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August 27[edit]

This Last Mars rover[edit]

It’s just me, or someone else thinks the Curiosity rover’s cameras are a bit low in resolution terms??
I mean, you can sum all the megapixels of all the onboard cameras and don’t even match the Mpix count of a Nikon d800 and this is just a consumer level camera
All the specs in this page are really “true”? and in this case… Why!? Iskánder Vigoa Pérez 13:45, 27 August 2012 (UTC) — Preceding unsigned comment added by Iskander HFC (talk • contribs)

One reason may be that the rover's instruments were selected and tested back in 2004: Curiosity_rover#Timeline. - Lindert (talk) 13:56, 27 August 2012 (UTC)[reply]

See: Mars rover camera project manager explains 2MP camera choice.--Aspro (talk) 14:00, 27 August 2012 (UTC)[reply]

(ec)Curiosity was designed in 2004 Mars Science Laboratory#History and build by 2008. Aerospace equipment, and particularly equipment that once deployed cannot be replaced, is very conservatively designed. The low-weight, low-power, vibration-tolerant, radiation-resistant, high-thermal-range requirements mean they always lag consumer technology by a long way, in terms of raw numbers. Check Comparison of embedded computer systems on board the Mars rovers and see how much better your phone is than what they run. -- Finlay McWalterჷTalk 14:02, 27 August 2012 (UTC)[reply]

There's also apparently a bandwidth issue - often we only get to see thumbnails until the next day, and some of the images from the first panorama were backlogged for quite a while before finally being transmitted. I guess they have lousy cell phone coverage on Mars. :) Wnt (talk) 14:08, 27 August 2012 (UTC)[reply]

Bandwidth is only a problem if you want the full resolution all the time which you never dare to ask.--Stone (talk) 14:10, 27 August 2012 (UTC)[reply]

The cameras are better than those in the MER rovers. The problem is and always will be that you need flight qualified parts to be allowed to get into a NASA mission. Qualification takes years and adds 3 to 5 zeros to the price and if NASA has to do it on their own it will be a million or several. To get a microchip for space is hard work. A FPGA cost you 50000 $ and than you have a chip capable to do not more than a newest calculator of your kids. Military quality is easy to get, but space qualified makes your life complicated. NASA would simply never consider to fly parts which might harm the success of the mission, even if the cameras are not the most important science tool they are the most important PR tool. --Stone (talk) 14:10, 27 August 2012 (UTC)[reply]

This article [1] discusses the issue. Acroterion (talk) 14:39, 27 August 2012 (UTC)[reply]

The CCD KAI-2020CM for MER and the Mitel (DALSA) doi:10.1029/2003JE002070 for MSL are not 2012 style, but still they were the best choice.--Stone (talk) 14:57, 27 August 2012 (UTC)[reply]

Another thing to consider is that consumer cameras are lower-resolution than they appear. A consumer camera uses a Bayer filter pattern for color: this gives you about N/2 megapixels of brightness resolution, and only about N/4 megapixels of color resolution, eg. a 16MP camera only has 8 megapixels of brightness resolution, and 4 megapixels of color resolution. The Mars rover cameras use broad-spectrum sensors and a collection of single-color filters, so a 2 megapixel sensor gives 2 megapixels of color resolution. --Carnildo (talk) 00:02, 1 September 2012 (UTC)[reply]

Stopping neutrinos[edit]

Since neutrinos have mass, does that mean they could, in theory, be slowed or even stopped relative to some stationery observer? Likewise, our article on photons mentions that they could as well have a very small amount of mass. If that turned out to be the case, would that mean that even light could be stopped? Goodbye Galaxy (talk) 15:26, 27 August 2012 (UTC)[reply]

Light is slowed, by literally every material it passes through. See Refractive index. Individual photons always travel at the speed of light, but this is not the same thing as the speed of the light (as a bulk phenomenon) itself. It sounds like you understand this, but I just wanted to clarify incase you, or others reading, don't. The speed at which light (as a phenomenon) travels depends greatly on exactly what is meant by the words "speed", "light" and "travels". --Jayron32 15:32, 27 August 2012 (UTC)[reply]

Yes, slow neutrinos are a possibility, but you can't really keep them in one place because of the uncertainty principle and the lack of any way to confine them (since they interact so weakly with everything else).

The standard model predicts that the photon is exactly massless, but experimentally you can never prove the mass is exactly zero; you can only establish better and better upper bounds. -- BenRG (talk) 15:56, 27 August 2012 (UTC)[reply]

And what if we presuppose that the photon has a slight mass? Goodbye Galaxy (talk) 16:55, 27 August 2012 (UTC)[reply]

In that case, you could create a Bose-Einstein condenstate of consisting of photons. The macroscopic description would be given in terms of a "classical" Proca action. Count Iblis (talk) 17:06, 27 August 2012 (UTC)[reply]

Would that be any different from an ordinary standing electromagnetic wave? Naively looking at the B-E transition temperature, it goes to infinity as the mass goes to zero, implying that this is the normal state for photons anyway. -- BenRG (talk) 17:53, 27 August 2012 (UTC)[reply]

Like the neutrino it would be hard to keep it in one place. You could confine it in a mirrored box, but you can confine a massless photon in a mirrored box too. Having a very tiny mass is not much different from having no mass at all. -- BenRG (talk) 17:53, 27 August 2012 (UTC)[reply]

Does the quantum foam (or Higgs field) slow observed photon velocities in vacuum to slightly less than the true speed of light? Hcobb (talk) 00:24, 28 August 2012 (UTC)[reply]

No, not in the standard model and not as far as experiments can tell. The vacuum is Lorentz invariant and the only Lorentz invariant thing it could do to the photon is give it a Lorentz invariant mass, so your question is equivalent to asking whether the photon has mass. In the standard model the photon is massless by construction: not all of the electroweak symmetry is broken, and the particle associated with the unbroken part is called the photon. I don't know what the story is in beyond-standard-model theories. -- BenRG (talk) 03:09, 28 August 2012 (UTC)[reply]

Finding Electric Field Magnitude and Direction at a Point Charge[edit]

Given three point charges positioned in a plane at the vertices of an equilateral triangle, is it possible to find the electric field magnitude and direction at the position of one of the point charges as a function of the charges of the three points and the length of the sides of the triangle? For example, given point charges with specified electric charges at points ABC located at the vertices of equilateral triangle ABC with side length l, could I determine the electric field magnitude and direction at point A?

—Dromioofephesus (talk) 17:02, 27 August 2012 (UTC)[reply]

No. The electric field direction is discontinuous at A, and the magnitude is singular at A. Red Act (talk) 20:01, 27 August 2012 (UTC)[reply]

Classical electromagnetism does not really admit of point-charges, as these would have infinite energy and therefore infinite mass (see classical electron radius). At any of the vertices, you can however calculate the contribution to the E-field from the point-charges at the other two vertices - and this immediately gives you the force on each of the three point-charges (under the assumption that a point-charge does not exert a force on itself). --catslash (talk) 00:16, 28 August 2012 (UTC)[reply]

How come my Prickly Pears won't bloom?[edit]

A few years ago I transplanted some Opuntia humifusa (Eastern Prickly Pear) from one property to another. They bloomed yearly at the old location. The six or so lobes I moved to the new location are doing spectacularly, and there are about 30 lobes now. They are in a well drained partially sandy area with strong direct sunlight all day until June, when they would normally bloom, and about 6 hours of direct sunlight once the oaks leaf out and partially shade them in the late afternoon. (They usually bloom around June 1, so I don't think the leaf cover is at all significant.) I have only had one single flower in all that time, and it was while the plant was still in the pot I used to move it in, before it was put back in the ground. The same with one I keep indoors. I planted it as a cutting and it bloomed once in the pot, and never since. Any suggestions as to why this may be and what to do about it? Google searches have been of no use at all. Thanks. μηδείς (talk) 18:12, 27 August 2012 (UTC)[reply]

FWIW (and I have flowered quite a few Opuntiae) I think the most likely problem is the roots are unconstrained and the soil is too rich, a bit like with (completely unrelated) figs. They tended to flower for me when they had no option to grow instead. I suggest stress them a bit; limit the roots and ensure the soil is a bit barren. Or even just cover the soil so they get less water. --BozMo talk 18:21, 27 August 2012 (UTC)[reply]

It did occur to me that they might need stress. The interesting thing is that the potted one indoors is very constrained and gets water only when it starts wilting, but still hasn't bloomed in three years. The ones outdoors may be a bit hard to stress, since they are going to receive a lot of water unless there is an east coast drought. Are you suggesting they will perhaps bloom once they get crowded? μηδείς (talk) 18:40, 27 August 2012 (UTC)[reply]

I would also suggest that they are more likely to bloom when crowded. Think of it in terms of investment in sexual reproduction vs. clonal growth. If it is easy to spread vegetatively, why invest precious resources in low-odds sexual reproduction? You could install a bed-guard around the perimeter like those used in landscaping, or even just increase competition by planting a different plant all around the perimeter. Lastly, the easiest thing might be to sever the below-ground connections, so that all the individuals in the middle are competing against their neighbors (if they are connected, competition between ramets is much much less). SemanticMantis (talk) 19:17, 27 August 2012 (UTC)[reply]

Chilling requirement? Count Iblis (talk) 19:20, 27 August 2012 (UTC)[reply]

One of my undergrad majors was in Biology with a focus on plant ecology, so I grok the above comments. I have also found a suggestion that 0/10/10 fertilizer helps with flowering, which is the only solution that seems doable without digging up the bed. The chilling requirement might be relevant for the indoor plant. It doesn't freeze, but it is in an unheated and unlit (except for sunlight) room that I use to force my Poinsettias to bloom. There would be no trouble putting the potted plant outside. Unfortunately we can't expect results till next June, long after the Mayan apocalypse. μηδείς (talk) 19:31, 27 August 2012 (UTC)[reply]

PS, this is obviously a clone of a plant that did bloom well before, and which was not really crowded, even by itself. The prior soil was a bit more sandy, and it was never watered. (These plants are wattered occasionally and indirectly, but not a lot.) The old soil probably had a much higher salt content (it was occasionally flooded by the ocean during storms). Perhaps covering the plants in a few inches of beach sand will help? μηδείς (talk) 19:37, 27 August 2012 (UTC)[reply]

Maybe beach sand will help. Another thing to keep in mind is the age of the genet. Some plants, especially succulents, will only bloom once they've reached some ratio of above ground/below ground mass, or even a critical threshold of total biomass. So, assuming you didn't also start the last colony from a few cuttings, I suspect doing nothing will generate plenty of blooms within a few years. SemanticMantis (talk) 14:40, 28 August 2012 (UTC)[reply]

Light source measurement in CRI/Ra[edit]

What does the "Ra" part of CRI/Ra stand for? Electron9 (talk) 22:29, 27 August 2012 (UTC)[reply]

The R seems to stand for "rendering", but the "a" I'm unsure of. According to [2] "General Color Rendering Index R_a" contrasts with "Special Color Rendering Indices R_i" (apparently the "i" is for index). Perhaps "a" was just chosen as the first variable in the alphabet (were there R_b and R_c values initially, too ?). StuRat (talk) 23:00, 27 August 2012 (UTC)[reply]

Ra = the general color rendering index. It's an average value, hence the "a." Zoonoses (talk) 03:45, 29 August 2012 (UTC)[reply]