WIP; better figures
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jverzani
@@ -680,9 +680,93 @@ For a general cone, we use this [definition](http://en.wikipedia.org/wiki/Cone):
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Let $h$ be the distance from the apex to the base. Consider cones with the property that all planes parallel to the base intersect the cone with the same shape, though perhaps a different scale. This figure shows an example, with the rays coming from the apex defining the volume.
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::: {#fig-generic-cone}
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```{julia}
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#| echo: false
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plt = let
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gr()
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rad(t) = 3/2 - t
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viewp = [2,-1/1.5,1/2+.2]
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empty_style = (xaxis=([], false),
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yaxis=([], false),
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framestyle=:origin,
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legend=false)
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axis_style = (arrow=true, side=:head, line=(:gray, 1))
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unitize(x) = x / norm(x)
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"""Orthogonal projection along the vector viewp"""
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function make_Pmat(viewp)
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a = unitize( [-viewp[2], viewp[1], 0] )
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b = unitize( [-viewp[3]*viewp[1],
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-viewp[3]*viewp[2],
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viewp[1]^2 + viewp[2]^2] )
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collect(zip(a,b))
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end
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#linear projection of R^3 onto R^2
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function proj(X, viewp)
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Pmat = make_Pmat(viewp)
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x=sum([Pmat[i][1]*X[i] for i in 1:3])
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y=sum([Pmat[i][2]*X[i] for i in 1:3])
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(x, y) # a point
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end
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proj(X) = proj(X, viewp)
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drawdiscF(t) = Shape([psurf(t, 2*i*pi/100) for i in 1:101])
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# discrete determinant of Jacobian; area multiplier?
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function jac(X, u, v)
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ϵ = 0.000001
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A = map((p,q) -> (p-q)/ϵ, X(u+ϵ/2, v), X(u-ϵ/2, v))
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B = map((p,q) -> (p-q)/ϵ, X(u, v+ϵ/2), X(u, v-ϵ/2))
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return A[1]*B[2]-A[2]*B[1]
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end
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# our surface
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R, r, rho = 1, 1/4, 1/4
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f(t) = (R-r) * cos(t) + rho * cos((R-r)/r * t)
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g(t) = (R-r) * sin(t) - rho * sin((R-r)/r * t)
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surf(t, θ) = (rad(t)*f(θ), rad(t)*g(θ), t)
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psurf(t,θ) = proj(surf(t,θ))
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plot(; empty_style..., aspect_ratio=:equal)
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for (i,t) in enumerate(range(0, 3/2, 30))
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plot!(drawdiscF(t); fill=(:gray,1), line=(:black,1))
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end
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θ = 0; plot!([psurf(0, θ), psurf(3/2, θ)]; line=(:black, 2))
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θ = pi/2; plot!([psurf(0, θ), psurf(3/2, θ)]; line=(:black, 1))
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θ = 3pi/2; plot!([psurf(0, θ), psurf(3/2, θ)]; line=(:black, 1))
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current()
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end
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plt
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```
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```{julia}
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#| echo: false
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plotly()
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nothing
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```
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A "cone" formed from the parameterized curve
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$r(t) = \langle
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(R-r) \cdot \cos(t) + \rho \cdot \cos((R-r)/r \cdot t),
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(R-r) \cdot \sin(t) - \rho \cdot \sin((R-r)/r \cdot t)
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\rangle$ with apex at the point $[0,0,3/2]$ and rays extending down through the origin following $3/2-z$.
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:::
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```{julia}
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#| echo: false
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#| eval: false
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plt = let
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h = 5
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R, r, rho = 1, 1/4, 1/4
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@@ -696,7 +780,7 @@ plt = let
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end
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current()
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end
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plot
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plt
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```
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A right circular cone is one where this shape is a circle. This definition can be more general, as a square-based right pyramid is also such a cone. After possibly reorienting the cone in space so the base is at $u=0$ and the apex at $u=h$ the volume of the cone can be found from:
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@@ -736,6 +820,11 @@ plt = let
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r(h,z) = (x(h,z), y(h,z))
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r1(h,z) = (2,0) .+ r(h,z)
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empty_style = (xaxis=([], false),
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yaxis=([], false),
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framestyle=:origin,
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legend=false)
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Nh=30
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heights = range(-1/2, 1/2, Nh)
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h0=heights[Nh ÷ 2]
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@@ -779,7 +868,8 @@ end
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plt
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```
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This figure shows the volume of a figure being comprised of slices. A discrete approximation would be found by estimating the volume of each slice by the cross sectional area times a small $\Delta h$.
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This figure shows the volume of a figure being comprised of slices. A discrete approximation would be found by estimating the volume of each slice by the cross sectional area times a small $\Delta h$. This leads to a formula
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$V = \int_a^b A(h)dh$, where $A$ computes the cross sectional area.
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(This figure was ported from @Angenent.)
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:::
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@@ -836,39 +926,197 @@ Illustration of Cavalieri's first principle. The discs from the left are moved a
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:::
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With the formula for the volume of solids based on cross sections, this is a trivial observation, as the functions giving the cross-sectional area are identical. Still, it can be surprising. Consider a sphere with an interior cylinder bored out of it. (The [Napkin](http://tinyurl.com/o237v83) ring problem.) The bore has height $h$ - for larger radius spheres this means very wide bores.
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With the formula for the volume of solids based on cross sections, this is a trivial observation, as the functions giving the cross-sectional area are identical. Still, it can be surprising.
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Consider a sphere with an interior cylinder bored out of it. (The [Napkin](http://tinyurl.com/o237v83) ring problem.) The bore has height $h$ - for larger radius spheres this means very wide bores.
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::: {#fig-napkin-ring-1}
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```{julia}
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#| echo: false
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plt = let
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# Follow lead of # https://github.com/SigurdAngenent/WisconsinCalculus/blob/master/figures/221/09surf_of_rotation2.py
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# plot surface of revolution around x axis between [0, 3]
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# best if r(t) descreases
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rad(t) = (t = clamp(t, -1, 1); sqrt(1 - t^2))
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rad2(t) = 1/2
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viewp = [2,-2,1]
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##
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unitize(x) = x / norm(x)
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"""Orthogonal projection along the vector viewp"""
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function make_Pmat(viewp)
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a = unitize( [-viewp[2], viewp[1], 0] )
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b = unitize( [-viewp[3]*viewp[1],
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-viewp[3]*viewp[2],
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viewp[1]^2 + viewp[2]^2] )
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collect(zip(a,b))
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end
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#linear projection of R^3 onto R^2
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function proj(X, viewp)
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Pmat = make_Pmat(viewp)
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x=sum([Pmat[i][1]*X[i] for i in 1:3])
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y=sum([Pmat[i][2]*X[i] for i in 1:3])
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(x, y) # a point
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end
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proj(X) = proj(X, viewp)
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# discrete determinant of Jacobian; area multiplier?
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function jac(X, u, v)
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ϵ = 0.000001
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A = map((p,q) -> (p-q)/ϵ, X(u+ϵ/2, v), X(u-ϵ/2, v))
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B = map((p,q) -> (p-q)/ϵ, X(u, v+ϵ/2), X(u, v-ϵ/2))
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return A[1]*B[2]-A[2]*B[1]
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end
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# surface of revolution about the z axis
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surf(t, z) = (rad(t)*cos(z), rad(t)*sin(z), t)
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surf2(t, z) = (rad2(t)*cos(z), rad2(t)*sin(z), t)
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# project the surface at (t, a=theta)
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psurf(t,z) = proj(surf(t,z))
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psurf2(t, z) = proj(surf2(t,z))
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bisect(f, a, b) = find_zero(f, (a,b), Bisection())
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_fold(t, zmin, zmax) = bisect(z -> jac(psurf, t, z), zmin, zmax)
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_foldz(z, tmin, tmax) = bisect(t -> jac(psurf, t, z), tmin, tmax)
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# create shape holding project disc
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drawdiscF(t) = Shape([psurf(t, 2*i*pi/100) for i in 1:101])
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drawdiscI(t) = Shape([psurf2(t, 2*i*pi/100) for i in 1:101])
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# project a line between two points
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pline!(p,q; kwargs...) = plot!([proj(p),proj(q)];
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line_style..., kwargs...)
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α = 1.0
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line_style = (; line=(:black, 1))
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plot(; empty_style..., aspect_ratio=:equal)
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# washer
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t0 = sqrt(3/4)
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Δ = .03
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δ = 0.785398 + 0.05
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x₀ = -.25
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plot!(drawdiscF(x₀-Δ); fill=(:black,), line=(:black,1))
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plot!(drawdiscF(x₀); fill=(:orange,), line=(:black,1))
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plot!(drawdiscI(x₀); fill=(:white,1.0), line=(:black,1))
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x₀ = 0.35
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plot!(drawdiscF(x₀-Δ); fill=(:black,), line=(:black,1))
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plot!(drawdiscF(x₀); fill=(:orange,), line=(:black,1))
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plot!(drawdiscI(x₀); fill=(:white,1.0), line=(:black,1))
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z0 = 3pi/2 - δ
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pline!(surf(t0, z0), surf(-t0, z0); line=(:black, 1))
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pline!(surf(t0, z0+pi), surf(-t0, z0+pi); line=(:black, 1))
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# boundary of sphere
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z0 = 3pi/2 - δ
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curve = [psurf(t, z0) for t in range(-t0, t0, 100)]
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plot!(curve; line=(:black,3))
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z0 = 3pi/2 - δ + pi
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curve = [psurf(t, z0) for t in range(-t0, t0, 100)]
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plot!(curve; line=(:black,3))
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# caps
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curve = [psurf(t0, θ) for θ in range(0, 2pi, 100)]
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plot!(curve, line=(:black, 2))
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curve = [psurf(-t0, θ) for θ in range(0, 2pi, 100)]
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plot!(curve, line=(:black, 2))
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# Shade lines
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δ = pi/6
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Δₜ = (4pi/2 - (3pi/2 - δ))/(2*25)
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for θ ∈ range(3pi/2-δ, 4pi/2, 25)
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curve = [psurf(t, θ) for t in
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range(-t0, max(-t0, -t0 + 1/2*sin(θ+δ+pi/2 + pi/2)), 20)]
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plot!(curve, line=(:black, 1))
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curve = [psurf(t, θ+Δₜ) for t in
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range(-t0, max(-t0, -t0 + 1/3*sin(θ+δ+pi/2 + pi/2)), 20)]
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plot!(curve, line=(:black, 1))
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end
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#=
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f1 = [[t, _fold(t, 0, pi/2)] for t in range(-0.5, -0.1, 26)]
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for f in f1
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plot!([psurf( f[1], f[2]-k*0.01*(6-f[1]) )
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for k in 1:21]; line=(:black, 1))
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end
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=#
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current()
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end
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plt
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```
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Figure showing sphere with interior cylinder bored out.
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:::
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This cross-sectional figure is used to better understand the key dimensions.
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::: {#fig-napkin-ring-2}
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```{julia}
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#| hold: true
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#| echo: false
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#The following illustrates $R=5$ and $h=8$.
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#The following illustrates $R=1$ and $h=2sqrt(3/4$.
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plt = let
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gr()
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empty_style = (xaxis=([], false),
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yaxis=([], false),
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framestyle=:origin,
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legend=false)
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R =5; h1 = 2*4
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R = 1; h = 2*sqrt(3/4)
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theta = asin(h1/2/R)
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thetas = range(-theta, stop=theta, length=100)
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ts = range(-pi, stop=pi, length=100)
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y = h1/4
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θ = theta = asin(h/2/R)
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thetas = range(-theta, stop=theta, length=100)
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ts = range(-pi, stop=pi, length=100)
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y = h/4
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p = plot(legend=false, aspect_ratio=:equal);
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plot!(p, R*cos.(ts), R*sin.(ts));
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plot!(p, R*cos.(thetas), R*sin.(thetas), color=:orange);
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plot(; empty_style..., aspect_ratio=:equal)
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plot!(R*cos.(ts), R*sin.(ts); line=(:black,));
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plot!(R*cos.(thetas), R*sin.(thetas), line=(:orange,1));
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plot!(p, [R*cos.(theta), R*cos.(theta)], [h1/2, -h1/2], color=:orange);
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plot!(p, [R*cos.(theta), sqrt(R^2 - y^2)], [y, y], color=:orange)
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plot!([R*cos.(theta), R*cos.(theta)], [h/2, -h/2]; color=:orange);
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plot!([R*cos.(theta), sqrt(R^2 - y^2)], [y, y]; line=(:orange,3))
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plot!(p, [0, R*cos.(theta)], [0,0], color=:red);
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plot!(p,[ 0, R*cos.(theta)], [0,h1/2], color=:red);
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plot!([0, R*cos.(theta)], [0,0], color=:red);
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plot!([ 0, R*cos.(theta)], [0,h/2], color=:red);
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annotate!(p, [(.5, -2/3, "sqrt(R²- (h/2)²)"),
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(R*cos.(theta)-.6, h1/4, "h/2"),
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(1.5, 1.75*tan.(theta), "R")])
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p
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x₀ = sqrt(R^2 - (h/2)^2)
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annotate!( [
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(x₀/2, 0, text(L"\sqrt{R^2- (\frac{h}{2})^2}",10, :top)),
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(x₀, h/4, text(L"\frac{h}{2}",:right)),
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(R/2*cos(θ),R/2*sin(θ), text(L"R", :bottom; rotation=rad2deg(θ)))
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])
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current()
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end
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plt
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```
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```{julia}
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#| echo: false
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plotly()
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nothing
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```
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Side view illustrating key dimensions of napkin ring problem with $R$ being the radius of the sphere and $h$ being the height of the resulting interior cylinder.
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:::
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The small orange line is rotated, so using the washer method we get the cross sections given by $\pi(r_0^2 - r_i^2)$, the outer and inner radii, as a function of $y$.
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