tears posted:yeah it doesnt really answer the question but i appreciate the effort

This is really something that many serious physicists have struggled with. "Is the wavefunction real?" is a question that has seen a decent amount of back and forth over the years by physicists studying quantum phenomena

toyotathon posted:yup let's do visual group theory. should i catch up to ch 6?

sounds good to me

A ⊆ A for every A

x ≤ x for every x,

A ⊆ B and B ⊆ A, therefore, A = B

x ≤ y and y ≤ x, therefore, x = y,

and,

A ⊆ B and B ⊆ C, therefore A ⊆ C

x ≤ y and y ≤ z, therefore, x ≤ z.

so it's like, A = {1,2,3}, B = {1,2,3}, C = {1,2,3,4,5,6,7}

drawing out the trig problem. all this machining stuff's just a lot of trig.:

you use the hypotenuse measured on a steel rule, and the known radius of a pin. then take a picture using a level camera, and use software to determine the angle of the rule. that lets you determine the X and Y distance of the feature on the lathe bed, to the lathe centerline.

example pictures:

it's nice that the skewed reflection kind of 'hones in' on the steel rule and pin contact point.

the measurement, the pin radius, and the angle, make two right triangles with a same angle, so you can get the distances. results were pretty good! the steel rule measures 1/64" or 0.5mm, and of the measurements that should be the same, the formula gives: 98.50mm and 98.62mm, 105.27mm and 105.20mm, and 104.43mm and 104.66mm.

it's to make an accurate jig with a V-channel to hold bar stock, which can slide along the lathe bed and kind of dock into the lathe chuck already along an eccentric centerline, then the 4-jaw chuck tightens down on it. otherwise it's basically hard to chuck something at a precise angle, and off-center in the lathe. this is so it sets a specified angle on a specific radius, to drill into the face of bar stock, except off-center. at least hopefully, who knows.

this is what the jig looks like:

fingers crossed that it more or less works...

Always inspiring to see an independent researcher come up with something new.

In this case the 'amateur' has a PHD in physics but it still gives me hope lol

On September 16, 2011, an anime fan posted a math question to the online bulletin board 4chan about the cult classic television series The Melancholy of Haruhi Suzumiya. Season one of the show, which involves time travel, had originally aired in nonchronological order, and a re-broadcast and a DVD version had each further rearranged the episodes. Fans were arguing online about the best order to watch the episodes, and the 4chan poster wondered: If viewers wanted to see the series in every possible order, what is the shortest list of episodes they’d have to watch?

In less than an hour, an anonymous person offered an answer — not a complete solution, but a lower bound on the number of episodes required. The argument, which covered series with any number of episodes, showed that for the 14-episode first season of Haruhi, viewers would have to watch at least 93,884,313,611 episodes to see all possible orderings. “Please look over for any loopholes I might have missed,” the anonymous poster wrote.

edit: also saw a shoutout today to the science of exact constraint: https://hackaday.com/2018/11/16/expanding-3d-printer-bed-stays-true-under-fire/

this guy who built it is a VERY irritating man, and a dentist. and he also picked the wrong coupling. Kelvin's 1-2-3 is known to be not very good w/ thermal expansion. if the distance between his 3-constraint (socket) ball and 2-constraint (vee) ball grows, like during thermal expansion, then the bed will rotate in the socket, along the path of the vee, until it compensates itself for the increased distance. unless he machined it exactly parallel to the line between them. part of the point of exact constraint is you don't need expensive stuff like 'exactly parallel' anymore... it's why maxwell made the 2-2-2 that everybody uses...

Edited by toyot ()

Edited by toyot ()

Edited by toyot ()

Belphegor posted:I'm working through Linear Algebra by Friedberg

how are you liking friedberg? we're slowly working thru halmos, but it might be too abstract for us..

It's more rigorous/theoretical than the type of linear algebra I learned in Uni, which was very much applied, whereas Friedberg covers generalized vector spaces, not just the 'spatial' ones in R^n

I'm going to go do some right now!

first is about curvature on a 2D surface. if you take the set of normal vectors at every point on a 2D surface (call it

**A**), and multiply them by a scalar r, it makes a new surface (call it

**B**). could you find the radius of curvature of

**A**, by looking at where

**B**intersects itself, for a given r? if you look at my old exact constraint notes maybe you can see what i'm getting at, and why...

another's about spheres. if you're standing on a sphere, holding a compass facing forward, is it possible to walk a path on the sphere and come back to where you started, w/o the compass doing a full rotation, either its needle, or spinning around its N/S axis, or E/W axis?

toyotathon posted:another's about spheres. if you're standing on a sphere, holding a compass facing forward, is it possible to walk a path on the sphere and come back to where you started, w/o the compass doing a full rotation, either its needle, or spinning around its N/S axis, or E/W axis?

just put your finger on the needle as you walk around

toyotathon posted:had a math-specific question, well 2.

first is about curvature on a 2D surface. if you take the set of normal vectors at every point on a 2D surface (call itA), and multiply them by a scalar r, it makes a new surface (call itB). could you find the radius of curvature ofA, by looking at whereBintersects itself, for a given r? if you look at my old exact constraint notes maybe you can see what i'm getting at, and why...

why would **B **necessarily intersect with itself? it sounds like you're just talking about rescaling the surface **A **by r, which wouldnt by itself introduce any sort of self-intersections that werent in **A **to begin with.

another's about spheres. if you're standing on a sphere, holding a compass facing forward, is it possible to walk a path on the sphere and come back to where you started, w/o the compass doing a full rotation, either its needle, or spinning around its N/S axis, or E/W axis?

theres nothing stopping you from holding the compass still, facing the same direction, while you walk in a very small circle. if you want to mark a point on the compass and align that with your velocity then you will always make a full rotation.

c_man posted:c_man posted:

why would B necessarily intersect with itself? it sounds like you're just talking about rescaling the surface A by r, which wouldnt by itself introduce any sort of self-intersections that werent in A to begin with.

no no not rescaling. i'm probably not explaining it properly. making a new surface (B) out of the tips of the surface normals of A, the normals all the same height, determined by some scalar r. B's the normal plane.

this example may or may not clarify what i mean.. If surface A is a flat plane, but has a half-sphere indentation in it of radius r, then the radius of curvature there is r (right?). all the normals of A will meet at a single point (looking down at A, if the normals are length r, it's the center of the circle that's formed by the equator... the rest of the normals are just above A making another plane, distance r away). greater than r and there will be a self-intersection of B (even if A is well-behaved), less than r and no intersection. so you can kind of imagine B changing as r changes, gradually growing off the surface like blades of grass, and when the grass inside the half-sphere is length r (or greater) they touch.

Like c-man says, B is not guaranteed to have intersections.

On which 'side' do we draw the blades of grass? The top or the bottom?

Also, imagine you have an elliptical indent. Which radius are we interested in, the minor one or the major one? What would the 'correct' radius of curvature be, as you call it?

he was interested about the theory of exact constraint... it's so obvious, that self-intersections of the normal bundle should equal the kinematic path. i just know i'm re-treading some 19th century head's work. sure do wish i wasn't living in a fascist state and could speak to kinematicians. every one i look up's eating military grants and making white weapons. real glad i got talking to this particular math guy. i've seen him biking around with a sign on his backpack that says "CLASS WAR NOW"

tears posted:maths is for harry potter nerds

life finds a way