The equation (X^2)/25+(Y^2)/25+(Z^2)/2=1 yields a 3D ellipsoid figure on WolframAlpha. However I am trying to find out how to enter an equation that will give me a form with the ellipsoid open at 2 ends. One way would be to restrict the domain of x such that x is between -3.5 and 3.5 . But I can't figure out the notation for that in the WolframAlpha text box. Any suggestions?
This is the code for Mathematica. It happens to work in Wolfram Alpha also. (It's not generally true that Mathematica code runs in WA.)
ContourPlot3D[x^2/25 + y^2/25 + z^2/2 == 1, {x, -3.5, 3.5}, {y, -5, 5}, {z, -5, 5}]
Related
I am trying to use demo.py in nkolot
/
GraphCMR | GitHub. I am interested in obtaining joints from the inferred SMPL image and visualize it similar to described in README of this project: gulvarol
/
smplpytorch | GitHub.
I also posted the issue here: https://github.com/nkolot/GraphCMR/issues/36.
What I tried that didn't work.
I changed https://github.com/nkolot/GraphCMR/blob/4e57dca4e9da305df99383ea6312e2b3de78c321/demo.py#L118 to
pred_vertices, pred_vertices_smpl, pred_camera, smpl_pose, smpl_shape = model(...) to get smpl_pose (of shape torch.Size([1, 24, 3, 3])). Then I just flattened it by doing smpl_pose.cpu().data.numpy()[:, :, :, -1].flatten('C').reshape(1, -1) and used the resulting (1, 72) pose params as input in pose_params variable of smplpytorch demo.
The resulting visualization doesn't look correct to me. Is this the right approach? Perhaps there is an easier way to do what I am doing.
How to get 3d joints from demo.py and visualize it | nkolot
/
GraphCMR
The problem is that
smpl_pose (of shape torch.Size([1, 24, 3, 3]))
is the SMPL pose parameters expressed as a rotation matrix.
You need to make a transformation from rotation matrix to axis-angle representation which is (72,1). You can use Rodrigues formula to do it, as claimed in the paper:
Get more information from the paper:
SMPL: A Skinned Multi-Person Linear Model
When using ol3-cesium and the map is in 3d mode, calling map.getView().getZoom() returns undefined. This might affect setZoom as well.
I understand we are in a 3d world, so there are no z-levels as in the tiled maps. On the other hand, Google Maps calculates a z-equivalent when coming back grom 3d to 2d.
How can I convert from height to a z-equivalent? Any formula, taking into account the latitude and altitude, to get the z equivalent?
There's no easy formula to get a 2D "Z" value from 3D, because the 3D camera can be tilted, can see different levels of tiles in the foreground vs the background, etc.
For individual tiles however, there are specific known "Level" values from the imagery quadtree. You can see these in Cesium Inspector by clicking the little + next to the word Terrain on the right side, and then put a checkmark on Show tile coordinates. The coordinates shown include L, X, and Y, where L is the tile's level (0 being the most zoomed-out, higher numbers more zoomed in), and X and Y are 2D positions within the imagery layer.
I posted an answer on GIS SE that shows how to reach in and grab these tiles, the same way Cesium Inspector does, along with a number of caveats involved. You could potentially look for the highest-level visible tile, and use that as your "Z" value.
I know this is not accurate, but sharing in case this is of use to anyone.
I have moved to several altitudes in Google Maps, switching between the 2D and 3D maps, writing down the z or altitude shown in the address bar:
z altitude (metres)
----- -----------------
3 10311040
4 5932713
5 2966357
6 1483178
7 741589
8.6 243624
11.35 36310
13.85 6410
15.26 2411
17.01 717
18.27 214
19.6 119
20.77 50
21 44
With the above correspondences, I have approximated the following function:
function altitudeToZoom(altitude) {
var A = 40487.57;
var B = 0.00007096758;
var C = 91610.74;
var D = -40467.74;
return D+(A-D)/(1+Math.pow(altitude/C, B));
}
Based on your formula, the reverse conversion should be:
altitude = C * Math.pow((A-D)/(zoomLevel-D) -1, 1/B);
I am currently trying to simulate ballistics on an object, that is otherwise not affected by physics. To be precise, I have a rocket-like projectile, that is following an parabolic arc from origin to target with a Lerp. To make it more realistic, I want it not to move at constant speed, but to slow down towards the climax and speed up on its way back down.
I have used the Mathf.Smoothstep function to do the exact opposite of what i need on other objects, i.e. easing in and out of the motion.
So my question is: How do I get an inverted Smoothstep?
I found out that what i would need is actually the inverted formula to smoothstep [ x * x*(3 - 2*x) ], but being not exactly a math genius, I have no idea how to do that. All I got from online calculators was some pretty massive new function, which I'm afraid would not be very efficient.
So maybe there is a function that comes close to an inverted smoothstep, but isn't as complex to compute.
Any help on this would be much appreciated
Thanks in advance,
Tux
Correct formula is available here:
https://www.shadertoy.com/view/MsSBRh
Solution by Inigo Quilez and TinyTexel
Flt SmoothCubeInv(Flt y)
{
if(y<=0)return 0;
if(y>=1)return 1;
return 0.5f-Sin(asinf(1-2*y)/3);
}
I had a similar problem. For me, mirroring the curve in y = x worked:
So an implementation example would be:
float Smooth(float x) {
return x + (x - (x * x * (3.0f - 2.0f * x)));
}
This function has no clamping, so that may have to be added if x can go outside the 0 to 1 interval.
Wolfram Alpha example
If you're moving transforms, it is often a good idea to user iTween or similar animation libraries instead of controlling animation yourself. They have a an easy API and you can set up easing mode too.
But if you need this as a math function, you can use something like this:
y = 0.5 + (x > 0.5 ? 1 : -1) * Mathf.Pow(Mathf.Abs(2x - 1),p)/2
Where p is the measure of steepness that you want. Here's how it looks:
You seem to want a regular parabola. See the graph of this function:
http://www.wolframalpha.com/input/?i=-%28x%2A2-1%29%5E2%2B1
Which is the graph that seems to do what you want: -(x*2-1)^2+1
It goes from y=0 to y=1 and then back again between x=0 and x=1, staying a bit at the top around x=0.5 . It's what you want, if I understood it correctly.
Other ways to write this function, according to wolfram alpha, would be -(4*(x-1)*x) and (4-4*x)*x
Hope it helps.
I have a csv file with the following format.
Label 1, 20
Label 2, 10
Label 3, 30
.
.
.
LabelN, 5
How do I plot the second column using the labels given in the csv file as labels on the x-axis?
(Something like this, where 1891-1900 is a label)
EDIT:
Found these questions which are quite similar to mine,
Plotting word frequency histogram using gnuplot
Gnuplot xticlabels with several lines
After trying the commands given in answer 1.
set xtics border in scale 1,0.5 nomirror rotate by -90 offset character 0, 0, 0
plot "data.txt" using 2:xticlabels(1) with histogram
I'm getting a not so clean histogram because the number of labels is quite large. I've tried the formatting given in answer 2. Can anyone suggest a way to get a cleaner histogram?
You have several options:
Plot only the important labels (extremes, mean etc. for example)
Skip every 5th label or so if labels form a series
Split your graph if you must plot every single label.
Seems like case 2) applies here, and thus skipping some of the labels before plotting will make the plot look better.
You can pre-process the file to skip every 5th label (say) using something like the following script:
line_number = 0
for line in open("d1.txt", "r"):
line_split = line.split(",")
if(line_number % 5 == 0):
print line,
else:
print ",",line_split[1],
line_number += 1
You can now plot with appropriate font size
set xtics border in scale 1,0.5 nomirror rotate by -90 offset character 0, 0, 0
set xtics font ",9"
plot "d2.txt" using 2:xticlabels(1) with histogram title "legend_here"
I'm on the FIRST robotics team at my high school, and we are working on developing a kiwi drive robot, where there are three omni wheels mounted in a equilateral triangle configuration, like this:
The problem is programming the robot to drive the motors such that that the robot moves in the direction of a given joystick input. For example, to move "up", motors 1 and 2 would be powered equally, while motor 3 would be off. The joystick position is given as a vector, and I was thinking that if the motors were expressed as vectors too, vector projection might be what I need. However, I'm not sure if this is right, and if it is, how I would apply it. I also have a feeling that there may be multiple solutions to one joystick position. Any help would be greatly appreciated.
I've built 9 robots during my time at school (1 FIRST, 8 RoboCup). We used the same omnidrive layout as you do. Beta's answer looks correct but add rotation to all wheels afterwards:
W1 = -1/2 X - sqrt(3)/2 Y + R
W2 = -1/2 X + sqrt(3)/2 Y + R
W3 = X + R
[This is Beta's formula with some added Rotation]
You need to think about the available ranges for your motors. I am guessing it can take a PWM signal of +/-255, so either the input or the output has to be adjusted somewhat. (It's not that hard...)
A good paper with details
To answer your specific questions: Vector projection is essentially what you are doing here. You apply it by having a matrix M, your input from the joystick I and your output to the motors O. Thus O = M * I;
M = [(-0.5 -sqrt(3)/2 +1)
(-0.5 +sqrt(3)/2 +1)
(1 0 +1)]
First let's define some terms. In keeping with the usual convention, the X axis will point to the right and the y axis will point up (so that the thrust of wheel 3 is along the X axis). We'll call the motion of the wheels W1, W2 and W3, each defined so that Wi > 0 means that the wheel rotates in the clockwise direction. In your example, if W1 < 0, W2 = W1 and W3 = 0, the robot will move in the +Y direction.
If all three wheels rotated at the same rate (W1 = W2 = W3) the robot would rotate in place. I'm guessing you don't want that, so the sum of the rotations must be zero: W1 + W2 + W3 = 0.
The motion of each wheel contributes to the motion of the robot; they add as vectors:
W1 = -1/2 X - sqrt(3)/2 Y
W2 = -1/2 X + sqrt(3)/2 Y
W3 = X
So if you know the desired X and Y from the joystick, you have W1, W2 and W3. As we've already seen, the difference between W1 and W2 is what drives Y motion. Their sum drives motion in X.
Though this system can be solved mathematically, in 2002, FIRST Team 857 chose to solve it mechanically. Our control system used three joysticks mounted with their X-axes forming an equilateral triangle, and handles replaced with ball-socket arms connected with a Y-shaped yoke. Map the X-axis of each stick directly to a motor speed, and the control system has been solved. As an advantage, this system is very intuitive for laypeople to run--push the yoke in the direction you want to go, rotate it to turn.
As you have recognized, the first part of this will be finding an appropriate equation to represent the resultant motion for any motor settings. Depending on the level of control and feedback you have on your motor speeds, I would suggest the process you go thorough should start with writing a vector equation: (define positive X as straight ahead)
-M1Cos(30)+M2Cos(30)=X (the negative is because 1 and 2 must be powered the same magnitude, but opposite polarities for forward motion)
M1Sin(30)+M2Sin(30)-M3 = Y (as anticlockwise motion on 1 and 2 will result in the robot moving left in the Y and anticlockwise motion on 3 will result in the robot moving to the right)
The other input that you need to add into this is the desired rotation of the robot, thankfully, M1+M2+M3 = W (Rotational velocity)
Your joystick input will give you X,Y and W, so you have 3 equations with 3 unknowns.
From here it is simultaneous equations, so you may end up with multiple solutions, but these can generally be restricted based on possible motor speeds and the like.
An example of this is the rec::robotino::com::OmniDrive Class - the source code for this method is available too...