A mathematical difficulty

[Sundog]

I have a small problem I'm hoping one of our resident math whizzes can help me with.

 

A friend of mine has proposed a sensor grid for the solar system, consisting of "tripwire" neutrino beamsgoing out to a shell of hub stations 1/2 a light year out and one light-second apart. I have enough of an appreciation of scale to know that's utterly ridiculous - I figure the mass of the hubs would probably outweigh the solar system - but I don't have the math to prove it.

 

Could someone work out how many hubs you'd actually need?

[lemming]

Re: A mathematical difficulty

 

The area of a sphere of that size would be for simplistic sake:

 

(60*60*24*365/2) = 1/2 light year (It's off a bit)

 

The area of the sphere would be that squared x 4 x pi, so

 

(60*60*24*365/2) * (60*60*24*365/2) * 4 * 3.14 = 3,122,790,589,440,000

 

which is the area of the sphere in number of light seconds. So yea, there would be a few of them...

 

Oh, a light second is 299,792,458 meters

[Lawnmower Boy]

Re: A mathematical difficulty

 

Convert to meters and rounding like mad, we have one light second equal to 3E8, hence half a light year is very roughly 1E16 (x3600x24x365.244/2). Surface area of a sphere is 4xpixradius cubed, or roughly 1E33 square meters. Flatten it out, divide into cells of 1E8 metres radius, or area 1E17.

You would need on the order of 100 trillion hub stations.

[Sundog]

Re: A mathematical difficulty

 

Thanks, guys. That should make my argument a little more convincing.

[Nyrath]

Re: A mathematical difficulty

 

My figures are a little different. Please check my math.

 

For a sphere 0.5 light years in radius. Google says that is equal to 4.73 x 10^15 meters. (go to Google and type in "0.5 light years in meters")

Surface area of a sphere = 4 x Pi x r^2

Surface area of a sphere = 4 x Pi x (4.73 x 10^15)^2

Surface area of a sphere = 4 x 3.1416 x 2.24x10^31

Surface area of a sphere = 2.81 x 10^32 square meters

 

Now, the sensors are spaced 1 light second apart. This means that the space each one takes up on the surface of the sphere is approximately a circle with a radius of 0.5 light seconds (if they are 1 light second apart, half belongs to one sensor and half belongs to the other).

 

Google says 0.5 light seconds is 1.49 x 10^8 meters

 

Surface area of a circle = Pi x r^2

Surface area of a circle = 3.1416 x (1.49 x 10^8)^2

Surface area of a circle = 6.97 x 10^16 square meters

 

So the total number of these circles you would need to totally cover the sphere is

 

number of sensors = sphere surface area / circle surface area

number of sensors = 2.81 x 10^32 / 6.97 x 10^16

number of sensors = 4.03 x 10^15

which is about 4 quadrillion sensors (using the US quadrillion, not the British).

[lemming]

Re: A mathematical difficulty

 

My figures are a little different. Please check my math.

Accounting for that we went for different methods, our areas worked out the similar. I would say you were more accurate.

 

Now, you went ahead and did the area's of the sensors, however, your method points out we all made a bit of an goof. A round circle will have to have overlap so that we don't wind up with gaps, so we'd have to expand your method a little. And probably just treat each sensor as a squared circles, windup with having to use a factor of Pi more sensors.

[Nyrath]

Re: A mathematical difficulty

 

Now' date=' you went ahead and did the area's of the sensors, however, your method points out we all made a bit of an goof. A round circle will have to have overlap so that we don't wind up with gaps, so we'd have to expand your method a little. And probably just treat each sensor as a squared circles, windup with having to use a factor of Pi more sensors.[/quote']

 

Agreed. But it is still a most outrageous number of sensors required.

[lemming]

Re: A mathematical difficulty

 

Agreed. But it is still a most outrageous number of sensors required.

 

Yes indeed.

[Maur]

Re: A mathematical difficulty

 

My figures are a little different. Please check my math.

 

For a sphere 0.5 light years in radius. Google says that is equal to 4.73 x 10^15 meters. (go to Google and type in "0.5 light years in meters")

Surface area of a sphere = 4 x Pi x r^2

Surface area of a sphere = 4 x Pi x (4.73 x 10^15)^2

Surface area of a sphere = 4 x 3.1416 x 2.24x10^31

Surface area of a sphere = 2.81 x 10^32 square meters

 

Now, the sensors are spaced 1 light second apart. This means that the space each one takes up on the surface of the sphere is approximately a circle with a radius of 0.5 light seconds (if they are 1 light second apart, half belongs to one sensor and half belongs to the other).

 

Google says 0.5 light seconds is 1.49 x 10^8 meters

 

Surface area of a circle = Pi x r^2

Surface area of a circle = 3.1416 x (1.49 x 10^8)^2

Surface area of a circle = 6.97 x 10^16 square meters

 

So the total number of these circles you would need to totally cover the sphere is

 

number of sensors = sphere surface area / circle surface area

number of sensors = 2.81 x 10^32 / 6.97 x 10^16

number of sensors = 4.03 x 10^15

which is about 4 quadrillion sensors (using the US quadrillion, not the British).

 

This is a very close estimate, but probably still a slight over estimate (I doubt it is off by even a fraction of an order of magnitude) of the number needed as the hubs wouldn't be creating a true sphere. Instead they would be creating a tessellated surface (the beams would travel straight from one hub to the next).

[Sundog]

Re: A mathematical difficulty

 

This is a very close estimate' date=' but probably still a slight over estimate (I doubt it is off by even a fraction of an order of magnitude) of the number needed as the hubs wouldn't be creating a true sphere. Instead they would be creating a tessellated surface (the beams would travel straight from one hub to the next).[/quote']

 

 

No, the beams come from the inner system. The Hubs merely detect whether the beam has gone through any status changes between the launch point and the array - such as having interacted with normal matter.

[Maur]

Re: A mathematical difficulty

 

No' date=' the beams come from the inner system. The Hubs merely detect whether the beam has gone through any status changes between the launch point and the array - such as having interacted with normal matter.[/quote']

 

Ahh, but that doesn't change that the surface area one is calculating for isn't quite what the stations would need to create, heheh.

[Basil]

Re: A mathematical difficulty

 

No' date=' the beams come from the inner system. The Hubs merely detect whether the beam has gone through any status changes between the launch point and the array - such as having interacted with normal matter.[/quote']

 

You do realize this means that the inner system's information will therefor be up to a year "out of date," because anything the hubs 'detect' will have to be transmitted to the inner system, don't you?

 

Why are the hubs so far out, anyway?

[Maur]

Re: A mathematical difficulty

 

You do realize this means that the inner system's information will therefor be up to a year "out of date," because anything the hubs 'detect' will have to be transmitted to the inner system, don't you?

 

Why are the hubs so far out, anyway?

 

The information might be a year old, but anything traveling past them would be moving slower than the speed at which the information was flowing giving someone more time to prep the further out they detect the intruder coming in from.

[Cancer]

Re: A mathematical difficulty

 

With 4e+15 detectors over the sphere ...

• How often are neutrinos being emitted from each station? With any reasonable sampling rate, there'll be a lot of power converging at the hub. And by definition, the hub has to be able to detect a power drop of 1 part in 4e+15 in order to be useful. I am not sure any physical quantity can be measured with that precision at this time.
  
• That requires truly prodigious angular resolution in your neutrino detector array. There are 4 pi steradians to a sphere, and I think 4e+15 points spread over a sphere means that making out one of them requires a resolution of much, much better than a second of arc. We can't do that yet for wide-field opticla detectors....
  

[Basil]

Re: A mathematical difficulty

 

The information might be a year old' date=' but anything traveling past them would be moving slower than the speed at which the information was flowing giving someone more time to prep the further out they detect the intruder coming in from.[/quote']

 

True, but if the incoming craft is travelling at, say, .5 c, and it doesn't "trip a wire" until it is 1/3 light years out (or ~21,100 Astronomical Units), Earth won't find out about it until the craft has arrived.

 

Remember: the original idea is for beams to be emitted from the inner solar system (let's say from Earth, for simplicity's sake), hit the "hubs" 9.5 light-year out, and then the hubs send information to Earth if the beams encountered anything. So, if a craft going half the speed of light gets to 1/3 light-year out (or closer!) without being detected, it will take less time to get to Earth than light (or light-speed neutrinos) will take to get from that point, out to the "hubs," and then back to Earth.

 

A craft travelling at .1 that gets to within ~5750 AU before "tripping a beam" will get to Earth before the warning will. And at that distance, the beams will be separated by more than twice the diameter of Earth.

 

Not only are there a ludicrous number of hubs, but it's not all that frickin' good a system. :thumbdown :thumbdown