## c ++ – Libnoise Bounding Box – from spherical to planar

I'm trying to convert a bounding box in a spherical map (equiangular image as a result of Libnoise / c ++) into a planar projection, but I can not find the right algorithm for it, and I'm not good at projections.

I'm using javascript and python, and I've tried to customize some of the libraries I found and convert the equiangular images into cubemaps. However, these work by processing each face separately, and I find it extremely complicated to customize such libraries to process only a bounding box of the equiangular result.

This is the library I use for reference.

And this is the diagram of what I'm trying to do with the resulting image.

I know that I will have some distortions and that the resulting image will be "narrower" as I get closer to the poles. Therefore, I plan to compute such offsets with respect to the bounding box to cover the desired area.

That way, I could edit the map with flat XY coordinates without spherical distortion.

## Equipment recommendation – Is there a digital camera that can be used to store bounding boxes for people?

My comparatively old Panasonic DMC-FZ200 (released in 2012) has this in the manual for the playback mode:

``````(Face Rec Edit)
You can delete or change the information relating to Face Recognition for the selected picture.
1 Select (Face Rec Edit) on the (Playback) menu. (P52)
2 Press 3/4 to select (REPLACE) or (DELETE) and then press (MENU/
SET).
3 Press 2/1 to select the picture and then press (MENU/SET).
4 Press 2/1 to select the person and then press
5 (When (REPLACE) is selected) Press 3/4/2/1 to
``````

(…)

``````Note
• When all Face Recognition information in a picture has been canceled, the picture will not be
categorized using Face Recognition in (Category Selection) in (Filtering Play).
``````

So it recognizes multiple faces (I think you can save about 10 faces that it should recognize by name) and prioritizes them when you're doing face-focussing. In addition, they are labeled during playback and images containing them can be selectively reproduced.

I did not use this feature, so I do not know how it's stored: I would suggest using some EXIF ​​tags rather than XMP (but this should be relatively easy to convert). Since this sounds like the kind of feature that deserves a check in camera reviews, and since it was already available in 2012, I should be surprised if it is not buried in more or less useful ways in the menus of most current deals.

## graphics3d – 3D plotting: Some edges of the bounding box are dashed

I'm trying to create a very simple 3D scatter plot

``````ListPointPlot3D[stratdatacleannew]
``````

With `stratdatacleannew` a list of 3-tuples. For clarity, however, I would like to have a bounding box with dashed lines instead of solid lines for edges that are not visible to the viewer. How can I do that?

## Find the IOU from Bird Eye View (BEV) and 3D Bounding Box

I have x, y, z, h, w, l, alpha, and rotaion_y of bounding_box_1 and bounding_box_2 of the polygon. To calculate the intersection of the union (iou) with pythons shapely Library?

## java – How do I get a bounding box from the accessibilty node information?

I'm trying to fetch screen boundaries of a particular node in my Accessibilty service with getBoundsInScreen (), but every time an empty response is returned, I just want to retrieve the boundaries of a node and draw into a rectangular bounding box at that position in my app
Here is my code:

``````private fun printAllViews(mNodeInfo: AccessibilityNodeInfo?) {
if (mNodeInfo == null) return
var log = ""
log +=  mNodeInfo.getBoundsInScreen(Rect(0,0,0,0))
Log.d("hmm", log)
if (mNodeInfo.childCount < 1) return
for (i in 0 until mNodeInfo.childCount) {
printAllViews(mNodeInfo.getChild(i))
}
}
``````

here, if I do it `nodeinfo.gettext()` It successfully gives me the text back, but not the limits. My question is, how can I get that?

## Algorithms – Maximum and minimum distance from the query point within the bounding box

I read an article about approaching sums with KD trees (similar to FMM).

As part of my efforts, I try to make this article, which is cited, understandable.

I'm having trouble understanding this part:

Calculation of the maximum variation of the weights over all points below the node
ND is easy. We know the place of xquery and we know the borders
Hyper rectangle of the current node. A simple algorithm calculation
O (number of tree dimensions) can calculate the shortest and largest
possible distances to any point in the node. From these two values
and the assumption that the weight function does not increase, we can
Calculate the minimum and maximum possible weights wmin and wmax of any
Data point below the node ND.

The only solution I can think of is to limit the deviation (wmax-wmin) from the top to the four corners of the bounding box, and then to calculate the distance from the query point to each of the four corners is actually O (i.e. ). However, this does not seem to be the intention of the author. Could someone point out to me what I miss here?

## 3d – including vs exclusive voxel bounding box

Suppose I have a few voxels at (1,1,1) and (2,2,2). I would like to know if it is better to consider that the smallest box containing both voxels is either (1,1,2,2). 1) to (2,2,2) or (1,1,1) to (3,3,3). That is, is it more correct if the box contains the bottom of the voxels or the voxels as cubes in the box?

## Legend – Adjusting the bounding box of PlotLegends in TimelinePlot

I want to align the elements of `PlotLegend` in one single horizontal row under the `TimelinePlot`because there is a lot of room for it (especially if I have the `size` be big). Instead, the internal algorithms pack the `PlotLegends` in three rows in this case.

How can I fix this?

``````TimelinePlot[
{

<|{Entity["Person", "LeonardoDaVinci::47w36"] ->
interval[{"1452", "1521"}]
} |>,

<|{Entity["Person", "CamilleCorot::vx57d"] ->
interval[{"1796", "1875"}]
} |>,

<|{
"Piet Mondrian" -> interval[{"1872", "1944"}]
} |>,

<|{"Thomas Gainsborough" -> interval[{"1727", "1788"}]
} |>,

<|{Entity["Person", "JanDavidszDeHeem::2gt75"] ->
interval[{"1606", "1684"}]
} |>,

<|{"Pablo Picasso" -> interval[{"1881", "1973"}]
} |>,

<|{"Hokusai" -> interval[{"1760", "1849"}]
} |>

}
,
PlotStyle -> {Red, Orange, Darker[Yellow], Green, Blue, Purple,
Black},
PlotLegends ->
placed[{Text[Style["Italian", 16, Italic, FontFamily -> "Times"]],
text[Style["French", 16, Italic, FontFamily -> "Times"]],
text[Style["American", 16, Italic, FontFamily -> "Times"]],
text[Style["British", 16, Italic, FontFamily -> "Times"]],
text[Style["Flemish", 16, Italic, FontFamily -> "Times"]],
text[Style["Spanish", 16, Italic, FontFamily -> "Times"]],
text[Style["Japanese", 16, Italic, FontFamily -> "Times"]]},
Below],
Aspect ratio -> 1/2,
PlotLayout -> "Packed",
Background -> LightGray,
ImageSize -> 600,
AxesOrigin -> center]
``````

## graphics3d – MMA does not display a bounding box for 3D plots

I am not sure if this is a problem with MMA or my system, but I have a strange problem with 3D plots.

MMA does not display the bounding box, it only displays the drawing. Do you know any of you the possible reason. I am using Ubuntu 18.04.2 LTS.
I've tried a reboot and even upgraded from MMA 11.2 to 11.3, but the anomaly persists.

``````Plot3D[Sin[x + y^2], {x, -3, 3}, {y, -2, 2}, Boxed -> True]
``````

## Trigonometry – Maximizes dimensions of the bounding box around the rotated rectangle

Given an orthogonal rectangle with sides `on` and `b`what twists $$theta$$ results in a bounding box with maximum height c.q. Width?
I hope that's right. With bounding boxes I mean the smallest orthogonal rectangle that contains each of the 4 corners of the inner rectangle after that inner rectangle was rotated. I am interested in the largest single pages, not in their section.

I think it is important to derive and maximize the width and height of the resulting rectangle as a function of the given variable.
Take an example:

The pages of the resulting bounding box are indicated by $$h = h_1 + h_2 = a sin theta + b cos theta$$,

I am not sure if there is a difference between the situations $$a or not, but I believe that the result is obtained in an analogous way.

How would I in this example with $$a maximize $$h$$?