Quake3World

If you smoke dope, the asteroids win!

I think we could make that work!

Fender wrote:If you smoke dope, the asteroids win!

Well I guess we're doomed

It's mardi grass day here ...no work and I'm smoking one as we speak getting ready to go see some homegrown tata's

Tsakali_ wrote:It's mardi grass day here ...no work and I'm smoking one as we speak getting ready to go see some homegrown tata's

you're walking the path, son...besides, it has been proven again and again that if you don't smoke weed, you're automatically a moron.

lots of morons in here...

Nightshade wrote:If a nuclear weapon detonated on the surface of an asteroid, we wouldn't care about shockwaves traveling through a vacuum, no would we?
Then you're faced with the remaining bits still traveling on pretty much the original path. So lots of small asteroids still headed towards Earth.

Keep shooting them so they break into smaller and smaller bits. Shoot the really small bits and they just vanish.

Obviously a triangular shaped ship with a nose mounted weapon would be best for this.

maybe they can reason with it

i didn't read all this junk, but you're wrong julez. you need to deflect it by a small angle... arctan(d/D) where d is the radius of earth's event horizon and D is distance from asteroid to earth when you deflect it. so if you try to deflect it when it's 10 event horizons away you'd only have to deflect it by about .1 radians the way i see it.

iambowelfish wrote:I think there's a little confusion here.

As far as I can see Julios is right if the ship flies alongside the asteroid until it definitively misses earth.

If the ship flies alongside for a limited time before that point, then I think mjrpes and feedback would be right. Even with the force removed the asteroid keeps moving along the y axis at the speed it's been accelerated to. In that case the ship wouldn't need (following julios's figures) 5 hours of time alongside the asteroid, provided it caught it early enough.

no - the way I'm thinking, you'd need 5 hours if you flew along side it all the time. If you only did it for a small amount of time, the asteroid wouldn't accelerate to a high enough speed along the y axis, and you'd need more distance.

but coming to think of it, there is something wrong with my analysis... coz if the asteroid was far enough, you wouldn't need a five hour tug...

denz had a point.

gonna try to digest menkent's post.

Also, i made another error - i shouldn't've divided the force between them - so actually it would only take 2.5 hours.

Ok, I'm fairly convinced that if you know the "time to impact" in advance, you don't need to use angles, and can still use distance.

It's just like trajectory motion in two dimensions - you can analyze each component independently.

So you could ask the question like this:

imagine you have a stationary asteroid some distance from the earth.

Now draw an imaginary line between the centre of the asteroid and the centre of the earth, and call that line the X-axis.

How much force would you need to displace the asteroid 6 thousand kilometres in one month?

I gotta go to the lab - will do some calculations later on tonight, and i'll use the figures ed lu implied, with the apollo sized tractor ship.

switch your thinking to polar coordinates and set the asteroid at the origin.

and if the asteroid is stationary you're fucked. you can shove it as far as you want and it will fall toward the largest gravitational force.
to paraphrase Ender - "the Earth is down"

menkent wrote:switch your thinking to polar coordinates and set the asteroid at the origin.

i don't know how to deal with that sort of stuff - i'm sure u could end up with the same answer, but i'm convinced it's not necessary (though it might be more efficient).

also - read up - i made a post at the same time u did

think of it as you in a car driving toward a house and your brakes are out. you don't have someone push you 20m laterally to miss the house, you turn the steering wheel and change your trajectory to an angle where you will drive past it to one side or the other.... which wouldn't work well anyway if the house is PULLING you toward it because the force will continualy pull you back toward a heading where you'll collide.

as i said earlier, i'm ignoring the gravitational pull of the earth.

as for the car analogy, read my last post again - you could use angle of deviation, but you can also use component forces.

If u had jet thrusters on the side of your car, you could calculate how much lateral force is required to nudge the car to the side by the width of the house (and we'd assume here that the wheels were spherical) in the time allowed.

In fact, later on, we can do the calculations both ways and see if we get the right answer.

am off for real now!

also keep in mind that because this is a force applied over time, it's not a neat triangle, but rather a "curved angle"

component analysis is probably way more efficient.

I'm gonna try to figure out how to incorporate earth's gravity into this thing too - thing is i don't know calculus so i may have to make some simplifications

AN ASTEROID IS GONNA HIT EARTH IN 6 MONTHS LOL

really omg i better get busy on my hair and nails

2036 and 100% chance huh? is this landing on the east coast or west coast of the US? oh, and i'm not worried about it. hardcore style. yeah.

[xeno]Julios wrote:also keep in mind that because this is a force applied over time, it's not a neat triangle, but rather a "curved angle"

component analysis is probably way more efficient.

I'm gonna try to figure out how to incorporate earth's gravity into this thing too - thing is i don't know calculus so i may have to make some simplifications

you won't be able to do it without calculus and it will be MASSIVELY (pun) more complicated if you don't do it in polar coordinates (where F of grav is simply in the -x direction at all times). this is a fairly comlpicated mechanics problem - not a ball on an inclined plane.
that said, what you call a "curved angle" is likely referred to as acceleration. now if you want to solve for the motion of the asteroid under the gravitational influences of the earth and the deflector ship simultaneously you've officially discovered the "three body problem" which is nonlinear and thus unsolvable. gg. if you don't account for both you're making up fake problems that don't matter anyway. trust nasa, save yourself a headache.

I say let the asteroid hit. Tell all the fundamentalists that they can 'meet their maker' halfway out in the pacific. We might sustain some losses of our own, but at least we will solve the bible thumping christian problem.

i bet all those dorks who drank coolaid to catch a ride on the last asteroid are feeling like tards now that this one is going to make a nice bus stop for people to board.

Grandpa Stu wrote: 100% chance huh?

no

menkent wrote: you won't be able to do it without calculus and it will be MASSIVELY (pun) more complicated if you don't do it in polar coordinates (where F of grav is simply in the -x direction at all times). this is a fairly comlpicated mechanics problem - not a ball on an inclined plane.

If we ignore earth's gravity, I think my way is extremely easy. Polar may be easy too, but component forces is very handy. Do you agree here?

if you don't account for both you're making up fake problems that don't matter anyway. trust nasa, save yourself a headache.

If we ignore earth's gravity, then it will require less tugging. The whole purpose of my post was to assess the feasibility - if it turns out that you'll require a 30 years of tugging without earth's gravity, then we can safely say that with earth's gravity, it'll require more time.

btw, i'm not convinced you understand what i mean by component forces, based on your reply to my stationary asteroid thought experiment.

let me ask you this: if we ignore earth's gravity, do you agree that the stationary asteroid example is identical in terms of figuring out how much force is required along the orthogonal axis?

Watching the arm chair physicist argue with the physics grad student is funny

iambowelfish wrote:

Nightshade wrote:If a nuclear weapon detonated on the surface of an asteroid, we wouldn't care about shockwaves traveling through a vacuum, no would we?
Then you're faced with the remaining bits still traveling on pretty much the original path. So lots of small asteroids still headed towards Earth.

Keep shooting them so they break into smaller and smaller bits. Shoot the really small bits and they just vanish.

Obviously a triangular shaped ship with a nose mounted weapon would be best for this.

Planning on setting up a nuclear machine gun in orbit?