Cool physics/navigation question.

[dogjones]

This is a cool question.

I am in Bermuda (33° 22' N 64° 41' W).

My girlfriend is in Durham, UK (54°47'N 01°34' W).

We have been describing the sky to each other (yeah, yeah, I know). Then she made this point - that it's all the same sky. I was baffled until she clarified that you can see the same stars in the UK that you can in Bermuda. Aha! So, what I want to know is:

1. What is (and how to work out) the minimum altitude of a single point that could be commonly seen by both parties (assuming for simplicity that we are both at the same level (sea I guess) with no obstructions on the horizon)

2. At what rate relative to altitude does this point 'expand' to a meaningful area?

(Bear in mind I have very little math so layman explanations of all squiggly bits would be much appreciated!)

Thanks!

D

 

[I Ratant]

Looks like this.
At any one time...

 

[Dancing David]

Your lattitude will have the most to do with it, as far as seeing the same sky at different times.

You will see about 22° worth of the southern stars she wont, she will be able to see more of the circum-polar stars than you.

Now what you can both see at the same time is another issue.

 

[Beady]

Why not buy an astronomy/planetarium program, and see for yourself?

However, from my days on the Coast Guard cutter, the distance to the horizon in miles is 1.221 multiplied by the square root of the height of the observer's eyes in feet. I have no idea how to write the formula on a computer keyboard, but I'll try:

D in miles = 1.221 X (sq rt of H in feet).

Therefore, assuming that you can stand on the water and that your eyes are therefore 6 feet off the water:

D = 1.221 X (sq rt of 6) = 1.221 X 2.45 = 2.99 ~ 3 miles.

The furthest spot you can see is ~3 miles away. Likewise, another person an additional ~3 miles away, or six miles from you, could see the same spot.

Now, say you and your friend are 2,000 miles apart, and you want to find the minimum altitude you both must be at to see a common point, do this:

1000 miles (remember, there are two of you) = 1.221 X (sq rt of "X")

1000/1.221 = sq rt of "X"

sq rt of (1000/1.221) = X squared

sq rt of 819 = X squared

28.6 = X squared

sq rt of 28.6 = X

5.35 miles altitude = X

I *think* I got that right. If not, I'm sure someone will point it out.

OOPS! I can alkready see I screwed it up. Well, I'll let one of the math geniuses fix it; I figure suplying the formula is enough.

 

[~enigma~]

Why not buy an astronomy/planetarium program, and see for yourself?

Agreed in full with everything except the word buy.

 

[dogjones]

OK, that's cool, but I'm thinking of lower points than stars, and I'm assuming it's exactly the same time so the rotation of the earth does not eventually reveal the same sky. So longitude does matter because it affects the distance between the two points on the sphere.

OK after some further research I think I can rephrase the question.

The earth's circumference is 40,075.16 km.

The distance from Bermuda to Durham is 5355 km. Point C is the exact midway point between these two points.

I start rising straight up at point C. What is my altitude when I first become visible from both both Bermuda and Durham?

 

[Ziggurat]

I have no idea how to write the formula on a computer keyboard, but I'll try:

D in miles = 1.221 X (sq rt of H in feet).

This message board supports a subset of LaTeX, a typesetting language designed to handle math. Quote this post to see the code I use to generate the following:

[latex]$D = 1.221 \times \sqrt{H}$[/latex]

 

[dogjones]

Why not buy an astronomy/planetarium program, and see for yourself?

However, from my days on the Coast Guard cutter, the distance to the horizon in miles is 1.221 multiplied by the square root of the height of the observer's eyes in feet. I have no idea how to write the formula on a computer keyboard, but I'll try:

D in miles = 1.221 X (sq rt of H in feet).

Therefore, assuming that you can stand on the water and that your eyes are therefore 6 feet off the water:

D = 1.221 X (sq rt of 6) = 1.221 X 2.45 = 2.99 ~ 3 miles.

The furthest spot you can see is ~3 miles away. Likewise, another person an additional ~3 miles away, or six miles from you, could see the same spot.

Now, say you and your friend are 2,000 miles apart, and you want to find the minimum altitude you both must be at to see a common point, do this:

1000 miles (remember, there are two of you) = 1.221 X (sq rt of "X")

1000/1.221 = sq rt of "X"

sq rt of (1000/1.221) = X squared

sq rt of 819 = X squared

28.6 = X squared

sq rt of 28.6 = X

5.35 miles altitude = X

I *think* I got that right. If not, I'm sure someone will point it out.

OOPS! I can alkready see I screwed it up. Well, I'll let one of the math geniuses fix it; I figure suplying the formula is enough.

Thanks! Am trying to work through it. Um, where do you think you screwed it up?

 

[Gribble]

I'm no mathematician but it seems you are needing to know how high you need to be to be able to see both places at the same time? You can calculate the distance between two positions given as lat/long if you google for "movable type geodesic" and enter your values in the boxes on the movable type page.
Try ???.movable-type.co.uk/scripts/latlong-vincenty.html - and the answer will come out at around 5356km. So you'll be halfway between those points and needing to know the height you are at when your horizon is half that distance away (in each direction)

Google for "Steve Sque horizon" and it should take you to
?ttp://newton.ex.ac.uk/research/qsystems/people/sque/physics/horizon/ and there's a calculator for you to enter 5356/2 = 2678km, (didn't work for me!) and also a graph where you can see that your horizon is around 2700km away when you are about 600 km above the Earth - which is the number you wanted.

Near enough? G

 

[ddt]

OK, that's cool, but I'm thinking of lower points than stars, and I'm assuming it's exactly the same time so the rotation of the earth does not eventually reveal the same sky. So longitude does matter because it affects the distance between the two points on the sphere.

OK after some further research I think I can rephrase the question.

The earth's circumference is 40,075.16 km.

The distance from Bermuda to Durham is 5355 km. Point C is the exact midway point between these two points.

I start rising straight up at point C. What is my altitude when I first become visible from both both Bermuda and Durham?

I don't quite get where Beady's square root fits in.

Consider the great circle of the earth covering Bermuda, C and Durham. Draw the triangle with points: the centre of the earth; Bermuda; and a point at height 'h' above point C. Call the angle at the centre of the earth 'phi'. Call the radius of the earth R (= 6,371 km).

ETA: height 'h' is such that the angle at Bermuda is 90 degrees.

Then basic goniometry gives:
cos phi = R / (R+h)
and therefore
h = (1 - cos phi) / cos phi * R

and the angle phi is given by:
phi * R = (distance C to Bermuda) / 2 = 5355 / 2

This gives approx. phi = 0.4203 (in radians)
and thus h = 607.24 (in km)

 

[dogjones]

I'm no mathematician but it seems you are needing to know how high you need to be to be able to see both places at the same time? You can calculate the distance between two positions given as lat/long if you google for "movable type geodesic" and enter your values in the boxes on the movable type page.
Try ???.movable-type.co.uk/scripts/latlong-vincenty.html - and the answer will come out at around 5356km. So you'll be halfway between those points and needing to know the height you are at when your horizon is half that distance away (in each direction)

Google for "Steve Sque horizon" and it should take you to
?ttp://newton.ex.ac.uk/research/qsystems/people/sque/physics/horizon/ and there's a calculator for you to enter 5356/2 = 2678km, (didn't work for me!) and also a graph where you can see that your horizon is around 2700km away when you are about 600 km above the Earth - which is the number you wanted.

Near enough? G

Yes, very cool and thank you! Um. It's a bit... higher than Beady's answer though?? Should I take an average?

 

[Gribble]

You could take an average of my "about 600" and ddt's "607.24" if you wish

And of course there's the little issue of the world not being perfectly round ...

ETA The Winter Solstice occurred just over two hours ago, so Happy New Year everyone! G

 

[dogjones]

I don't quite get where Beady's square root fits in.

Consider the great circle of the earth covering Bermuda, C and Durham. Draw the triangle with points: the centre of the earth; Bermuda; and a point at height 'h' above point C. Call the angle at the centre of the earth 'phi'. Call the radius of the earth R (= 6,371 km).

Then basic goniometry gives:
cos phi = R / (R+h)
and therefore
h = (1 - cos phi) / cos phi * R

and the angle phi is given by:
phi * R = (distance C to Bermuda) / 2 = 5355 / 2

This gives approx. phi = 0.4203 (in radians)
and thus h = 607.24 (in km)

Nice one. This is interesting. The space shuttle cruises at approx 300km. So even at that height they can't see Bermuda & the UK at the same time. I always thought they could see pretty much all of one side of us. Wow! It's not such a small world after all!

 

[ddt]

Yes, very cool and thank you! Um. It's a bit... higher than Beady's answer though?? Should I take an average?

I'm quite sure my formula is right. It is also corroborated by Gribble's. Beady said his formula came from the Coast Guard. I suspect it's a simple approximation that works on small distances.

Nice one. This is interesting. The space shuttle cruises at approx 300km. So even at that height they can't see Bermuda & the UK at the same time. I always thought they could see pretty much all of one side of us. Wow! It's not such a small world after all!

Compared to the radius of the Earth, 300 km is pretty small. Still, plugging this in into my formula gives them a horizon of 1918 km. That means that they can easily see the whole of Europe and then some, when on the right spot.

 

[Ziggurat]

I don't quite get where Beady's square root fits in.

From the small angle approximation:
[latex]$1-cos(x) \approx x^2$[/latex]

solve for x, and you get a square root on the other side.

 

[I Ratant]

I'm quite sure my formula is right. It is also corroborated by Gribble's. Beady said his formula came from the Coast Guard. I suspect it's a simple approximation that works on small distances.



Compared to the radius of the Earth, 300 km is pretty small. Still, plugging this in into my formula gives them a horizon of 1918 km. That means that they can easily see the whole of Europe and then some, when on the right spot.

.
Not over Yurp, but the typical viewing points that can see the ISS look like this.
For points more distant between each other, the altitude gets higher, and the object would need to be larger.
There's geosynchronous satellites that may be simultaneously "visible" electronically, but they're 22000 miles up above the equator.

 

[ddt]

From the small angle approximation:
[latex]$1-cos(x) \approx x^2$[/latex]

solve for x, and you get a square root on the other side.

Yes. I think you're missing a factor 1/2 in there, btw, as the power series for the cosine is
[latex]\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots[/latex]

 

[I Ratant]

A 3D view.
No "squiggly bits" needed, just line of sight.

 

[Ziggurat]

Yes. I think you're missing a factor 1/2 in there

Oops, you're correct.

 

[jsfisher]

Why not buy an astronomy/planetarium program, and see for yourself?

Stellarium is a fantastic program....and you will really like the price.