Three models of Gravity

Weight is a fictitious force in the same way that centrifugal force doesn't really exist. The reason an object in your hand feels as though there is a downward force on it is because you are accelerating up with the Earth, and the object is resisting this acceleration.

What causes that resistance? If I were traveling in a pressurized vehicle that was accelerating at the same rate as the earth I would feel no resistance if I tried to move forward inside the vehicle would I not?

Quote from: Robosteve on August 20, 2008, 06:44:19 PM

The mass of air above a point ten kilometres above the ground is less than the mass of air above the surface.

And why is that?

Because the mass of air above the surface is equal to the mass of air above a point ten kilometres above the ground plus the mass of air between the surface and a point ten kilometres above the ground. Unless you want to argue that the mass of air between yourself and the cruising altitude of a passenger jet is zero or negative, I think we can agree that my above statement was correct?

What causes that resistance? If I were traveling in a pressurized vehicle that was accelerating at the same rate as the earth I would feel no resistance if I tried to move forward inside the vehicle would I not?

You would. You feel resistance to the Earth's acceleration, too. People who don't fully understand the phenomenon call it "gravity".

How high are the ice walls again?

The one keeping in the atmoplane is thought to be about a hundred kilometres high.

Enjoying the discussion but have to put the kids to bed. Will resume tommorrow if it's still going at work.

You would. You feel resistance to the Earth's acceleration, too. People who don't fully understand the phenomenon call it "gravity".

If I were travelling 90 miles per hour in my vehicle and reached forward to change the radio station what resistance would I feel? Would this resistance differ at various speeds? Would if differ even if I continued to accelerate constantly?

In the same way why, in a pressurized atmosphere, would my body or any other object resist the upward motion?

Sorry, couldn't resist.

Because the mass of air above the surface is equal to the mass of air above a point ten kilometres above the ground plus the mass of air between the surface and a point ten kilometres above the ground. Unless you want to argue that the mass of air between yourself and the cruising altitude of a passenger jet is zero or negative, I think we can agree that my above statement was correct?

Wasn't arguing. Just trying to understand.

Quote from: oldsoldier on August 20, 2008, 06:55:07 PM

How high are the ice walls again?

The one keeping in the atmoplane is thought to be about a hundred kilometres high.

I think this is too high. I think if you take a plane with walls hundred kilometres high and add atmosphere till full and accelerate to 9.8m/s/s you'll get more air pressure at the surface than we experience.

No math yet, probably never from me as PV=nrT is not my strong suit and besides, I doubt it'd be conclusive to any FEer but you robosteve.

At any rate, I gotta imagine the density profile of the atom"sphere" is different in all 3 models of gravity. I'd be very surprised if they were all the same.

Wasn't arguing. Just trying to understand.

Fletch, don't worry about it, I had the same trouble. I was trying to get my head around his initial statement that "The mass of air above a point ten kilometres above the ground is less than the mass of air above the surface" before I realized that he was essentially saying "6 inches of ruler weighs less than the whole 12 inches of ruler." 

If I were travelling 90 miles per hour in my vehicle and reached forward to change the radio station what resistance would I feel? Would this resistance differ at various speeds? Would if differ even if I continued to accelerate constantly?

Velocity and acceleration are not the same thing. At constant velocity, you would not feel any acceleration. However, I'm sure you'll have noticed that when braking (which causes negative acceleration), you feel pulled towards the front of the car? That is because your body is resisting the acceleration.

In the same way why, in a pressurized atmosphere, would my body or any other object resist the upward motion?

"Motion" is an ambiguous term in physics. To be more specific, it is the acceleration you are resisting, acceleration being the rate of change of velocity. The pressure in the air has nothing to do with your own inertia (inertia is the phenomenon whereby you tend to resist acceleration), because air is all around you and so applying an upward force from underneath as well as a downward force from above. As the Earth accelerates upwards, you feel as though you are being pulled downwards.

Sorry, couldn't resist.

No need for apologies. Discussion is what I am here for.

I think this is too high. I think if you take a plane with walls hundred kilometres high and add atmosphere till full and accelerate to 9.8m/s/s you'll get more air pressure at the surface than we experience.

No math yet, probably never from me as PV=nrT is not my strong suit and besides, I doubt it'd be conclusive to any FEer but you robosteve.

At any rate, I gotta imagine the density profile of the atom"sphere" is different in all 3 models of gravity. I'd be very surprised if they were all the same.

Well, the RE atmosphere extends about that high into the sky. I was just assuming that the FE atmoplane would be similar. Also, I could try figuring out some mathematics regarding how high up the atmoplane should extend. I met pV=nRT last semester, and I don't see any reason why the maths would be beyond me. I still need to get around to answering the question about the speed of the Earth in the other thread too, even though it's a pointless question.

Quote from: Robosteve on August 20, 2008, 06:57:25 PM

You would. You feel resistance to the Earth's acceleration, too. People who don't fully understand the phenomenon call it "gravity".

If I were travelling 90 miles per hour in my vehicle and reached forward to change the radio station what resistance would I feel? Would this resistance differ at various speeds? Would if differ even if I continued to accelerate constantly?

To use the same analogy, take your car to a drag strip and try to change the radio station while you're at full throttle. 

Its not your speed...its the rate at which you're changing your speed which is defined as acceleration.

I think if you take a plane with walls hundred kilometres high and add atmosphere till full and accelerate to 9.8m/s/s you'll get more air pressure at the surface than we experience.

I have trouble picturing the geological forces that could build and support the mass of those walls.

I have trouble picturing the geological forces that could build and support the mass of those walls.

They aren't so much walls as gently sloping mountains.

I think this is too high. I think if you take a plane with walls hundred kilometres high and add atmosphere till full and accelerate to 9.8m/s/s you'll get more air pressure at the surface than we experience.

You have the order a little wrong. First you take a disk with walls 62 miles tall (I refuse to go metric... all those zeros ), then you accelerate it upwards at 9.8 m/s/s, then you fill it with air until the atmoplanic pressure at sea level is around 15 psi.

At any rate, I gotta imagine the density profile of the atom"sphere" is different in all 3 models of gravity. I'd be very surprised if they were all the same.

You're right, the density profile would be different, but our starting point is the same: Sea level air pressure of 15 psi. You could easily construct a "Halo model" which follows these criteria, provided you have those 62-mile tall walls to contain the atmotorus.

We'd also need a centripetal acceleration of 9.8 m/s/s to account for apparent "gravitation." Assuming our Halo is the same diameter as our current round earth (to give us enough room for all our continents and stuff), what would our tangential velocity need to be to give us the right gravitation? The answer is 7906 m/s, which is 17 times faster than our current equitorial velocity. This means one day in our Halo world would be 1 hour 25 minutes long!

Of the three models (Flat, Round, and Halo), personally I'm rooting for the Halo model.

Quote from: mayhem on August 20, 2008, 07:31:59 PM

I have trouble picturing the geological forces that could build and support the mass of those walls.

They aren't so much walls as gently sloping mountains.

Either way, we're talking about RET planetary mass here.  The base must be spectacularly huge to support a vertical like that. 

Either way, we're talking about RET planetary mass here.  The base must be spectacularly huge to support a vertical like that. 

Apparently it is, since air pressure doesn't seem to be decreasing in anything of a hurry.

Rather than a 62-mile tall mountain range that is 78,000 miles long, it would be much better (and lighter) to have plexiglass shield to hold in the atmosphere. But at that size plexiglass would probably be too weak, so maybe something like Transparent Aluminum like in the movie Star Trek 4.

Rather than a 62-mile tall mountain range that is 78,000 miles long, it would be much better (and lighter) to have plexiglass shield to hold in the atmosphere. But at that size plexiglass would probably be too weak, so maybe something like Transparent Aluminum like in the movie Star Trek 4.

It's much longer than that. That's the length of the Inner Ice Wall, which is only around fifty metres high. The Greater Ice Wall is a long way further out, and as such is a lot longer, probably millions of kilometres long.

Of the three models (Flat, Round, and Halo), personally I'm rooting for the Halo model.

I'm rooting for it too and have been ever since I read "Rendezvous with Rama". But I'm afraid it's gonna have to fail.

As soon as you jump in a halo you're not subject to any force, not even fictitious forces. You take whatever the tangent velocity you had the moment of the jump along with the vector sum of your velocity in your jump direction and... you're moving at constant speed. So... you CAN distinguish a halo inside a room w/o windows. Sad sad day for HE'ers.

I'm rooting for it too and have been ever since I read "Rendezvous with Rama". But I'm afraid it's gonna have to fail.

As soon as you jump in a halo you're not subject to any force, not even fictitious forces. You take whatever the tangent velocity you had the moment of the jump along with the vector sum of your velocity in your jump direction and... you're moving at constant speed. So... you CAN distinguish a halo inside a room w/o windows. Sad sad day for HE'ers.

You aren't subject to any force when you jump on a round or a flat Earth, either.

Quote from: oldsoldier on August 20, 2008, 11:27:11 PM

I'm rooting for it too and have been ever since I read "Rendezvous with Rama". But I'm afraid it's gonna have to fail.

As soon as you jump in a halo you're not subject to any force, not even fictitious forces. You take whatever the tangent velocity you had the moment of the jump along with the vector sum of your velocity in your jump direction and... you're moving at constant speed. So... you CAN distinguish a halo inside a room w/o windows. Sad sad day for HE'ers.

You aren't subject to any force when you jump on a round or a flat Earth, either.

Really? You're gonna do this to me? Did you miss my inclusion of "fictitious forces" in the paragraph above? Additionally, are you taking exception to dispute my statement about how HE works? or just to be argumentative in general?

EDIT: Hey, 100th post!!!!

As soon as you jump in a halo you're not subject to any force, not even fictitious forces. You take whatever the tangent velocity you had the moment of the jump along with the vector sum of your velocity in your jump direction and... you're moving at constant speed. So... you CAN distinguish a halo inside a room w/o windows. Sad sad day for HE'ers.

Sorry, but you misunderstand your trajectory. When you jump in a halo as I described (the one  going almost 8000 m/s) your "jump" velocity is very tiny compared with your tangent velocity (provided by your inertia from when you were standing on halo-firma, if you'll pardon my Latin). What happens then? The halo "rises up" to meet you! (It doesn't actually rise up, it's been going in a circle the whole time.) Remember the whole point of the spinning halo is to provide a centripetal acceleration of 9.8 m/s/s, which is indistinguishable from Earth's Gravity given a big enough Halo.

What you feel with your body is not gravity, but centrifugal force (which is a fictitious force).

But then it's a circular, and not parabolic arc, isn't it?

And if you were somehow able to measure your "arc velocity" you'd measure it as a constant, wouldn't you?

Really? You're gonna do this to me? Did you miss my inclusion of "fictitious forces" in the paragraph above? Additionally, are you taking exception to dispute my statement about how HE works? or just to be argumentative in general?

EDIT: Hey, 100th post!!!!

You aren't subject to any fictitious forces when in freefall in either of the other two models either.

I'm working on the theory that the earth is slightly porous. The earth has reached it's maximum speed, but the vacuum beneath the earth sucks air through the earth, and therefore sucks objects down to ground. The bigger the object, the bigger the suck. this gives the air pressure. The formula for measuring the "suck weight" of an object is E= mC² where E= sucking Energy. m=mass of object and C= Circumference of the object. I know odd shaped objects won't have a round circumference, but you measure round it at it's fattest (F) point. so C= pi *F. I hope this is not too technical for you folks, please look aout for my book "Flat Math" coming soon.

^^ You'll have to figure out how the air is being continuously replentished.  When it gets sucked through the earth's pores, does it circulate through the greater ice wall?  This could perhaps account for coreolis and hurricanes!

But then it's a circular, and not parabolic arc, isn't it?

And if you were somehow able to measure your "arc velocity" you'd measure it as a constant, wouldn't you?

Nope. Jumping in a halo appears to you to seem like jumping on Earth. You jump in the air and seem to come back down. In reality you jumped up, then the halo hit you on your feet, while you're still moving up!

Remember, there's no gravity, and thus no parabolas. Your trajectory (from an observer standing next to you in the halo) is straight up, then straight down, just like on Earth. But an observer standing near you, but floating outside the halo, will see something completely different. Your airborne trajectory will be a perfectly straight line! It sounds freaky, but it works, thanks to the fact that centrifugal force is fictitious. But you sure do feel it!

And yes, your arc velocity is constant (aka tangent velocity).

I want a halo to play with.

I'm working on the theory that the earth is slightly porous. The earth has reached it's maximum speed, but the vacuum beneath the earth sucks air through the earth, and therefore sucks objects down to ground. The bigger the object, the bigger the suck. this gives the air pressure. The formula for measuring the "suck weight" of an object is E= mC² where E= sucking Energy. m=mass of object and C= Circumference of the object. I know odd shaped objects won't have a round circumference, but you measure round it at it's fattest (F) point. so C= pi *F. I hope this is not too technical for you folks, please look aout for my book "Flat Math" coming soon.

You are an idiot. If I didn't know you were trolling, I'd list the many things that are wrong with that theory, but I won't bother.

Quote from: oldsoldier on August 20, 2008, 11:59:53 PM

But then it's a circular, and not parabolic arc, isn't it?

And if you were somehow able to measure your "arc velocity" you'd measure it as a constant, wouldn't you?

Nope. Jumping in a halo appears to you to seem like jumping on Earth. You jump in the air and seem to come back down. In reality you jumped up, then the halo hit you on your feet, while you're still moving up!

Remember, there's no gravity, and thus no parabolas. Your trajectory (from an observer standing next to you in the halo) is straight up, then straight down, just like on Earth. But an observer standing near you, but floating outside the halo, will see something completely different. Your airborne trajectory will be a perfectly straight line! It sounds freaky, but it works, thanks to the fact that centrifugal force is fictitious. But you sure do feel it!

And yes, your arc velocity is constant (aka tangent velocity).

I want a halo to play with.

I want one to play with too, but I beg to differ with your math.

I just did some back of the envelop calculations. To keep this simple, let's assume a halo 500 meters in radius spinning so that it's tangential velocity is 70 mps, this gives the 9.8 acceleration we're looking for on the surface. Now let's take a pole 9.8m long and drop a ball from the top of it. Looks like there's a 2/100ths of a second difference in how long this takes to fall in the halo vs RE/FE. Here's the math:

First RE:
At 9.8 meters high, and d = 1/2 * a * t² we can solve for t and get sqrt(2) = 1.414 seconds

Second HE:
part 1: how fast is it "falling"
At 9.8 meters high, the radial velocity is a little slower than at the surface. The surface is 70m/s while at this height it's 68.628 m/s (Justification: 70 mps at 500 meters radius gives rotational speed of .14 radians per second w = v/r, now w is constant, so when r is 490.2 v is then 68.628)

part 2: how far does it "fall"
When we "drop" the ball it starts moving along the chord from 9.8m high. How long is this chord? It's pretty easy to calculate from Pythagorean theorem, as the full radius is the hypotenuse, the semi-chord length (what we're interested in) is one leg and the (radius - the pole) is the other leg. Anyway, do the math and this length is 98.51m.

Part 3: how long does the "fall" take
98.51m / 68.628 m/s gives 1.435 seconds 2/100ths of a second longer than in FE/RE

Part 4: Where does it land?
Fortunately there are 2 different equations for this. But each equation is easier if you think about the angle that the halo turns through in both cases instead of linear distances. Let's start with the easier one.
The ball took 1.435 seconds to fall and we've already established that the halo rotates at 0.14 radians/second so in 1.435 seconds the pole that the ball was dropped from will have rotated by 0.20096 radians. On the ground of the halo this is 100.48 meters. That is to say that after 1.435 seconds the halo rotates 100.48 meters.
The other equation is where the chord intersects with the halo. That's an inverse trig function from the triangle in part 2. Here it's the inverse cosine of 490.2/500 which is .1983 radians. On the ground of the halo this is 99.16 meters.

Conclusion:
At this scale, the HE ball drops .02 seconds slower than RE/FE and "mysteriously" lands about 1.3 meters behind the pole from which it was dropped.

EDIT: I just did the calculation at an earth sized halo, approximately 5,000 kmeters in radius and the answers are much closer to each other, though at this scale the ball still falls behind the pole that it was dropped from by a little more than a centimeter.

But isn't this because 9.8m is nearly 2 percent of the radius of the Halo you used? The larger the Halo, the smaller percentage difference in the velocity at any height above the ground. And obviously an Earth Halo is dramatically larger than 1 km across.