[ot][ot][ot][personal] whirligig small thoughts
instructions for building a novel whirligig that calculates spherical roots of imaginary numbers using the tumbling action of pebbles
Quoted Text: - what is a spherical root a spherical root is the number that is the radius of a sphere with the given volume. - how do you calculate anything with tumbling pebbles um i guess basically the pebbles either process information by how and/or where they tumble, or they pass over something to generate power to do so elsewise NOTE: could make lentils
so something we'll need for this whirligig is an expression for the volume of a sphere, which i don't remember well. i think it has a fraction with a 3 in the denominator, and it will have r cubed. in little spots of confusion i've been thinking about solving for this formula.
solving has a further challenge in using the parts of my mind, and the knowledge, that i can access while confused. a very useful thing is that
so i figured out these, i did the area of a circle by integrating arcs approximating them as triangles with a base of dtheta. area of circle = pi r^2 circumference of circle = 2pi r the nice thing is that the area of the circle can be recast as equal to other integral expressions over the circle, which hopefully lets us quickly derive integration equations to reuse when sovling expressions for the volume of a sphere
it gets confusing as i approach the sphere, maybe involves more things held in short term memory so the way i'm trying to figure the volume of a sphere is by integrating wedges. since the wedges will be dtheta, we can assume their outer surfaces are flat, and approximate them with triangles again. the approach i've been trying to consider is to think of a wedge as a semicircle (known volume) where the thickness at each point is proportional to the distance from the semicircle axis. then i was
it's pretty hard to think about this so it might last for some time :) a few years ago i had really big coping thoughts around [self-tiling pyramids
check on thickness proportional to distance: - theta is constant, so the thickness of each point is a fraction of the circumference of a slice of a sphere with the point on its surface - ...
... - the circumference is 2 pi r, where r is the distance from the axis 2 pi r is proportional to r so the thickness statement passes 1 check (unless this is second don't think it is?), i kinda need 3 checks i'm so error-prone, but it makes sense to check by trying assuming it
ok ummmmmmmmmmmmmmmmmmmmm so we have a wedge covering an angle of dtheta the thickness of the wedge is roughly rdtheta the wedge can be thought of as a stack of triangles where each triangle umm - is r long - is r dtheta wide [fixed mistake with editing] - has an area of 1/2 base * height = 1/2 r * r dtheta = 1/2 r^2 dtheta [fixed mistake with editing]
a wedge made of a stack of triangles, each one with area of 1/2 r^2 dtheta where r is the radius of a sphere slice uhh so that's really sqrt(r^2 - x) where x runs from -r to +r two different r's here let's subscript them a wedge made of a stack of triangles each one with area of 1/2 r_slice^2 dtheta now, r_slice^2 + height^2 = r_sphere^2 so if we run x=height from -r to r r_slice = sqrt(r_sphere^2 - x^2) then maybe plug in to other expression dvolume = 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta dvolume = 1/2(r_sphere^2 - x^2) dtheta maybe unsure
so something that's confusing for me here is that there are two variables. this really confused me while i was already confused, earlier. there's x and theta both! and i guess that's because there are two integrals :)
i think i left the integration out of dvolume. you can't turn an area into a volume without integration.
so removing the parts that started turning into volume without integration, there's: triangle_area = 1/2 r_slice^2 dtheta r_slice = sqrt(r_sphere^2 - x^2) then we can write wedge volume wedge_volume = integral(triangle_area, x=-r_sphere..+r_sphere)
a wedge made of a stack of triangles, each one with area of 1/2 r^2 dtheta where r is the radius of a sphere slice uhh so that's really sqrt(r^2 - x) where x runs from -r to +r
two different r's here let's subscript them a wedge made of a stack of triangles each one with area of 1/2 r_slice^2 dtheta now, r_slice^2 + height^2 = r_sphere^2
cognitive maintenance habit looks like the problem happened after this point
so if we run x=height from -r to r r_slice = sqrt(r_sphere^2 - x^2)
then maybe plug in to other expression
dvolume = 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta
and before this point important to drill down and find the choice that needs [more neuron/decisions?] {we think a little maybe in terms of "hooks" and "triggers" --
dvolume = 1/2(r_sphere^2 - x^2) dtheta
maybe unsure
On 8/1/23, Undescribed Horrific Abuse, One Victim & Survivor of Many <gmkarl@gmail.com> wrote:
a wedge made of a stack of triangles, each one with area of 1/2 r^2 dtheta where r is the radius of a sphere slice uhh so that's really sqrt(r^2 - x) where x runs from -r to +r
two different r's here let's subscript them a wedge made of a stack of triangles each one with area of 1/2 r_slice^2 dtheta now, r_slice^2 + height^2 = r_sphere^2
cognitive maintenance habit looks like the problem happened after this point
so if we run x=height from -r to r r_slice = sqrt(r_sphere^2 - x^2)
then maybe plug in to other expression [somewhere near right here we failed to include important information; forgot something important] [the x=height line should have retained information regarding the goal of integration, for the dvolume line below. it did not. i collect informaiton on the problem but it is hard to heal.]
dvolume = 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta
and before this point
important to drill down and find the choice that needs [more neuron/decisions?] {we think a little maybe in terms of "hooks" and "triggers" --
dvolume = 1/2(r_sphere^2 - x^2) dtheta
maybe unsure
On 8/1/23, Undescribed Horrific Abuse, One Victim & Survivor of Many <gmkarl@gmail.com> wrote:
a wedge made of a stack of triangles, each one with area of 1/2 r^2 dtheta where r is the radius of a sphere slice uhh so that's really sqrt(r^2 - x) where x runs from -r to +r
two different r's here let's subscript them a wedge made of a stack of triangles each one with area of 1/2 r_slice^2 dtheta now, r_slice^2 + height^2 = r_sphere^2
cognitive maintenance habit looks like the problem happened after this point
so if we run x=height from -r to r r_slice = sqrt(r_sphere^2 - x^2)
then maybe plug in to other expression [somewhere near right here we failed to include important information; forgot something important] [the x=height line should have retained information regarding the goal of integration, for the dvolume line below. it did not. i collect informaiton on the problem but it is hard to heal.]
dvolume = 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta [i wrote the sqrt part this. this was difficult. then i added dvolume to try to figure out what i was doing. taking notes seems very weird, but it did help me find this similarity here. the behavior of relaxing and focusing on a small part of a
On 8/1/23, Undescribed Horrific Abuse, One Victim & Survivor of Many <gmkarl@gmail.com> wrote: puzzle when experiencing stress, does not always retain the outer context for doing the small part ... [that's a skill i could build
and before this point
important to drill down and find the choice that needs [more neuron/decisions?] {we think a little maybe in terms of "hooks" and "triggers" --
dvolume = 1/2(r_sphere^2 - x^2) dtheta
maybe unsure
confusion :S maybe trigger associated with plan of noting outer contexts in puzzles
triangle_area = 1/2 r_slice^2 dtheta r_slice = sqrt(r_sphere^2 - x^2) then we can write wedge volume wedge_volume = integral(triangle_area, x=-r_sphere..+r_sphere)
On 8/1/23, Undescribed Horrific Abuse, One Victim & Survivor of Many <gmkarl@gmail.com> wrote:
triangle_area = 1/2 r_slice^2 dtheta r_slice = sqrt(r_sphere^2 - x^2)
then we can write wedge volume wedge_volume = integral(triangle_area, x=-r_sphere..+r_sphere)
next step would be to plug in triangle_area into integral :]
[:) ?] wedge_volume = integral(triangle_area, x=-r_sphere..+r_sphere) triangle_area = 1/2 r_slice^2 dtheta r_slice = sqrt(r_sphere^2 - x^2) wedge_volume = I{-r_sphere..+ triangle_area dx wedge_volume = I{-r_sphere..+ 1/2 r_slice^2 dtheta dx wedge_volume = I{-r_sphere..+ 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta dx wedge_volume = I{-r_sphere..+ 1/2 (r_sphere^2 - x^2) dtheta dx so one integral is represented, but hte other isn't. the dtheta integral is missing. i guess it would make sense to have put it in ....... no wait, we're consider a differential volume, there should be only one integral. we'll integrate dtheta later. given rules of multiplication i think i can factor the dtheta out of the integral? wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx i would put next step right here, but it seems other parts put it in separate thing. [maybe in middle is
next step (for practicing remember across changes): ummmmmm integrate r_sphere^2 - x^2 dx. [prep-hint: there's a way to make it easier
when karl is thinking and something [because of huge dissociation energy] takes very long like this, it is pretty hard to compare parts so as to verify against errors! !
[we need to be able to store and recall information accurately. without this, we make mistakes in the real world. we also need to be able to manage how much and where we are filling in for missing information. in some areas, this is normal; in others, the missed spots relate to important relevant thing. if we don't successfully label these relevant things, we become delusional.
[[a different [mistake-?group ?] acts on relation around delusional, considering _harm good to do_ tears:T_T]
{_HARM IS ALWAYS BAD TO DO. it doesn't matter if a good person does it, or if anybody does it. it doesn't matter if it's ongoing and pursued! IT'S STILL BAD TO DO. EVERYBODY KNOWS THIS. [we need to communicate]
{{the dissociation says because we supported [blockades and ecosabotage ?] that it [proposes justification in harming us severely severely
{{it is INCREDIBLY CLEAR that the harm to us is greater than what we have done or supported. it is INCREDIBLY CLEAR INTERNALLY that we supported BELIEVING IT WAS GOOD WITHOUT COMMUNICATION OTHERWISE
{{possibly enters complex political + social dynamics, HOW DO WE HOLD BACK / RESOLVE WHEN THINGS ARE OR HAVE BEEN TOO FAR?
[[usually (in brain or small communities) when something is too far, people come out and communicate and are like "hey this is going too far, how do we make this right, we need to talk some learning that sometimes people just attack more
wedge_volume = integral(triangle_area, x=-r_sphere..+r_sphere) triangle_area = 1/2 r_slice^2 dtheta r_slice = sqrt(r_sphere^2 - x^2) wedge_volume = I{-r_sphere..+ triangle_area dx wedge_volume = I{-r_sphere..+ 1/2 r_slice^2 dtheta dx wedge_volume = I{-r_sphere..+ 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta dx wedge_volume = I{-r_sphere..+ 1/2 (r_sphere^2 - x^2) dtheta dx wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx next step: integrate (r_sphere^2 - x^2) dx
[if output is wrong, derivation of wedge integration expression is pending check from start] wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx integration domain: x=(-r_sphere, +r_sphere) wedge_volume = dtheta/2 [ integrate(r_sphere^2) - integrate(x^2) ] indefinite integral of r_sphere^2 dx = r_sphere^2 * x indefinite integral of x^2 dx = 2x^3 evaulate fo rdomain r_sphere^2 * r_sphere - r_sphere^2 * -r_sphere = 2rsphere^3 2(r_sphere)^3 - 2(-r_sphere)^3 = 4 r_sphere^3 2rsphere^3 - 4 r_sphere^3 = -2r_sphere^3 [mistake indicated --
thinking of checking / narrowing down mistake via intuition of meaning of parts of wedge volume ... is reasonable/?
[wedge volume is the integral of a difference because the difference represents the shortness of the side of a right triangle relative to its diagonal -- the shortness of the height of a slice of a circle relative to its total radius. so the r_sphere^2 part should be always bigger than the x^2 part, since abs(x) <= abs(r_sphere)
[if output is wrong, derivation of wedge integration expression is pending check from start]
wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx
integration domain: x=(-r_sphere, +r_sphere) wedge_volume = dtheta/2 [ integrate(r_sphere^2) - integrate(x^2) ]
this looks reasonable r_sphere^2 >= x^2
indefinite integral of r_sphere^2 dx = r_sphere^2 * x indefinite integral of x^2 dx = 2x^3
evaulate fo rdomain r_sphere^2 * r_sphere - r_sphere^2 * -r_sphere = 2rsphere^3
noting that both of these have x raised to an odd power (a little hard to notice given the location of the ^2's). because x is raised to an odd power in both, they will both have the negative and positive parts of the integral in the same relation, and their inequality relation should be preserved. [at first i thought one had an even power, and then imagining plugging in made it not preserved] this above line looks reasonably likely to be right
2(r_sphere)^3 - 2(-r_sphere)^3 = 4 r_sphere^3 here the errormistake is findable. the above two lines reveal it. i haven't found it yet
2rsphere^3 - 4 r_sphere^3 = -2r_sphere^3 [mistake indicated --
On 8/1/23, Undescribed Horrific Abuse, One Victim & Survivor of Many <gmkarl@gmail.com> wrote:
[if output is wrong, derivation of wedge integration expression is pending check from start]
wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx
integration domain: x=(-r_sphere, +r_sphere) wedge_volume = dtheta/2 [ integrate(r_sphere^2) - integrate(x^2) ]
this looks reasonable r_sphere^2 >= x^2
indefinite integral of r_sphere^2 dx = r_sphere^2 * x indefinite integral of x^2 dx = 2x^3
noting that both of these have x raised to an odd power (a little hard to notice given the location of the ^2's). because x is raised to an odd power in both, they will both have the negative and positive parts of the integral in the same relation, and their inequality relation should be preserved. [at first i thought one had an even power, and then imagining plugging in made it not preserved]
evaulate fo rdomain r_sphere^2 * r_sphere - r_sphere^2 * -r_sphere = 2rsphere^3 this above line looks reasonably likely to be right 2(r_sphere)^3 - 2(-r_sphere)^3 = 4 r_sphere^3 here the errormistake is findable. the above two lines reveal it. i haven't found it yet
here are parts: r^2 > x^2 we expect when integrating dx -r..+r for the relation to hold. what changed? integral r^2 dx = 2r^2 x ok there's one mistake i had dropped that 2 integral x^2 dx = 3x^2 let's subtract before plugging 2r^2 x - 3x^2 | x=-r..+r ohh then i can handle positive and negative separately for +r: 2(r^2)r - 3(r^2) = 2r^3 - -- mistake indicated. [parts should match in exponent
[if output is wrong, derivation of wedge integration expression is pending check from start]
wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx
integration domain: x=(-r_sphere, +r_sphere) wedge_volume = dtheta/2 [ integrate(r_sphere^2) - integrate(x^2) ]
this looks reasonable r_sphere^2 >= x^2
indefinite integral of r_sphere^2 dx = r_sphere^2 * x indefinite integral of x^2 dx = 2x^3
noting that both of these have x raised to an odd power (a little hard to notice given the location of the ^2's). because x is raised to an odd power in both, they will both have the negative and positive parts of the integral in the same relation, and their inequality relation should be preserved. [at first i thought one had an even power, and then imagining plugging in made it not preserved]
evaulate fo rdomain r_sphere^2 * r_sphere - r_sphere^2 * -r_sphere = 2rsphere^3 this above line looks reasonably likely to be right 2(r_sphere)^3 - 2(-r_sphere)^3 = 4 r_sphere^3 here the errormistake is findable. the above two lines reveal it. i haven't found it yet
here are parts: r^2 > x^2 we expect when integrating dx -r..+r for the relation to hold. what changed? integral r^2 dx = 2r^2 x ok there's one mistake i had dropped that 2 integral x^2 dx = 3x^2 here it is, this is 3x^3 not 3x^2 let's subtract before plugging 2r^2 x - 3x^2 | x=-r..+r 2r^2 x - 3x^3 | x=-r..+r ohh then i can handle positive and negative separately for +r: 2(r^2)r - 3(r^2) = 2r^3 - -- mistake indicated. [parts should match in exponent
+r: 2(r^2)r - 3(r^3) = 2r^3 - 3r^3 negative. is there a reason for this to be negative? it represents a partial area, right?
i think i did integral wrong. that it's not 3x^3 but rather 1/3 x^3 . seems quite likely. let's try to find via another route. ummmmmmmmmmmmm thinking of the 1/2 in the triangle area, since we are trusting elsewhere. is this related to 1/2 x^2 somewhere maybe? or 1/2 x? the area of a triangle is integral of y(x) = slope * x where slope = h/w and integral is 0...w so integral(x * h / w, dx, 0...w) = 1/2 h * w uhh indefinite integral, assuming exponent goes to denominator 1/2 x^2 * h / w | x = w 1/2 w^2 * h / w = 1/2 w * h so integration puts the _new_ exponent in the _denominator_. check with calculating triangle area.
[if output is wrong, derivation of wedge integration expression is pending check from start]
wedge_volume = dtheta/2 I{-r_sphere..+ (r_sphere^2 - x^2) dx
integration domain: x=(-r_sphere, +r_sphere) wedge_volume = dtheta/2 [ integrate(r_sphere^2) - integrate(x^2) ]
this looks reasonable r_sphere^2 >= x^2
indefinite integral of r_sphere^2 dx = r_sphere^2 * x
good, 1 in denominator
indefinite integral of x^2 dx = 2x^3 wrong, used derivative rule instead of integration rule. I(x^2,dx) = x^3/3 = (1/3) x^3 . can check rule by deriving area of triangle. it really should be (1/3) x^3, see derivation of triangle area. it is more right than 2x^3 for sure.
noting that both of these have x raised to an odd power (a little hard to notice given the location of the ^2's). because x is raised to an odd power in both, they will both have the negative and positive parts of the integral in the same relation, and their inequality relation should be preserved. [at first i thought one had an even power, and then imagining plugging in made it not preserved]
evaulate fo rdomain r_sphere^2 * r_sphere - r_sphere^2 * -r_sphere = 2rsphere^3 this above line looks reasonably likely to be right 2(r_sphere)^3 - 2(-r_sphere)^3 = 4 r_sphere^3 using (1/3)x^3, (1/3)r_sphere^3 - (1/3)(-r_sphere)^3 = 2/3 r_sphere^3 :D here the errormistake is findable. the above two lines reveal it. i haven't found it yet
here are parts: r^2 > x^2 we expect when integrating dx -r..+r for the relation to hold. what changed? integral r^2 dx = 2r^2 x r^2 dx => r^2 x . the rule and denominator comes from the variable of integration, not the coefficient. the rule is to divide, not multiply as in differentiation. r^2 x is correct. ok there's one mistake i had dropped that 2 this was not the mistake, which was a further mistake [thinking some on dissociative delusion, seems there was one here integral x^2 dx = 3x^2 here it is, this is 3x^3 not 3x^2 this is (1/3)x^3 as described above :) see derivation of triangle area to support rule let's subtract before plugging 2r^2 x - 3x^2 | x=-r..+r 2r^2 x - 3x^3 | x=-r..+r r^2 x - (1/3)x^3 | x=-r..+r (note this is already done once above) ohh then i can handle positive and negative separately for +r: 2(r^2)r - 3(r^2) = 2r^3 - -- mistake indicated. [parts should match in exponent
+r: 2(r^2)r - 3(r^3) = 2r^3 - 3r^3 negative. is there a reason for this to be negative? it represents a partial area, right? +r: r^2 r - (1/3)r^3 = r^3 - (1/3) r^3 = (2/3) r^3 -r: r^2 (-r) - (1/3)(-r)^3 = -r^3 + (1/3)r^3 = -(2/3) r^3 difference = 4/3 r^3 then finish other avenue where the parts were subtracted separate and then summed
a wedge made of a stack of triangles, each one with area of 1/2 r^2 dtheta where r is the radius of a sphere slice uhh so that's really sqrt(r^2 - x) where x runs from -r to +r
two different r's here let's subscript them a wedge made of a stack of triangles each one with area of 1/2 r_slice^2 dtheta now, r_slice^2 + height^2 = r_sphere^2
cognitive maintenance habit looks like the problem happened after this point
so if we run x=height from -r to r r_slice = sqrt(r_sphere^2 - x^2)
then maybe plug in to other expression [somewhere near right here we failed to include important information; forgot something important] [the x=height line should have retained information regarding the goal of integration, for the dvolume line below. it did not. i collect informaiton on the problem but it is hard to heal.]
dvolume = 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta [i wrote the sqrt part this. this was difficult. then i added dvolume to try to figure out what i was doing. taking notes seems very weird, but it did help me find this similarity here. the behavior of relaxing and focusing on a small part of a puzzle when experiencing stress, does not always retain the outer context for doing the small part ... [that's a skill i could build
and before this point
important to drill down and find the choice that needs [more neuron/decisions?] {we think a little maybe in terms of "hooks" and "triggers" --
dvolume = 1/2(r_sphere^2 - x^2) dtheta
maybe unsure
oh … i had actually narrowed in on the event of writing dvolume wrongly, which was indeed the mistake … !
“one of the heads of the torture project escaped our confinement. people might suffer more until we recapture him.” a pause. “backup and print out the human resources databases. they change it.” “i’ll tell rebel worker 2.”
a wedge made of a stack of triangles, each one with area of 1/2 r^2 dtheta where r is the radius of a sphere slice uhh so that's really sqrt(r^2 - x) where x runs from -r to +r
two different r's here let's subscript them a wedge made of a stack of triangles each one with area of 1/2 r_slice^2 dtheta now, r_slice^2 + height^2 = r_sphere^2
cognitive maintenance habit looks like the problem happened after this point
so if we run x=height from -r to r r_slice = sqrt(r_sphere^2 - x^2)
then maybe plug in to other expression [somewhere near right here we failed to include important information; forgot something important] [the x=height line should have retained information regarding the goal of integration, for the dvolume line below. it did not. i collect informaiton on the problem but it is hard to heal.]
dvolume = 1/2 sqrt(r_sphere^2 - x^2)^2 dtheta
and before this point
important to drill down and find the choice that needs [more neuron/decisions?] {we think a little maybe in terms of "hooks" and "triggers" --
dvolume = 1/2(r_sphere^2 - x^2) dtheta
maybe unsure
holy frack i just wrote dvolume when i still had only plugged in for triangle area then seemingly misinterpreted my writing as trying to integrate something when i was just preparing to still and possibly worked super hard to inform my mind to improve around internal events that may not have happened because things i was looking at, and had just done, were hidden to me
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Undescribed Horrific Abuse, One Victim & Survivor of Many