Jump to content

Recommended Posts

  • Replies 117
  • Created
  • Last Reply

Top Posters In This Topic

Posted
It is simply not possible draw a line tangent to the pole that will intersect any point at the equator.  All line tangent to the pole have to lie on a plane tangent to the pole and normal to the axis of the sphere.

 

Your analogy fails beause as soon at Tom leans to one side, the plank  will no longer be in contact with the pole but would be in contact with (and tangent to) some other point on the ball.

598768[/snapback]

It will at a point above the equator on a high enough platform

Posted
So if Tom leans the plank will be in contact with a point above the equator?  :doh:

604308[/snapback]

point in space.

 

Yes, it’s possible to create a platform high enough all you need is a telescoping shaft with a with reflective media on the end

Posted
It will at a point above the equator on a high enough platform

604305[/snapback]

 

Isn't that what I said back on page 1? :doh::P:)

 

Now the question is, how high must the platform be to track an object at 1000 km altitude? Does it have to be higher than a house? Does it have to be higher than the Empire State Building? Does it have to be higher than Mt Everest? How feasible is it to build a platform of such a height at the South Pole?

Posted

STOP THE INSANITY!

 

I'm not doing this. If you don't understand the concept of parallel lines and the tangent of a circle, but insist you have some sort of greater insight into orbital mechanics than a friggin' physicist, it is in no way my problem.

Posted
Isn't that what I said back on page 1?  :lol:  :lol:  :lol:

 

Now the question is, how high must the platform be to track an object at 1000 km altitude?  Does it have to be higher than a house?  Does it have to be higher than the Empire State Building?  Does it have to be higher than Mt Everest?  How feasible is it to build a platform of such a height at the South Pole?

604326[/snapback]

 

When we get done with this discussion, can we proceed to the effects of atmospheric attenuation on shooting straight out on a tangent vs. shooting straight up? :doh::P:)

Posted
When we get done with this discussion, can we proceed to the effects of atmospheric attenuation on shooting straight out on a tangent vs. shooting straight up?  :doh:  :P  :)

604328[/snapback]

 

Don't we have to figure out how to protect the telescoping platform from being knocked over by all the satellites in polar orbit several hundred km below the platform first?

Posted
Don't we have to figure out how to protect the telescoping platform from being knocked over by all the satellites in polar orbit several hundred km below the platform first?

604336[/snapback]

 

 

It's not a problem. If the platform extends out far enough to reach the equator, it'll be so far from the Poles that anything in a polar orbit can't reach it.

 

Unless a clown tilts the platform, of course...

Posted

You know, I'm getting really tired of us Poles being drug into this.

 

We have had a bad rap undeserved for a long time now, and I thought things were getting better.

 

Just when it seems it has - we're thrust to the forefront of another conversation about stupidity.

Posted
You know, I'm getting really tired of us Poles being drug into this.

 

We have had a bad rap undeserved for a long time now, and I thought things were getting better.

 

Just when it seems it has - we're thrust to the forefront of another conversation about stupidity.

604348[/snapback]

You are a Pole in the northern hemisphere. It is those southern hemisphere Poles ruining it for the rest of you. They are running around with planks on them, teatering with leaning clowns and telescoping platforms. Shameful behavior, but clearly less dignified than those that follow the laws of geometery.

Posted

Geo orbiter can go out to 60,000 miles (approx 100,000KM). You basically have a right triangle. Side 1= 60,000 miles side 2= 3963 miles (straight down through the earth to the equator) hypotenuse = 60,131 miles. Angle to geo obiter at 60,000 miles out = approx 86 degrees from South Pole. 90-86= 4 degrees below the horizon. The platform would have to be about 276 miles high. So I guess it’s not realistic, unless you bounce it off a polar obiter. So the biggest advantage is that you can see or destroy every polar obiter within one rotational orbit around the earth.

Posted
Geo orbiter can go out to 60,000 miles (approx 100,000KM).  You basically have a right triangle. Side 1= 60,000 miles side 2= 3963 miles (straight down through the earth to the equator) hypotenuse = 60,131 miles.  Angle to geo obiter at 60,000 miles out = approx 86 degrees from South Pole.  90-86= 4 degrees below the horizon.  The platform would have to be about 276 miles high.  So I guess it’s not realistic, unless you bounce it off a polar obiter. So the biggest advantage is that you can see or destroy every polar obiter within one rotational orbit around the earth.

604433[/snapback]

 

Wow. That's almost right. Almost.

Posted
Geo orbiter can go out to 60,000 miles (approx 100,000KM).  You basically have a right triangle. Side 1= 60,000 miles side 2= 3963 miles (straight down through the earth to the equator) hypotenuse = 60,131 miles.  Angle to geo obiter at 60,000 miles out = approx 86 degrees from South Pole.  90-86= 4 degrees below the horizon.  The platform would have to be about 276 miles high.  So I guess it’s not realistic, unless you bounce it off a polar obiter. So the biggest advantage is that you can see or destroy every polar obiter within one rotational orbit around the earth.

604433[/snapback]

 

Geosynchronous orbits are nowhere near 60,000 miles. And supersynch's are transfer orbits only. Wrong again, Euclid.

Posted
Geosynchronous orbits are nowhere near 60,000 miles.  And supersynch's are transfer orbits only.  Wrong again, Euclid.

604549[/snapback]

Tom, I didn’t say anything about Geosynchronous in my example because they have to be about 23,000 miles out for a 23 hr 56 min rotation. For the example I gave myself the widest berth to be right. The furthest earth or geo obiter that I read about was 50,000 to 60,000 miles out. Turns out at the very least the pole or platform would have to be over 250 miles high to observe a obiter around the equator from the South pole.

Posted
Tom, I didn’t say anything about Geosynchronous in my example because they have to be about 23,000 miles out for a 23 hr 56 min rotation.  For the example I gave myself the widest birth to be right. The furthest earth or geo obiter that I read about was 50,000 to 60,000 miles out. Turns out at the very least the pole or platform would have to be over 250 miles high to observe a obiter around the equator from the South pole.

604640[/snapback]

 

And I never said you were talking about geosyhcnronous orbits, Einstein. :doh:

×
×
  • Create New...