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Physics behind Halo, the game

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  1. Mar 14, 2012 #1
    Now I've played this awesome game recently, called Halo: Combat Evolved ( I'm sure a lot of people have played it already). It's based on intergalactic warfare.

    A lot of adventures of Master Chief (the main guy) happen to be around on this ring shaped planet-sized thing called the Halo. And the character is actually able to walk around on the inner surface of this. The gravity acts downwards.

    So is it possible to be able to analytically measure the magnitude of the gravity acting on the inner surface provided we have all the required details? I've managed to find the gravitational field intensity at any point on the axis of the ring. But how do I find it near the surface?
     
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  3. Mar 14, 2012 #2

    sophiecentaur

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    Which direction do you define as "downwards"? Do you mean 'towards the local bit of floor'? And is this halo a 'tube'?
     
  4. Mar 14, 2012 #3
    I'm not sure I understand the question since the question implies a non-zero "gravitational field intensity at any point on the axis of the ring."

    Otherwise, this looks like simple Physics 101 question on centrifugal force. I don't know the game, but assume that Halo is a ring somewhat similar to Larry Nivens' Ringworld, with centrifugal force providing the outward force that is felt like gravity. In that case, assuming you know the radius "r" in feet and the velocity of rotation "v" in feet per second then the outward acceleration "a" in feet per second squared is computed as:

    a=r/v^2

    Obviously if you don't know the velocity, but do know the radius and the time for one complete rotation (a day?) in seconds, then:

    v=2*pi*r/t.

    If you know that the "gravity" on Halo is equivalent to one Earth gravity (a=32 feet/sec^2) or any other value (Moon gravity, Mars gravity, etc.), then you can also reverse the computation and calculate "r" and "v" from "t" or vice versa.


    Keep in mind this is computing centifugal force, not surface gravity. A cute exercise left to the reader is the difference in gavity felt at a single point on Earth at Noon vs. Midnight due to orbital centrifugal force. I'd love to see some replies with that calculation.
     
    Last edited: Mar 14, 2012
  5. Mar 14, 2012 #4

    Ryan_m_b

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    Halo: CE, now there's a blast from the past! If you liked that wait until you play reach :wink: Essentially the Halo in the setting is a stanford torus which is similar in principle to an O'Neill cylinder.

    According to the halo wiki the diametre of a ring is 10,000km. I'm assuming it's pretty much Earth gravity so using the equations provided we get:

    v = √(r*a)

    v = √(5e6*10)

    v = 7071mps​

    (hoping there's no basic maths fail here)
     
  6. Mar 14, 2012 #5

    A.T.

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    Is the gravity produced by rotation of the ring, its mass or by both?
     
  7. Mar 14, 2012 #6
    Thanks for the extra info on "r" and "a". I concur with the estimation for "v", so I guess one "day" is 2*pi*r/v = 4443 seconds or about 1.23 hours. :)
     
  8. Mar 14, 2012 #7
    I recall confirming the calculus in Phys101 that gravity everywhere inside a hollow sphere is zero regardless of mass. I expect the same would be true on the inside surface of a ring, aside from a purely lateral component pulling one sideways towards the ring "equator".
     
  9. Mar 14, 2012 #8
    Now Halo is a ring that basically looks like a flat strip curved to form a circle. Indeed, the concept was inspired from Ringworld (I'm looking forward to play the second game). Here's a picture:

    Delta_Halo.jpg


    Downwards meant away from the centre and towards the inner surface of the ring. I'm assuming that all the gravity is due to actual mass of the ring and rotation isn't playing any part (because any such rotating ring would eventually stop or slow down after all the rotational energy is lost in pulling things down).

    Since I noticed that this discussion wasn't really going as I expected, here's my attempt to the problem. I've used Wolfram Integrator to Integrate because so far I've learned only very basic level of lntegration (I'm in eleventh grade, India).

    Point out mistakes if you feel, I'd love to be corrected.


    8KuuY.jpg


    Although, if it isn't due to actual gravity and is just centripetal force, the answer is already there. I don't debate that. I am interested in the 'gravity' rather than rotation.


    But I do debate wkrasl. He says it should be zero. That's not true. Gravity field is zero only because of a special consequence of the Shell theorem.

    It might be true that in the fictional universe the gravity is indeed because of centripetal force. But can't we just assume it isn't for the sake of a nice discussion? Because then it isn't even a problem.
     
  10. Mar 14, 2012 #9

    sophiecentaur

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    No, the zero g situation only happens for a spherical shell. If the ring is (truly!) massive then objects would fall 'down' vertically to the ground (the inside surface). If the radius were large enough, you could even treat the thing as a 'flat earth', the same as you can on a planet sized planet, over smallish distances (artillery trajectories etc). But the mass would need to be extreme to give g at the surface and I doubt that there would be a strong enough material to prevent the whole thing collapsing into a shere in the middle. The structure would be much more stable if it were spinning. It would need comparatively low structural strength.
     
  11. Mar 14, 2012 #10

    sophiecentaur

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    No, the zero g situation only happens for a spherical shell. If the ring is (truly!) massive then objects would fall 'down' vertically to the ground. If the radius were large enough, you could even treat the thing as a 'flat earth', the same as you can on a planet sized planet, over smallish distances (artillery trajectories etc). But the mass would need to be extreme to give g at the surface and I doubt that there would be a strong enough material to prevent the whole thing collapsing into a shere in the middle. The structure would be much more stable if it were spinning. It would need comparatively low structural strength.
     
  12. Mar 14, 2012 #11
    well, actually the Halos are spinning, and that is the cause of the apparent gravity on their inner surface

    HOW are they spinning? magic!
     
  13. Mar 14, 2012 #12
    Hmmm... At least two people, including the originator of the thread, think I'm wrong about the shell theorem applying to a ring. Thanks for all the math in the reply.

    For a little more clarity, see the diagram and the math on
    http://www.saddleback.edu/faculty/bhubbard/documents/SHELLTHEOREMPROOF1_000.pdf.

    With apologies for not having an exact solution, I'll stand by my assertion, which should be intuitively obvious from the diagram. This is a trivial thought experiment. The point of the shell theorem is that, inside the sphere, no matter where you are, the greater attraction of nearer mass is exactly canceled by the lesser attraction of a greater amount of mass pulling towards the opposite side.

    Look at it this way: Any three points including (1) the sphere center, (2) you on the inside surface, and (3) an arbitrary third point elsewhere on the surface, defines a corresponding arbitrary great circle. The net gravitational pull is the infinite sum of the net pull from all of the infintely many infinitely thin great circles, each of which cancels out when inside the sphere.

    The ring is simply one of those great circles, but of non-zero width. If you are exactly on the centerline of that non-zero width ring, the math calculating the shell theorem proves that gravity would be zero if the ring were drawn in ink but the mass still there. That would be true if you relocated yourself any arbitrary distance from your starting point.


    It is clear to me that removing the mass symmetrically outside the parallel ink lines centered on your current position therefore cancels out. All that remains is the mass of the remaining non-zero width ring. Now if you move horizontaal to the surface some distance away from the centerline, there will then be greater mass towards the centerline resulting in a sideways gravitational pull perpendicular towards that centerline. (Discounting the ring thickness placing the actual centerline subsurface.)

    Peace.


    By the way. I see a large planet near the ring. All of the above excludes that consideration. Otherwise I expect the gravity from that planet will be a substantial varying vector added to the centrifugal force vector, resulting in a continuously varying pull similar to amusement park rides that rotate in a plane not parallel to the ground. Barf city!
     
    Last edited: Mar 14, 2012
  14. Mar 14, 2012 #13

    sophiecentaur

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    "The Math", which proves what you say about the ring would need to be spelled out (by you). The situation with a sphere is entirely different from the situation with a ring. The integration is over a solid arc which balances nearby effects of a small area of nearby mass with the effects of a large area of distant mass. Inverse square law doesn't compensate for a ring.
     
  15. Mar 14, 2012 #14
    The Saddlebrook proof actually didn't include the entire sphere. If it had, it would have included a double integral over "theta" (which only traverses a single great circle) plus a separate rotation angle "phi" about the z line axis (to sweep the infinitely many great circles comprising the entire sphere). My plain English reply simply spelled out the obvious point that the double integral, regardles whether "phi" covers a sphere or a subset, it is still an integration of nothing but zeroes.
     
  16. Mar 14, 2012 #15
    I think the problem is that the Halo does not seem to be curved along its shorter length. Or rather than it's like a cylinder that is very short and very wide.

    small height, large radius: this is not the same thing as a great circle of a sphere with infinitesimally small size

    at least that is my understanding
     
  17. Mar 14, 2012 #16

    sophiecentaur

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    That Saddlebrook proof of the shell theorem integrates over the whole sphere.
    The integral over all those circles only becomes zero because complete circles are included, whose mass varies. If you do the same thing, formally, using 'points' along the ring, the contributions of mass for all the points is the same so the integral is different and the sum doesn't go to zero.

    Just consider - it's called the 'Shell Theorem' and not the 'Ring Theorem'. Why do you think that is? You need to consider that your appreciation of that bit of maths may not be quite right. (i.e. assume that you can't be right then see why)

    The detailed shape of the section of the halo is not too relevant to the overall result.
     
  18. Mar 14, 2012 #17
    Alright, so if it's non-zero, how do we calculate it? I've tried doing it above, but failed miserably (no one seems to notice).

    Would someone tell me what went wrong with my integration?

    I've made some corrections, but the answer is still indeterminate.

    u1r52.jpg
     
    Last edited: Mar 15, 2012
  19. Mar 15, 2012 #18

    sophiecentaur

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    Should it not be cot(θ/2) in that final result? (Still indeterminate, though)
    Also, when you put h=0, you may be introducing a 0/0 situation, which will affect the subsequent conclusion.

    It would be 'triv' to do it numerically, wouldn't it? (Not a satisfying method, I admit but it would give you an answer).
     
  20. Mar 15, 2012 #19

    Okay, I'll try putting h=1000 m (which is small compared to 10000 km)
    Putting R=10000000 m
    Putting G= the usual value
    Putting M=1.7x10^17 kg (from wikipedia)

    Hmmmm. This gives me 0.000275 as the answer. That's really less.
     
    Last edited: Mar 15, 2012
  21. Mar 15, 2012 #20

    sophiecentaur

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    Here's a bit of arm waving.
    Divide your ring into equal segments. (small as you like).
    With the test mass at the centre, they are all equidistant from the mass so no net force.

    Move towards the edge. Each segment on ' your side' of the the ring will have a corresponding piece on the other side of the ring. You will be nearer to each of the pieces on your side than to the mirror image segments on the other side. The effect of each segment will be proportional to 1/d2 so the sum of contributions from each of the nearer pieces will be far greater than the sum (same number) of all contributions from their equivalent pieces. But the number of segments involved is only a linear function of your position on the radius. This has to result in a finite sum and not zero. In the limit, for a large ring and a small measurement frame, you would have a 'flat earth' situation.

    This contrasts with the shell situation in which there are progressively more elements on the 'opposite wall' - a 'square law, in fact - which compensates the inverse square law for the force from each.
     
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