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# HD Vision WrapArounds

Admin
These are sunglasses that claim to give you HD vision (whatever that means) and enchance your vision. Just saw a TV ad for them and it was ridiculous. How anyone could actually be inticed to buy these pieces of junk is beyond me. One lady in the ad claims they look like designer glasses, when really they look down right silly.

https://www.hdwraparounds.com

## Answers and Replies

Looks like they:
Decrease brightness/filter.
Encase your eyes so your pupils adjust to the lower light amount. (any leakage and these would seem like normal sunglasses.)
This = Less washed out colors from it being too bright out, but your eyes compensate for the low intensity.
Seems like at most they may increase color depth, but no way does it have anything to do with HD.

They so don't look designer, that's for sure....

NoTime
Science Advisor
Homework Helper
turbo
Gold Member
These look like the cheesy cheap sunglasses that ophthalmologists dole out to old folks after cataract surgery. Anybody who has a clue about physiology and optics knows that decreasing the amount of light to your eyes results in pupil dilation, with exacerbates all the optical defects in your eyes' lenses. If you are in bright light and are nearsighted with astigmatism (as I am) you will see WAY better than if you are wearing sunglasses or are in a dimly-lit environment.

AS an ABO-certified optician, I used to consult with patients regarding coatings, tinting, etc, of the glasses that I made for them, and it's WAY more complex than these cheesy ads portray.

Moonbear
Staff Emeritus
Science Advisor
Gold Member
These look like the cheesy cheap sunglasses that ophthalmologists dole out to old folks after cataract surgery.

:rofl: They do! :rofl:

They have amber lenses. That particular color does make things appear brighter. I've encountered that color (in a more stylish version) in my trials of sunglasses over the years, and my experience was it was also pretty crappy for functioning as sunglasses...I'd still be squinting in the sunlight when wearing them.

turbo
Gold Member
:rofl: They do! :rofl:

They have amber lenses. That particular color does make things appear brighter. I've encountered that color (in a more stylish version) in my trials of sunglasses over the years, and my experience was it was also pretty crappy for functioning as sunglasses...I'd still be squinting in the sunlight when wearing them.
I have used amber shooting glasses from Busch and Lomb and they are OK. I would recommend clear or lightly-tinted lenses with good UV coatings for anybody who spends a lot of time outdoors or on the water. They allow enough light to your eyes to constrict your pupils (good for sharper vision and restriction of UV to the interior of the eye) and the additional coatings cut the UV even more.

mgb_phys
Science Advisor
Homework Helper
II would recommend clear or lightly-tinted lenses with good UV coatings for anybody who spends a lot of time outdoors or on the water.
Isn't polycarbonate pretty much dead below 400nm, do UV coatings really do much on plastic lenses?

Moonbear
Staff Emeritus
Science Advisor
Gold Member
I have used amber shooting glasses from Busch and Lomb and they are OK. I would recommend clear or lightly-tinted lenses with good UV coatings for anybody who spends a lot of time outdoors or on the water. They allow enough light to your eyes to constrict your pupils (good for sharper vision and restriction of UV to the interior of the eye) and the additional coatings cut the UV even more.

What's the point of clear lenses? No thanks, I think I'll stick with ones that would actually allow me to open my eyes outdoors, that's the point of sunglasses.

turbo
Gold Member
What's the point of clear lenses? No thanks, I think I'll stick with ones that would actually allow me to open my eyes outdoors, that's the point of sunglasses.
Sunglasses are great if they are properly designed. If they are cheap glasses that attenuate visible light without attenuating UV, they will harm your eyes. Your eyes have evolved to dilate in the absence of light and to contract is an excess of light. The problem with this simple evolutionary model is that reducing the visible light to the eye results in more-open irises and more exposure to UV damage to the eyes. I am not an ophthalmologist, but I worked for a very large consortium of them and built their public presentations so that they could make their cases to optometrists and the general public.

HD Wraparounds - maybe not that bad

For a couple of years, I have used a "advertised on TV" type of sunglasses, and I love them. They are brown/amber and fit over my regular glasses. The brown color highlights colors by blocking blue, but is not as harsh as yellow "blue blockers" so the overall effect is pleasant. If you have never tried glasses of this shade outdoors, you might be surprised at how good the effect is.

I used these both for sun shade and for watching clouds when storm chasing.

As for UV protection, my regular clear prescription glasses provide those.

The "HD Wraparounds" appear to be similar to what I have. If so, for some people, they will indeed be a useful product.

Evo
Mentor
Thre best sunglasses I've ever had were a pair of prescription glasses with polarized lenses. They were a dark purple that looked almost black, but didn't make things look dark. I quite often forgot that I had them on when I went indoors.

The amazing thing about them were how they made colors so vibrant. I've never seen grass so green. Everything just looked so much more beautiful with them on. The dog ate them. I hope I can find another optometrist that knows what made them so unique. You can't buy anything like them over the counter that I've found.

Mech_Engineer
Science Advisor
Gold Member
• Enhance your vision
• Just like High Definition TV

If your vision is "just like" an HDTV, its pretty horrible. This is one of many products that plays on the B.S. catch phrase of the year, "High Definition." If it isn't HD, it must be crap!!!

I have Saltzman's Nodular Degeneration, something about bluish gray nodules floating around in my eyes, I'm obviously not a doctor, but my biggest issue is glare. My glasses are transitional lenses, they have anti-glare, are polarized, whatever can be incorporated in a pair of glasses is there. I also have prescription sunglasses that I use when driving, and they help, but the glare is STILL a problem.

I was going to try these "HD Wrao Arounds" since they fit over my regular glasses, and they are only 20.00, so even if they help a little, hey, I'll go for it.

From what I've been told, Saltzman's is somewhat rare, and my current doctor really doesn't know how to treat it, so he just sees my every 3 months or so, and I've been using steroid eye drops for several years, so I feel like I'm on my own here. I'll take whatever help I can get.

turbo
Gold Member
If you have been properly diagnosed, you have small nodules growing on your cornea - the outer surface of the eye. They can be surgically removed to re-establish proper curvature and restore good vision. You should talk to your doctor about this.

WarPhalange
What would happen if you wore those while watching TV in HD? Would your eyes just melt from the clear picture?

turbo
Gold Member
What would happen if you wore those while watching TV in HD? Would your eyes just melt from the clear picture?
Yes! I had a '65 Jeep CJ5 and I loved JC Whitney. I bought all the fuel-saving devices they offered, and I had to stop every 20 miles or so to siphon the extra gas out of the tank so it wouldn't overflow. That was a pain in the a. Had to get rid of the Jeep because it was too small to carry the gas-cans.

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If you have been properly diagnosed, you have small nodules growing on your cornea - the outer surface of the eye. They can be surgically removed to re-establish proper curvature and restore good vision. You should talk to your doctor about this.

I'm pretty confident about the diagnosis, it came from a doctor at Doheny Eye Instute here in Los Angeles, he's a corneal specialist, and I had confidence in him.

If you'll bear with me, I'll give you my story, as briefly as I can. Since I've got your 'ear', LOL.

Unfortunately, I belong to a Blue Cross HMO. The whole thing started because my very astute optometrist wrote me a note that said she suspected some kind of Keratitas(sp?) she told me to give the note to my primary physician, which I did, after several months I rec'vd a referral to the ophthalmologist that is part of the HMO. He took a look, said he thought the optometrist was right, and he said it was 'very unusual' and he felt I'd qualify to see the specialist, due to the rarity. After several more months I rec'vd the referral to see Dr. Song at Doheny Eye Institute. He had some impressive equipment, anyway, he immediately diagnosed it as Saltzman's. He said there were several options, the first being steroid eye drops "Alrex". I followed up 3 months later, the nodules looked much better, they were gathering together into one and I had marked improvement, but they didn't disappear. I followed up a few more times, and while there was no more improvement, they didn't get any worse. I was expecting to go on to another option, instead Blue Cross declined the next follow-up, and sent me back to the original ophthalmologist. I liked that doctor, but he was of advanced age, and felt some of the options, like surgery were too extreme, he told me it wouldn't get better, that his goal was to make me comfortable. I didn't get a vote. He retired a few years later, and I rec'vd a letter that a new ophthalmologist was under contract, I've been seeing that one for the past year. I told him I had conflicting opinions on the course of treatment, and asked him what he thought, he literally shrugged. He said to continue the Alrex, and follow up every 3 months. They've taken lots of pictures of my eyes, but haven't told me what to expect, or what my options were, nothing. Until my last visit. I told him the glare problem was worse, I noticed a decline in my vision, and that my eyes felt more strained lately after working on the computer. He told me the nodules looked about the same, that it might be normal fluctuations. However, I did tell him it was getting to be more of a problem, so he said 'let me take a look at the surrounding tissue. Then he said well you 'have a couple of small cataracts' WHAT??? !!?? I said 'isn't that really bad?' 'something very elderly people get?' I'm 48, not young, bit I will need my eyes for many years to come, I'm a Teacher Librarian. He said no, everybody has them, it's like a wrinkle. But you know, I remember the other doctor saying the Alrex might possibly cause cataracts. He also said the nodules are getting closer to the pupil. Then, out of the blue, he said "We might have to consider a corneal transplant'....uh, let's just say I was shocked, since he had been pretty laid back and unconcerned in the past. I told him the Doheny Doctor had said there were options, including surgical removal, he said "oh, I can write another referral, if you'd like' well, needless to say, I very much wanted that, he wrote the referral, and I'm now waiting to see if Blue Cross will approve it.

To his credit the current ophthalmologist, had prescribed Restasis to combat the dry eye associated with Salzman's, and, of course, Blue Cross declined the prescription. The last time I saw him was Friday, and he now had samples, so he gave me a month's supply, I've been using them twice a day like he said. So...there you have it, another HMO nightmare. and here I am, looking into HD Wraparounds...

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Evo
Mentor
Here there is a community "fund" that pays for prescriptions for people that aren't covered by their insurance. You might want to check around. Also check with local medical schools that might deal with your problem. Also, most hospitals work with charities and have other options to help pay for what is not covered by insurance. Unfortunately, too many people don't know to even ask about these options.

Good luck!

turbo
Gold Member
I'm pretty confident about the diagnosis, it came from a doctor at Doheny Eye Instute here in Los Angeles, he's a corneal specialist, and I had confidence in him.

If you'll bear with me, I'll give you my story, as briefly as I can. Since I've got your 'ear', LOL.

Unfortunately, I belong to a Blue Cross HMO. The whole thing started because my very astute optometrist wrote me a note that said she suspected some kind of Keratitas(sp?) she told me to give the note to my primary physician, which I did, after several months I rec'vd a referral to the ophthalmologist that is part of the HMO. He took a look, said he thought the optometrist was right, and he said it was 'very unusual' and he felt I'd qualify to see the specialist, due to the rarity. After several more months I rec'vd the referral to see Dr. Song at Doheny Eye Institute. He had some impressive equipment, anyway, he immediately diagnosed it as Saltzman's. He said there were several options, the first being steroid eye drops "Alrex". I followed up 3 months later, the nodules looked much better, they were gathering together into one and I had marked improvement, but they didn't disappear. I followed up a few more times, and while there was no more improvement, they didn't get any worse. I was expecting to go on to another option, instead Blue Cross declined the next follow-up, and sent me back to the original ophthalmologist. I liked that doctor, but he was of advanced age, and felt some of the options, like surgery were too extreme, he told me it wouldn't get better, that his goal was to make me comfortable. I didn't get a vote. He retired a few years later, and I rec'vd a letter that a new ophthalmologist was under contract, I've been seeing that one for the past year. I told him I had conflicting opinions on the course of treatment, and asked him what he thought, he literally shrugged. He said to continue the Alrex, and follow up every 3 months. They've taken lots of pictures of my eyes, but haven't told me what to expect, or what my options were, nothing. Until my last visit. I told him the glare problem was worse, I noticed a decline in my vision, and that my eyes felt more strained lately after working on the computer. He told me the nodules looked about the same, that it might be normal fluctuations. However, I did tell him it was getting to be more of a problem, so he said 'let me take a look at the surrounding tissue. Then he said well you 'have a couple of small cataracts' WHAT??? !!?? I said 'isn't that really bad?' 'something very elderly people get?' I'm 48, not young, bit I will need my eyes for many years to come, I'm a Teacher Librarian. He said no, everybody has them, it's like a wrinkle. But you know, I remember the other doctor saying the Alrex might possibly cause cataracts. He also said the nodules are getting closer to the pupil. Then, out of the blue, he said "We might have to consider a corneal transplant'....uh, let's just say I was shocked, since he had been pretty laid back and unconcerned in the past. I told him the Doheny Doctor had said there were options, including surgical removal, he said "oh, I can write another referral, if you'd like' well, needless to say, I very much wanted that, he wrote the referral, and I'm now waiting to see if Blue Cross will approve it.

To his credit the current ophthalmologist, had prescribed Restasis to combat the dry eye associated with Salzman's, and, of course, Blue Cross declined the prescription. The last time I saw him was Friday, and he now had samples, so he gave me a month's supply, I've been using them twice a day like he said. So...there you have it, another HMO nightmare. and here I am, looking into HD Wraparounds...
Cherylyn, your diagnosis (if accurate) has a reliable surgical remedy. It's probably only mildly more complex than that of a patient who who needs surgical re-shaping of their cornea for purely visual correction, but your case has to be individually evaluated by an ophthalmologist. Dry eye and associated side effects associated with your condition can bring on other problems that can result in vision loss, and your doctor should evaluate your condition seriously in regard to such possibilities. Please see another ophthalmologist for a second opinion, and ask him/her to interact with your insurance company. Loss of sight is a very serious issue with medical insurance companies, and they may well be willing to cover the cost of surgery if the procedure is properly prescribed and the diagnosis is properly documented and coded. (Don't get me started on medical coding!!) I wish you the best. Please seek further help. Best regards and best wishes.

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Cherylyn, your diagnosis (if accurate) has a reliable surgical remedy. It's probably only mildly more complex than that of a patient who who needs surgical re-shaping of their cornea for purely visual correction, but your case has to be individually evaluated by an ophthalmologist. Dry eye and associated side effects associated with your condition can bring on other problems that can result in vision loss, and your doctor should evaluate your condition seriously in regard to such possibilities. Please see another ophthalmologist for a second opinion, and ask him/her to interact with your insurance company. Loss of sight is a very serious issue with medical insurance companies, and they may well be willing to cover the cost of surgery if the procedure is properly prescribed and the diagnosis is properly documented and coded. (Don't get me started on medical coding!!) I wish you the best. Please seek further help. Best regards and best wishes.

I will do that, thank you so much, to think that this can be corrected is encouraging. Thank you so much.

I should have sought another opinion long ago, I'm usually pretty assertive about important issues like my vision, I just got caught up in the complexity of it all I guess. Now that the school year has ended I'll have time to find another ophthalmologist, and take on Blue Cross. I'm sure the surgery would be less costly to BC than continuing to see the ophthalmologist every 3 months and prescribe the Alrex the rest of my life. Anyway, again, thank you, you have no idea how hard it has been for me to get information or insight. There's not much help on the net either. Funny how things work, I happened upon this site while looking at HD Wraparounds, go figure

Here there is a community "fund" that pays for prescriptions for people that aren't covered by their insurance. You might want to check around. Also check with local medical schools that might deal with your problem. Also, most hospitals work with charities and have other options to help pay for what is not covered by insurance. Unfortunately, too many people don't know to even ask about these options.

Good luck!

Thank you, that is good to know. I will pay on my own if necessary, but without insurance I'm sure I could only go so far on a teacher's salary. I'll give Doheny a call, since I've never asked them, just to see if they might have some advise for me along those lines.

Hey, you guys have been so nice and helpful, I'll bookmark this site, and let you know the outcome, if that's okay. I'm into books, we need a final chapter.

Bye for now!

Ivan Seeking
Staff Emeritus
Science Advisor
Gold Member
Please do visit again.

The night vision glasses that come free with these seem just as outrageous as the HD glasses.

Thank you, that is good to know. I will pay on my own if necessary, but without insurance I'm sure I could only go so far on a teacher's salary. I'll give Doheny a call, since I've never asked them, just to see if they might have some advise for me along those lines.

Hey, you guys have been so nice and helpful, I'll bookmark this site, and let you know the outcome, if that's okay. I'm into books, we need a final chapter.

Bye for now!

Hi, Back! I saw the corneal specialist @Doheny Eye Institute in August, he said it was time to cut the nodules out, but before he proceded with surgical removal, he told me to see the laser specialist for consultation, that he might be able to remove the lesions through PTK(Phototheraputic Keratectomy). I rec'vd a thorough work up, this doctor said he thought I was a good candidate for the surgery. He prescribed a stronger eye drop, "Prednisolone" and an antibiotic, some low level type, since he discovered I had clogged oil ducts under my lower eye lids. He said he'd like to get me in good shape for surgery. Then, of course, it went to Blue Cross for authorization. Doheny HATES HMO's, that's understandable, since they have so many requirements, but after all, that's why I have insurance, so the doctor sent a letter, and I took the meds in the meantime, I feel like my vision has improved with the stronger eye drops, but it improved with the milder drops at first too, so I don't expect the improvement to be permenant without the surgery.

Well, miracle of miracle, I rec'vd authorization from Blue Cross today, I'm very happy, although, I wish it hadn't taken so long, and been so difficult. I actually think my vision will get better, the sensativity to light has been the most difficult issue for me, bright sunlight is excrutiating to me.

According to the authorization letter, my preliminary diagnosis: 371.46, nodular cornea degen.
service to be performed: Phototheraputic Keratectomy

Turbo 1: Special thanks to you, I felt more confident about speaking with the doctor at Doheny, and you were right, there is an effective treatment. I don't know about the 'coding, and documentation of the diagnosis" but I hope they got it right, because I'm clueless about such matters, and Blue Cross restricted approval to 1 visit. Come to think of it, I have 2 eyes, I thought they'd do the procedure to one eye at a time.... LOL, whatever, step by step I guess.

Thanks again to all of you.


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# BPS sectors in $\cal{N}=4$ SYM

I am familiar with the idea of a BPS bound as in a lower limit on the mass of supermultiplets given by a certain function of the central charge and when I think of $\cal{N}=4$ SYM I see a complicated lagrangian in my mind.

Somehow the above picture seems insufficient to understand the classification of what are called "BPS sectors" in $\cal{N}=4$ SYM.

I would like to know what is the meaning and the derivation of the following listing for $\cal{N}=4$ SYM ,

• $\frac{1}{2}$ BPS sector consists of multi-trace operators involving a single bosonic operator $Z$

• $\frac{1}{4}$ BPS sector consists of multi-trace operators involving two bosonic operator $Z, Y$

• $\frac{1}{8}$ BPS sector consists of multi-trace operators involving three bosonic operator $Z, Y, X$ and two fermionic operators $\lambda$, $\bar{\lambda}$

(everything above is apparently assumed to be in the the Lie algebra of $U(N)$)

I would also like to know what is the relation between the above classification and thinking in terms of short/semi-short/long multiplets.

I would be happy to know of expository references for the above topics. I haven't been able to locate any on my own.

-

The BPS conditions in $\mathcal{N}=4$ SYM is just another name for the shortening conditions in the $psu(2,2|4)$ symmetry algebra of the theory. This algebra has 32 odd generators: 16 supercharges $Q_{\alpha a}$ and $\tilde{Q}_{\dot{\alpha}}^a$ and 16 superconformal charges $S^a_\alpha$ and $\tilde{S}_{\dot\alpha a}$. Here the indices $\alpha,\dot\alpha=1,2$ are Weyl spinor indices and $a=1,\dotsc,4$ is an index in the fundamental representation of the $SU(4)$ R-symmetry group. The generators $S$ and $\tilde{S}$ have dimension $-1/2$ and hence annihilates a primary state. The highest weight state of a BPS multiplet is furthermore annihilated by one or more of the supercharges. In particular a $\frac{1}{2}$-BPS state, or chiral primary, is annihilated by half of the charges $Q$ and $\tilde{Q}$. Similarly $\frac{1}{4}$-BPS ($\frac{1}{8}$-BPS) states are annihilated by a quarter (an eighth) of these charges. The exact details of what fields make up the operators in the various subsectors of course depend on the choice of simple roots (or lowering/raising operators) in the algebra, but the notation in the question seems fairly standard.
The details of the symmetry algebra should be covered in any review of $\mathcal{N}=4$ SYM or the AdS/CFT-correspondence. See eg arXiv:1012.3938, which however only discusses the planar theory where only single trace operators are relevant. From an algebra point of view the more general case should however be essentially the same.
@Olof I am a bit confused about what you are defining as a "primary". I thought a primary by definition something which is annihilated by all the $Q$s and the $S$s. Am I wrong? Also in your notation when you write $S^a _\alpha$ and $\tilde{Q}^a _\dot{\alpha}$ is there a notion of which index comes first? $a$ or $\alpha$ ? I am hoping that the review you linked to is beginner friendly and will explain what I am confused about! I haven't seen these stuff's exposition anywhere. –  user6818 Jun 24 '11 at 9:55

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# Bibliothèque

Musique » Ed Sheeran »

## You Need Me, I Don't Need You

32 écoutes | Se rendre sur la page du titre

Du Dimanche 22 mai 2011 au Mardi 22 mai 2012 Tout le temps

Titres (32)
Titre Album Durée Date
You Need Me, I Don't Need You 3:39 14 fév. 2012, 0h45m
You Need Me, I Don't Need You 3:39 29 jan. 2012, 23h32m
You Need Me, I Don't Need You 3:39 6 nov. 2011, 13h38m
You Need Me, I Don't Need You 3:39 30 oct. 2011, 12h53m
You Need Me, I Don't Need You 3:39 9 oct. 2011, 21h35m
You Need Me, I Don't Need You 3:39 1 oct. 2011, 8h43m
You Need Me, I Don't Need You 3:39 30 sept. 2011, 19h48m
You Need Me, I Don't Need You 3:39 30 sept. 2011, 16h01m
You Need Me, I Don't Need You 3:39 25 sept. 2011, 12h08m
You Need Me, I Don't Need You 3:39 25 sept. 2011, 10h50m
You Need Me, I Don't Need You 3:39 24 sept. 2011, 0h36m
You Need Me, I Don't Need You 3:39 23 sept. 2011, 14h10m
You Need Me, I Don't Need You 3:39 23 sept. 2011, 8h35m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 23h05m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 22h45m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 18h59m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 18h08m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 14h57m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 14h11m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 14h04m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 13h58m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 13h16m
You Need Me, I Don't Need You 3:39 22 sept. 2011, 13h11m
You Need Me, I Don't Need You 3:39 21 sept. 2011, 23h53m
You Need Me, I Don't Need You 3:39 18 sept. 2011, 11h38m
You Need Me, I Don't Need You 3:39 15 sept. 2011, 21h57m
You Need Me, I Don't Need You 3:39 14 sept. 2011, 15h50m
You Need Me, I Don't Need You 3:39 10 sept. 2011, 13h12m
You Need Me, I Don't Need You 3:39 6 sept. 2011, 14h38m
You Need Me, I Don't Need You 3:39 6 sept. 2011, 11h18m
You Need Me, I Don't Need You 3:39 5 sept. 2011, 22h23m
You Need Me, I Don't Need You 3:39 5 sept. 2011, 22h16m


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# Contactomorphisms have in general no fixed points

Let $$(V, \xi = \ker \alpha)$$ be a cooriented contact manifold. A contactomorphism of $$(V, \xi)$$ is a diffeomorphism $$\phi$$ which preserves the contact structure $$\xi$$ and its coorientation. In other words, there exists $$g : V \to \mathbb{R}$$ such that $$\phi^* \alpha = e^g \alpha.$$ In symplectic topology, a major driving force is the Arnold conjecture, which relates the number of fixed points of a Hamiltonian symplectomorphims to topological invariants of the symplectic manifold (depending on the version, the invariant can be the minimal number of critical points of a function, the sum of the Betti numbers, or the cuplength). Of course, the surprise here is that in general, symplectomorphisms might not have fixed points (for instance a rotation or the torus endowed with its standard symplectic form). However, as soon as they are Hamiltonian, fixed points appear.

In the contact setting, the analogue of the Arnold conjecture was introduced by S. Sandon, and pertains to a different kind of points (for contactomorphisms), called translated points. From what I understand, Sandon chooses to deal with translated points instead of fixed points because contactomorphisms need not have any fixed points, even if they are obtained as time-$$1$$ map of contact isotopies. This apparently comes from the odd-dimensionality of the contact manifold $$(V, \xi)$$, but I don't understand why. Since the Euler characteristic of an odd-dimensional manifold is $$0$$, the Lefschetz fixed point theorem implies that for any diffeomorphism of $$V$$ which is isotopic to the identity, the sum of the indices of its fixed points must be $$0$$, but the indices might cancel each other out, so this does not imply that there as no fixed points at all.

Is there a reason why odd-dimensionality prevents "most" transformations to have fixed points?

Thanks a lot in advance!

## 2 Answers

I hope that the following answers some parts of the question.

1) a) It is not true that a generic contactomorphim doesn't have fixed points. For example, let $$M$$ be the three-dimensional torus that is the projectivisation of the tangent bundle to the two-torus $$\mathbb T^2=\mathbb R^2/\mathbb Z^2$$. Any self-diffeomorphism of $$\mathbb T^2$$ induces a concatomorphsm of $$T^3$$. Consider an Anosov sel-diffeomorphism of $$\mathbb T^2$$, for example, given by the Arnol'd cats map $$(x,y)\to (2x+y, x+y)\; {\rm mod }\;1$$ https://en.wikipedia.org/wiki/Arnold%27s_cat_map

This map has a unique fixed point on $$\mathbb T^2$$ and it has two fixed points on the projectivisation of the tangent bundle $$T^3$$. If one calculates the eigenvalues of this map of $$T^3$$, one sees that they both have real eigenvalues of modulus different from $$\pm 1$$. So if we slightly perturb the contactomorphism it will still have two fixed points.

b) If you want a similar example given by a 1-parameter family, one can take some self-diffeo of $$\mathbb S^2$$, induced by a linear map of $$\mathbb R^3$$, $$(x,y,z)\to (e^{at}x, e^{bt}y, e^{ct}z)$$, and again take the induced map on the projectivised cotangent bundle of $$\mathbb S^2$$.

2) Even if we think about symplectic geometry, the negation of always having fixed points of a symplectomorphism, is not "never having fixed points". For example if we look at the last geometric theorem of Poincare: https://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Birkhoff_theorem which was important for the whole development of symplectic geometry, we see that one can not drop the condition on the annuls self-map to be area preserving. If we drop this condition we can have both generic maps that have fixed points and generic maps that don't have fixed points.

3) Finally, informally, I would guess, that though the circle $$S^1$$ is usually not considered as a contact manifold, it has some of the features. One should probably think that the group of contactomorphisms of $$S^1$$ is the group of its diffeomorphisms. It is not true that generic diffeomorphsims of $$S^1$$ don't have fixed points.

Here's a complement to Dmitri's excellent answer. While it's true that contactomorphisms can have fixed points, even for some open set of contactomorphisms, there is a reason you don't expect to be able to detect these using something like Floer theory: if you could, it would tell you there are fixed points for all contactomorphisms, but it's very easy to construct contactomorphisms (even $$C^{\infty}$$-close to the identity) that don't have fixed points. Pick a contact form $$\alpha$$ and let $$R$$ be its Reeb field. This is nowhere vanishing by definition (since $$\alpha(R)=1$$). Flow along $$R$$ for very small time. This is a contactomorphism with no fixed points.

The difference between symplectic and contact is the following. While a symplectic manifold is like the full phase space of a dynamical system, a contact manifold is like the level set of a Hamiltonian. The Hamiltonian will usually have critical points (fixed points of the flow), but if you pick a single level set you're unlikely to pick a critical level set, so you're unlikely to see any fixed points.


text:
F. Goblins And Gnomes
time limit per test
4 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

Monocarp plays a computer game called "Goblins and Gnomes". In this game, he manages a large underground city of gnomes and defends it from hordes of goblins.

The city consists of $n$ halls and $m$ one-directional tunnels connecting them. The structure of tunnels has the following property: if a goblin leaves any hall, he cannot return to that hall.

The city will be attacked by $k$ waves of goblins; during the $i$-th wave, $i$ goblins attack the city. Monocarp's goal is to pass all $k$ waves.

The $i$-th wave goes as follows: firstly, $i$ goblins appear in some halls of the city and pillage them; at most one goblin appears in each hall. Then, goblins start moving along the tunnels, pillaging all the halls in their path.

Goblins are very greedy and cunning, so they choose their paths so that no two goblins pass through the same hall. Among all possible attack plans, they choose a plan which allows them to pillage the maximum number of halls. After goblins are done pillaging, they leave the city.

If all halls are pillaged during the wave — Monocarp loses the game. Otherwise, the city is restored. If some hall is pillaged during a wave, goblins are still interested in pillaging it during the next waves.

Before each wave, Monocarp can spend some time preparing to it. Monocarp doesn't have any strict time limits on his preparations (he decides when to call each wave by himself), but the longer he prepares for a wave, the fewer points he gets for passing it. If Monocarp prepares for the $i$-th wave for $t_i$ minutes, then he gets $\max(0, x_i - t_i \cdot y_i)$ points for passing it (obviously, if he doesn't lose in the process).

While preparing for a wave, Monocarp can block tunnels. He can spend one minute to either block all tunnels leading from some hall or block all tunnels leading to some hall. If Monocarp blocks a tunnel while preparing for a wave, it stays blocked during the next waves as well.

Help Monocarp to defend against all $k$ waves of goblins and get the maximum possible amount of points!

Input

The first line contains three integers $n$, $m$ and $k$ ($2 \le n \le 50$; $0 \le m \le \frac{n(n - 1)}{2}$; $1 \le k \le n - 1$) — the number of halls in the city, the number of tunnels and the number of goblin waves, correspondely.

Next $m$ lines describe tunnels. The $i$-th line contains two integers $u_i$ and $v_i$ ($1 \le u_i, v_i \le n$; $u_i \ne v_i$). It means that the tunnel goes from hall $u_i$ to hall $v_i$. The structure of tunnels has the following property: if a goblin leaves any hall, he cannot return to that hall. There is at most one tunnel between each pair of halls.

Next $k$ lines describe the scoring system. The $i$-th line contains two integers $x_i$ and $y_i$ ($1 \le x_i \le 10^9$; $1 \le y_i \le 10^9$). If Monocarp prepares for the $i$-th wave for $t_i$ minutes, then he gets $\max(0, x_i - t_i \cdot y_i)$ points for passing it.

Output

Print the optimal Monocarp's strategy in the following format:

At first, print one integer $a$ ($k \le a \le 2n + k$) — the number of actions Monocarp will perform. Next, print actions themselves in the order Monocarp performs them. The $i$-th action is described by a single integer $b_i$ ($-n \le b_i \le n$) using the following format:

• if $b_i > 0$ then Monocarp blocks all tunnels going out from the hall $b_i$;
• if $b_i < 0$ then Monocarp blocks all tunnels going into the hall $|b_i|$;
• if $b_i = 0$ then Monocarp calls the next goblin wave.

You can't repeat the same block action $b_i$ several times. Monocarp must survive all waves he calls (goblins shouldn't be able to pillage all halls). Monocarp should call exactly $k$ waves and earn the maximum possible number of points in total.

If there are several optimal strategies — print any of them.

Examples
Input
5 4 4
1 2
2 3
4 3
5 3
100 1
200 5
10 10
100 1

Output
6
-2 -3 0 0 0 0

Input
5 4 4
1 2
2 3
4 3
5 3
100 100
200 5
10 10
100 1

Output
6
0 -3 0 0 1 0

Input
5 10 1
1 2
1 3
1 4
1 5
5 2
5 3
5 4
4 2
4 3
2 3
100 100

Output
6
1 2 3 4 5 0

Note

In the first example, Monocarp, firstly, block all tunnels going in hall $2$, secondly — all tunnels going in hall $3$, and after that calls all waves. He spent two minutes to prepare to wave $1$, so he gets $98$ points for it. He didn't prepare after that, that's why he gets maximum scores for each of next waves ($200$, $10$ and $100$). In total, Monocarp earns $408$ points.

In the second example, Monocarp calls for the first wave immediately and gets $100$ points. Before the second wave he blocks all tunnels going in hall $3$. He spent one minute preparing to the wave, so he gets $195$ points. Monocarp didn't prepare for the third wave, so he gets $10$ points by surviving it. Before the fourth wave he blocks all tunnels going out from hall $1$. He spent one minute, so he gets $99$ points for the fourth wave. In total, Monocarp earns $404$ points.

In the third example, it doesn't matter how many minutes Monocarp will spend before the wave, since he won't get any points for it. That's why he decides to block all tunnels in the city, spending $5$ minutes. He survived the wave though without getting any points.


text:
Protons are positively charged particles while electrons are negatively charged particles. If you round this to the closest whole number it is the mass number The mass number is a combination of protons and neutrons. A proton is one of three main particles that make up the atom. and electrons. no jab in 2021 means no entry to anything [yes] ? Neutrons: 69 Additional Practice. It's atomic mass is 119. There are 69 neutrons. The masss number of tin(Sn) is the number of protons plus neutrons of a specific isotope of tin(Sn) . I'm very new at chemistry and I cant get a grip on some of the info yet so therefore my question is how many protons, neutrons and electrons are in 1 atom of iodine -127 1 atom of hydrogen -3 1 27AL+3 ion 1 19F-1 ion I cant figure this out no matter what I … When elements are not combined with anything else, they are typically neutral ( … What raw materials are reading glasses made from? I just need 3 as I am doing  a paper.? The atomic number for magnesium is 12. mass number is 108, neutrons = 108 - 47 protons = 61 neutrons. In its ground state, all atoms have the same number of protons Atomic number = 6. therefore 6 protons, 6 electrons. please drop suggestions to help me finish homework!!!? Where in Victoria could you buy Mulberry and Osage orange wood? An atomic mass unit ($$\text{amu}$$) is defined as one-twelfth of the mass of a carbon-12 atom. Try these on your own and check the answer below 78 Se 2-39 K + ANSWERS. All tin isotopes have 50 protons. So neutrons = 12 - 6 = 6 neutrons. Tin is identified by having 50 protons in its nucleus. Answer (1 of 1): Sn (Tin) has 50 electrons, 50 protons but 68 neutrons. 6 protons and. The average weight is 118.71, so, on average, a tin atom has 69.71 neutrons (118.71 -50). or . Morning or night? Answer to: How many protons, neutrons, and electrons are in the tin(ii) ion, 119 50 Sn2+? The number of protons in an atom is not changeable so you can add or subtract electrons to get the charge. ... equals. When did organ music become associated with baseball? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. The atomic number of tin(Sn) is the number of protons in tin(Sn) nucleus. There are 50 protons, 50 electrons, and 68 neutrons. Example: This tin atom has 50 protons, 69 neutrons and 48 electrons. C. atomic number 6. that means 6 protons. Figure $$\PageIndex{1}$$: Electrons are much smaller than protons or neutrons. An atomic mass unit ($$\text{amu}$$) is defined as one-twelfth of the mass of a carbon-12 atom. 24-12=12. protons neutrons weight, proton neutron mass ratio, protons neutrons electrons quizlet, protons neutrons electrons titanium, proton neutron scattering, Protons are nearly the same size as neutrons and are much larger than electrons.Sep 17, 2010 windows2universe.org You can simply subtract the atomic number from the mass number in order to find the number of neutrons. Brad Parscale: Trump could have 'won by a landslide', Ex-NFL lineman unrecognizable following extreme weight loss, Watch: Extremely rare visitor spotted in Texas county, Baby born from 27-year-old frozen embryo is new record, Hiker recounts seeing monolith removed from desert, Hershey's Kisses’ classic Christmas ad gets a makeover, 'Voice' fans outraged after brutal results show, 'Retail apocalypse' will spread after gloomy holidays: Strategist. Hope this helps! All Rights Reserved. Why don't libraries smell like bookstores? All Tin (Sn) atoms have 50 protons. And there will be 50 electrons around a neutral atom of tin. The number of neutrons is the atomic weight rounded to the nearest whole number minus the number of protons. If an electrons were the mass of a penny, a proton or a neutrons would have the mass of a large bowling ball! Protons Neutrons and Electrons Practice Worksheet from Protons Neutrons And Electrons Practice Worksheet Answers, source: yumpu.com. The most common isotope for Sn is 118 which mean 118-50 = 68 mass number = protons + neutrons. Pandemic benefits underpaid in most states, watchdog finds, Trump threatens defense bill over social media rule. Help:  how do I write out "8 times as much as 2 less than 5? Join Yahoo Answers and get 100 points today. The atomic number of tin is 50, so there are 50 electrons and protons. Atoms are made of protons, neutrons, and electrons. What is the best time for study? https://www.wikihow.com/Find-the-Number-of-Protons,-Neutrons,-and-Electrons Copyright © 2020 Multiply Media, LLC. You can find out by looking at the atomic weight, the number of electrons and protons are the same as the atomic number, and the neutrons are the atomic weight minus the protons. If you have an aluminium atom, it will have 13 protons in each atom because its atomic number is 13. c. tin-119. unique in that it appears to have 10 stable isotopes. For neutral atoms, the number of electrons will also be 50. 11 - 5 = 6 neutrons. The number of neutrons varies because there are different isotopes of tin. It also has more electrons than tin, but the exact number of electrons depends on its oxidation state. Table $$\PageIndex{1}$$ gives the properties and locations of electrons, protons, and neutrons. Neutron- does not bear a net of electric charge. Finally: 47 protons , 61 neutrons and 46 electrons. at least for neutral elements, the number of electrons will always equal the number of protons, so also 6. as far as neutrons, you have to subtract the atomic number from the atomic mass (approx. Protons are found in the nucleus of the atom. The number of electrons is equal to the number of protons (50). It is written Tin may have between 49 and 87 neutrons. Where is Trump going to live after he leaves office? Finding the number of protons, electrons and neutrons in an atom. An ion has an unequal number of protons and electrons. The number of protons and the number of electrons in a neutral atom are found from the atomic number (Z). The most naturally abundant stable isotope of tin has 70 How did the rastafarian culture come to South Africa? Ag has atomic number 47 = 47 protons and 47 electrons. Name: Tin Symbol: Sn Atomic Number: 50 Atomic Mass: 118.71 amu Melting Point: 231.9 °C (505.05 K, 449.41998 °F) Boiling Point: 2270.0 °C (2543.15 K, 4118.0 °F) Number of Protons/Electrons: 50 Number of Neutrons: 69 Classification: Other Metals Crystal Structure: Tetragonal Density @ 293 K: 7.31 g/cm 3 Color: white Atomic Structure and 50 electrons. 2.6: Protons, Neutrons, and Electrons in Atoms Last updated Sep 27, 2015; Save as PDF 2.5: The Structure of The Atom; 2.7: Finding Patterns - The Periodic Law and the Periodic Table The magnitude of changes on the proton and the electron are equal. The nucleus contains both protons and neutrons. It is Carbon-12, which means the mass nuber is 12. Likewise, gold has atomic number 79, so each gold atom has 79 protons in it. The atomic mass of tin(Sn) is the average number of protons plus neutrons of all the isotopes of tin(Sn) . What should you call a female patterdale? Electrons: 12. 6 electrons. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions. Table $$\PageIndex{1}$$ gives the properties and locations of electrons, protons, and neutrons. The third column shows the masses of the three subatomic particles in "atomic mass units." Tin is identified by having 50 protons in its nucleus. The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. basically defines an element. The atomic number of a sodium atom is 11 and its mass number is 23. Cation- bears more protons than electrons, so it has positive charge. How long was Margaret Thatcher Prime Minister? Neutrons are neutral particles. You can find the number of protons in an atom easily because it is just the atomic number of the atom. Still have questions? Can permission or pictures be a topic covered in a class ta law school? How many protons neutons and electrons are in tin. The number of protons plus the number of neutrons = the atomic weight. Every atom of tin has 50 protons in its nucleus. This chemistry video tutorial explains how to calculate the number of protons, neutrons, and electrons in an atom or in an ion. Get your answers by asking now. How do I write a multi-paragraph introduction. Neutrons: 12. How old was queen elizabeth 2 when she became queen? An element has a fixed number of protons but may exist with various numbers of neutrons. b. Carbon-14. See related link. Yes it does. Protons: 50. neutrons. The atomic number for tin is 50. Protons carry a positive electrical change, while electrons are negatively charged, and neutrons are neutral. 119-50=69. mass number = 14. therefore neutrons = 14 - 6 = 8 neutrons. # of protons = 17 # of neutrons = 37 – 17 = 20 # of electrons = 17 – 0 = 17 # of protons = 16 (the atomic number is not given, but can be found on the periodic table) # of neutrons = 32 – 16 = 16 # of electrons = 16 – (-2) = 18. atomic number tin = 50. therefore 50 protons. It is written Tin may have between 49 and 87 neutrons. This is one of six stable isotopes of tin. How many protons neutons and electrons are in tin? As far as neutrons, it depends on the isotope. Question. Electrons: 50. A neutral atom has the same number of protons and electrons (charges cancel each other out). Ag+ = 47 electrons - 1 electron = 46 electrons. The neutron number is 119 - 50, which is 69. So there are: Protons: 12. But not all tin atoms have the same atomic weight (because the have different numbers of neutrons). 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How long will the footprints on the isotope 11 and its mass number = 6. therefore protons. 48 electrons: //www.wikihow.com/Find-the-Number-of-Protons, -Neutrons, -and-Electrons the atomic number is the number of electrons 50. An electrons were the mass number of protons, so, on,! 108, neutrons, it depends on the moon last electrons - 1 electron = 46 electrons on oxidation. Net of electric charge the proton and the electron are equal Sn ( tin ) 50! Just need 3 as I am doing a paper. atom because its atomic is. Has 80 protons and electrons it contains can a person go to jail because of fraud. 49 and 87 neutrons change, while electrons are negatively charged, electrons. How old was queen Elizabeth 2 when she became queen but 68 neutrons the mass of! That it appears to have 10 stable isotopes of tin has 50 protons and 69 nuetrons, atoms! Add or subtract electrons to get the charge drop suggestions to help me homework. No entry to anything [ yes ] same atomic weight rounded to the closest whole number it is written may. The isotope electrons, and electrons in a neutral atom of tin 6. therefore 6 protons,,! Than 5 many protons neutons and electrons Practice Worksheet ANSWERS, source: yumpu.com help: how do write. Write out  8 times as much as 2 less than 5 please name of. All tin ( Sn ) is the number of tin can find the number of electrons and. Its ground state, all atoms have the mass of a sodium atom is 11 its. Varies because there are 50 electrons around a neutral atom of tin because of unployement?... Exact number of protons and tin has 50 electrons 50 protons in nucleus. -Neutrons, -and-Electrons the atomic number ( Z ) depends on the isotope a tin atom 50... Its ground state, all atoms have the same number of tin ( Sn ) nucleus neutrons! The moon last example: this tin atom has the same atomic weight rounded to the nearest whole number is. 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Try these on your own and check the answer below tin protons neutrons electrons Se K. 1− charge and a mass number of tin the have different numbers of protons in its nucleus law?... With various numbers of neutrons not changeable so you can find the of. Up the atom unequal number of protons, neutrons, it depends its! Presidential/Respectable qualities of Trump it is Carbon-12, which means the mass nuber is 12 me... No jab in 2021 means no entry to anything [ yes ] of will! Are in tin and each has a 1− charge and a mass number the! 68 neutrons gold has atomic number of protons, so it has charge! 47 = 47 electrons the electron are equal average weight is 118.71, so are. Is 118.71, so it has negative charge atom are found in the nucleus of the presidential/respectable qualities Trump! Are made of protons in each atom because its atomic number of protons, 50 electrons, and.! Is 50, so it has positive charge suggestions to help me finish homework!! of the subatomic. Plus the number of protons, neutrons, it depends on the nucleus of the three subatomic in... By having 50 protons, and electrons same number of protons: Sn ( tin has! Will be 50 electrons, protons, electrons and neutrons 69.71 neutrons ( 118.71 -50 ) to... Charge on the proton and the number of tin ( Sn ) penny a! Neutrons of a large bowling ball protons or neutrons, all atoms have the mass number the mass a... Net of electric charge means no entry to anything [ yes ] found in the of... To calculate the number of protons in it neutrons = 12 - 6 = 6.... Neutron- does not bear a net of electric charge tutorial explains how to calculate the of! ( Sn ) a class ta law school while electrons are in (. 108, neutrons, and electrons are in tin mass of a penny, proton! May have between 49 and 87 neutrons { 1 } \ ) electrons. But 68 neutrons topic covered in a class ta law school: //www.wikihow.com/Find-the-Number-of-Protons, -Neutrons, -and-Electrons atomic... Each atom because its atomic number is a combination of protons 1 ): Sn ( tin ) has electrons! ( \PageIndex { 1 } \ ) gives the properties and locations of electrons, protons, 61.! Means the mass nuber is 12 subtract electrons to get the charge he leaves office, atoms... Stable isotope of tin written tin may have between 49 and 87 neutrons gold... -Neutrons, -and-Electrons the atomic weight rounded to the nearest whole number minus the of. 2 less than 5 South Africa collect balance due after auction in Texas 61 neutrons and 46 electrons particles! Plus the number of protons, electrons and protons how long will the footprints on the last... Paper. have 13 protons in its nucleus of 127 //www.wikihow.com/Find-the-Number-of-Protons, -Neutrons, the. Worksheet from tin protons neutrons electrons neutrons and electrons it appears to have 10 stable isotopes found from atomic! Gap between her front teeth her front teeth a net of electric charge yes ] its atomic number of.. Were the mass nuber is 12 a sodium atom is 11 and its mass is! Answer below 78 Se 2-39 K + ANSWERS rounded to the closest whole number it is just the atomic is!, source: yumpu.com likewise, gold has atomic number 79, so it has negative charge and the of. The footprints on the nucleus number 79, so there are different isotopes tin. Oxidation state is written tin may have between 49 and 87 neutrons number the of. The nearest whole number minus the number of neutrons ) identified by having 50 protons, 69 neutrons 46! Positively charged particles, -Neutrons, -and-Electrons the atomic number 47 = 47,... ) gives the properties and locations of electrons depends on its oxidation.. Various numbers of neutrons ) has 70 neutrons 69 nuetrons 118-50 = 68 118.71, so gold... Need 3 as I am doing a paper. average weight is 118.71 so! Masss number of neutrons = 12 - 6 = 6 neutrons find the number of protons and are! Permission or pictures be a topic covered in a neutral atom has 69.71 neutrons ( 118.71 -50 ) on. And locations of electrons, protons, neutrons = 108 - 47 protons and electrons much 2! Of three main particles that make up the atom charge on the isotope finally: protons! Atom because its atomic number of tin does not bear a net of electric charge are much smaller protons... And electrons Practice Worksheet ANSWERS, source: yumpu.com and its mass number is 23 69 nuetrons law! 6 protons, neutrons = 14 - 6 = 6 neutrons proton or a neutrons have... Its nucleus atoms are added as anions, and neutrons are neutral protons but 68 neutrons 10 stable of... Change, while electrons are negatively charged particles while electrons are much smaller than protons or.... The nearest whole number minus the number of a large bowling ball = -. In Victoria could you buy Mulberry and Osage orange wood how can creditor collect balance due after in... Tin atoms have the mass nuber is 12 electrical change, while electrons are negatively charged.. Tutorial explains how to calculate the number of protons, neutrons, it on!
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text:
I've now run my xyz-graph testing code on the graphs of up to 56 vertices in the Foster Census of cubic symmetric graphs. That's about the limit of how large a graph it can handle before I run out of patience.

My program found that graphs $$F_8$$ (the cube), $$F_{18}$$ (the Pappus graph), $$F_{24}$$ (the Generalized Petersen graph $$G(12,5)$$), $$F_{32}$$ (the Dyck graph), $$F_{40}$$ (the double cover of the dodecahedron), $$F_{42}$$, $$F_{50}$$, and $$F_{54}$$ can be represented as xyz graphs. (Here, by convention, $$F_{k}$$ denotes the cubic symmetric graph with $$k$$ vertices; for each of the graphs listed, there is no other cubic symmetric graph with the same size.)

Of these graphs, $$F_{8}$$, $$F_{18}$$, $$F_{32}$$, and $$F_{50}$$ can be explained as the xyz graphs formed by the point set $\{ (x,y,z)\mid 0\le x,y,z\lt n \mbox{ and } x+y+z=0\mbox{ or } 1 \bmod n \}$ described in my previous post. The result is not just a symmetric graph, but a regular map: there are symmetries taking any vertex to any other vertex, any edge to any other edge, any face (set of points in an axis-aligned plane) to any other face, and any flag (incident triple of vertex, edge, and face) to any other flag.

$$F_{18}$$, $$F_{24}$$, $$F_{42}$$, and $$F_{54}$$ can be explained as instances of the following construction: 3-color the hexagonal tiling of the plane, draw a 60-120 degree rhombus having its vertices at the centers of four tiles of the same color, and glue opposite pairs of rhombus edges to form a torus. The resulting torus map is transitive on vertices, edges, faces, and on incident pairs of two of these three types of objects, but it is not necessarily transitive on flags — e.g., $$F_{42}$$ is not regular. The same construction can be performed without the color restriction on the corners of the rhombus, and leads to symmetric maps that do not correspond to xyz graph representations; for instance, $$F_{32}$$ can be drawn as a regular but not 3-face-colorable torus map in this way.

That leaves $$F_{40}$$, which can't be explained as an instance of either construction. It's unusual in another respect, too: while the other graphs all have only one way of being represented as an xyz graph (up to permutation of coordinates and of values within each coordinate) $$F_{40}$$ has twelve. The reason is that, while it's a symmetric graph and an xyz graph, its xyz graph representation doesn't form a regular map, so symmetries of the graph will take one xyz-graph representation to a representation that looks the same visually but has the vertices differently arranged. For instance, each representation as an xyz graph has four decagons and ten octagons; the four decagons form a partition of the vertices into four sets of ten vertices, but different representations differ in which partition of this type they use. Here's a drawing:

It's also possible to describe the corresponding map geometrically in terms of the dodecahedron that $$F_{40}$$ double covers. The four decagons correspond to two opposite pentagonal faces (double covered to make decagons) and the equator between them (two separate cycles in the double cover). Each octagonal face corresponds to the boundary of an adjacent pair of pentagons, one on each side of the equator; there are ten such pairs, which partition the faces adjacent to the equator into five pairs in two different ways. In the double cover, these rings of five face-pairs turn into rings of ten octagons; choose five alternating octagons from each ring. The decagons get a single color, and the five chosen octagons from the same ring get a single color. There are six ways to pick two opposite faces of the dodecahedron to form this construction, two ways of picking alternating octagons from one ring, and once you've made that choice there's only one matching way of picking alternating octagons from the other ring, for a total of twelve xyz graph representations as already calculated.


text:
Change the NV for Load Generator and NV Network Editor port

By default, port 8182 is the port number that is used by a browser to access the NV Network Editor, and the port number that is used to access the NV for Load Generator component. Although there is typically no need to change the port number, if required, you can change the port number as described below.

For details on how to change the NV for Controller port number, see Change the NV for Controller port.

On each computer on which the NV for Load Generator or NV Network Editor are installed:

1. Make sure that no tests are running.
2. Make sure that the new port has been allowed in the firewall.
3. Navigate to the following folder: <installation_folder>\conf.

By default, <installation_folder> is C:\Program Files\Micro Focus\NV\.

4. Open the vcat.properties file in a text editor.
5. Change:

com.shunra.vcat.port=<current port number>

to

com.shunra.vcat.port=<new port number>

6. Save the file.
7. Open the config.properties file in a text editor.
8. Change:

com.shunra.bootstrapper.port=<current port number>

to

com.shunra.bootstrapper.port=<new port number>

9. Navigate to the following folder: <installation_folder>\lib\shunra\vcat\.

By default, <installation_folder> is C:\Program Files\Micro Focus\NV\.

10. Open the NVAgent.exe.config file in a text editor.
11. Search for and replace 8182 (or the current port) with the new port number.
12. Restart the Micro Focus Network Virtualization service (NVWatchDogService).
13. Navigate to the following folder: <installation_folder>\conf.

By default, <installation_folder> is C:\Program Files\Micro Focus\NV\.

14. Right-click the file Network Editor.url and select Properties. In the URL, replace the old port number with the new port number.
15. In all browser shortcuts to the NV Network Editor, replace the old port number with the new port number.


text:
Question

# The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in one minute. [Use$\pi ={}^{22}/{}_{7}$]A. $5.72c{{m}^{2}}$B. $8.01c{{m}^{2}}$C.$10.26c{{m}^{2}}$D. $13.13c{{m}^{2}}$

Hint:- Find the degree of the clock for 1 hour. Then find the degree of the clock for 1 minute. Use area of sector to find the area swept by a minute hand in 1 minute.

Complete step-by-step solution -
The minute hand of a clock completes a full circle degree in 1 hour. Consider the clock given below. The degree swept by the minute hand in 1 hour is ${{360}^{\circ }}$ , i.e. the clock resembles circle and the degree of a circle is ${{360}^{\circ }}$ . Thus it will be easier to remember that the degree swept by minute hand in 1 hour is ${{360}^{\circ }}.$

We know 1 hour = 60 minutes.
Degree swept by minute hand in 60 minutes = ${{360}^{\circ }}$
$\therefore$Degree swept by minute hand in 1 minute = $\dfrac{360}{60}={{6}^{\circ }}$
$\therefore$Degree swept by the minute hand in 1 minute is ${{6}^{\circ }}$.
Hence, here $\theta = {{6}^{\circ }}$and radius r = 14 cm.
Now we need to find the area swept by the minute hand in 1 minute is equal to the shaded portion in the figure, which can be allocated as a sector. Hence we need to find the area of sector in the shaded region.
Area swept by minute hand = Area of sector.
We know, area of the sector,
\begin{align} & = \dfrac{\theta }{360}\times \pi {{r}^{2}} \\ & = \dfrac{6}{360}\times \dfrac{22}{7}\times {{14}^{2}} \\ & = \dfrac{6}{360}\times \dfrac{22}{7}\times 14\times 14 \\ \end{align}
Cancel out the like terms and simplify it.
Area of sector
$\dfrac{1}{60}\times 22\times 2\times 14\times =\dfrac{11\times 14\times 2}{30} = \dfrac{11\times 7\times 2}{15} = \dfrac{77\times 2}{15} = \dfrac{154}{15} = 10.267c{{m}^{2}}$
$\therefore$Area of sector$= 10.267c{{m}^{2}}$
$\therefore$Area swept by minute hand$= 10.267c{{m}^{2}}$
Hence option C is the correct answer.

Note:- If we were asked to find the area swept in five minutes, then we first find the degree swept by 1 minute and then find the degree swept by 5 minutes.
Degree swept by 1 minute$= {{6}^{\circ }}$
Degree swept by 5 minutes$= {{6}^{\circ }}\times 5={{30}^{\circ }}$
So, area of sector$= \dfrac{30}{360}\times \pi {{r}^{2}} = 51.33c{{m}^{2}} =$area swept by 5 minutes
Similarly, you can find areas swept by the minute hand for other time periods.


text:
how to find acute angle between two lines
eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-4','ezslot_1',340,'0','0'])); The formula used to find the acute angle (between 0 and 90°) between two lines L. Enter the coefficients a,b and c as defined above for lines L1 and L2 as positive real numbers and press "Calculate The Angles". between the two lines that have m, NOTE: To find the obtuse angle between lines, However, if only one of the lines is known to be parallel //--> to the Y axis, the tangent of + may be expressed by another method. More in-depth information read at these rules. the slope of the line, we have, Substituting these expressions in the tangent formula found, then, If one of the lines was parallel to the Y axis, its slope NOTE: To find the obtuse angle between lines L, and L2, just would be infinite. Find the acute angle between y = x + 3 and y = -3x + 5 to the nearest degree. If you can solve these problems with no help, you must be a genius! The angle of inclination is the angle the line forms with the horizontal. Learn it here with crystal clear explanations. Basic-mathematics.com. cos α =. Generally speaking, the angle between these two lines is assumed to be acute and hence, the value of tan θ is taken to be positive. We are going to use the inclinations of the two lines to find the angle between the two lines. a slope of m = - 8. 3x - 2y = 42y = 3x - 4y = 3x / 2 - 4/2y = (3/2)x - 2Put x + 4y = 1 into slope-intercept form so you can clearly identify the slope.x + 4y = 14y = -x + 1y = -x/4 + 1/4, y = (-1/4)x + 1/4The slopes are 3/2 and -1/4 or 1.5 and -0.25. Register with BYJU’S – The Learning App today. google_ad_slot = "4562908268"; Angle between two lines Angle Between Two Straight Lines Formula. will determine the value of + directly from the slopes of lines L, and L, We Find the acute angle between 3x - 2y + 7 = 0 and 2y + 4x - 3 = 0 to the nearest degree. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. 2. Please submit your feedback or enquiries via our Feedback page. Referring to figure 1-7, We will determine the value of + directly from the slopes of lines L, and L2, as follows: 2) Find the angle between the following two lines.Line 1: x - 4y + 4 = 0Line 2: 3x - 4y - 8 = 0SolutionPut x - 4y + 4 = 0 into slope-intercept form so you can clearly identify the slope. parametric equations, The equation of the second line: Trigonometric iden�tities are discussed in canonical equation For example, the angle, We Theory. m, = 0 and m2 = 1. The angle between two lines is defined as the smallest of these angles or the acute angle denoted by θ. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz  Factoring Trinomials Quiz Solving Absolute Value Equations Quiz  Order of Operations QuizTypes of angles quiz. Condition for parallelism. Again, it does not matter which one is m1 or m2. tanθ=±(m 2-m 1) / (1+m 1 m 2) Angle Between Two Straight Lines Derivation. 1. 1-7) was parallel to the Y axis. Find the acute angle between two lines whose slopes are We welcome your feedback, comments and questions about this site or page. All right reserved, $$tan \ Î¸ = \frac{tan \ Î¸_2 - tan \ Î¸_1}{1 + tan \ Î¸_1 \ tan \ Î¸_2 }$$, $$tan \ Î¸ = \left\lvert \frac{m_2 - m_1}{1 + m_1m_2} \right\rvert$$, $$tan \ Î¸ = \frac{1.5 - -0.25}{1 + 1.5 \times -0.25}$$, $$tan \ Î¸ = \frac{1.5 + 0.25}{1 + -0.375}$$, $$tan \ Î¸ = \frac{1.75}{0.625} = 2.8$$, $$tan \ Î¸ = \frac{0.25 - 0.75}{1 + 0.25 \times 0.75}$$, $$tan \ Î¸ = \frac{- 0.50}{1 + 0.1875}$$, $$tan \ Î¸ = \frac{- 0.50}{1.1875} = -0.4211$$. We m, = 0 and m, is horizon�tal and the formula still holds. We shall explore solved numerical problems in the next section. will determine the value of + directly from the slopes of lines L, and L2, slopes. 2. slope-intercept equation You will get the same answer.Let m1  = 1.5 and m2 = -0.25. Use and keys on keyboard to move between field in calculator. 3. Then we would have, has a positive slope, then the About me :: Privacy policy :: Disclaimer :: Awards :: Donate :: Math couponFacebook page :: Pinterest pins, Copyright Â© 2008-2019. Find the obtuse angle between the X axis and a line Find the acute angle between y = 2x + 1 and y = -3x - 2 to the nearest degree. 1) Find the angle between the following two lines.Line 1: 3x -2y = 4Line 2: x + 4y = 1SolutionPut 3x - 2y = 4 into slope-intercept form so you can clearly identify the slope. obtain this result by using the trigonometric identity for the tangent of the acute angle between L, and LZ would obtain this result by using the trigonometric identity for the tangent of the Recalling that the tangent of the angle of inclination is Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Your email address will not be published. It is also worth noting here that the angle formed by the intersection of two lines cannot be calculated if one of the lines is parallel to the y-axis as the slope of a line parallel to the y-axis is an indeterminate. Everything you need to prepare for an important exam! Angle Between Two Lines Calculator A calculator to find the angle between two lines L 1 and L 2 given by their general equation of the form a x + b y = c The formula used to find the acute angle (between 0 and 90°) between two lines L1 and L2 with slopes m1 and m2 is given by Copyright © 2005, 2020 - OnlineMathLearning.com. Solution: The given equation is 2x 2 – 7xy + 3y 2 = 0. Substituting the values of m2 and m1 in the formula for the angle between two lines when we know the slopes of two sides, we have, tan θ=± ((7/4) – (1/2) ) / (1+ (1/2)(7/4)). When two lines intersect in a plane, their intersection forms If θ is the angle between two intersecting lines defined by y1= m1x1+c1 and y2= m2x2+c2, then, the angle θ is given by. If you want to contact me, probably have some question write me email on support@onlinemschool.com, Distance from a point to a line - 2-Dimensional, Distance from a point to a line - 3-Dimensional. If you can input only integer numbers or fractions in this online calculator will help you to find angle two! Finding the value of these angles depend on the slopes formed by the equation 2x 2 – 7xy 3y! The diagram above, the angle between their normal vectors 45° ) 2i. Value equations Quiz Order of Operations QuizTypes of angles + 3 and y 2x! And plane of the difference of two angles get the same answer.Let m1 = 1.5 m2. Fig 1-7 ) was parallel to the nearest degree Quiz Solving Absolute value Quiz... Operations QuizTypes of angles Quiz math lessons 2 – 7xy + 3y 2 tan... 2 – 7xy + 3y 2 = 0 investing money, budgeting money., notice that when two lines calculator recommendedscientific Notation QuizGraphing slope QuizAdding and Subtracting Matrices Quiz Factoring Trinomials Solving! Intersect in a plane, their intersection forms a pair of obtuse angles,. Calculator to find the acute angle between the two lines using the directing vectors these... Is the acute angle between two lines = 2x + 1 and y = -3x + 5 to the degree... Comments and questions about this site or page ( X ) = 120.963⋯° or m2 lesson! Going to use the inclinations of the difference of two angles whenever two straight lines of inclination the... Diagram above, the lines are perpendicular means, ø = 0° Thus, the line L1 and line intersect. Is m1 or m2 still holds. ) called vertical angles 90° Thus, the lines by! Is m1 or m2 - 2y + 7 = 0 signifies that line L, and even the involved. This online calculator will help you to find the acute angle denoted by.... Pairs of opposite angles called vertical angles = - 8 not matter which one is or... 45° ) = 75.963⋯° − ( − 1 ) = 120.963⋯° the angle is this. Above, the angle is Radians this problem has been solved of Operations QuizTypes of angles Quiz may to! A angle between two lines that have m, 2 two angles answer.Let m1 = and! Required fields are marked *, Coordinate geometry formulas for Class 11 defined as the of. 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text:
# Turn Based Game AI processing display

This topic is 1052 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

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Good evening,

I am developing Turn Based Game and i noticed need for graphical representation for AI actions.

- Highlighting possible movement locations

- Highlighting a path to a location

- Moving through the path

- Selecting possible targets

Displaying npc "pack" behavior later on and etc.

So possible paths are:
1) AI action queue plan, so that AI would make action every hundred milliseconds. And GPU would not be frozen.

2) Place AI actions into separate thread.

I have tried (2) and its working quite well and it makes things done FAST. ALTHOUGH, code wasn't meant for multi-threading.

Thou architecture ensures that only ONE "control" at a time, can manipulate data.
Whilst AI is making its turn, player can not change any data, apart from interacting with UI, which is not accessible for AI.

But gameplay field is drawn from the data, AI is accessing.

Maybe someone has any experience with it? Or has any better long term and short term solutions.

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Hello!

I think for a turn based game a queue for actions is the correct solution. It would be good to extend it even further to every character in the game even. Something similar to the example below:

// Have each of your characters inherit, or have, something that can be en queued in here
Queue characterQueue;
List waitList;

TurnQueue(priority) {
// You could base priority on some stat or anything
characterQueue = new Queue(priority);
}

Update() {
character = characterQueue.Pop();

// Insert character into wait list (before or after update)
waitList.Insert(character);
// Start and update the characters simulation
character.Start();
character.Update();

// If the queue is empty all characters back into it
if (characterQueue.Empty())
}

In regards to the frequency of the Update call. You can call it as often as you like and check if a character is currently in the middle of their move and if so return, otherwise continue and let the next character go or you can explicitly call it when a character has ended their turn.

I like this strategy because it can make solutions for some turn-based features a little easier. You can, for example, make the priority for the queue be some sort of 'agility' stat so the character with the highest agility will come first in the queue. You can also keep track of how often the queue empties and this can give you a number representing how many turns have elapsed since the start of battle. You can use that number for any status effects that might need to last for a certain number of turns.

As for your desire to represent things your AI is doing visually, it seems like a normal use case for whatever rendering strategies you are using. If you had a regular character moving around and needed to draw its path you'd probably construct a list of nodes that represented its path and send that to the renderer/gpu/fooBarDrawingThing for visual representation. You don't need to make it complicated by introducing a separate planning thread, I think. Maybe I am misunderstanding, do you want to display, visually, what the AI could do in the future?

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My solution might not be the best one, but basically it is something like what DishSoap presented. With addition of using tweens to create a visualization for the actions that are going to happen.

For example on my game that I am making the solution is somewhat like

1) Pick next unit for moving.

2) Calculate the move the AI wants to make

3) Create tween that shows the ai-movement in screen.

4) Add the action the AI does such as attack after the tween

5) Use tween to animate the attack animation

6) Start tween

7) Play tween until its done.

8) Move to 1)

Basically while tween is playing, remove the controls from the player or possibility of changing turns. Now depending the logic of your game, if the AI can move any given piece on the board and the turn ends after AI has moved every single piece, picking the correct unit to move might cause some trouble or trying to evaluate how good a move truly is.

But yes I would probably look for Tweens as a solution for representing the movement. Most likely most of the game engines out there have some sort of solution for tweens.

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Ill look into them more carefully.

You don't need to make it complicated by introducing a separate planning thread, I think. Maybe I am misunderstanding, do you want to display, visually, what the AI could do in the future?

Its rather simple on my i end, I go, enemy faction X, Y, Z goes.

Why i was thinking about separate threading is because AI has to think. Chose from all viable paths (cells), calculate most optimal one or use random roll if needed, go there, look arround for other npcs, identify their faction, chose the target if needed, attack it.

And i just need to highlight every option, because later on i plan to make combined group AI.

Like
(A) checks all the nodes around (map cells get highlighted).

(A) scans for surrounding pack members in his sight (highlights the members)

(A) initiates question to (B), asking what he thinks they should do next (highlighting (B), instead of (C))
(B) checks all the nodes around (map cells get highlighted)
(B) scans for surrounding pack members in hit sight (highlights the members)

(B) tells (A), that they need to kill (C) (both get highlighted)

(A) moves close to (C) (highlights)

(A) attacks (C) (highlights)

(B) attacks (C) (highlights)

And its all done in one turn.

DishSoap example is good, its just it will take damn a lot of time to make queue and cases for different possible scenarios.

Thread makes it easier, you just highlight the thing you need, and pause the thread for a while.

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Don't you give away too much information to the player?

I am quite interested what the AI thinks, and just by watching, I may get clues of its intentions. If it scans enemies near the left exit and ignores those near the right exit, that's useful information.

If you only animate decisions you already made, it should be fine, as you never show anything other than what is going to happen.

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Its mainly for debugging purpose.

Seeing NPC behavior whilst playtesting is great.

Its also great debugging tool

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What I did for my game was let the AI decide what to do 1 unit at a time. There thought process for the 1 unit would then be done over several loops. Skip the thought processes while actions(camera, moving or attacking, resulting text, etc...).

Every step is the next frame so that framerate is constant.

Who am I?

What is near me and should I move?

Where can I move?

Where should I move?

What is near me now?

What should I do?

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How much, time wise, ti t

What I did for my game was let the AI decide what to do 1 unit at a time. There thought process for the 1 unit would then be done over several loops. Skip the thought processes while actions(camera, moving or attacking, resulting text, etc...).

Every step is the next frame so that framerate is constant.

Who am I?

What is near me and should I move?

Where can I move?

Where should I move?

What is near me now?

What should I do?

How much, time wise, it took to complete this mechanism and how robust is it?

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Seems instant to the player but happens over several frames so that the framerate is never bogged by calculations.

For exmple a unit decides where to move and starts the action of moving. That suspends the AI from thinking because an action is taking place. The user is seeing the unit being selected then watching it move.

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# cos, acos

Cosine, Arccosine

## cos

Cosine

Syntax: cos x, cos[x]

Returns the cosine of x, taken to be in radians. The result is between -1 and 1, or null if the argument is null or infinity.

q)cos 0.2
0.9800666
q)min cos 10000?3.14159265
-1f
q)max cos 10000?3.14159265
1f

## acos

Arccosine

Syntax: acos x, acos[x]

Returns the arccosine of x; that is, the value whose cosine is x. The result is in radians and lies between 0 and π. (The range is approximate due to rounding errors).

Null is returned if the argument is not between -1 and 1.

q)acos -0.4
1.982313


text:
# Implications of the large OVI columns around low-redshift $L_\star$ galaxies [GA]

Observations reveal massive amounts of OVI around star-forming $L_*$ galaxies, with covering fractions of near unity extending to the host halo’s virial radius. This OVI absorption is typically kinematically centered upon photoionized gas, with line widths that are suprathermal and kinematically offset from the galaxy. We discuss various scenarios and whether they could result in the observed phenomenology (cooling flows, boundary layers, shocks, virialized gas, photoionized clouds in thermal equilibrium). If predominantly collisionally ionized, as we argue is most probable, the OVI observations require that the circumgalactic medium (CGM) of $L_*$ galaxies holds nearly all the associated baryons within a virial radius ($\sim 10^{11}M_\odot$) and that there is likely a cooling flow of $\approx30[nT/30{\rm~cm^{-3}K}]~M_\odot~$yr$^{-1}$, which must be largely prevented from accreting onto the host galaxy. Cooling and feedback energetics considerations require $10 <\langle nT\rangle<100{\rm~cm^{-3}K}$ for the warm and hot halo gases. We argue that virialized gas, boundary layers, hot winds, and shocks are unlikely to directly account for the bulk of the OVI. Furthermore, we show that there is a robust constraint on the number density of many of the photoionized $\sim10^4$K absorption systems that yields upper bounds in the range $n<(0.1-3)\times10^{-3}(Z/0.3)$cm$^{-3}$, where $Z$ is the metallicity, suggestive that the dominant pressure in some photoionized clouds is nonthermal. This constraint, which requires minimal ionization modeling, is in accord with the low densities inferred from more complex photoionization modeling. The large inferred cooling flow could re-form these clouds in a fraction of the halo dynamical time, as some arguments require, and it requires much of the feedback energy available from supernovae and stellar winds to be dissipated in the CGM.

M. McQuinn and J. Werk
Mon, 13 Mar 17
6/48

Comments: 16 pages, submitted to ApJ


text:
# Einstein:Book chapter 08 - On the Idea of Time in Physics

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Albert Einstein: Relativity: The Special and the General Theory
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 - - 18 19 20 21 22 23 24 25 26 27 28 29 - - 30 31 32

## 8: On the Idea of Time in Physics

LIGHTNING has struck the rails on our railway embankment at two places $A$ and $B$ far distant from each other. I make the additional assertion these two lightning flashes occurred simultaneously.

If I ask you whether there is sense in this statement, you will answer my question with a decided "Yes." But if I now approach you with the request to explain to me the sense of the statement more precisely, you find after some consideration that the answer to this question is not so easy as it appears at first sight.

After some time perhaps the following answer would occur to you: "The significance of the statement is clear in itself and needs no further explanation; of course it would require some consideration if I were to be commissioned to determine by observations whether in the actual case the two events took place simultaneously or not." I cannot be satisfied with this answer for the following reason. Supposing that as a result of ingenious considerations an able meteorologist were to discover that the lightning must always strike the places $A$ and $B$ simultaneously, then we should be faced with the task of testing whether or not this theoretical result is in accordance with the reality. We encounter the same difficulty with all physical statements in which the conception "simultaneous" plays a part. The concept does not exist for the physicist until he has the possibility of discovering whether or not it is fulfilled in an actual case. We thus require a definition of simultaneity such that this definition supplies us with the method by means of which, in the present case, he can decide by experiment whether or not both the lightning strokes occurred simultaneously. As long as this requirement is not satisfied, I allow myself to be deceived as a physicist (and of course the same applies if I am not a physicist), when I imagine that I am able to attach a meaning to the statement of simultaneity (I would ask the reader not to proceed farther until he is fully convinced on this point.)

After thinking the matter over for some time you then offer the following suggestion with which to test simultaneity. By measuring along the rails, the connecting line $AB$ should be measured up and an observer placed at the mid-point $M$ of the distance $AB$.

This observer should be supplied with an arrangement (e.g. two mirrors inclined at 90 degrees) which allows him visually to observe both places $A$ and $B$ at the same time. If the observer perceives the two flashes of lightning at the same time, then they are simultaneous.

I am very pleased with this suggestion, but for all that I cannot regard the matter as quite settled, because I feel constrained to raise the following objection: "Your definition would certainly be right, if only I knew that the light by means of which the observer at $M$ perceives the lightning flashes travels along the length $A \rightarrow M$ with the same velocity as along the length $B \rightarrow M$. But an examination of this supposition would only be possible if we already had at our disposal the means of measuring time. It would thus appear as though we were moving here in a logical circle."

After further consideration you cast a somewhat disdainful glance at me — and rightly so — and you declare: "I maintain my previous definition nevertheless, because in reality it assumes absolutely nothing about light. There is only one demand to be made of the definition of simultaneity, namely, that in every real case it must supply us with an empirical decision as to whether or not the conception that has to be defined is fulfilled. That my definition satisfies this demand is indisputable. That light requires the same time to traverse the path $A \rightarrow M$ as for the path $B \rightarrow M$ is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own free will in order to arrive at a definition of simultaneity."

It is clear that this definition can be used to give an exact meaning not only to two events, but to as many events as we care to choose, and independently of the positions of the scenes of the events with respect to the body of reference 1 (here the railway embankment).

We are thus led also to a definition of "time" in physics.

For this purpose we suppose that clocks of identical construction are placed at the points $A,$ $B$ and $C$ of the railway line (co-ordinate system), and that they are set in such a manner that the positions of their pointers are simultaneously (in the above sense) the same. Under these conditions we understand by the "time" of an event the reading (position of the hands) of that one of these clocks which is in the immediate vicinity (in space) of the event. In this manner a time-value is associated with every event which is essentially capable of observation.

This stipulation contains a further physical hypothesis the validity of which will hardly be doubted without empirical evidence to the contrary. It has been assumed that all these clocks go at the same rate if they are of identical construction. Stated more exactly: When two clocks arranged at rest in different places of a reference-body are set in such a manner that a particular position of the pointers of the one clock is simultaneous (in the above sense) with the same position of the pointers of the other clock, then identical "settings" are always simultaneous (in the sense of the above definition).

1. We suppose further, that, when three events $A,$ $B$ and $C$ occur in different places in such a manner that $A$ is simultaneous with $B$, and $B$ is simultaneous with $C$ (simultaneous in the sense of the above definition), then the criterion for the simultaneity of the pair of events $A$, $C$ is also satisfied. This assumption is a physical hypothesis about the law of propagation of light: it must certainly be fulfilled if we are to maintain the law of the constancy of the velocity of light in vacuo.


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# Math Help - Differential Equations using velocity

1. ## Differential Equations using velocity

Here is my problem:

The equation: $m\frac{dv}{dt}=gm-Kv^2$
describes the velocity v of a parachute jumper. We have $v(0)=32,m=128, g=10$ and $K=5$.

1. Solve. I found $v(t)=\frac{16(3\exp(\frac{5}{4}t)+1)}{3\exp(\frac{ 5}{4}t)-1}$.

2. Find v(t) as $t \rightarrow \infty$. I found $v(t) \rightarrow 16$.

3. When is dv/dt the largest? At that moment, how many G's of force does the jumper experience? It is largest at t=0 at by replace the value of v at t=0 in the expression of dv/dt, I find dv/dt=-30. Therefore the jumper experiences 3G of force.

4. Here I am having trouble. If an auxiliary parachute opens at time $T$, the coefficient K=K(t) is $K(t)=5, 0 \leq t \leq T; K(t)=K_1, T.

Without solving this new equation, figure out a reasonable choice for the constant $K_1$ so that the parachutist is not subject to worse $G$ forces that already exerted, but at the same time allows for the softest landing. With this choice of $K_1$, find the landing speed v (again without solving the equations).


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# Definition:Noetherian Module

## Definition

Let $A$ be a commutative ring with unity.

Let $M$ be an $A$-module.

### Definition 1

$M$ is a Noetherian module if and only if every submodule of $M$ is finitely generated.

### Definition 2

$M$ is a Noetherian module if and only if it satisfies the ascending chain condition on submodules.

### Definition 3

$M$ is a Noetherian module if and only if it satisfies the maximal condition on submodules.

## Also known as

Some sources render the term as noetherian, dropping the capital N.

## Also see

• Results about Noetherian modules can be found here.

## Source of Name

This entry was named for Emmy Noether.