## optics – Why is the image plane at Z = f in pinhole camera models?

Don’t confuse formulas meant to be used with refractive optics (i.e., lenses) with projection mapping functions (such as the pinhole projection model).

The thin lens formula only applies for refractive lenses, such as glass elements that bend light. The thin lens formula is really just an idealization of a refractive element described by the lensmaker’s equation with negligible thickness of the lens (hence the name, thin lens). Concepts such as depth of field are defined and derived from applications of the thin lens formula.

The pinhole model is a projection model. That is, it describes the mapping from the field of view to the image plane. The pinhole model is a 1:1 mapping — every ray entering the pinhole leaves the pinhole at the same angle, for the entire field of view. Many refractive lenses (i.e., subject more-or-less to the thin lens model) have a pinhole projection mapping function. But not all. Wide angle lenses, and especially fisheye lenses, do not follow the pinhole projection formula.

This is easy to understand in the degenerate case: how can a circular fisheye lens with a 180° angle of view in all directions project onto the camera’s image plane, unless there is some sort of angular distortion such that the further away from the optical axis the subject is, the more the image rays are bent to project within a confined cone? That’s impossible to do with a pinhole projection model. But it’s not difficult with a series of concave lenses in front of the lens to bend the incoming light into the “funnel” of the lens’s collection area and project it onto the camera’s image plane.

It appears your first image came from a slide deck PDF (or one of it many copies online) for a senior-level undergraduate class in computer science. Unfortunately, the slide deck could have used one more very simple image to demonstrate the pinhole projection model:

Pinhole camera model, from Wikipedia Commons. Public domain.

Here is easy to see the relation between the real-word subject (tree), and its image formation inside the pinhole camera. The depth of the pinhole camera is the focal length, ƒ. The two red rays, the bounding rays of the subject tree, enter the pinhole, and leave (continue towards the image plane) at the same angle. Thus, simple similar-triangle geometry describes the pinhole projection formula.

## dnd 5e – Is it possible to Gate from the ethereal plane back to the material plane?

The basics of this seem fine; Etherealness takes you from the Material to the Ethereal, Gate lets you travel to a specific point on a Plane that isn’t your current one. It seems the Ethereal Plane and the Material Plane aren’t the same, even though they overlap.

The only thing that might be a problem, but it’s up to the DM, is that Etherealness has this clause:

When the spell ends, you immediately return to the plane you originated from in the spot you currently occupy. If you occupy the same spot as a solid object or creature when this happens, you are immediately shunted to the nearest unoccupied space that you can occupy and take force damage equal to twice the number of feet you are moved.

I don’t think this spell was written to deal with the consequences of using it and then using something else to plane-shift further. I would rule that if you’re not on the Ethereal Plane anymore, nothing happens, but check with your DM.

Alternatively, you could use Plane Shift. It’s also on the Cleric list, also takes you to a different Plane to Gate from and has no weird automatic return clauses that might cause issues.

## dnd 5e – Is it possible to Gate from the ethereal plane back to the material plane in 5e?

dnd 5e – Is it possible to Gate from the ethereal plane back to the material plane in 5e? – Role-playing Games Stack Exchange

## 3D Plane in a discrete domain.

I have a 3D plane defined by 3 floating points. I determine the plane equation parameters (ax+by+cz+d=0). I want to test all integer points (x,y,z) of the domain if them are in the plane.
I have to verify the equation a
x+by+cz+d=0. But because (x,y,z) are discrete the equation could not be fulfill by any point. How can I find “delta” in order that ‘|ax+by+c*z+d| < delta’ in order to find a plane with only one layer of points.

Thanks,

Luís Gonçalves

## air travel – I would like to book single round trip ticket from USA to China, but would like to have my friend in the same plane in my return trip back to USA

Me: USA to China (round trip)
my partner China to USA only.

But I would like to have my partner in same flight and next seat.

or should I book like, single one way ticket from USA to China and 2 one way ticket from China to USA for both of us.

which is less expensive? any thought or suggestions?

Thanks

## Can I use a personal satellite phone/Internet connection as a passenger in a plane?

I am reading some contradictory statements regarding the use of a personal satellite phone/Internet connection in a plane. E.g.:

https://www.bluecosmo.com/satellite-phones

Hardened structures, buildings, mountains, and heavy tree cover can all negatively affect your signal. Pilots often leave a satellite phone or an Iridium GO! on the dashboard of their plane’s cockpit to receive a signal.

You can absolutely use a satellite phone indoors, whether in a building, car, boat, airplane, or any other enclosed space as long as you install an unobstructed antenna on the outside of the vehicle or structure. Outfitter Satellite carries kits for in-building, in-vehicle, in-aircraft, and marine satellite phones. The idea of always needing to step outside or lean dangerously out a car window to use your satellite phone is incorrect.

Can I use a personal satellite phone/Internet connection as a passenger in a plane? By "can", I mean both legally and technically (= will it have a usable connection, assuming that I cannot plant an antenna on the outside of the plane prior to the flight).

## calculus – Linear Differential Equation on the Complex plane solution has zero of order n

If $$f(z)$$ is a solution of $$y^{(n)}(z) = a_{n-1}(z) y^{(n-1)}(z)+cdots+a_{0}(z) y(z)$$ having a zero of order $$n$$ then $$f equiv 0, ; ; forall z in R in mathbb{C}$$ where $$a_k forall k in {0,1,2,..n-1}$$ are continous functions in the region $$R$$ which is open and connected in the complex plane.

I need help understanding the proof of this Lemma as stated by this : Paper. After trying out with $$n = 1$$ I do not understand why this lemma would imply that the trivial solution only exists of a zero of infinite order.

Counter Example:
$$n = 1$$ Then Let us saying we have the following equation when $$n = 1$$ then we have

$$y^1 = a_{0} y, ; a_{0} = g(z)$$

Now this is easily solvable and the general solution is given by
$$y = Ce^{int a_{0} dz}, forall C in R$$

Now note that the exponential function has no zeros $$iff$$ no poles hence there does not
$$exists z_0,; text{such that } e^{z_0} = 0$$.
Therefore
$$not exists z_0 text{ s.t } e^{int a_{0} dz} = 0$$

But the key lies in the fact that if we let $$a_0 = 1/z$$ then we get $$y = z + C$$ as a solution and clearly at $$z = -C$$ we have $$f(-C) = 0$$ but $$f'(-C) = 1$$ which implies that the solution $$f(z)$$ has order $$1$$, but this is clearly not the trivial solution and hence why I cannot understand why this lemma works. I was thinking maybe it’s because $$1/z$$ is not defined at 0 and hence not continuous in the region R ?

Regarding the general proof : From the hypothesis of our lemma and after substituting $$f(x)$$(which is a solution) in our original equation ie
$$y^{(n)}(z) = a_{n-1}(z) y^{(n-1)}(z)+cdots+a_{0}(z) y(z)$$ it is clear that $$f$$ has a zero of order atleast $$n+1$$. $$f(z_0) = 0, ; z_0 in mathbb{R}$$.
Now we prove this by contradiction. Suppose that $$f$$ is not identically $$0$$ ie
$$f notequiv 0$$ then $$exists p geq 1$$ such that $$f$$ has a zero of order $$n+p$$ at $$z_0$$.

Why is this true? Why does would a solution have a zero of order at least $$n + 1$$ when it already has a zero of order $$n$$ doesn’t that contradict the definition of a zero of order $$n$$? Then how do we go about assuming it has order $$n+p$$? If anyone would be so kind as to explain the general proof and how this lemma is proved I would be grateful as I have tried a lot but still cannot understand why this works.

Thank you in Advance

## plotting – Drawing a straight line in complex plane

I have these point

`p = {1.4048301787418027` + 0.13589831076513692` I, 1.2937864859657142` + 0.09826442126033227` I, 1.2439169436341522` + 0.08152085090585755` I, 1.2112251127756706` + 0.07057066521001881` I, 1.1867797359572914` + 0.06239213624408477` I, 1.1671819237029237` + 0.05584007186977612` I, 1.1507678762649645` + 0.050355104251339586` I, 1.1365972662436867` + 0.04562153217223474` I, 1.124085953294084` + 0.041443428931470704` I, 1.1128457075115394` + 0.037690672840169796` I, 1.1026049923874517` + 0.034272306091684974` I, 1.0931660902037226` + 0.031122129476503592` I, 1.0843802257181003` + 0.028190350543236032` I, 1.076132312976041` + 0.025438465710119672` I, 1.0683311706306955` + 0.02283597890972666` I, 1.060903003028386` + 0.020358216403672635` I, 1.053786913489549` + 0.017984823404026413` I, 1.0469317264511402` + 0.015698699656082248` I, 1.0402936772643052` + 0.013485225924340693` I, 1.0338346910665444` + 0.011331687936209292` I, 1.027521069271944` + 0.009226836941693186` I, 1.021322461994918` + 0.007160546106256114` I, 1.0152110424227831` + 0.0051235346034142525` I, 1.0091608233679767` + 0.003107139398530981` I, 1.0031470719276647` + 0.0011031199836812119` I, 0.9971457882879402` - 0.0008965152800982101` I, 0.9911332209734355` - 0.0028996704469104913` I, 0.9850853942726508` - 0.004914318965930066` I, 0.9789776246938359` - 0.006948664028432134` I, 0.9727840022953297` - 0.009011310260944524` I, 0.9664768094358419` - 0.011111455803159562` I, 0.9600258433964858` - 0.013259115782829277` I, 0.9533975994565845` - 0.015465391402985825` I, 0.9465542556662075` - 0.017742803834833825` I, 0.9394523769004093` - 0.02010571977953537` I, 0.9320412189796319` - 0.022570907472936778` I, 0.9242604553685143` - 0.025158280726805794` I, 0.9160370544286163` - 0.027891919050845516` I, 0.9072808772098193` - 0.030801502627903685` I, 0.8978782922954255` - 0.03392438838789599` I, 0.8876826109570178` - 0.037308710478383814` I, 0.8764992119353089` - 0.041018184124120435` I, 0.864061351522674` - 0.045139880920104784` I, 0.8499886226970275` - 0.04979749980583017` I, 0.8337105619927074` - 0.05517557183546196` I, 0.814313066911355` - 0.06156755652317412` I, 0.7901896657264539` - 0.0694831242611486` I, 0.7580949362489926` - 0.07993065003390927` I, 0.7096897586662804` - 0.09538398209765551` I, 0.6071597345880032` - 0.1250060626682752` I}`

Now I did

``````sol = FindFit(p, a x + b, {a, b}, x);
eqnR(x_) := a x + b /. sol;
eqnR(x)
``````

Which gives me the output

`(1.24601 + 0.0818429 I) - (0.00962482 + 0.00319108 I) x`

I want to plot the list plot by doing

`ListPlot(Transpose({Re(p),Im(p)})) `

and also plot the straight line from the find-fit given by `eqnR(x)` on the same plot.

But I am not sure how to plot a straight line with complex parameters.
Any suggestions?

This is the list plot. What I would like to have is a plot of straight-line fitting all these points.

## air travel – How can I approximate ahead of time the wifi speed and robustness of some wifi on the plane?

While other focus on what may be wrong with the connection, I would take the opposite approach. Assume the best case scenario (given the parameters of your trip) and then try to figure out what that scenario looks like. Needless to say that you can follow these steps when you are still considering different airlines and routes.

I’d propose a simple checklist that you will go through for your situation. You may view this analysis as a risk assessment which commonly involves risk and uncertainty. In many cases, you’re not going to have a certain answer, but you will be able to get an indication based on information on the internet.

### 1. Find reviews of the wifi covering your route and airline combo

As other have said, you’re going to have different experiences with different airlines and their connection may vary with different routes. Just search for reviews that contain the airline name and their (non-hub) destination (that often captures the route).

This step is meant to get an overall indication of the wifi on flights similar to the one you will be taking. If the connections are bad, there will probably be reviews covering that.

### 2. Find out about obstacles that will surely interfere with wifi

If your route faces specific obstacles (the Atlantic Ocean, some large countries that may restrict on-board wifi) you use those as a search term as well. For example, the wifi on my Qatar Airways flight to Bangkok cut out when flying over India. That’s not a coincidence, that’s something you can take as a fact and you can plan around that if you research ahead of time.

### 3. (more advanced) Research the airline fleet and cross reference with route

A plane that’s not retrofitted with wifi capabilities isn’t going to give you internet access. Since you know what airline you will be flying with, it makes sense to research how many of their planes actually have wifi capabilities. Do all their planes have wifi access or do only specific models have that technology? Depending on the answer you may have to research which planes are actually used on your route. Some of the flight tracking websites keep records of historical flights which may or may not be available for free.

### 4. Be prepared for disappointment

This is basically covered by the other answers. Despite your own best research, there may be technical difficulties that prevent you from using the wifi. The previous steps are meant to set your expectations given the flight that you will be taking. For a significant part, those other steps will provide good indicators.

In some cases, the other answers will be correct. In some places you can be sure that the wifi will be slow and spotty. But there is no universal law of crappy wifi, I’ve had some reliable connections flight after flight on specific routes and that’s not a coincidence. It simply depends on the quality of the service and how seriously the airline takes their wifi product.

## ag.algebraic geometry – Some possible implication of existence of a \$g^r_d\$ on a smooth plane curve

Let $$X$$ be a smooth plane curve of degree $$d$$ and genus $$g$$ (over complex numbers). For example we can take for the time-being $$d=6$$ and $$g=10$$. Let’s also assume that there exists a divisor on $$X$$ having degree $$d^{‘}$$ and exactly $$r+1$$ sections. Then my question is the following:

Question : With these informations at hand, is it legitimate to expect that $$X$$ admits a morphism $$pi_i: Xto C_i$$, where $$C_i$$ is a curve of genus $$h_i$$( which depends on $$g, r, d, d^{‘}$$) for $$i=1,2$$ and the map $$(pi_1,pi_2): X to C_1 times C_2$$ is birational onto its image?

Is there a concrete counterexample? or do we need to impose some more condition to relate these two phenomenon?

Any insight from anyone is appreciated.