What is the gravity on a “partial” ringworld?

Let's say I have a part of a Ringworld (see link for specifications). Specifically, I have $$\theta$$ (in radians) of the complete ring; full width, partial circumference ($$\theta = 2\pi$$ would be the full ring).

Questions:

• What is the gravity at various points, at ground level? In particular:
• In the center
• In the middle of each edge
• At a corner
• What is the gravity at the center of gravity (and how high is that above ground level)?

For simplification, assume that the ring is a radial section of a tube having 1 AU radius (at this scale, inner vs. outer radius is negligible), 1,600,000 km tall (i.e. axial dimension), and 70 m thick, with a uniform density of 20,000 kg/m³. (Although "more accurate" answers, and/or answers that estimate how much error this assumption introduces, are welcome!)

Note: Obviously, the gravity at the midpoints of opposite edges will be the same, so the first question is asking about the four points of one quadrant of the segment.

Edit: I believe the general form of the answer comes from integrating $$a = d_{u} (\frac{G m}{|d|^2}) = \frac{G m d}{|d|^3}$$ ($$d_u = \frac{d}{|d|}$$, where $$d$$ is the vector from where we want to determine gravity to the point of the ring section being integrated) over the volume of the ring section, but calculus was never my strong suit... (Also, my attempt at throwing FEA at the problem causes me to wonder if I've botched the formula somehow. It did, however, suggest that my intuition is correct that gravity is never away from the ring section, i.e. the typical simplification of a planet[oid] → its center of mass is inappropriate here.)

• Note that in Ringworld the "gravity" is artificial as a result of the rotation of the ring. – StephenG Jul 10 '19 at 5:46
• Yes, I'm aware. I'm not asking about the pseudo-gravity; I'm specifically interested only in the actual gravity arising from the ring's mass. – Matthew Jul 11 '19 at 15:45

Since the gravitational field has power law of $$r^{-2}$$, distant mass effectively contributes to the field if it scales as $$r^2$$. For example, if you consider a cone and want to calculate the contribution to the field in the vertex from different sections, then section that is twice further away will have 4 times less field from each point, but since the section in 4 times larger, the contribution will be approximately the same.
With that being said, the main factor determining the gravity in this ringworld will be the thickness $$t$$ of the ring. $$g = 2\pi\rho Gt,$$ where $$\rho$$ is the density of the ring material.
The gravity will be approximately the same everywhere and pointing downwards (the difference won't be noticeable to human) and only at $$\approx 70~\mathrm{m}$$ zones near the edge there will be noticeable change in the direction and magnitude of the field, since at the exact edge the gravity is pointing at $$45^\circ$$ to the surface (However, the edge is likely to be filleted).