Oh my, you really need another bit of spherical trig, namely the measure of an angle in "radians"
there are 2 pi radians in a circle of 360 degrees
so the circumference is
C = 2 pi radians * R
the handy thing is for any angle in radians now
arc length along circle = T R
where T is the T = angle in radians
to get from degrees to radians
radians= T = angle in degrees
* 2 pi/360
If your angle is one minute that is 1/60 degrees
so
T = 2 pi/(60*360) radians
so your arc length in light years is
2 pi(8.611)/(60*360)
= .0625048 light years
which is 3.67*10^11 miles
(those are land, not nautical miles) To get it go to Google and type .0625 light years = miles and hit enter or use calculator and speed of light.
367,000,000,000
hit
Damon, Gloucester MA
I posted yesterday but feel I confused the issue with an incorrect calculation of an angle, I have just realised my mistake!. I also think I should come clean about my circumstances. I am a mature (very) student, in fact a retired commercial fisherman and am studying celestial navigation. Apart from celestial navigation being closely related to my life at sea I have a very strong family history of dementure with 7 close family members having died or currently suffering. I am advised to keep my brain engaged which may help delaying the inevitable. my memory is already noticeably affected.
I went to school 50 years ago and it was when it was optional and I spent much time on the golf course eventually becoming a golf professional before going to sea. About the only thing I can remember about maths from school is Pi 22 over 7 which I now discover is a hopeless approximation.
Regarding the celestial. People like you (mathematicians) have made it very simple for people like me (simpletons) with the introduction of tables. All I need to do is to be able to add and subtract to do celestial. I am getting on with it nicely.
I am making my own notes in my own language (what makes sense to me) and when I read parallax needs to be taken into account for Sun and Moon but not for stars I thought I would do a calculation to put into my own notes to confirm why parallax is not an issue for stars.
I chose the star Sirius as it so comparatively close to Earth you can almost smell it and I lie in bed and watch it cross my bedroom window every morning (at this time of the year).
I had to decide how to do the calculation and decided to make Sirius 8.611 lyrs away the centre of a circle with Earth on the circumference. Calculate the distance round the circumference and divide to determine the distance along the arc of 1 minute. I determined this to be 14,730,933,173.9.
It is best practice while navigating to confirm positions with additional information so to confirm my calculation above I decided to create a triangle with 1 angle at Sirius and 2 angles on the circumference of the circle. This will yield a straight line instead of an along the arc distance which will be a little shorter.
We know 3 angles and 2 sides and it is extremely frustrating (losing sleep) to realise I am so stupid I am unable to determine the length of the 1 side we do not know. I suppose I should have spent less time on the golf course but then I would have forgotten everything learned by now!.
Additionally I know I am up against it with the large numbers involved. I have googled solving a triangle but am only able to put 4 or 5 digits into the formulas.
So to reiterate
My triangle has 3 known corners
1 minute/and 2 of 89 degrees 59 minutes 30 seconds
2 of the 3 sides are known and are equal at 14,730,933,173.9 miles
Can someone please help this old sea dog.
Thanks
Mike
10 answers
Hello Damon, Thank you but you have grossly over estimated my capabilities. I have done all you suggested and the first result furnished a ridiculous number with an e in it. The second site I tried was only interested in calculating my BMI. I did get a result from another site that I could understand but it was about 352 billion miles more than I was expecting. I am looking for a figure of a bit less than 14,730,933,173.9. If the figure of 367 billion is correct then my original calculation is pants although I do not believe this to be the case as I have confirmed it using rounded figures and mental arithmetic. Look forward to further help. Thanks Mike PS I should have mentioned in my first post as well as being stupid I struggle to use computers
for that pesky e notation:
http://www.mathsisfun.com/numbers/scientific-notation.html
look at example at the end:
Example: 3.12E4 is the same as 3.12 × 104
3.12E4 = 3.12 × 10 × 10 × 10 × 10 = 31,200
http://www.mathsisfun.com/numbers/scientific-notation.html
look at example at the end:
Example: 3.12E4 is the same as 3.12 × 104
3.12E4 = 3.12 × 10 × 10 × 10 × 10 = 31,200
You are right, I have an arithmetic error
I estimate that 8.6 light years ia about 5.06 * 10^13 or 5.06e13 miles
one minute is 1/21,600 of the circumference [ that is 1/(360*60) because 360 degree and 60 minutes per degree]
circumference = 2 pi r
= 2 pi * 5.06e13
= about 31 e13 or 3.1 e14 miles around
divide that by 21,600
(31,000/21,600)e10
= 1.4 e10
= 14,000,000,000
I estimate that 8.6 light years ia about 5.06 * 10^13 or 5.06e13 miles
one minute is 1/21,600 of the circumference [ that is 1/(360*60) because 360 degree and 60 minutes per degree]
circumference = 2 pi r
= 2 pi * 5.06e13
= about 31 e13 or 3.1 e14 miles around
divide that by 21,600
(31,000/21,600)e10
= 1.4 e10
= 14,000,000,000
Clearly parallax is not a problem at this distance :)
found error
so your arc length in light years is
2 pi(8.611)/(60*360)
= .0025048 light years
which is 1.47e10 miles
(those are land, not nautical miles) To get it go to Google and type .0025048 light years = miles and hit enter or use calculator and speed of light.
so your arc length in light years is
2 pi(8.611)/(60*360)
= .0025048 light years
which is 1.47e10 miles
(those are land, not nautical miles) To get it go to Google and type .0025048 light years = miles and hit enter or use calculator and speed of light.
Hello Damon, Thank you. The answer obtained is now much more in line but I have just realised having logged on again that you are helping me calculate the length along the arc and what I posted for is to solve the triangle. ie Find the length in a straight line. Please re read my original post where I hope I have made this clear. I have always accepted that parallax is not a problem but doing the calculations for my notes is just the way I decided to go. Thanks Mike
LOL,
If you really want to do it as an isoceles triangle with equal legs of length 5.06 e13 miles
then
sin .5 minutes = (x/2)/5.06e13
x = 5.06 e13 * 2 sin .5 min
sin .5 min = sin (.5/60) degrees = .000145441
times 10.12 e13 = .00147186292 e13
= 1.47 e10
again:)
If you really want to do it as an isoceles triangle with equal legs of length 5.06 e13 miles
then
sin .5 minutes = (x/2)/5.06e13
x = 5.06 e13 * 2 sin .5 min
sin .5 min = sin (.5/60) degrees = .000145441
times 10.12 e13 = .00147186292 e13
= 1.47 e10
again:)
By the way
for small angles
most of us assume
sin theta = tan theta = angle in radians
where angle in radians = angle in degrees * pi/180
then the arc length = chord length = R theta
for small angles
most of us assume
sin theta = tan theta = angle in radians
where angle in radians = angle in degrees * pi/180
then the arc length = chord length = R theta
Thanks Damon but please do not waste any more of your time proving your superior intelligence. It is not becoming. You have simply not managed to grasp the situation. I am an old man bordering dementure and have no comprehension whatsoever of what you are on about. All I ever wanted was the answer. Mike
4 answers