Understanding negative g-force? I was reading this on reddit:

If you were in an elevator accelerating upwards which, you might
  experience a force of +2g. And if the elevator was accelerating
  downwards very quickly, you might actually feel an upwards force of
  -0.5g. That's what a negative g-force is, when it feels like you are falling up.

So I understand that the case scenario when an elevator is accelerating upwards, the net force on the person is in the up (positive) direction so the force applied by the person in reaction is in the down (negative) direction, which is positive g-force. 
But I don't at all understand how you will feel an upwards force of -0.5g when an elevator is accelerating downwards. Because when in an elevator, accelerating downwards at theoretically $4.9m/s^2$, the force normal will still be upwards (as it's preventing free fall), but will be less than if there was no acceleration (less weight). But the reaction force therefore is downwards. So this is still just like the scenario when the elevator is accelerating upwards! 
Or am I misreading this? Does this person actually mean if the elevator is somehow accelerating downwards at 1.5 times the acceleration of gravity? In that case, I don't see how this would make any sense.
 A: When we say you are experiencing an acceleration this means something must be exerting a force on you, because force and acceleration are related by Newton's second law. In a stationary elevator it is the floor of the accelerator that exerts an upwards force on you, and this force is just $mg$ giving you an acceleration of $g$.
If the elevator is accelerating downwards, for example $-0.5g$, the force exerted on you by the floor of the elevator is decreased and your total acceleration is decreased (in this case to $+0.5g$). If the downward acceleration of the elevator is $-1g$ then the force on you decreases to zero and you become weightless i.e. your acceleration is zero $g$.
If the acceleration of the elevator becomes greater (more negative) than $-1g$ you will find yourself standing on the roof of the elevator so it is now the roof that exerts a net downwards force on you. This is how your acceleration can become negative. Instead of the elevator floor accelerating you upwards the elevator roof accelerates you downwards.
A: Are you sure the paragraph is correct, because I can't make out which direction is positive. So we are going to take downward as positive and upwards as negative.
When you are going downwards, what the elevator is doing is actually by a little counter acting the force of gravity. That is, the g-force(gravitational force) acts downwards, acceleration of g. Now the elevator according to your paragraph provides an upward acceleration of -0.5g. So the net acceleration of the elevator now is g-0.5g=0.5g in the downward direction. Now that means you in the elevator is also moving at 0.5g acceleration. This is obvious since if you had a net acceleration you wouldn't be standing still inside the elevator. Now coming back to your body, your has a downward acceleration of 0.5g. But your body will have an acceleration of g if you were to jump of a building, given you measure it before you die ;-) . Now when you are travelling in an elevator you only have an acceleration of 0.5g, so we can conclude that the elevator gives you a force of -0.5g in the upward direction, thereby reducing your weight.
A: The text you're reading is slightly ambiguous, and there are two possible readings of what it means:


*

*The more reasonable one is that the elevator is accelerating downwards at $a=\tfrac12 g=4.9\:\mathrm{m/s^2}$, which already counts as "accelerating downwards very quickly". In this case, you will feel a net force from the elevator floor of $g-a=\tfrac12g=4.9\:\mathrm{m/s^2}$, or in other words you will feel as if there was a mysterious force of $-ma=-\frac12 mg$ acting on all your body.

*Alternatively, the elevator could be accelerating downwards at $a=\tfrac32 g=14.7\:\mathrm{m/s^2}$, i.e. faster than vacuum freefall, so you will be stuck to the ceiling, pulled there with a net force of $m(a-g)=\tfrac12mg$, which is offset by an equal and opposite normal force exerted by the ceiling.
Frankly, I think it's much more likely that the author meant the first scenario and wasn't very precise in their language.
