I derived my won formula

I realise I can derive my own formula...

V - A = 2 (lg648000D-lg32.6KB) / lg2.512


this formula is only applicable to Astrophysic but then again for those physic you can help me check out... I out all my explanation in this forum http://www.astronomy.com/asy/community/forum/topic.asp?TOPIC_ID=14225


My acc if you wan to know in this forum is Berserk...
 

waqas

Oh, nevermind...
I have to say I'm impressed with Smith's efforts. In fact it's inspired me to create my own formula; one for having a girlfriend! Here I go;

Having Life + Social Skills - Time spent on net forums - Knowledge of maths/physics/Computing = G.Q. (Girlfriend Quotient)

Eureka!

Remember, the higher your girlfriend quotient, the greater the chances of you having a girlfriend. Happy dating!

PS. Take away a million points if you have an internet blog.
 

Majin_Tenshi

The can opener went bye-bye...
waqas said:
I have to say I'm impressed with Smith's efforts. In fact it's inspired me to create my own formula; one for having a girlfriend! Here I go;

Having Life + Social Skills - Time spent on net forums - Knowledge of maths/physics/Computing = G.Q. (Girlfriend Quotient)

Eureka!

Remember, the higher your girlfriend quotient, the greater the chances of you having a girlfriend. Happy dating!

PS. Take away a million points if you have an internet blog.

/stab

Now smith, I'm lazy, so explain your formula yerself.
 
Majin Tenshi said:
/stab

Now smith, I'm lazy, so explain your formula yerself.

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]

Since V = A + M then i came up with


V - A = 2(lgX-lg32.6) / lg2.512

and 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


It can also be simplified as


V - A = 2 (lg648000D-lg32.6KB) / lg2.512


Lastly with further touch up i guess the above formula can also be used to calculate the size of our own solar system objects with just apparent diameter and distance. Astronomical measure like lightyear can be neglected as well


648000 D = X.K.B



D is the diameter in km (Note the different?)
B is the apparent diameter in second arc (same so remember must convert)
X is the distance in km
K is a constant for pie or 3.1416



Notes: Units of D is always constant with X, so if D is km then X must be km or vice versa...


When calculating large objects in the sky such as Sun and moon there tend to be slight error due to the problem with small increment but i guess it can be neglected (since normally we round off to 3 significant figures)

for eg like the moon, we know that the average distance (X) of moon is 384400 km, with the diameter (D) of 3476.6 km and also an mean apparent diameter (B) of 31' 06" arc... so show what i mean by error...

648000 D = X.K.B
648000 D = 384400 x 3.1416 x (31x60 + 6)
D = 384400 x 3.1416 x 1866 / 648000
= 3477.5 km (slightly larger)

Other than that i guess there isnt any problem, just hope that mistake can be negligible...


In case if you didnt realise this formula can be very handy for parallax calculation and it is very very fast... substitute D as radius of Earth Orbit and the parallax angle (B) it shift in six month, straightaway you will get the distance (X) of the object ...
 

ShinHell9

I started on here when I was like 14...
but wait, our planets orbit elliptically, wouldn't our solar syatem be an ellipse? Also, we still don't know the exact number of PI, in simple little pre-calc problems of finding linear velocity it doesn't make much of a dent, but with numbers that huge, knowing the trillionth place value of PI or not can make a significant difference, wouldn't you say? However, I'm no astronomer, so I may be wrong as hell...and I do believe this is not your work, because I could come on here and say they dicovered
e^(PI*i)+1=0 and everyone would probably believe they discovered it.

(I had to use PI because I can't get the greek symbol on here.)
 
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.
 
Eddie said:
everyone would probably believe they discovered it.


I guess this what i am expecting... But looking at the post i have in the other forum i can only tell you this is my work... hard to believe and yes i got no back up to uphold my stand


All this best i can do is, i can modified my acc -or the "berserk" you c- in that forum in whatever way u want to prove that that acc my mine...


Anyway to Saiya: Not unbelievable when u spent nearly one whole day typing that post and edited all the spelling errors in that forum, and anyway u are right i cut and paste that stuff becz it's tired to explain everything again...
 
E

ELEKTROFUNK

Guest
Eddie said:
but wait, our planets orbit elliptically, wouldn't our solar syatem be an ellipse? Also, we still don't know the exact number of PI, in simple little pre-calc problems of finding linear velocity it doesn't make much of a dent, but with numbers that huge, knowing the trillionth place value of PI or not can make a significant difference, wouldn't you say? However, I'm no astronomer, so I may be wrong as hell...and I do believe this is not your work, because I could come on here and say they dicovered
e^(PI*i)+1=0 and everyone would probably believe they discovered it.

(I had to use PI because I can't get the greek symbol on here.)

God eddie shut up who cares what you think in a topic where there is math involved.
 

DarkBlademaster

Jesus cries when he looks at me.
ELEKTROFUNK said:
God eddie shut up who cares what you think in a topic where there is math involved.

Kowabunga to that... not just to eddie but to everyone who is trying to show off their 1337 math skillz.
 
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