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How to calculate g-forces on pilots in climbs, dives, outer space

  1. Jun 12, 2012 #1


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    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.

    Attached Files:

  2. jcsd
  3. Jun 12, 2012 #2


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    Well, lets say Luc skywalker is doing loops in his starfighter in outer space with negligable gravity. The normal force he would experience on the seat would be mv^2/r at all times. We will can this force Cf, for centripital force.

    If Luc now comes down to the planet so Princess Leia can see better and does circular loops ( quite a proficient pilot he is ) then at the bottom of the loop the force he feels is Cf + mg. At the top of the loop Luc would feel Cf-mg.

    Now whether at the top of the loop Luc feels the force on his butt or on his shoulder harness depends on v^2/r.

    Does that help you out?
  4. Jun 12, 2012 #3


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    Almost. Let's say that the maximum g-force that Luke can take is 9g's for a force of 9mg. This would be the maximum value the Normal force would have. Now, on Earth he can feel Cf + mg = N at the bottom of the loop and crank his velocity up (v^2 in Cf) until Cf+mg = 9mg. Out in space, since mg=0, Cf can go even higher, and Luke can take the Cf until it equals 9mg. Thus, Luke can take the bottom of the turn with a higher speed in space than he would be able to on Earth.

    At the top of the Loop Luke feels a force of N=Cf-mg and the max value of N is again, 9mg. It seems that this is the reverse of the bottom of the loop. Cf goes up but it always subtracts out the mg so it can get really big (and V can get big). But if I go into space, the mg again disappears, now Cf can't go as high becuase the -mg is gone.

    Thus, in summary, it seems that in deep space, Luke can go much faster at the bottom of a turn than he could on Earth, but not as fast as the top of a loop on Earth.

    Is this correct?
  5. Jun 13, 2012 #4


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    Staff: Mentor

    In deep space, there is no "bottom" and "top".
    He can fly in a circle with the same radius faster than at the bottom on a planet and slower than at the top on a planet.
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