Hi all, I was working on same basic physics with calculate the apparent weight of a pilot in a turn. Let's have the plane be oriented such that the cockpit faces the sky, the belly faces the ground to start off with. Now, "Positive G-forces" are for a plane that begins to climb, or to put the nose on an incline. "Negative G-forces" would be a plane that dives, tips the nose down, or under the horizontal. In the image I attached, step 1) is my Newton's Laws for a "positive G scenerio", where the plane begins to climb and I only worry about the lowest point in the turn, for simplicity. I wrote at the bottom that: N=mg+mv^2/r where the N is the Normal force (the weight the pilot "feels") and I mention that the extra mv^2/r is what adds weight to the pilot. In step 2) I show my analysis for a "negative g-forces", which I conclude with N =mg-mv^2/r and this time the pilot feels like he weighs less because the mv^2/r term detracts from the true weight. I read the following statement: Negative g-forces are harsher than positive g-forces so what a pilot can do, is rather than dive where you tip the nose down, roll the plane over (now, the cockpit faces the ground and the belly the sky) and pull the stick back. You will now experience positive g-forces". In my diagram, I have that situation in the bottom right corner and I conclude with: N=mv^2/r-mg. However, this expression does not look at all like N= mg+mv^2/r. In fact, if the pilot were to roll over, it seems like he would still experience weightlessness. What would the force diagram look like for a plane rolling over that would make the pilot experience postive g-forces? On my diagram, the floor is the bottom of the plane or the seat of the plane. The Normal force is the force of contact between pilot and floor.