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Gravity in Zorlond & Kiya's Sphere

The Sphere's construction was performed in 152645. In that episode it was specified that the Sphere's gravity came from two sources: a set of all-around grav plating which provided a uniform half-gee everywhere on the surface, and the rotation of the Sphere which produced an effective two-gee pull at the Sphere's equator.

In 281671 and 282234, Zorlond made adjustments to the situation by angling and strengthening the grav plating, and by slowing the rotation of the Sphere, in an effort to fix the unrealized verticality problems. (The former episode also contained details of the Sphere's central star.)

If I've made a mistake in the math, please let me know...

The original setup

The pull of the grav plating is perpendicular to the surface of the Sphere, while the rotational pull would be perpendicular to the Sphere's axis. The resulting gravity effect would therefore be at an angle which is a combination of those two vectors. Because of this effect, someone standing on the inside of the sphere will always perceive a slope to the Sphere's surface unless they're at the equator or poles.

The intensity and direction of gravity in the Sphere depends on the viewer's distance from the Sphere's equator. For convenience, I'll designate the distance in terms of degrees of latitude -- for example, a person standing one-third of the distance from the equator to one of the poles would be at 30 degrees latitude.

Skip the calculations and go to the sample results

Calculating the gravity due to rotation

I'm guessing here (I could be wrong), but I think that effective gravity of a point on a rotating object is proportional to the angular velocity of the point. By my calculations, the ratio of the angular velocity at latitude a to the angular velocity at the equator is cos(a). Here are the calculations Since we know the effective rotational gravity at the equator (2G), the effective rotational gravity at any latitude will be 2 cos(a).

Calculating the resultant gravity

The following section references this diagram. Note that this is crudely drawn, so don't tell me that the vectors aren't the right length, or don't point at exactly the right angles -- this is just meant to give you the general idea.

The length of P is always 0.5, and its angle is the latitude "a". The length of R is 2 cos(a), as stated above, and its angle is always 0.

I calculated the x/y dimensions of F using x = P sin(a) and y = R + P cos(a).

The angle "c" of F (that is, the angle between R and F) is c = arctan(x/y), and the slope angle "b" is b = a - c.

I then calculated the length of F, which is the total force of gravity, using F = x / sin(c). Alternatively, I could have used F = sqrt(x^2 + y^2), but the other way seemed easier.

Sample results

With regards to slope, "downhill" will always be towards the equator. To find the angle the sun will appear to be in the sky, subtract the slope angle from 90 degrees.

The adjustments

A terrace solution was considered at first, but was determined to be insufficient. So in 281671 Zorlond angled the gravity plating to counterbalance the rotational gravity vector and hopefully bring gravity back to vertical throughout the Sphere.

However, in 282234 it was realized that the plating was not strong enough to totally overcome the slope -- that in the middle latitudes there would still be some angle to the gravity. So Zorlond doubled the strength of the grav plating to 1G, and then mounted additional plating like gear teeth on the outside of the Sphere to slow the rotation down so that the equator would be at 1.5G (ignoring the effect of the grav plating).

The rotational speed needed to a achieve a desired gravity is v = sqrt ( a * r ), where a is the desired gravity and r is the radius of the circle. The desired gravity at the equator being (originally) 2G and (later) 1.5G, and the radius of the Sphere being 1AU, then:

• 2G = 19.6133 m/s^2
• 1AU = 149,597,870 km (multiply by 1000 to convert to meters)
• v (starting) = sqrt( 19.6113 * 149,597,870,000 ) = 1,712,923 m/s, which is about 5% of the speed of light!
• 1.5G = 14.709975
• v (ending) = sqrt( 14.709975 * 149,597,870,000 ) = 1,483,435 m/s
• So the Sphere's rotation needs to be reduced by 229,488 m/s, a decrease of 13%.

In 282234 Zorlond elected to slow the rotation gently, at a rate of .4 km/h/s, or 400 m/h/s, or about 0.111 m/s^2. Since t = v / a, this should take about 2,065,392 seconds, or 24 days.

The new setup

Skip the calculations and go to the sample results

This diagram has been oriented so that the surface of the sphere at the sample latitude is downards -- the previous diagram was oriented so that the equator was downwards. This is the third way I tried of calculating this, and it's the one that seems to work best.

Symbols on this diagram --

• Q is not on the diagram, but represents the force of rotational gravity at the equator, which is now 1.5.
• R = the force of gravity due to rotation (perpendicular to the axis). The length of this line is Q * cos(c),
• P = the force of gravity due to the gravity plating (here rather severely angled). The length of this line is now 1.0.
• F = the total force of gravity, perpendicular to the surface of the sphere.
• c = the angle between the surface and rotational gravity, which is equal to the latitude
• b = the angle of the grav plating
• D = a line drawn from the end of R, perpendicular to F
• G = the segment of F between the surface and D
• E = the segment of F from D to the endpoint of F
• m = the angle between D and P

We start out knowing Q, P, and c, and we need F and b.

I used the following sequence of formulae:

• Q = 1.5
• P = 1.0
• R = Q * cos(c)
• D = R * sin(c)
• G = R * cos(c)
• m = arccos(D/P)
• E = P * sin(m)
• F = G + E
• b = 90 - m (assuming m is in degrees)
  b = PI / 2 - m (assuming m is in radians)

Or, if you want to combine everything into two formulae:

• F = Q * cos(c) * cos(c) + P * sin( arccos( Q * cos(c) * sin(c) / P ) )
• b = 90 - arccos( Q * cos(c) * sin(c) / P )

Here's the final results