Xx_Chaos_xX Posted March 15, 2005 Share Posted March 15, 2005 Recently during my free time, with nothing better to do i begin trying to derive my own formula... So far i have comes out with this, the formula for calculating the brightness of stars in term of magnitudes in relation to distance: M = [2lg(x) - 2lg(y)]/ lg2.512 M: Magnitude change, if ans is +ve it appear brighter if -ve it will look dimmer X: Any distance Y: Distance of 10 parsec or 32.6 LY(The standard distance for absolute magnitude) Therefore: V = A + M or A = V - M where V is the Visual magnitude, A is Absolute magnitude and M is magnitude change For those astrophysic experts here.... Can this formula work Well here is the eg. of how i used my formula to calculate star brightness, the best eg is of course would be Sun... As you well know (For those Astronomy expert) Sun had a visual magnitude(V) of -26.8 and Absolute Mag.(A) of +4.83, to show how this formula work... Eath-Sun Distance: 1.496 x 10^8 km No. of seconds per year: 60 x 60 x 24 x 365.25 = 31557600 Speed of light (c): 299792.458 km/s Sun Dist in Light-year (X): 1.496 x 10^8 / (299792.458 x 31557600) = 1.581273248 x 10^ (-5) LY Distance of 10 parsec for Absolute Mag. (Y): 32.6 LY To use formula: M = (2lgX - 2lgY) / lg2.512 M = {2[lg(1.581273248) + (-5)lg10 ) - 2lg32.6} / lg2.512 = [2(-4.80) - 2(1.51)] / lg 2.512 = -31.56950371 Mag It show that the sun will shine -31.6 magnitude (or 4.25 trillions) brighter on Earth than on 32.6 LY away Since V = A + M Then V = +4.83 + (-31.57) = -26.74 Mag (Ans) Oops! My formula show that the visual mag. of the sun is -26.74, not the exact answer but i guess the answer came quite close... =P [EDITED] Anyway here is another typical eg whether it work or not (It can also come in handy in determining whether a particular website is reliable or not) look at this webpage for eg.... [url]http://en.mimi.hu/astronomy/beta_orionis.html[/url] It claimed that Rigel had a Visual Mag (V) of +0.12 and an Absolute Mag (A) of -7.1, and its located at over 900 years away.. To prove its reliability... We can find the distance from the given Visual and Absolute Mag V - A = 2 (lgX - lg32.6) / lg2.512 Sub V = 0.12 and A = -7.1 +0.12 - (-7.1) = 2 (lgX-lg32.6) / lg2.512 +3.61 x lg2.512 = lgX - lg32.6 therefore: X = 10^ ( 3.61lg2.512 + lg32.6 ) = 906.3344389 Distance of Rigel = 906 light-years (3 S.F) The source is quite reliable lol... Link to comment Share on other sites More sharing options...

Lennex3 Posted March 15, 2005 Share Posted March 15, 2005 :animestun [color=darkred]Edit: This post is complete spam. Only posting one emotion is not a good example of the quality of posting we like to see here on theOtakuBoards. Please be aware that we have a rules page that all members are required to read before posting. The [URL=http://www.otakuboards.com/rules.php?"][U]Rules[/U][/URL] and [URL=http://www.otakuboards.com/faq.php?][U]FAQ[/U][/URL] can also be found on the left navigation menu. Failure to follow the rules will result in the banning of your account. The reason I am leaving this post is to warn you about your post quality and to keep the author of this thread from having a double post. If you have any questions please feel free to PM myself or any of the other moderators. ~Panda [/COLOR] Link to comment Share on other sites More sharing options...

Xx_Chaos_xX Posted March 16, 2005 Author Share Posted March 16, 2005 [quote name='Lennex3']:animestun[/quote] I know its hard to read, and harder to understand, unless you study in Astronomy and Astrophysic, Well i guess i am not clever in anyway, its just that i am too obsessed in Astronomy and Astrophysic that I find every bits of information regarding this subjects... You just need to know 2 things to derive this formula also... 1st: The order of Magnitude and the different of each units. A 1st Magnitude star is always 2.512 times brighter than a 2nd Mag object, 2nd Mag brighter another 2.512 times than 3rd Mag. and so on... Therefore you will realised than an 1st Mag star is always 100 times brighter than an 6th Mag star [2.512^(6-1) = 100], Note: The lower the Mag the brighter the object get... And the brightness will increase or decrease in an exponential rate 2nd: The ratio in relation to distance and brightness is D : B^2 respectively, so if the same object is place 2 times the distance away it will appear 2^2 or 4 times dimmer, the object at 3 times distance away will be 9 times dimmer and so forth... So with this 2 stuff in mind you will slowly realise: (X/Y)^2=2.512^M Where M is the unknown variable in Mag. change... Where X and Y is the positions of 2 different points at different distance from the object of focus... Using Logarithm rules u will derive that M = (2lgX-2lgY)/lg2.512 [My formula] Well whether it really work that well I'm not very sure, but i do hope someone can help me to correct or even modified it to make it better like for eg. to make it more convenient it can be written as: V - A = 2(lgX-lg32.6) / lg2.512 Notes: Distance of X must be in light-year ... Anyway i am quite sure also I am not the 1st one to thought of this... It just that it is really that interesting to derive your own formula and use it in your own calculation Link to comment Share on other sites More sharing options...

Guest Novapup Posted March 16, 2005 Share Posted March 16, 2005 Wow, that's some interesting calculations! I'm not sure I understand all of it but did you take into account the star's spectral class? For instance: Polaris an F7 I and is 132 parsecs away. if a star such as Tau Ceti, a G class star was the same distance away, you probably wouldn't see it at all. That would be an important factor in calculating distance. If you're interested, I've created a calculator that plots locations of stars in X Y Z coordinates. You need to enter the star's Right Asention, Declanation and distance. The calculator will give you the stars location in three dimensions from our solar system. The calculator was made using flash 5. I could forward it to you if you want. A good book to pick up if you don't have it is Peterson's guide to stars and planets. If you're really ambitious, try the Sky Catalogue. Hope that helps. Link to comment Share on other sites More sharing options...

Xx_Chaos_xX Posted March 16, 2005 Author Share Posted March 16, 2005 Anyway this is a very basic formula of mine, I need to incorperate with another formular to make it work better (Dont u guys realised my formula is virtually useless other than trying to prove a reliabilty of an particle scource?) Here is what i am trying to think hard, just by determing the visual, absolute and the apparent diameter of a star (Too bad not many telescopes are that powerful) , i am able to derive the diameter of the star and thus, the surface temperature and which catagories it is in (for eg. M1 or G2 etc) This is getting more and more ridiculous... But through a very complicated way i did manage to come out with a long formula to derive a star diameter with the given Visual, Absolute and Apparent diameter (In second arc) knowing that apparent diameter of a star form a right triangle with angle, opposite and hypotenuse, I can derive the diameter of the star for eg from Tan (B/3600) = D / X so that D = X Tan (B/3600) Where D is the real diameter of the object (can be a Star, Nebulas or even Galaxies) B is the apparent diameter in second arc units X is the distance of the object So from my formula, using only Visual and Absolute Magnitude with distance unknown V - A = 2(lgX-lg32.6) / lg2.512 I can make the X as a subject 1st [(V - A) x lg2.512 ] = 2 (lgX-lg32.6) lgX = {[(V - A) x lg2.512] / 2 + lg32.6} therefore X = 10^ [ 0.5(V - A)lg2.512 + lg32.6 ] Combining 2 formula I got a long chain of messy number and symbols D = 10^ {(V - A) x lg2.512] / 2 + lg32.6} Tan(B/3600) and can written as D / Tan (B/3600) = 10^ [ 0.5(V - A)lg2.512 + lg32.6 ] or lg D - lg Tan(B/3600) = 0.5(V - A)lg2.512 + lg32.6 D is the diameter in light-year B is the apparent diameter in second arc V is the Visual Mag. and A is the Absolute Mag However let find out whether it works... Betegeuse is a very good eg as scientist have successfully find out the apparent diameter (B)of the supergiant, to be around 0.054 degree arc... From Space.com, [url]http://www.space.com/scienceastronomy/brightest_stars_030715-10.html[/url] It mention that the star sun at a Visual Mag (V) of 0.5 and from this solstation [url]http://www.solstation.com/x-objects/betelgeuse.htm[/url] it mention that the star has a Absolute Mag (A) of -5.1 and located at 430 lightyear (X) away (Space.com also mention that) ... To show both source are correct at the same time to find the true size of this super giant... we have D / Tan (0.054/3600) = 10^ { 0.5[0.5 - (-5.1)]lg2.512 + lg32.6 } D / Tan (0.054/3600) = 10^ ( 2.8lg2.512 + lg32.6 ) D / Tan (0.054/3600) = 429.8061033 light year As you can see from above D = X Tan (B/3600) X = 430 LY (3 s.f.) Which mean both website are correct and to continue D = 429.8061033 Tan [1.5x10^(-5)] = 1.12x10^(-4) light year (3 sf) The exact diameter of betelgeus is = 1.12x10^(-4) x 299792.458 x 31557600 = 1,064,549,536 km or 764.8 times the diameters of sun... Big ar isnt it Btw that is not just it... The formula can be used to determine the separation distance of an binary star system base on the apparent separation, visual and apparent magnitude of the 2 stars in the system, of course that require more complex calculation as it involved 3D vector diagram... but as long as you get the both the star distance away using this formula and the apparent separation, I guess you can calculate the exact separation of the 2 binary stars by pythagorus therom.... Link to comment Share on other sites More sharing options...

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