Shown here is the first roller coaster in the US to feature a vertical loop. At the top, the car is going about 45 mph. Can we estimate the radius of curvature, and the resulting centripetal acceleration at the top?

I can see people in the cars, sticking out their arms. (Why do they do that?) Looks to me like the radius of the circle is roughly ten "arm lengths" (maybe about 10 meters?) That would put the top of the circle at 20 meters=60 feet, which doesn't seem crazy. A speed of 45 mph is, very roughly, 20 m/s. So the centripetal acceleration at the top is about v^2/r = 400 m^2/s^2/ 10 m = 40 m/s^2, about 4 g's. (The web site says 4.9 g's, so we're not too far off)

I'm glad it came out bigger than g. If the TOTAL centripetal acceleration was LESS than g up there, then people would start falling out of the car (see the next calculation - the normal force would come out negative!)

At the top, the Normal force of the seat is down, as is your weight, so these two forces ADD together to form the net force, which by N-II must be mv^2/r = 4 mg (that's what we just estimated). So, at the top, using my numbers, N+mg = 4mg, or N=3mg. You are squished into your seat and "feel" three times heavier (upside down) than usual.

If you bring a water bottle with you, and you leave it open, will the water spill out at the top of the loop? I don't see why it should. The normal force of the bottle on the water is downward, the water is undergoing uniform circular motion just like you. It's like the "water in the bucket" demo in class. The water IS accelerating down, towards the center of the circle (as are you!) but the sideways motion means that it ends up lower, but also sideways, staying in the bottle (and you in your seat) as you go around.