well i don't know how much do i need to know to consider is a lot nor do i know how much is little but i can tell you is the highest math class i ever finished is precalculus. Pendragon5 (talk) 01:55, 21 February 2012 (UTC)[reply]
All you will need is an understanding of the definition of sine and cosine (and the intuition to use them properly). -- ToE 11:34, 21 February 2012 (UTC)[reply]
So there are two different kinds of z? The second z is the radial velocity, how about the first z? What is the first z stands for? And i do know that the wavelength of light is in nm. Anyway basically i have to put them all in the same unit when i calculate right?Pendragon5 (talk) 01:59, 21 February 2012 (UTC)[reply]
Uh oh! I don't know what I said to create this confusion. We've only been discussing one z. It is a unitless parameter which is a measure of the redshift, it is calculated from the observed wavelength, and it is used to calculate the velocity. This same z appears in both equations. -- ToE 11:34, 21 February 2012 (UTC)[reply]
And to calculate the z, radial velocity. The wavelength needs to be in nm and the speed of light needs to convert to meter per second right? 186,282 miles X 1.609344 = 299,791.819008km X 1000 = 299,791,819.008 m/s. So 656.5386/299,791,819.008 = approximately .0000021899. I have no idea what is that number suppose to mean and how it is going to help me solve the problem.Pendragon5 (talk) 02:05, 21 February 2012 (UTC)[reply]
Keep your complete units in place and do math to them. You may not have been required to do so up to this point, but it will help you now, and it will be required at some point in the near future. Have you learned the rules for this? You can only add or subtract like units, and they yield the same unit. For instance, you can write 123 minutes - 180 seconds = 2 hours (all units of time), but it doesn't make any sense to add or subtract dissimilar units; for instance, I can't compute 3 days + 3000 miles. Note that you can compute 1 meter - 1 foot, as they are both lengths, but as with the time calculation, you have to scale the values appropriately. You can multiply or divide any units and they are treated the same way as the numbers. So 25 meters/second ÷ 5 meters/second2 = 5 seconds, but 25 meters/second ÷ 5 meters/second = 5, a unitless value. Do you see how that works? You can also square or take the square root of units just as with numbers, so that sqrt(25 meters2/second2) = 5 meters/second. Note that logarithms and exponentiation are only done to unitless values, so that Apparent magnitude#Calculations, for instance, has the formula where the argument of the logarithm is the ratio of two fluxes, so that the units will cancel out.
Applying this to the work you just did, you should have written: 656.5386 nm / 299,791,819.008 m/s = 0.0000021899 ns, and then said, "What?" Not only does your number not make any sense (which you recognized), the units don't either. What are we getting a time for? We'll figure out what's going on with the formulas later (hint: z is not the radial velocity), but first ...
Use Units conversion by factor-label properly. I am glad to see you effortlessly convert c from miles/second to meters/second, but your work should look like this: 186,282 miles/s X 1.609344 km/mile = 299,791.819008 km/s X 1000 m/km = 299,791,819.008 m/s. Notice that when you multiply out miles/s X km/mile X m/km = m/s the units cancel out to give you what you want. Always put the complete units on your conversion factors. Your mathematical justification of multiplying a speed in miles/second by 1.609344 km/mile is that because 1 mile = 1.609344 km, 1.609344 km/mile = 1, and you can multiply any term by 1 and not change its value. Also, if you are heading into astronomy or physics, you should get used to using c in SI units, both mks and eventually cgs. (All the general relativists love cgs -- when they are not working in natural units .)
Do you understand redshift and Doppler effect well enough to see how measuring the shift in wavelength of a spectral line can be used to determine the relative radial velocity, and do you understand why the radial velocity has the dominant effect on redshift, and the transverse velocity is not a significant factor until the velocities approach relativistic speeds? You don't need to understand the relativistic part yet, but you should be able to look at Q16 and immediately know whether Star C, at the moment of its spectroscopic analysis, was moving towards the earth or away from the earth. Which is it and why?
OK. Let me know what we need to do to make the above points clear. Once they are, we will figure out what the two redshift formula do for us: and . -- ToE 05:45, 21 February 2012 (UTC)[reply]
Unfortunately i'm not familiar with redshift and doppler effect well enough (i think i understand them little to nothing, not even close to well enough). Basically the entire paragraph second to the last one is completely something i'm totally clueless on. And i got the points above clear enough now i think. So basically i have to put them in the same conversion unit to cancel them out to get the z. So 656.5386 nm X 10^-9 = 6.565386 X 10^-7 m. So 6.565386 X 10^-7/299,791,819.008 m/s = 2.1899 X 10^-15. I also have a question. and why there are two formula for z? What are the differences? I got to admit this is the hardest thing i have been working on in astronomy, so confusing (maybe cuz i'm dumb?). Can you explain which formula individually? And how am i going to plug the information i got from the question into the formula? Thanks!Pendragon5 (talk) 22:22, 21 February 2012 (UTC)[reply]
Don't get too discouraged; you simply have not yet developed the level of mathematical sophistication that is expected from a student tackling problems like these, and this sophistication comes from experience. The confusion is bi-directional, as I've often not understood where your misunderstandings lay, but I think I'm starting to catch where you are coming from. In the context of this problem, and are not two different formula for z, but one formula which lets you calculate z from your spectral information, and another which lets you use z to calculate the relative radial velocity (here v). You probably wouldn't have had any confusion if the second formula had been presented at , or if the two had been combined as , would you? Well, don't expect to be hand fed such pat formulas in the future. You are expected to combine and manipulate equations so that you can solve for your desired value in terms of known values. The reason z is broken out separately isn't just to give you a hard time, but because this redshift factor z can be computed several different ways and used in several places, and combining them all would yield way to many overly complicated equations. Note also that you may drive some redshift problems in reverse, where you know the velocities and need to compute the wavelength of some spectral line. -- ToE 23:59, 21 February 2012 (UTC)[reply]
Oh, and stop dividing λ / c for this problem. I harped on it earlier not just because it was the wrong thing to do here, but because it was the wrong thing to do and you were doing it badly. Notice that when computing , the numerator is the difference of two wavelengths (wavelengths just being lengths) and thus is a length, and the denominator is a wavelength (a length), so the quotient is unitless, and as long as all your wavelengths are in the same units, (whether nanometers or picofurlongs) no unit conversion is required. Then use you z in the second equation to compute the relative radial velocity. Back to dividing λ / c, this result which has units of time (do you understand how length / speed = time ?) is the inverse of the frequency of the light with that particular wavelength. c = λ f. -- ToE 00:20, 22 February 2012 (UTC)[reply]
And again noting that there is no reason for us to calculate λ / c, you are still doing it badly:
You wrote: So 656.5386 nm X 10^-9 = 6.565386 X 10^-7 m.
Don't write it that way because 656.5386 nm X 10^-9 = 6.565386 X 10^-7 nm.
Instead write either 656.5386 nm X 10^-9 m/nm
or 656.5386 nm X 10^-9
You wrote: So 6.565386 X 10^-7/299,791,819.008 m/s = 2.1899 X 10^-15.
Where did the units go? It should be: 6.565386 X 10^-7 m / 299,791,819.008 m/s = 2.1899 X 10^-15 s. We didn't need to calculate that, but since your did, we might as well see what it means. Because c = λ f, you computed λ / c = 1/f, so f is the reciprocal of your answer. 1/2.1899 X 10^-15 s = 4.5664 x 10^14 s^-1 = 4.5664 x 10^14 Hz which is the frequency of the Hα band received from your Star C. BTW, have you taken High School Chemistry yet? That is where the factor-label method is usually introduced, but it is used throughout science. -- ToE 01:17, 22 February 2012 (UTC)[reply]
Oh wow. It turned out i messed up pretty badly. λ observed is 656.5386 nm right? And λ emit is 656.3 nm? So it that so then (656.5386 - 656.3)/ 656.3 = approximately .00036 for z X 186,282 miles/s = approximately 67.7234 mile/s, so it's the radial velocity right here right? If the λ emit is 656.3 nm then what i don't understand is why the laboratory value is the λ emit? What the λ emit even means?Pendragon5 (talk) 00:32, 22 February 2012 (UTC)[reply]
Good question, and it is important to understand redshift as it is used a great deal in astronomy. You should read our redshift and Doppler effect articles to the best of your ability. I know that many of our physics articles are not written for beginners -- it is hard to balance completeness and accuracy with ease of understanding -- so you may want to seek out other sources as well. When we study hydrogen in the laboratory we see characteristic spectral lines, one being the H-alpha line with a wavelength of 656.281 nm. When we look at light from stars we see the same pattern of spectral lines, but their wavelengths are not always quite the same -- in the case of this problem we are seeing an Hα of 656.5386 nm. Our assumption is that the Hydrogen in the star is the same as the hydrogen on earth, so the emitted wavelength is the same a our laboratory measured wavelength, and that the wavelength is redshifted or blueshifted due to the Doppler effect. Note that this is not a blind assumption. Young physicists are all the time trying to figure out if the laws of physics are different in different parts of the universe (see our article Variable speed of light, for instance), but there appears to be no reason to doubt that λ_emitted = λ_laboratory. I have no idea why Q16 gives you the observed wavelength to 7 significant figures, but rounds the laboratory value to 4 significant figures, but your should probably go ahead and use the value they gave you. -- ToE 01:05, 22 February 2012 (UTC)[reply]
Alright λ_emitted = λ_laboratory is all i need to know for future uses. I didn't know λ_emitted = λ_laboratory at all since the beginning, no wonder why i was like so lost. I still don't understand the concept of λ laboratory. Why is it 656.3 in the problem as they state but not 656.28 nm? Or is .02 is too small and they want us to not even consider about it? Oh wait a minute, they rounded it on purpose right? Alright we are at 67.7234 mile/s then what? What is next? What do i do with the 67.7234 mile/s radial velocity to find out the angle, and sorry that i'm pretty stupid on this. Until now i still can't imagine up what the angle am i trying to solve T_T. Thanks!Pendragon5 (talk) 01:40, 22 February 2012 (UTC)[reply]
Whoever wrote the problem should not have rounded λ_laboratory, but they did, so that is what you should use. I assume that you know the speed of light to be 186,000 miles/second or 186,282 miles/second. Learn 3 x 10^8 meters/second. That is accurate to better than one part in a thousand and is the value you will be expected to use in most physics problems. (Use higher precision where necessary, such as during actual experiments, but carrying the extra digits doesn't forward the purpose of most exercises.) The magnitude of 300,000 km/second can be easy to remember if you know 186,000 miles/second and that there is a little more than one and a half km per mile. That gives you c = 3 x 10^5 km/s = 3 x 10^8 m/s. (In graduate school you might hang out with the wacky cgs crowd and use 3 x 10^10 cm/s, but will eventually become a naturalist and use c = 1 with no units at all!) So, what is the radial velocity of Star C in km/s? -- ToE 02:27, 22 February 2012 (UTC)[reply]
To answer the question above you asked, i haven't take chemistry class yet. As my plan, i will take it in my senior year as my junior i will be taking physic class. Approximately 109.065 km/s radial velocity.Pendragon5 (talk) 04:11, 22 February 2012 (UTC)[reply]
"Approximately 109.065 km/s radial velocity." How does that compare to the orbital velocity? What does that mean? Will something look familiar when you work out questions E, F, & G in the model question below? -- ToE 05:22, 22 February 2012 (UTC)[reply]
So it's slower than the orbital velocity which is 218 km/s. About 2 times slower, what a coincidence? I have no idea what is that tells me.Pendragon5 (talk) 22:20, 22 February 2012 (UTC)[reply]
Is Star C moving 109 km/s toward you or away from you, and why? The author of the question set up the numbers so that the radial velocity was half that of the orbital velocity so that the next part would be easy (once you visualize it). You should see the answer as soon as you figure out sub-questions E, F, and G. -- ToE 00:00, 23 February 2012 (UTC)[reply]
File:Astronomystuff9.JPGWhy are we assuming that it is at the straight angle from Earth? I just put 4 as an example, there could be so many locations the binary system could locate compare to the Earth. How do we know which one is which? I still can't understand the angle concept. Since the star is ALWAYS moving then isn't the angle will just always changing? This is so CONFUSING! GOD!Pendragon5 (talk) 03:52, 23 February 2012 (UTC). Note the line on the paper says All four =... And the line to those four ellipse are the line of sight when we look at them if that where the binary system isPendragon5 (talk) 03:57, 23 February 2012 (UTC)[reply]
(outdent)Since it's the red shift then it's moving toward me with the speed of 109 km/s. That doesn't makes any sense to me. The star is simply just rotating around itself, it will never move toward me an inch.Pendragon5 (talk) 02:33, 23 February 2012 (UTC)[reply]
Two problems. First: "Since it's the red shift then it's moving toward me". No. What is the difference between redshift and blueshift? (Read the blueshift article!) Second: "The star is simply just rotating around itself". No. Remember problem Q15. Star C is one half of binary system with Star D. The binary system rotates, but Star C is revolving (or orbiting) around their center of mass at 218 km/s. See the difference? Star C is moving like the tip of our second hand (where the center of the clock is the center of mass of the system). Star C is not represented by the entire second hand itself, but by the very tip. If you wish, you can glue a second second hand to the first that points in the opposite direction. (So one points at 9 while the other points at 3.) Then the tip of one second hand represents Star C while the tip of the other second hand represents Star D. That more fully represents the binary system, but all they are asking about in Q16 is the motion of Star C, so one second hand could do. -- ToE 02:51, 23 February 2012 (UTC)[reply]
Ops, redshift = moving away. Anyway i just found out the different is positive = moving away, negative = moving toward. So it must be moving away from us. And yea by rotating i meant since it is the binary system. I already know the 2 stars are rotating around each other in the ellipse orbit. What i don't understand is, they will stay in that orbit forever until they explode so therefore they actually never move any closer to us.Pendragon5 (talk) 03:42, 23 February 2012 (UTC)[reply]
I think i can picture the binary system now but not the angle we're trying to find. How do we know which way the binary system is rotating compare to our Earth perspective.
Well, from question 15, you know nothing about the orientation of their orbital plane from our perspective, but the fact that they asked you for Star C's orbital velocity implies that its orbit is pretty circular. If its elliptical orbit were highly eccentric then the orbital velocity would vary greatly, being highest at periapsis and lowest at apoapsis, so by asking for the orbital velocity, they are making an unstated assumption that the orbit is roughly circular.
Question 16 gives some more information, which, along with a couple of other unstated assumptions, lets us figure something, but not everything about the binary system's orientation toward earth. First, they clearly want us to use the ratio of the orbital velocity to the observed radial velocity to determine the angle of the orbital plane wrt (with respect to) our line-of-sight. There is an unstated assumption here that the radial velocity is entirely due to the orbital velocity at some angle to the line of sight. If we were told that the binary system as a whole had some net velocity toward or away from the earth, we could take it into account, but since we were not given such information, we can't do anything with this problem unless we assume that there is no net radial velocity of the entire binary system.
Up to this point we haven't assumed anything about orientation of the binary system with respect to us, the observers. It could be squarely facing us (90°) so that the circular orbit actually appears circular to us, and could be rotating either clockwise or counter-clockwise as viewed by us. Or it could be edge on (0°) so that the stars don't trace out a visible circle or ellipse from our point of view, but just appear to move back and forth. Or it could be anywhere between 0° and 90°. But wait, if it was at 90° (square on to us) then we would never detect any radial velocity (just like the tip of the second hand on the clock aimed squarely at you which never comes any closer nor moves any farther away, even momentarily), and the math you've done with the redshift shows a radial velocity of half the orbital velocity. So we know that the orbital plane is not at 90° to our line-of-site, that it is not squarely facing us. It must be inclined with respect to our line-of-sight, sufficiently far from 90° for the for orbital velocity of Star C to, at some point in its orbit, manifest itself as a radial velocity of the value you calculated.
But, as you have described in the clock face edge on, the tip of the second hand is at times moving towards you, at times moving away from you, and at times doing neither. So with an orbital inclination of 0°, the radial velocity (wrt you, the observer) is constantly changing. This exposes another unstated assumption of the question, that the redshift they gave was not just from a random measurement. Had they measured the redshift for a full ten days (the binary system's period), they would have found at times a blueshift, at times a redshift, and at times neither. Since they are asking for "the angle between the plane of the systems orbit and the line of sight of the observer", and this does not change as Star C orbits -- the star will be at different places in the orbital plane at different times, but the orbital plane remains the same -- they they must have given you a particular redshift, and the only one which make sense to give is the maximum one. So the radial velocity you calculated must be the maximum Star C experiences (wrt to you, the observer) anywhere in its orbit. Have you visualized the orbit sufficiently to see where that occurs? Then you just need too determine the inclination where that maximum radial velocity matches the one you calculated.
Finally, there are a couple of things we will never know about the orientation of the orbit because not enough information was given. We don't know if the star's orbit is clockwise or counter-clockwise as viewed from the earth, and we don't know in what direction the orbital plane is inclined. This is equivalent to not knowing if the clock started out facing you or facing away from you, and not knowing which edge your assistant tilted toward or away from you. Did your assistant tilt the "6" up toward you, the "12" down toward you, or the "3" or "6" to the side toward you, but that is OK, because we weren't asked that information; we only have to figure out the angle of inclination. For ease of visualization, without loss of generality, you may assume that the clock started facing you and that you assistant tilted the "6" up toward you and the "12" down away from you. -- ToE 21:34, 23 February 2012 (UTC)[reply]
You have also expressed confusion regarding the "the angle between the plane of the systems orbit and the line of sight of the observer". Perhaps you should consider a model. Imagine a clock with a second hand which extends just about 13.66 inches from the center of the clock, so that the circle drawn by the tip of the second hand has a circumference of 2180 mm. Now "juice up" the clock so that it spins the second hand around once every 10 seconds instead of once per minute. Do you see the analogy? Orbital period of 10 seconds vs 10 days & orbital velocity of 218 mm/s vs. 218 km/s. The face of the clock represents the plane of our binary system's orbit and the tip of the second hand represents Star C. Have someone hold the clock across the room from you at eye level, pointing squarely at you so that you can clearly see the face. A) what is the angle between the clock face and your line of sight?, and B) what is the radial velocity of the tip of the second hand with respect to you, the observer -- that is, how fast is the tip of the second hand moving toward or away from you? Now have your assistant tip the clock 90° so that you are observing the face edge-on. C) what is the angle between the clock face and your line of sight?, and D) what is the radial velocity of the tip of the second hand with respect to you, does this velocity change with time, and how would you characterize it? -- ToE 01:36, 22 February 2012 (UTC)[reply]
Visualize is never my strong point, especially in this case since i'm completely unknown to the concept. The answer for A should be 0 degree of inclination right? And for B the hand doesn't move closer and way from me because it is just rotating around the clock, and the clock is just stay where it is. I have a feeling that i probably visualize things wrong so far. God, wish you have a camera and just draw it and upload it here, it would be a life saver for me. I having problem visualize this since yesterday even after countless hours of efforts. Sometimes there is something i just don't get it and it remains like that for a while.Pendragon5 (talk) 02:00, 22 February 2012 (UTC)[reply]
Don't get what you mean by "juice it" either.Pendragon5 (talk) 02:03, 22 February 2012 (UTC)[reply]
In other word, do whatever is necessary to make it run faster. -- ToE 02:27, 22 February 2012 (UTC)[reply]
Re answer A, No. If the clock face is perpendicular to your line of sight, what angle does the plane of the face make with your line of sight? -- ToE 02:39, 22 February 2012 (UTC)[reply]
Re: "Visualize is never my strong point" -- Then that is something you better work at. It will be quite important for calculus and analytic geometry as well as vector analysis courses. The flip side is that learning the math will help you visualize better. -- ToE 03:00, 22 February 2012 (UTC)[reply]
Re: "wish you have a camera and just draw it". The camera is not the problem. I am clean out of four dimensional sheets of paper. -- ToE 03:06, 22 February 2012 (UTC)[reply]
Once you answer question D -- and I don't need a mathematical formula; a short description will suffice -- then questions E, F, and G are where the very simple trigonometry (though perhaps not so simple visualization) comes in to play. What is the radial velocity of the tip of the second hand with respect to you when the clock is tilted as E) 30°, F) 45°, G) 60°? At that point the answer to Q 16 will pop out! -- ToE 03:15, 22 February 2012 (UTC)[reply]
Alright i just can't picture it. I don't think i got line of sight either? Is the line of sight is just the line coming out from my eye that goes straight in any direction i look?Pendragon5 (talk) 04:28, 22 February 2012 (UTC)[reply]
Alright let say i clock at this clock from my laptop as i sat exactly in front of my laptop. and let assume this clock is as same as my eye level. Is this the scenario were you describing? Is my line of sight will make 90 degree angle at the plane of the clock?Pendragon5 (talk) 04:25, 22 February 2012 (UTC)[reply]
Re: "Is the line of sight is just the line coming out from my eye that goes straight in any direction i look?" Well, yes, but in the context of the problem they expect you to be looking in the right direction, that is, in the direction of the binary system. In the context of our model, look at the center of the clock face, and if the clock is aimed "squarely at you", then your line of sight will be perpendicular to the clock face, right? Thus your line of sight makes a 90° angle with the plane of the clock face. If it is tilted edge on, where you can't read the numbers at all, then your line of sight makes a 0° angle with the plane of the clock face. At 30° the numbers would be visible, but fairly distorted (foreshortened by a factor of 1:2 because sin(30°) = 0.5). (Note that in analytic geometry, it is common to measure angles from the surface normal, so what we are calling 90° would be 0°, but they are explicitly asking for "the angle between the plane of the systems orbit and the line of sight of the observer".) -- ToE 05:22, 22 February 2012 (UTC)[reply]
So, are you ready to answer question D? -- ToE 15:50, 22 February 2012 (UTC)[reply]
I don't think i ever even hear about analytic geometry before not until you mentioned it to me before. Don't understand this part "(Note that in analytic geometry, it is common to measure angles from the surface normal, so what we are calling 90° would be 0°, but they are explicitly asking for "the angle between the plane of the systems orbit and the line of sight of the observer".)"
Introductory Calculus courses and textbooks are often titled "Calculus and Analytical Geometry", and George B. Thomas's text by that name is particularly well known. The only reason I brought up the concept of the surface normal was that at one point when I asked what angle your line of sight made with the plane of the clock face when it was pointed squarely at you (so that you could read it easily), you replied 0°. The answer was wrong -- when the plane is perpendicular to your los (line-of-sight), it makes a 90° angle with your los (that's what perpendicular means, right), and that's what you need to know for this problem -- but I wanted to point out that your 0° answer made sense in some context. I could expound on the uses of surface normals in astronomy, and discuss why the orbital angular velocity vector is perpendicular to the instantaneous orbital velocity vector, but that is tangential (so to speak) to the discussion of this problem, so if you want to discuss such vectors, ask after we get an answer to this question. The important part is that you understand that when the clock face is aimed squarely at you (so that you can read it easily), the plane of the clock face makes a 90° angle with your los, and when it is turned edge-on (so that you can't see the numbers at all), it makes a 0° angle with your los. Are you seeing that? -- ToE 00:39, 23 February 2012 (UTC)[reply]
The angle between the plane, is it this one? Is it the angle between the blue disc and the red ellipse? And the angle between the line of sight and the angle between the plane as i'm looking directly to to that picture? Or from the left or from the right?Pendragon5 (talk) 22:46, 22 February 2012 (UTC)[reply]
That image is showing a dihedral angle between the intersecting planes, but I don't see how to use it to illustrate our problem. The image File:Angle of incidence.svg which appears at Angle of incidence might help. θ is not the angle we are talking about -- θ is measured against the surface normal as discussed above. What we do want is the unlabeled complement of θ, that is 90° - θ, what they call the grazing angle in that article. -- ToE 00:59, 23 February 2012 (UTC)[reply]
So, if the angles make sense, what is the answer to sub-question D? Specifically, when the clock face is edge on (0° with your los), with the bottom of the clock (six o'clock ) closest to you, what is the radial velocity of the tip of the second hand (with respect to you, the observer) as the second hand passed the 12, the 3, the 6, and the 9? Recall that we constructed the clock so that the orbital speed of the tip of the second had was 218 mm/s. -- ToE 01:10, 23 February 2012 (UTC)[reply]
So the angle we are looking for is 90 degree - angle of incidence = ? angle, correct? As it passed from 12 to 6, it is moving toward me and from 6 to 9 is away from me with the speed of 109 nm/s? What can that information going to help us to solve the problem? Still can't picture the angle we're trying to solve.Pendragon5 (talk) 02:44, 23 February 2012 (UTC)[reply]
Yes, the angle between the line-of-sight and the plane is 90° - θ_incidence, but there is no particular reason to think of it in terms of the angle of incidence here; I was just going after an illustration. Re: "109 nm/s", I'm sure you meant 109 mm/s, but where did that figure come from? What is the instantaneous radial velocity (with respect to you, the observer) of the tip of the second hand as it passes by the 12, the 3, the 6, and the 9? Yes, it will all average out to zero because the clock has no net motion towards or away from you, but the tip of the second hand is at times moving away from you, at times moving towards you, and at times doing neither. If you can characterize that, then you can consider what happens when the clock face is canted at a 45° angle to your line-of-sight, and then the behavior of this binary system may start making sense. -- ToE 04:28, 23 February 2012 (UTC)[reply]
Yea i meant mm, i saw nm for some reasons. This is a similar model of the binary system but in smaller scale, it's in mm instead of km. So the clock has the inclination of 45 compare to my line of sight, and my line of sight is directly point at the center of the clock? I don't know if it's relevant. Then i would guess that would make the speed goes up to 109 divided by square root of 2 mm/s because i just use the triangle 45-45-90. Pendragon5 (talk) 04:49, 23 February 2012 (UTC)[reply]
But where are you getting 109 mm/s from? The tip of the second hand has an "orbital velocity" of 218 mm/s. -- ToE 21:40, 23 February 2012 (UTC)[reply]
I got it my friend! 218 mm/s is the hypotenuse since the whole orbital plane is 45 inclination from the plane. So the 2 legs are 109 due to 45-45-90 angle, one of them is correspond to line of sight that's why we saw its speed 109 instead of 218 because we see it as perspective of the line of sight. I already talked to you some more at the bottom of this page! Everything is all CLEAR NOW!! HAAHOOO!Pendragon5 (talk) 03:20, 24 February 2012 (UTC)[reply]
You do "get it" as is made clear by your answers below, but there appears to be a confusion of numbers immediately above. I doesn't really matter now, but you first appeared to mention 109 mm/s as an answer for the 0° case, where the maximal radial velocity of tip of the second hand is 218 mm/s. Then you seemed to be addressing the 45° case with (218 / sqrt(2)) mm/s ≈ 154 mm/s but again stating 109 mm/s = (218 / 2) mm/s, which, as you clearly now understand, is the answer for the 60° case. -- ToE 00:52, 25 February 2012 (UTC)[reply]
Shoot, something just crossed my mind, not sure it's correct but i have a feeling i got it eventually. No time to do it now, i will figure it out tomorrow. For now it's time to sleep lol.Pendragon5 (talk) 04:49, 23 February 2012 (UTC)[reply]
And let me be clear on something, on number 26, they didn't not ask about orbital angular velocity nor do they ask about observed angular velocity right?Pendragon5 (talk) 21:09, 20 February 2012 (UTC)[reply]
I'm pretty sure they were not asking about observed angular velocity (proper motion), but I do think that they were asking about orbital angular velocity of the stars orbiting their respective clusters. Isn't that what you were thinking when you calculated the 23/2 ratio? (Perhaps we are using the term differently. For a circular orbit the orbital angular velocity is 2π/period.) I've also tacked one more coda to Wikipedia:Reference desk/Science#Orbits of stars within globular clusters. -- ToE 22:30, 20 February 2012 (UTC)[reply]
Yea i was think they asked for orbital angular velocity too.
Re: "Alright λ_emitted = λ_laboratory is all i need to know for future uses."
If you say so. If I were writing these questions (and I could see asking this one, but would have made explicit the couple of unstated assumptions I mentioned at WP:RD/S#Spectrum of stars), I would follow it up with:
Q 17: The spectrum of Star D was measured at the same time that Star C's Hα line was measured as 656.5386 nm. What was the wavelength of Star D's Hα line?
With an understanding of redshift and the behavior of your binary system, this should be a pretty easy question to figure out. -- ToE 07:05, 22 February 2012 (UTC)[reply]
I suppose an even easier question would be:
Q 18: The spectrum of Star D was measured 120 hours after Star C's Hα line was measured as 656.5386 nm. What was the wavelength of Star D's Hα line?
While I wouldn't be surprised if the actual Q 17 matched my suggestion above, there is now way they would use my Q 18 "gimme" question. You see the answer? -- ToE 07:12, 22 February 2012 (UTC)[reply]
I don't even know how to do number 16 yet lol. I have no knowledge of the information i have found and have no idea how to connect them together to find the angle, and again i'm still very blurry about the angle i'm trying to find. This is pretty frustrating. For all problems i have done and my life there is always some pictures were given with them so i never have to visualize things up. Plus it's harder to communicate like this, if you are my teacher in my life then it would be a billion times easier (exaggerated).Pendragon5 (talk) 22:42, 22 February 2012 (UTC)[reply]
File:Astronomystuff8.JPGWhich edge am i looking at when i tilted the clock 90 degree? And tell me edge number as i already labeled on them.Pendragon5 (talk) 22:38, 22 February 2012 (UTC) - - For ease of discussion, have your assistant tilt the lower edge (six o'clock) up toward you. -- ToE 02:02, 23 February 2012 (UTC)[reply]
The beauty of my Q17 is that you don't need to have done any more of Q16 that you have already completed, and of my Q18 is that you don't need to have done any of Q16 at all, other than visualizing the orbital arrangement. But at the moment they are distractions, so let's concentrate on 16. -- ToE 01:28, 23 February 2012 (UTC)[reply]
Alright finally i got everything down now, it would be very easy if you just draw it as the triangle but well by doing that i won't understand the whole picture so i have to learn it in the hard way (so many hours of trying to figuring out what you meant and visualize it to make sense out of it) and it is worth it. Since it, the orbit plane, is tilted 60 degree from out line of sight so the actual velocity is 218 km/s but when we're looking at it by the line of sight it is only 109 km/s. It all made sense now, to find the angle i just have to do arccos: 109/218 = here we go 60 degree! Thank you a lot for your patient and countless effort to explain lengthen lectures. And to do number 17 i need to be clear on something. At the point of Star C is moving 109 km/s away from us, the star D is moving 109 km/s toward us.
So according to the formula and since it's moving toward us it will be (-109,000 m/s)/3 X 10^8 m/s. So the red shift is approximately -.000363. And to inverse the z formula, i got appropriately 656.06154 (note i used 656.3 as the emit).Pendragon5 (talk) 22:13, 23 February 2012 (UTC)[reply]
Congratulations! I'll write up some final thoughts on this problem tonight. -- ToE 03:43, 24 February 2012 (UTC)[reply]
Real-life intervenes; it may be several days. -- ToE 00:26, 26 February 2012 (UTC)[reply]
And forgot to answer number 18, the answer for number 18 is 656.5386 nm haa.Pendragon5 (talk) 19:21, 24 February 2012 (UTC)[reply]
HAAA the nomad planets are pretty weird. I kind of don't understand how it all works out. If they don't orbit any star then how can they form into the planets? And if they don't orbit around any stars what are their movements? Or are they just stand still where they are?Pendragon5 (talk) 04:43, 26 February 2012 (UTC)[reply]
Hi! You know the formula for this? Like if they give me the distance of the stars from Earth or its apparent then i can plug in the formula and find out the other missing information. Thanks!Pendragon5 (talk) 01:25, 1 March 2012 (UTC)[reply]
Do you have a particular problem in mind? Specifically, what missing information are you seeking? The formula given at Absolute magnitude#Computation
comes directly from the inverse-square falloff of radiation (such as light) with distance and the definition of absolute magnitude as being equivalent to the apparent magnitude of an object at the standard luminosity distance of 10 parsecs. If that is what you are trying to determine, then I'd be happy to work throught the formula with you. (Have you done anything with logarithms yet? They are a simple but very helpful tool, especially for working with numbers which vary over a large range.) If the missing information includes more than than, then they are presumably expecting you to use the absolute magnitude along with more given information (star type, for example) to compute additional values (star mass, for example). -- ToE 02:16, 1 March 2012 (UTC)[reply]
Well yea i have done log before. I don't think i can post pictures on Commons anymore since i don't own the copyright of it. I just post it here the last time for temporary use. Alright the problem i'm trying to solve is problem 2 of section C. The number on the x and y axis are not readable in my copy so i just make up some numbers for them. Well it doesn't matter what numbers on there as long as i know how to do it then i can apply it to any numbers.Pendragon5 (talk) 03:09, 1 March 2012 (UTC)[reply]
It would appear that you need to "use your resources and knowledge of variable stars" to determine, from the graph they gave you, what the absolute magnitude of the star is. Once you have that, then the formula above will give you the distance. Is there something characteristic about that intensity curve? You might want to read our articles Cepheid variable and Standard candle. -- ToE 03:39, 1 March 2012 (UTC)[reply]
Hello! Thanks a lot for your countless effort of teaching me about astronomy. You are my first teacher and the best so far in astronomy. I'm really appreciated your hard teaching. This Barnstar stands for your effort of teaching me! It's the only thing I can do for you now. Thanks A LOT again. Hope you have a great day!! (I probably will contact you sometimes again next year) Pendragon5 (talk) 00:34, 4 March 2012 (UTC)[reply]
Some little extra things i want to tell you. I just attended the state competition today. I was very excited about it and expected to get like top 3 or something. I turned out to be 12th rank out 33. I am really disappointed at myself, i don't know what happened. They didn't return the tests back so nobody knows what did they do wrong on. I thought i did really well and expected me to get top 3... But I somehow ended up with 12th T_T. I still don't understand what happened. I think they should have given the tests back so people can learn from the mistakes they made. I have no idea what kind of mistakes i made, i thought i got almost all of them correct. So anyway i think it could have been worse without you. Well on the positive side, i still have 2 more years to try so I guess I need to try harder next year. Alright thanks and cya! Take care!Pendragon5 (talk) 00:46, 4 March 2012 (UTC)[reply]
Almost forgot, i have the last problem i want to ask you for this year. What is not true about molecular cloud?
Actually there is another one problem. Alright i was given 4 pictures looking through different wavelength or electromagnetic radiation of one star and the problem asked me to identify which picture is correspond to each of the following options: Optic, X-ray, radio wave, Ultraviolet. So how am i suppose to know which one is which?Pendragon5 (talk) 03:16, 4 March 2012 (UTC)[reply]
Interesting. Can you describe, in general terms, how the pictures differed? -- ToE 03:48, 4 March 2012 (UTC)[reply]
They are different in colors as far as i can tell.Pendragon5 (talk) 14:53, 4 March 2012 (UTC)[reply]
Colors would be artificial for all but the optic (and optical images are sometimes rendered in false colors), so that would only tell you something if there were some standard convention used. These were just "fuzzy blob"-like images, not detailed images of solar surface structure as found in Sun, right? If they were all of the same shape, were they also all of the same size and intensity? -- ToE 00:49, 5 March 2012 (UTC) (I'm offline for several days while traveling.)[reply]
4 pictures are of the same object. 2 of them look the same but the other 2 look different. I don't have them with me, it was part of the test. I only have it somewhere in my memories. I have no ideas about their intensity, they are just pictures with no labels or anything. Different colors and little in shapes that's all i can tell.Pendragon5 (talk) 22:59, 5 March 2012 (UTC)[reply]
(The following is a response to questions asked here. -- ToE 00:34, 12 March 2012 (UTC))[reply]
Physic or chemistry first is up to the student's choice. I can do chemistry before physic but unfortunately it's too late to change now. The registration was last month, students are expected to stay with classes they have signed up which means they don't allow any change after the registration time.
About the question you asked me. I can't do it in my head. I used the formula that used apparent mag, absolute mag and distance. According to my calculation the answer is -3. And like i said before, they don't expect students to memorize formula. It's up to the students, it's one own's advantage to memorize formula due to the time limit. Pendragon5 (talk) 21:53, 9 March 2012 (UTC)[reply]
Well I don't know how chemistry classes could have prepared me for my physic classes but in my school, the prerequisite is precalculus, which is the math i need in physic. Well in my school this is how physic classes work. There is honor physic then AP physic. I will take both next year consecutive of course and take this exam in May Advanced Placement Physics B (My AP physic class is designed for that exam). My math classes in my junior year is Calculus courses all year and i will take AP calculus BC exam in May too. And in my senior year as my plan for now i'm going to take multivariable calculus and linear algebra.
About the preparations for astronomy next year. Let me tell you the timeline first though. It started around October every year and then in February is the regional competition (not to brag but it is very easy, all multiple choices). Beside, you may not know but astronomy is not the only thing i need to worry about (it is by far the one i'm interesting in the most and spent the most time but not the only thing). I have to get ready to some other events in SO too. There are like 20 of them (astronomy is 1 of them), each students need to choose at least 2 or 3 or up to 5 if you think you can handle it. I had 3. Well i do spend sometimes every week from October to work on SO stuffs but well not as much as since January. Since January i worked a lot harder because part of it is it is closer to the competition but another big reason is i don't really have much time in first semester, which ended in February. Beside SO, i have my classes and a bunch of other activities which consumed a holy cow of time everyday of my life. My first semester, i had about average 4-5 hours of homework everyday not to consider other activities and clubs i'm in and home chores. I was so exhausted during my first semester due to the lack of sleep and so happy that it is over now. My second semester is a lot easier now but doesn't mean i have time to rest. I just have more time to sleep but well it's life. Always have something for me to do. One more thing about second semester is i need to study hard for my AP exams coming up in May. I have 2 this year.
The state one is much harder due to the format and only first place got to go to national (so i have 11 more ranks to go, which i know for that the fact that i have a long way to go or in other word i need to improve a lot). Technically i have 2 more years but i'm think this way, on my first time at the national competition, chances are i won't be first place. So if i made it in my senior year = no more IAO for me. So i need to make it to national in my junior = doesn't guarantee an IAO spot for me but higher chance for sure. So yea i have 1 more year is what i should be thinking. Thanks for letting me know about the true fact, to be honest i was underestimate everyone at the state competition until the results came back to me. I still can't believe it and still wonder what kind of mistakes i could have made (one thing that i think they did not do a good job at is they didn't give back the tests so students can learn from their mistakes for later years, i don't know why but i'm sure they have their own reasons to not give them back) So anyway right at the beginning i know for the fact that i have little chance even i have tried my best. Well plus and to be the first place in national or international, i think i need some luck too. I know that one self knowledge is the main thing to determine how well you do but let put them this way. Knowledge is infinite so there will always be some chances that you encounter some problems you haven't learn before but someone else will know how to do it. It doesn't mean you study less than them but luck may not favor you that time by the fact that they gave you the problem you never done before but someone else has done it. Or some people may have been a lucky guesser once. Well in general term, the more you know = the better chance to win but that doesn't mean you will win, you just got a higher chance. Let say person A study 1 hour a day compare to person B study like 5 hours a day. There is a chance that person A will win but a lot less compare to person B. I guess i have made my point clear, so effort is majority yet it's not everything i needed to win. I also need some luck so my IAO dream can come true. Of course i'm not going to rely on luck (i'm just trying that i probably need some). I still try to best effort to reach my goal and yep even though i may not make it. But it's ok. The fact that i know i have tried my best is all it matters. I will be little upset but well i just have to move on. I'm an optimist guy btw, at least i try to be haa.
Anyway enough for side talking. Let get to the main point. I have all the astronomy tests in the national of the past few years. I'm going to just spend a great time in the summer to try to solve for them. I have completed one of them. I meant like i have done the entire test of a single year. The one i have been asking on here is that one i completed. The problems i did at state were indeed easier because those problems we have been doing are from national. But well there are some problems at state i didn't know how to do like the pictures as an example and few others but the rest i thought i did correctly. Anyway perhaps, those problems that i didn't know how to do was the reason why i was 12th rank. Or maybe i just made quite a bit of stupid mistakes which i will never find out what mistakes were they. (WOW i just noticed i just wrote a very long one lol, i don't know why i told you all of the above but well teacher should know quite a bit about the student, shouldn't they :D? I guess so you can have a perspective at my life and therefore = better for both of us somehow lol) Pendragon5 (talk) 22:29, 12 March 2012 (UTC)[reply]
