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July 16

Telescope mount

Back in the day, fancy amateur telescopes used equatorial mounts, which were mechanically complicated but straightforwardly tracked the stars' apparent motion as the earth rotated. X-Y mounts were (and are) mechanically simpler, but figuring out how to point them required intricate calculations.

Now that there are computers in everything, are X-Y mounts displacing equatorial ones, say in the midrange amateur segment? Is XY used at all in the high/professional segment? Thanks. 67.164.113.165 (talk) 05:45, 16 July 2019 (UTC)[reply]

According to our telescope mount article, computerized positioning has enabled the simpler design of an altazimuth mount to be used for all levels up through professional. DMacks (talk) 06:17, 16 July 2019 (UTC)[reply]

Thanks! Yes that is what I meant by XY mount. It looks like the right thing. This stuff is very expensive ready-made but it seems more DIY-able than an equatorial mount. 67.164.113.165 (talk) 06:33, 16 July 2019 (UTC)[reply]

I remember trying to aim a heavy small field of view scope 0.7 degrees north of Polaris in the day without even a protractor, making small adjustments by picking it up and putting it back down and trying to get a nondescript axis of tiny weightlifting weights to point my latitude degrees up and 28.9 degrees left of the street grid which is 28.9 degrees right of real north. The Sun drifted where it shouldn't every half hour or hour cause this is very hard. Fun times. (at night you can look through a monocular in the rod the weights are on with a map of the Polaris area they make for this and aim to within 0.1 degrees in a snap) Sagittarian Milky Way (talk) 16:59, 16 July 2019 (UTC)[reply]

Figure skating jump energies

Out of curiosity, I derived a minimum lower bound on the amount of energy needed to do a figure skating jump with n rotations.

Specifically, the additional kinetic energy needed to both jump into the air and spin is the sum of the necessary translational and angular kinetic energies,

$K={\frac {1}{2}}(I\omega ^{2}+mv_{vert}^{2})$ .

where $v_{vert}$ denotes the vertical component of the takeoff velocity. The minimum angular velocity needed is ${\frac {\pi ng}{v_{vert}}}$ (as derived from the kinematics equation for constant acceleration and the fact that these n rotations must be done in the time between takeoff and landing).

Under these assumptions, by differentiating K with respect to the vertical takeoff velocity, the needed kinetic energy injection is found to be $K=\pi ng{\sqrt {Im}}$ . This is only a lower bound, since additional air time is needed for the needed angular acceleration due to the snap and body inefficiencies (the body doesn't convert all food energy to mechanical energy).

I've heard of a study that says a quintuple jump is the maximum theoretical limit. However, the sky seems to be the limit in practice, and the linear growth of this formula in weight and number of rotations suggests that the limit will really be the force needed, since all this extra kinetic energy needs to be injected in just two stages – the takeoff impulse, and the snap (where conservation of angular momentum is used for getting the rotations needed).--Jasper Deng (talk) 08:46, 16 July 2019 (UTC)[reply]

Dunno about jumping harder, but maybe they could wear weighted gloves to get more angular momentum during the arms-out pre-jump spin. 67.164.113.165 (talk) 09:21, 16 July 2019 (UTC)[reply]

Using equipment other than skates on the ice is frowned upon in this sport except for training purposes, such as ankle weights which help the legs snap together on takeoff.—Jasper Deng (talk) 10:46, 16 July 2019 (UTC)[reply]

The limit is certainly not the sky, rather, the energy best athletes can muster in explosive feat like high jump, or sprint. Or, even better, pole vault, which probably sum up the max energy (aka height, in this instance) that can be turned into something. Then you need calculate the best split of the energy into the jump (determines the time up) and the spin (given the time up, the number of spin). (edit). Long jump will also help, for the maximum jumping time. Gem fr (talk) 10:28, 16 July 2019 (UTC)[reply]

The ideal split calculated in my model is about half angular and half translational. It’s really an apples to oranges comparison to cite other sports since I know of virtually no other sport that uses angular mechanics as much. The phenomenon of angular acceleration by reduction of the moment of inertia is pretty unique to figure skating. What can be known based on weight distribution is the maximum proportional change in it, since one’s moment of inertia about the vertical axis is bounded below by that of the air position. It may not look like it but skaters virtually never have zero angular momentum while moving, especially in jump takeoff situations. Every jump takes off from an edge which is on the circle. This is not the only source of angular momentum, though.—Jasper Deng (talk) 10:48, 16 July 2019 (UTC)[reply]

Not surprised at all by that equipartition (see related equipartition theorem). Both long jump and pole vault involve angular mechanics (although their axis is horizontal, as opposed to vertical in spinning), but that is not my point. The point is, the skater cannot build up significant spinning energy before the jump, so it has to come from the jump impulse (be taken from translation) (edit: while he wont be able to zero the useless horizontal movement energy), meaning, the max energy of vault jumper is a pretty good approximation of the max spinning+ translational energy of the skater. Gem fr (talk) 11:31, 16 July 2019 (UTC)[reply]

Not so. The angular acceleration does significant work: when decreasing the moment of inertia, the angular velocity rises by a factor of the multiplicative inverse of the ratio of the new and previous moments of inertia, and thus so does the angular kinetic energy. Also, especially on toe jumps (the toe loop, flip, and lutz), a very significant amount of energy is injected by the tap on the ice, so it’s not really just the conversion of horizontal translational motion to vertical. If you watch skaters’ toe jumps, even quadruple jumps diminish skaters’ horizontal linear momentum by little.—Jasper Deng (talk) 18:13, 16 July 2019 (UTC)[reply]

The record breakers at pole vault are able to rise their center of mass up to ~5 m, which translate into a maximum vertical speed of (2gh)^0.5= ~10 m/s; which, not strangely at all, is close to the horizontal maximum speed of top running sprinters, meaning, pole vault is pretty efficient at turning horizontal speed into vertical; not sure skaters can be as efficient.Gem fr (talk) 11:51, 16 July 2019 (UTC)[reply]

The pole must store energy. Thus when more bendy better energy storing poles are invented the world record rises. Sagittarian Milky Way (talk) 22:34, 16 July 2019 (UTC)[reply]

Skaters can achieve significantly higher on-ice speeds than the best runners simply due to having less resistance, so they also have much more translational energy to work with.—Jasper Deng (talk) 18:13, 16 July 2019 (UTC)[reply]

OK, you are right, then, you probably should rely on speed skating max speed as a better approximation of max energy. List of world records in speed skating mention 54.40 km/h (15 m/s, which is slightly less than 50% more speed than sprint, but more than 2x as much available energy; it is also a speed at which the air is a significant drag). However, if the skater cannot really tap on the horizontal energy (as per your just above added observation) and must rely on his jumping impulse, then for the really available energy you are back at the same as long jump, that is, sprint, that is, ~10 m/s; or even high jump, where the vertical max speed will be ~5m/s. In any case 15m/s is an upper bound. Gem fr (talk) 22:35, 16 July 2019 (UTC)[reply]

Even if we use 10 m/s, my model predicts an upper bound significantly greater than five rotations – well over a dozen rotations. That is, if no translational energy is converted. Edge jumps like the salchow and loop jump do convert significant translational energy into rotational (by tightening up the edge, that is, lowering the turning radius), at the cost of not being able to inject as much energy anew (a significant amount is still injected new by the legs themselves, though not as much as with toe jumps).--Jasper Deng (talk) 22:55, 16 July 2019 (UTC)[reply]

Intuitively, I think the skater will be able to transform significant part of the horizontal speed/energy into rotation speed/energy, but not into vertical motion. While equipartition would be the best, I doubt the skater will actually achieve it. He will be limited by vertical impetus, which will be of the same magnitude, at best, as a high jumper (~5 m/s; this value is reasonable since I found ref for 4.5m/s). This would make flying time the strongest limit to the number of spin. Gem fr (talk) 00:50, 17 July 2019 (UTC)[reply]

A takeoff velocity of 5 m/s does not seem like a reflection of real limits; air time is given by

{\frac {2v_{vert}}{g}}

which yields only a bit more than a second of air time for that vertical velocity, but I have seen skaters in the air for significantly longer. But yes, real skaters are currently not that close to equipartition. Instead, the limit probably is the maximum possible angular acceleration, since the moment of inertia is bounded below and above. Then that links into the amount of angular momentum one can have while still on their edge (remember, skaters basically never have zero angular momentum even while "gliding straight"). --Jasper Deng (talk) 01:36, 17 July 2019 (UTC)[reply]

one second is pretty long, actually. It can be extended by a small fraction by ending lower than started. And I just watched Javier Sotomayor record breaking jump; I cannot precisely figure his flying time, but it seems shorter than 1.5 s. And skaters do no better; this hints at air time only slightly longer than 1 second. Gem fr (talk) 07:41, 17 July 2019 (UTC)[reply]

This is not the best resource for maximum airtime; skaters known for their air time can get almost two seconds. —Jasper Deng (talk) 08:18, 17 July 2019 (UTC)[reply]

The phenomenon of angular acceleration by reduction of the moment of inertia is pretty unique to figure skating. -- it's a defining feature of the pirouette in ballet, and is probably used in other forms of dance as well. Also in diving and gymnastics, it seems to me. In the "tuck" it's done by pulling the knees towards the chest. 67.164.113.165 (talk) 16:13, 16 July 2019 (UTC)[reply]

Debateable whether ballet counts as a sport as much as I enjoy it. Diving and gymnastics don’t have even close to the same magnitude of angular acceleration; of those, only ballet has comparable angular acceleration.—Jasper Deng (talk)

Since a coin flip is in the realm of Lagrangian mechanics, I think this would be too, though I have no experience applying it myself.—Jasper Deng (talk) 18:39, 16 July 2019 (UTC)[reply]

Honeybee subspecies - A. m. carnica or A. m. carpatica?

Moved from Wikipedia:Teahouse § Honeybee subspecies - A. m. carnica or A. m. carpatica?

https://link.springer.com/article/10.1134/S1022795416020058 https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5872275/ https://link.springer.com/article/10.1134/S1022795418030109

The carpatica subspecies seems to be discussed by the russian federation and carpathian arc, but not accepted by the overall scientific western community. How much higher is the genetic variance threshold for naming a subspecies in the scientific community and what would A.m.carpatica have to show to be included in the categorization? — Preceding unsigned comment added by Anbu95 (talk • contribs) 12:37, 16 July 2019 (UTC)[reply]

Subspecies#Monotypic_and_polytypic_species looks relevant. The variations have to be nonrandom AND distinct geographically, so there may be too much geographical overlap between carnica and carpatica to consider them subspecies. 70.67.193.176 (talk) 14:39, 17 July 2019 (UTC)[reply]

Rotating black hole

If the gravity field around the black hole is spherically symmetrical, how can the rotation of the bh drag the surrounding spacetime around? It is exerting the same radially directed force in all direction, yes? — Preceding unsigned comment added by 86.8.200.199 (talk) 19:21, 16 July 2019 (UTC)[reply]

Short answer: No it’s not radially directed because it’s spinning.—Jasper Deng (talk) 20:30, 16 July 2019 (UTC)[reply]

So it is not radially directed because of the rotation? Can you quote a reference for that assertation,? — Preceding unsigned comment added by 86.8.200.135 (talk) 23:14, 16 July 2019 (UTC)[reply]

Motion around a rotating, uncharged black hole is modeled by the Kerr metric, which is rotationally but not spherically symmetric. I couldn't give you a straightforward equation for the non-radial acceleration from the perspective of a distant observer, but maybe someone else can, or you can find one in one of the references to that page. Someguy1221 (talk) 23:22, 16 July 2019 (UTC)[reply]

Can you explain why? In traditional Newtonian gravity, a symmetrical rotating mass can be modeled the same as a non-rotating mass. I assume that's what inspired the question. ApLundell (talk) 23:18, 16 July 2019 (UTC)[reply]

In general relativity, gravity is a manifestation of spacetime curvature, which is itself a function of mass/energy and momentum. The gravity of a rotating object is thus not even expected to be the same as that of a non-rotating object. Someguy1221 (talk) 23:27, 16 July 2019 (UTC)[reply]

General topic: frame-dragging. --47.146.63.87 (talk) 03:14, 17 July 2019 (UTC)[reply]

can this be explained by vector field theory at all?86.8.201.96 (talk) 19:44, 17 July 2019 (UTC)[reply]

It can expressed using a tensor field, of which a vector field can be seen as a special case. Someguy1221 (talk) 20:02, 17 July 2019 (UTC)[reply]

Moon umbra lighter around the edges

I'm watching the lunar eclipse tonight and it seems that the part of the Moon in the umbra is a little brighter around the edges. Is this due to eye's perception of contrast or just a chance alignment of the maria, or is it a real optical effect? I've observed it several hours ago when the eclipse began and now at a different part of the Moon, which leads me to think it's not the placement of the maria. 93.136.63.164 (talk) 22:21, 16 July 2019 (UTC)[reply]

That part is being lit by brighter parts of twilight. The edge is lit by sunrise and sunset which is why the penumbra is so bright, part of the Sun is above the horizon there. Sagittarian Milky Way (talk) 22:39, 16 July 2019 (UTC)[reply]

Oops I wasn't precise enough. I meant the lunar limb (didn't know the word for it). I agree with your reason for the edges of the umbra, that's also because Sun's radius is non-zero, and because of Earth's atmospheric refraction etc. For instance see here or here, the contrast is weaker than what I perceived visually this night but you can clearly see that none of the darkest parts are at the limb. 93.136.63.164 (talk) 03:47, 17 July 2019 (UTC)[reply]

I'm not sure if libration can ever make the limb a mare from Earth. The limb is generally Maria-free. The umbra radius shrinks slightly from middle of moon to limb with most of that near the limb which should help sharpen the horns a bit. Since the limb is a half percent further. Even an uneclipsed crescent also has sharper horn tips than you'd expect, one would assume a photograph of a crescent phase on a sphere would look like 2 perfect circles of different sizes overlapping but the horn tip would actually be slightly sharper than that. I can't visualize the geometry well enough to see if the geometric reason for this would also help sharpen the horns in a lunar eclipse. Sagittarian Milky Way (talk) 04:24, 17 July 2019 (UTC)[reply]

Thanks, that makes sense. I was kind of hoping for a more esoteric explanation, kind of like opposition surge, but then again it seems opposition surge is also mostly just shadows of a not-quite-2D lanscape. 93.136.58.135 (talk) 05:41, 18 July 2019 (UTC)[reply]

I've remembered something which should cause the horns to be sharper, the umbra is elliptical. The sharpest part of the umbra transition has been experimentally determined to be most closely correlated with the altitude of the tropopause (inside gets cloud blocked, scattered, refracted and dispersed too much) and this is more meters above sea level at Earth's equator than its poles which are already closer to the center of the Earth. Even Alaskan mountains have noticeably thinner air than Himalayan ones of the same altitude, this starts below the tropopause. This horn sharpening subportion is generally unsymmetrical though of course. Sagittarian Milky Way (talk) 12:19, 18 July 2019 (UTC)[reply]