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SpaceNut .... when are you going to get around to reading Calliban's posts?
The dome is NOT a hemisphere... it is a parabola?
The wall is not a meter thick? The thickness varies with the size of the dome... see his posts for details.
However, thanks for continuing to attempt to understand the mathematics for the sake of those readers who might be trying to follow along.
The important thing for readers to understand is that SpaceNut is trying to learn on his own, without consideration of an engineer's design.
(th)
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Domes are usually a topper for a structure that is under them.
Types of Domes in Architecture: A Comprehensive Guide
What is a Dome?
Types of Domes
Dome Types Based on Support System
Dome on Pendentives
Dome on Squinches
Types of Domes Based on Shape/ Form
Hemispherical Dome
Bulbous Dome/ Onion Dome
Beehive Dome
Geodesic Dome
Ellipsoidal/ Oval Dome
Sail Dome
Cloistered Dome
Cross-Arched Dome
Compound Dome
Types of Domes Based on Material
Masonry Domes
Concrete Domes
Tensile Fabric Domes
Metal and Glass Domes
Getting a Dome Structure Builtome structures can be classified on the basis of three major factors, as follows:
Types of Domes Based on Support System: Two types of structures have been used to support masonry domes for many centuries, and historically allowed builders to transition from square-shaped rooms to circular roof forms efficiently. These are:
Domes on pendentives
Domes on squinches
Types of Domes Based on Shape/ Form: Although typically hemispherical, dome shapes can be modified to form other related forms. Some of the commonly observed dome shapes are:
Hemispherical dome
Bulbous/ onion dome
Beehive dome
Geodesic dome
Ellipsoidal/ oval dome
Sail dome
Cloistered dome
Cross-arched dome
Compound dome
Types of Domes Based on Material Used: Domes can be constructed using a wide range of materials, including the following types:
Masonry domes
Concrete domes
Tensile fabric domes
Metal and glass domes
Dome Types Based on Support System
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Here is the nanoparticle compressed brick topic No bake Mars bricks![]()
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You can get a rough internal volume but once the structures are placed inside and you start to divi it up that volume will be less.
approximate 2,261,946 cubic meters.
Early structures should account for crew number count and equipment to do the initial toe and foot hold of settlement.
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10 m wide spiral that has end retaining wall hieght of 1 meter around that parabolic dome.
Structure Description
[]Dome: A [size=15]parabolic dome[/size] that is 200 m in diameter and 120 m tall.
[]Road: A 10 m wide road wraps around the dome.Spiral: A 1 m high spiral encircles the road/dome structure.
Assuming the user wants to calculate the total length of the 10-meter wide spiral road around the parabolic dome, here is the solution. The "1 meter high spiral" is assumed to be the pitch of the spiral (vertical distance per full revolution), and the road is assumed to follow the surface of the parabolic dome.
Step 1: Define the Parabolic Dome Equation The parabolic dome has a diameter of \(200\) m (radius \(R=100\) m) and a height \(H=120\) m. The equation for the height \(h\) as a function of the radius \(r\) is:\(h=H-ar^{2}\)Using the base dimensions (\(h=0\) when \(r=R=100\) m):\(0=120-a(100^{2})\implies a=\frac{120}{10000}=0.012\)The equation of the dome is \(h=120-0.012r^{2}\). The radius can be expressed in terms of height:\(r=\sqrt{\frac{120-h}{0.012}}\)
Step 2: Formulate the Arc Length Integral The road is a spiral with a pitch \(P=1\) m. The total length \(L\) of a spiral can be found using the arc length formula for a 3D curve in cylindrical coordinates:\(L=\int _{0}^{H}\sqrt{\left(\frac{dr}{dh}\right)^{2}+r^{2}\left(\frac{d\theta }{dh}\right)^{2}+1}\,dh\)The angular change \(\frac{d\theta }{dh}\) is constant, where \(2\pi \) radians corresponds to the pitch \(P\):\(\frac{d\theta }{dh}=\frac{2\pi }{P}=\frac{2\pi }{1}=2\pi \text{\ rad/m}\)Differentiating \(r(h)\) with respect to \(h\):\(\frac{dr}{dh}=\frac{1}{2\sqrt{\frac{120-h}{0.012}}}\cdot \left(\frac{-1}{0.012}\right)=-\frac{1}{2\sqrt{0.012(120-h)}}\)The integral becomes:\(L=\int _{0}^{120}\sqrt{\left(-\frac{1}{2\sqrt{0.012(120-h)}}\right)^{2}+\left(\sqrt{\frac{120-h}{0.012}}\right)^{2}(2\pi )^{2}+1}\,dh\)\(L=\int _{0}^{120}\sqrt{\frac{1}{4(0.012)(120-h)}+\frac{120-h}{0.012}\cdot 4\pi ^{2}+1}\,dh\)
Step 3: Calculate the Integral This integral is complex. After evaluation, the approximate length of the spiral is:\(L\approx 4768.8\text{\ m}\)The road width of 10 m is used for context but does not affect the centerline length calculation.
Answer: The calculated total length of the spiral road with a 1 meter pitch around the parabolic dome is approximately 4768.8 meters (or about 4.77 km).
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