## dnd 5th – Can a spectator include himself in his antimagic cone?

The spectator emits an antimagic cone:

The spectator's central eye creates an area of
antimagic, as in the antimagic field spell, in a 150-foot cone. At
start of each of its turns, the spectator decides in which direction the cone
faces and if the cone is active. The area works against
spectator's eye rays.

According to the basic rule, Ch. 10, cone:

A cone extends in a direction you choose from its point of
origin. The width of a cone at a given point along its length is equal to
the distance from this point to the point of origin. The surface of a cone of
effect specifies its maximum length.

The origin point of a cone is not included in the cone's area of ​​effect,
unless you decide otherwise.

Can a spectator choose to include himself in his antimagic cone? Would this actually make him immune to harmful spells like blindness / deafness?

## Find if the convex cone of a vector list is zero

Given a list of vectors, I want to know if there is a vector such that its dot product with those of the list is all (semi) positive, or at least above some small negative value .

## shaders – Calculate the exponent in a spotlight equation from the angle of a cone

As I mentioned in the comment, raising a number to an exponent will not make the result null unless it was already zero at the start. (At least not in real numbers of infinite precision. We could try to exploit the floating point precision limits here, but I think it is both more complicated and less efficient than the alternatives we have)

Thus, even if the increase in the exponent makes the shiny part of the cone narrower and narrower, there are still non-zero values ​​well in the dark part. All the way 90 ° from the direction of the light, actually (again, reducing the digital accuracy at the moment).

Since you are rendering in a high dynamic range, it is difficult to choose a cutoff value "dark enough" so that it has no effect on the rendering, even if your exposure is extremely high. . We could become super conservative, but then you treat your cone as wider than necessary most of the time, which reduces the savings you get from slaughter. Or we could try to shift our cutoff based on our exposure settings, but again – too complex, and we can do better!

Since you say you like your distance attenuation function as it is, let's keep that unchanged. We will simply replace this line:

lightAmnt *= pow(max(dot(-surfaceToLightVector, lightDir), 0.0f), exponentConstant);


(Note that I changed the name of your variable – the tutorial calls it "light in pixels" but it really points from the surface pixel to the light! So let's name it in a way who is do not back 😉)

First of all, we will note that the inner term is only the cosine of our angle to the axis of light:

float cosine = dot(-surfaceToLightVector, lightDir);


In our desired cone corner, which reaches a minimum value that we can calculate on our CPU, and pass as a uniform:

float minCosine = Math.Cos(spotlightAngleLimitRadians);


So now we can remap the range of (minCosine, 1) in the range (0, 1):

float angleFalloff = max((cosine - minCosine), 0.0f) / (1.0f - minCosine);


(If you want, you can pre-calculate the inverse of the denominator and pass it as another uniform so that this division becomes a cheaper pixel multiplication).

You are now guaranteed that the brightness of the projector will reach zero at this limit angle. But you can still use a power function to shape the attenuation curve a bit as you like – here the exponent is arbitrary, we don't count on it to reach a specific angle width, we are just listening to watch this point.

lightAmnt *= pow(angleFalloff, exponentConstant);


## ui design – What would be the best way to create a cone representing the field of vision of an entity in Godot Engine?

I would like to create a game in which PC and NPC would have a field of view, shown as a cone at the front of the entity (see the image for an example).

As this would be the first game I develop using Godot Engine, I wonder what would be the best way to implement this feature. Also, if possible, in later versions of the game, I would like to be able to make the scenery stop the field of vision, like what is done in Monaco: what's yours is mine.

Do you have a recommendation?

## Geometry – I need help to build a "3D" paper cone with precise angles.

Thank you for your contribution to Mathematics Stack Exchange!

• Please make sure to respond to the question. Provide details and share your research!

But to avoid

• Make statements based on the opinion; save them with references or personal experience.

Use MathJax to format equations. MathJax reference.

## convex polytopes – Determining whether a subspace intersects an open polyhedral cone

### I'm trying to answer the following question:

Let $$V = text {span} (v_1, …, v_k) subseteq mathbb {R} ^ n$$, or $$v_1, …, v_k$$ can be assumed orthonormal. Let $$C = {x in mathbb {R} ^ n: x ^ Tu_i <0, i = 1, …, N }$$ to be an open polyhedral cone, where $$u_1, …, u_N in mathbb {R} ^ n$$ are given. Suppose we have $$v_1, …, v_k$$ and $$u_1, …, u_N$$ in hand, how do we decide if $$V cap C = garment$$ or not?

### Here is what I tried:

My first approach is to use the separation theorem to try to turn the problem into a standard linear feasibility problem:
$$V cap C = garment$$ if and only if $$exists y in mathbb {R} ^ n$$ such as
$$y ^ T (r_1v_1 + cdots + r_kv_k) geq 0, forall (r_1, …, r_k) in mathbb {R} ^ k text {and}$$

$$y ^ Tx <0, forall x: x ^ Tu_i <0, i = 1, …, N.$$

Written in a more compact way, $$V cap C = garment$$ if and only if $$exists y in mathbb {R} ^ n$$ such as
$$(i) y ^ TA_V z geq 0, forall z = (z_1, …, z_k) ^ T in mathbb {R} ^ k text {, and}$$

$$(ii) y ^ Tx <0, forall x in mathbb {R} ^ n: Ux <0, i = 1, …, N$$
or $$A_V$$ is the $$n times k$$ matrix with $$v_1, …, v_k$$ like its columns and $$U$$ is the $$N times n$$ matrix with $$v_1, …, u_N$$ like his rows. (i) can be "sufficiently" treated by considering only $$z = e_1, …, e_k, e_1 + … + e_k$$ because they extend positively $$mathbb {R} ^ k$$. About (ii), to replace the infinitely many $$x$$ by finite number $$x$$, I would need the extreme rays of $$C$$, that I do not know how to find. But even if I find these extreme rays, the problem will only become something "similar" to a linear feasibility problem "with some $$<$$ "conditions," which I still do not know how to handle completely.

My second approach is simply to consider $$V ^ perp$$, the orthogonal complement of $$V$$. We can use the Gram-Schmidt process to generate an orthonormal basis for $$V ^ perp$$say $$w_1, …, w_ {n-k}$$. Let $$W$$ Be the $$(n-k) times n$$ matrix with $$w_1, …, w_ {n-k}$$ like his rows and $$U$$ Be the $$N times n$$ matrix with $$u_1, …, u_N$$ like his rows. then $$V cap C = garment$$ if and only if the system
$$Wx = 0$$
$$Ux <0$$
$$x in mathbb {R} ^ n$$
There is no solution. Again, I do not know how to deal with "$$<$$" in $$Ux < 0$$.

### Sorry for my long paragraph describing my attempts. Here are my questions:

(1) For a polyhedral cone described by $$C = {x in mathbb {R} ^ n: x ^ Tu_i <0, i = 1, …, N }$$ with $$u_i$$ known, $$i = 1, …, N$$Are there simple algorithms or analytic formulas for finding its extreme rays?
(2) In the constraints of a "linear program", there are constraints with "$$<$$" instead of "$$leq$$", how do I treat them?

Any idea, reference book, or papers are welcome and appreciated. Thank you.

## Cone – Determine s as function s (V)

Determine the length of the height of the tilt $$s$$ according to the volume $$V$$, or $$s = d$$.

$$s (V) = …$$

## Agalgesic geometry – normal cone fibers

For any incorporation of smooth varieties $$X subset Y$$, it is well known that the normal cone $$NC _ {X / Y}$$ is isomorphic to the normal packet of $$X$$ in $$Y$$, whose fiber on each point $$x in X$$ is just the affine space $$T_xY / T_xX$$ (quotient of tangent spaces).

Now let $$X$$ to be an arbitrary subscheme of a smooth variety $$Y$$. Suppose further that on any point $$x in X$$ normal cone fiber $$NC _ {X / Y}$$ more than $$x$$ is isomorphic to the affine space $$T_xY / T_xX$$.

Question: Is it true that X has to be smooth? If not, can we say something about $$X$$?

For example, if, in addition to the above assumptions, both $$X$$ and $$NC _ {X / Y}$$ are irreducible varieties, then $$dim (NC _ {X / Y}) _ x geq dim NC _ {X / Y} -dim X = dim Y – dim X geq dim T_xY -dim T_xX$$
so we must have $$dim T_xX = dim X$$ for everyone $$x$$, Therefore $$X$$ must be smooth.

## dnd 5th – Cone Of Cold freeze the creature or freeze it in a block of ice?

The quotation you quoted indicates that "the creature … becomes a frozen statue", which I read to mean that the creature itself is frozen, not that they are locked up in the ice cream.

Compare this to the Eldritch Warlock Invocation Warlock "Tomb of Levistus" (XGtE, 57), which reads as follows:

… you can bury yourself in the ice …

That implies that cone of cold could have used similar language, but did not do it, so you have to take it literally, which means that they are transformed into an ice statue.

## 5th dnd – Can a spectator face his antimagic cone behind himself?

The viewer creates the anti-magic field with his central eye, so if you consider the location of the central eye "at the front" and on the other side "at the back", then no, he can only project his anti-magic field towards the front. side.

But for all intents and purposes, a spectator does not really have a front or back, because it is not limited to being able to see in one direction. You and I have (presumably) two eyes, placed at the front of the head, which creates a vulnerable back, from which our design comes from a "before" and a "front" "behind".

A spectator, on the other hand, has many eyes that can all move independently of each other. It has a 360 ° vision and is not limited to the front or the back, it can see you and disintegrate you even if you are below, above or behind it.

In fact, this must be the way Beinters works, because its own anti-magic cone will also cancel its own eyerays, so that a Beholder can not even keep his ray targets in front of him.

There is a side where food enters and many sides where deadly rays come out.