Celestial Nav?

[StuM]

Originally Posted by Mike1956

Fully accept and fully aware of all points raised.

Thing is I simply do not have a calculator to handle the figures I am dealing with. Thought someone on the forum might have.

However pointless my question seems to you I happen to be interested to determine what I asked. How far away from Earth would I need to go to make a difference of 1 minute of arc in the rays of light from Sirius?

I am sure this sort of calculation is bread and butter to many people on here with a decent calculator.

Mike

OK, Sirius is 8.6 light years away which is 2.64 parsecs. (see earlier post about parsecs)

So you would need to go 2.64 x mean earth orbit to subtend 1 second.

So you would need to go 2.64 x mean orbit x 60 to subtend 1 minute.

That's 2.64 x 18693,000,000 x 60 which is approx 317 158 Astronomical Units (AU) or 29.5 15 billion miles.

To put that into perspective, that's very roughly ten 3-5 times further than the average distance to Pluto.


Edit: Numbers changed above to correct error.

[StuM]

Originally Posted by Kelkara

As far as celestial nav ... your question is indeed pointless.

But you asked nicely, so you deserve an answer:

it's a simple trig calculation y=x.tan(A)

x=distance to target (Sirius)= 8.6 ltyr * 9.5e+12 km/ltyr
A=angle of parallax (1 minute) = 2.9e-4 radians
tan(A)~=A for small A = 2.9e-4 (no calculator needed)

y=distance needed to move to create parallax =
(8.6*9.5*2.9) *10^(0+12-4)
~=9*9*3 *10^8 = 243*10^8 = 2.43e+10km (quill and parchment optional)


or to put into perspective that is about three times as far away as the orbit of Pluto. If you were interested in 1 second of arc you would still need to go most of the way to Jupiter.

Hmmm, one of us it out by a factor of two or three. I'll check.

[StuM]

Originally Posted by StuM

Hmmm, one of us it out by a factor of two or three. I'll check.

Kelkara is correct,

I used diameter instead of radius of Earth's orbit above .

I've corrected my figures.

[Kelkara]

I've only just now worked out what calculation Mike was trying to do in his original post. And after noticing his factor of 2 error, it seems he was right all along.

Unfortunately all he gave us was:

5878625373183.6 x 8.611 x 22 divide by 7 divide by 360 divide by 60

With no explanation.

In hindsight I could guess what the 360 and 60 were for, and then three posts later he tried to explain the 8.611 and 5878625373183.6 (but I had to google them anyway to confirm the units). But the 22 and 7 seemed to have no relevance.

Back when I was in school (when calculators were not allowed in maths class) I learned to look for the numbers 22 and 7 in math questions hoping to cancel them out with an approximation for pi ... but that was a long time ago.

[StuM]

Originally Posted by Kelkara

I've only just now worked out what calculation Mike was trying to do in his original post. And after noticing his factor of 2 error, it seems he was right all along.

Unfortunately all he gave us was:

With no explanation.

In hindsight I could guess what the 360 and 60 were for, and then three posts later he tried to explain the 8.611 and 5878625373183.6 (but I had to google them anyway to confirm the units). But the 22 and 7 seemed to have no relevance.

Back when I was in school (when calculators were not allowed in maths class) I learned to look for the numbers 22 and 7 in math questions hoping to cancel them out with an approximation for pi ... but that was a long time ago.

Possibly an approximation of pi used in converting degrees to radians (which would also explain the factor of 2) ?

Which sort of makes the precision of his 5878625373183.6 miles/light year a bit excesssive

[Kelkara]

Originally Posted by StuM

Possibly an approximation of pi used in converting degrees to radians (which would also explain the factor of 2) ?

Which sort of makes the precision of his 5878625373183.6 miles/light year a bit excesssive

I think he was trying to calculate the circumference of a circle around Sirius with a radius of the distance to earth. Then divide by 21600 to get the length of one minute of arc - a perfectly valid approach to get the same answer.

Using 14 digits of precision for a back-of-an-envelope calculation could be why his calculator couldn't cope most give up after about 8.

The factor of two was mixing up radius and diameter ... can happen to anyone

[StuM]

Originally Posted by Kelkara

The factor of two was mixing up radius and diameter ... can happen to anyone

Go on, rub it in!

[Benz]

Originally Posted by charliehows

Considering a; the best result you can expect from celestial nav is a matter of miles in inaccuracy and b; gps calculations are so accurate that calculations of time differentials between satellites have to take into account the light bending effects of gravitational waves - - - what exactly is the point you are trying to make? In the best tradition of naïve scientists you start with the entirely self evident statement that “I am no mathematician but...”

It is only the ignorant who claim that the best accuracy of Celestial has an error of miles. That's completely absurd! Far, far better accuracy is not only possible, but common among those who bother to practice. In perfect conditions, a position can be pinpointed by a skilled sextant user.
And musings about the whys and wherefores are not a waste of time: not only does using the mind keep it sharp, but understanding how celestial navigation works is of great usefulness in troubleshooting fixes.
This discussion is timely for me, since I was just yesterday explaining to my children the navigational triangle. Coincidence? yes, but a cool one.
Ben
zartmancruising.com

[Mike1956]

Originally Posted by Kelkara

I've only just now worked out what calculation Mike was trying to do in his original post. And after noticing his factor of 2 error, it seems he was right all along.

Unfortunately all he gave us was:

With no explanation.

In hindsight I could guess what the 360 and 60 were for, and then three posts later he tried to explain the 8.611 and 5878625373183.6 (but I had to google them anyway to confirm the units). But the 22 and 7 seemed to have no relevance.

Back when I was in school (when calculators were not allowed in maths class) I learned to look for the numbers 22 and 7 in math questions hoping to cancel them out with an approximation for pi ... but that was a long time ago.

Sorry, I did not realise I would need to explain the method, just seemed obvious.

Yes, need to create a circle with the centre Sirius and Earth/our Sun on the circumference. Calculate the total distance round the circumference then divide into degrees then minutes. Secondary school stuff.

I googled distance of 1 lyr and distance to Sirius, when you realise the figures involved (in terms of miles) it suddenly becomes obvious why the need to use lyrs as a unit of measure. Similarly confusing using distance to measure temperature in the atmosphere until the penny drops!

Someone put a link to an online calculator but I fear it will be beyond my comprehension. Most things online are! If anyone can boss it I will be pleased to receive the answer.

Mathematicians & scientists no doubt (my sister is one of them) will contrive all sorts of complex methods for sorting this calculation to bamboozle the rest of us and warrant their fancy salaries.

In reality it is not beyond any of us to complete the calculation longhand but speaking only for myself I have become idle. I do estimate the calculation would take a couple of hours and I have just realised I have probably spent a couple of hours on here trying to get someone to do the calculation for me.

I am aware the distance along the arc will be different to the straight line distance. I have no idea how I would start to calculate the distance in a straight line but along the arc is perfectly OK to satisfy my curiosity.

Once 1 minute of arc is determined to be 1 billion miles (or whatever) it will be clear why parallax is ignored for stars.

Mike

[Kelkara]

Originally Posted by Mike1956

Sorry, I did not realise I would need to explain the method, just seemed obvious.

There are many ways to approach even simple calculations ... being told what we're looking at always helps ... anyone can be bamboozled by numbers ... no need to insult my career choice.

Someone put a link to an online calculator but I fear it will be beyond my comprehension. Most things online are! If anyone can boss it I will be pleased to receive the answer.

I tried ... honestly I did.

Once 1 minute of arc is determined to be 1 billion miles (or whatever) it will be clear why parallax is ignored for stars.

Both StuM and I have already given you the answer, I also confirmed that your calculation gives the same result ... for the fourth time: it is 15 billion miles or 24 billion km.

you're welcome

[Mike1956]

Originally Posted by Kelkara

There are many ways to approach even simple calculations ... being told what we're looking at always helps ... anyone can be bamboozled by numbers ... no need to insult my career choice.

I tried ... honestly I did.

Both StuM and I have already given you the answer, I also confirmed that your calculation gives the same result ... for the fourth time: it is 15 billion miles or 24 billion km.

Thank you for your help. Much appreciated. Sorry for any offence.

Regarding my point about determining a straight line (as opposed to along the curve) to a position 1 minute of arc away from Earth, it occurred to me that it might be a simple calculation for some sort of triangle where 1 angle is known, (one sixtieth of a degree), and we have 2 sides of equal length which are also known, although they are pretty long.

Mike

Mike

[Adelie]

Originally Posted by Mike1956

I understand the light from stars is considered to strike the Earth at the same angle irrespective of our location on Earth. I have no problem with this whatsoever but have become curious to discover how far from Earth we need to venture before there is a difference of say 1 minute. Obviously it depends on the star but take for example SIRIUS.



I reckon the calculation would look something like this but I am no mathematician.......



5878625373183.6 x 8.611 x 22 divide by 7 divide by 360 divide by 60



This should give the distance from Earth to give a 1 minute difference



An additional calculation would determine what difference exists from various parts of the Earth.



Perhaps a mm or 2



I attempted the calculation on my calculator from the £1 shop and is started to smoke



Mike

For the nearest stars at about 4lyr, the difference in location of these stars against the background stars and galaxies is about 1.5 arcsec when photographed 6mo apart and in the right direction relative to the sun. This is where the term parsecs comes from. The limit of this method for measuring distance to a star was about 40lyr, don't know about currently.

[Kelkara]

Originally Posted by Mike1956

Regarding my point about determining a straight line (as opposed to along the curve) to a position 1 minute of arc away from Earth, it occurred to me that it might be a simple calculation for some sort of triangle where 1 angle is known, (one sixtieth of a degree), and we have 2 sides of equal length which are also known, although they are pretty long.

Yup, that's exactly how it is done. Your question didn't actually specify which direction from earth you would be moving ... I assumed a straight line perpendicular to the star and used the tan of the angle ... you assumed a circular arc around the star using the angle itself ... but the shortest distance is actually 1/60 degree away from the line I used, and is calculated by the sin of the angle. But for very small angles like this, all three answers are so close to each other that it really makes no difference which you use.

Traditional celestial nav was all about keeping the calculations simple enough so that they could be done without calculators. Hence ignoring tiny errors like parallax of stars, but do the same calculation for the moon, and it will become obvious why a parallax correction is needed there.

As a tip for next time you try to get your calculator to handle numbers too big for it ... work in trillions of miles: your calculation then becomes 6 x 8.6 x 6.28 / 21600 = 0.015 trillion miles, which any calculator can handle.

good luck, celestial nav is a great excuse for staring up at the stars, which is something we should all do more of.

[Mike1956]

Have just found an online calculator and did the calculation (twice) and my answer is different to 8 billion miles. As I intimated these things are generally beyond me.

Calculation as follows

5878625373183.6 x 8.611 =

x 2 =

x 22 =

divide by 7 =

divide by 360 =

divide by 60 =

14730933173.9 MILES

Any idea what I have done wrong?

Thanks

Mike

[Kelkara]

Originally Posted by Mike1956

14730933173.9 MILES

Any idea what I have done wrong?

You've done nothing wrong. That is the correct answer ... 15 billion miles.

now you're starting to make me doubt myself.