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For simplicity: Standard, 50,000km space elevator from equator to the counterweight. Acceleration, followed by constant velocity, followed by deceleration, and a stop past geostationary orbit.
What physical sensations do I experience (e.g. - not specific numbers, but general sensations: ".. then, you're pressed against the ceiling," and ".. now 'up' is the Earth.")? It has always been vague to me.
Also, I'm not interested in emotional sensations. Well, I am, but I can figure that out.
I will try to answer without using math, because I don't know how enter math notation in these answers.
So let's break it down. The first force to consider is gravity, and for this, let's assume you are stationary at whatever altitude. At sea level, you experience one g of gravity of course (I will use "force" when the more correct term would be "acceleration", but with a constant mass, the two are equivalent enough), and that force drops off as the inverse square of the distance from the Earth's center. So at one Earth radius (i.e., the surface), you feel 1 g, at 2 radii or 4,000 miles up, you feel 1/4 g, and at 3 radii or 8,000 miles up, you feel 1/9 g, etc.
But that is just gravity. There is also centripetal force, since you aren't just hovering motionless as the Earth spins below you, you are being slung around like a rock on a string. That force goes up linearly with your distance from the center of the Earth, so at 4,000 miles altitude, your centripetal force has doubled, and at 8,000 miles, it has tripled, etc. It is also in the opposite direction as the gravitational force.
As you rise from the surface, the gravitational force starts out much stronger than centripetal force (otherwise people on the ground would be slung out into space), but falls off rapidly, while the centripetal force increases steadily. The place where the two are exactly balanced is the geostationary altitude. For the Earth it happens to be about 22,200 miles. Past that, the centripetal force pulling outward exceeds the force of gravity pulling inward.
So if you were moving at a steady speed up the elevator, you would feel yourself get lighter rather rapidly. At 22,200 miles up, you would feel weightless, and then past that, you would feel a tug "upwards," towards the ceiling. You would settle your feet on what used to be the ceiling, and you could look "up" and see the Earth. That pull would increase until you reach the end of the tether, and it could be less than the 1 g you feel on the surface, or much more.
But there is still another force acting on you, and that is simple linear acceleration, from getting up to speed. Travel up and down the elevator would have to be pretty zippy, maybe a few times the speed of sound, so that the trip doesn't take weeks or months. The problem is if you accelerate straight up from the ground, the Earth's gravity will add its force to that acceleration force. (Fighter pilots know that their planes and their bodies can take more acceleration downwards than upwards.) If I were a clever designer, I would build long acceleration tunnels running along the ground that gradually curve upwards to meet the elevator trunk, so that some or most of the acceleration could take place without adding gravity. That would take some strain off the passengers' bodies. (And now that I think about it, those tunnels should be east of the main shaft.) I would also try to avoid much acceleration during the early parts of the vertical climb, where Earth's gravity is strongest.
And braking at the far end would also add to the centripetal force outward (which is increasing as you go, remember.) A clever designer would call for early braking early to spare the passengers some g forces. In fact, I imagine few people would actually travel to the far end, because braking would be take so long, and require so much energy (not to mention the return trip.) Most traffic to the far end would be outbound spacecraft, and for them there would be no braking at all. The craft would be slung off like David's stone that slew Goliath. The passengers in that craft would feel some side force as their lateral velocity (i.e., perpendicular to the radial velocity) increases. But that depends on the rigidity and design of the elevator trunk. (Imagine slinging a lead ball with a pipe. A rigid pipe works pretty well. A flexible pipe does not.) The instant they leave the elevator, they would be in free fall.
So a lot depends on the design choices made in the elevator. If the payload were allowed to accelerate all the way from 22,200 miles out, that would maximize the muzzle velocity (so to speak), but would place some demands on the bending strength of the elevator. On the other hand, if the payload came to a dead stop at the far end, and then was released, that would put no bending force on the tip of the elevator, but would waste some velocity. A compromise would be an elevator with a bent tip, to reduce bending stress on the tip, while still giving the departing spacecraft plenty of push. (See Larry Niven's "Integral Trees" for an analogous configuration.)
So I can imagine this. A passenger gets on a train car perhaps 50 miles from the actual elevator. It accelerates briskly, at perhaps 2 g's. The track gradually curves upward to meet the main trunk. Acceleration slows or stops, to maintain the 2 g's on the passengers. Then, as the car rises and gravity weakens, it begins to accelerate again, maintaining the net 2 g's. (How long could passengers tolerate 2 g's? Hours? Days? That might be a limiting factor.) At a certain point, braking begins, and at another, possibly different point, there is free-fall, and then ceiling and floor flip. There are more hours of 2 g, this time from deceleration plus centripetal force (minus the weakening gravity). The car comes to a stop somewhere above 22,200, where there is enough centripetal force to make things like walking and eating sort of ordinary, without requiring training. Then they stop and watch the actual spacecraft scream past them on the way to the other planets. :-)
