I'm using ROS and a RPLidar A2 and apparently the sensor turns clockwise which means its reference frame (see image below) isn't the same as a normal cartesian frame (x axis on 0°, y axis on 90°). So i need to transform the frame via tf to make it easier to understand. Thus, i guess i just need to make a +90° rotation around axis z to match the normal cartesian frame.
How can i make this happen? Do i need to add the static TF transformation information to the launch file?
You are correct that a transform needs to be applied. This can usually be done one of two ways, depending on your overall application. The easiest, like you mentioned, would be to just add a static transform; it can also be done via launch file:
<node pkg="tf" type="static_transform_publisher" name="link1_broadcaster" args="0 0 0 0 0 0.7071 0.7071 parent_frame_id child_frame_id 100" />
Note that the parameters for this node are as follows: x y z qx qy qz qw parent_frame_id child_frame_id period_in_ms
This works quite well for quick and easy transforms, however, it doesn't scale when many sensors and frames get added in. If you expect more sensors to be added that all need their own transforms, you should instead use a URDF file. It will allow you to easily layout relational data about position/orientation of features on your robot.
Related
I try to understand how to input material properties using engineering constants for VTI material ( vertical transverse isotropic) with direction 3 being vertical direction. Does anyone has inp example showing how to input these material properties? For example, I don't know how to input nu12, n13? Thanks
enter image description here
As for transversely isotropic materials you have to provide 5 different parameters (I don't know what the vertical means, I never read this before).
For the convention on the poissons number please see Abaqus online documentation - conventions.
For your example. You should usually have one "master" direction (englisch is not my first language). And you have a plane perpendicular to this which has rotational symmetry. If your 3-direction is this (maybe the direction in which fibers are oriented) then your E11 and E22 values should be identical. Furthermore two of your possion numbers in your case 31 and 32 (prependicular/parallel sometimes called). Also the same directions of shear modules have to be identical.
The last shear modulus can be obtain via
G = E/2(1+nu) where E must be prependicular E modulus (2,1) and the poisson number must be the prependicular one (21)
Your constants: E33, E22=E11, nu31=nu32, nu12, G31=G32 , calculate G21 from above equation
For clarification
from H. Schürmann - Konstruieren mit Faser-Kunststoff-Verbunden (construction with fiber reinforced materials)
NOTE: In this picture the 1 direction is the parallel direction where fibers are oriented, adjust it accordingly or follow what i wrote above.
I'm looking to create a Q-Q plot within Rascal using the Vis library. I have been told there is no positional system. Is this true? If true, how would I go about plotting this or any scatterplot? Does anyone have an example of this in use?
That's an excellent question. Certainly Rascal's Vis library is "point free" in the sense that its layout mechanism has no absolute coordinate system. However, there are certain Figure kinds which have a relative coordinate system wrt their own "origin". When you combine several of those using horizontal, vertical or overlay boxes (and align them properly), you can create the effect of bar charts, scatterplots and whatever you desire.
In particular the overlay Figure composition is interesting: http://tutor.rascal-mpl.org/Rascal/Libraries/Vis/Figure/Figure.html#/Rascal/Libraries/Vis/Figure/Figures/overlay/overlay.html
Figure point(num x, num y){ return ellipse(shrink(0.05),fillColor("red"),align(x,y));}
coords = [<0.0,0.0>,<0.5,0.5>,<0.8,0.5>,<1.0,0.0>];
ovl = overlay([point(x,y) | <x,y> <- coords]);
render(ovl);
Produces this (both code and image taken from the documentation linked above):
Each point is an ellipse which is aligned at the (x, y) position relative to the origin of the enclosing overlay box.
The origin by default of this overlay seems to be the upper-left corner, when no other FProperty's are given to the overlay. It's possible other alignment options for the overlay Figure also change the position of its origin.
With the help of Jurgen Vinju I wrote this code, hope it helps someone: https://gist.github.com/rlmhermans/c9e82a6a623b65f0c6957ab3ff2742cf
I am trying to make a new tool for the tabletop simulator community based on my "pack up bag". The Packup Bag is a tool that remembers world position and rotation of objects you place inside it, so you can then place them back in the same positions and rotations they came from when "unpacking the bag".
I have been trying to modify this so it spits things out in a relative position and rotation to the bag, instead of using hardcoded world coordinates. The idea here is that players can sit at any location at the table, pick the faction bag they wish to play.. drop it on a known spot marked for them and press the place and it will populate contents of the bag relative to its location.
Now I have gotten some of this worked out... I am able to get the bag to place relative in some ways .. but I am finding it beyond my maths skills to work out the modifications of the transforms.
Basically I have this part working..
The mod understands relative position to the bag
The mod understands relative rotation to the bag
BUT.. the mod dose not understand relative position AND rotation at the same time.... I need someway to modify the position data relative to the rotational data... but can not work out how.
See this video....
https://screencast-o-matic.com/watch/cFiOeYFsyi
As you can see as I move the bag around the object is placed relative to it.... but if I rotate the bag, the object has the correct rotation but I need math to work out the correct position IF it is rotated. You can see it is just getting placed in the same position it was as if there was no rotation... as I haven't worked out how to code it to do this.
Now I have heard of something called "matrix math" but I couldn't understand it. I'm a self taught programmer of only a few months after I started modding TTS.
You can kinda understand what I mean I hope.. In the video example, when I rotate the bag, the object should be placed with the correct rotation but the world position needs to be changed.
See this Example to see relative rotation ....
https://screencast-o-matic.com/watch/cFiOeZFsyq
My code dose this by remembering the self.getPostion() of the bag and the obj.Position() of the object getting packed up.. it then dose a self - obj and stores that value for the X and Y position. It also remembers if it is negative or position and then when placing it uses the self.postion() and adds or subtracts the adjustment value. Same for rotation.
Still I do not know what ot go from here.. I have been kinda hurting my head on this and thought maybe some of you math guys might have a better idea on how to do this.
: TL;DR :
So I have
bag.getPosition() and obj.getRotation()
bag.getRotation(0 and obj.getRotation()
These return (x,y,z}
What math can I use to find the relative position and rotation of the objects to the bag so if I rotate the bag. The objects come out of it in a relative way...
Preferably in LUA.. thank you!
I'd hope you've found the answer by now, but for anyone else finding this page:
The problem is much simpler than what you're suggesting - it's basic right triangle trigonometry.
Refer to this diagram. You have a right triangle with points A, B, and C, where C is the right angle. (For brevity, I'll use abbreviations opp, adj, and hyp.) The bag is at point A, you want the object at point B. You have the angle and distance (angle A and the length of the hyp, respectively), but you need the x,y coordinates of point B relative to point A.
The x coord is the length of adj, and y coord is the length of opp. As shown, the formulas to calculate these are:
cos(angle A) = adj/hyp
sin(angle A) = opp/hyp
solving for the unknowns:
adj = hyp * cos(angle A)
opp = hyp * sin(angle A)
For your specific use, and taking into account the shift in coordinate system x,y,z => x,z,y:
obj_x_offset = distance * math.cos(bag.getRotation().y)
obj_z_offset = distance * math.sin(bag.getRotation().y)
obj_x_position = bag.getPosition().x + obj_x_offset
obj_z_position = bag.getPosition().z + obj_z_offset
Diagram source:
https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-modeling-with-right-triangles/a/right-triangle-trigonometry-review
I want to add force to the grenade according to the touch positions of the user.
--this is the code
physics.addBody(grenade1,"dynamic",{density=1,friction=.9,bounce=0})
grenade1:applyForce(event.x,event.y,grenade1.x,grenade1.y)
Here more the x and y positions are the lower the force is. But the force here is too high that the grenade is up in the sky.
You must calibrate the force applied. Remember that F=ma so if x=250 then F=250, if the mass of the display object (set when added body, based on material density * object area) is 1 then acceleration a = 250, which is very large. So try:
local coef = 0.001
grenade1:applyForce(coef*event.x, coef*event.y, grenade1.x, grenade1.y)
and see what you get. If too small, increase coef until the response is what you are looking for. You may find that linear (i.e., constant coef) doesn't give you the effect you want, for example coef=0.01 might be fine for small y but for large y you might find that coef=0.001 works better. In this case you would have to make coef a function of event.y, for example
local coef = event.y < 100 and 0.001 or 0.01
You could also increase the mass of the display object, instead of using coeff.
Recall also that top-level corner is 0,0: force in top level corner will be 0,0. So if you want force to increase as you go up on the screen, you need to use display.contentHeight - event.x.
How to make a 2d world with fixed size, which would repeat itself when reached any side of the map?
When you reach a side of a map you see the opposite side of the map which merged togeather with this one. The idea is that if you didn't have a minimap you would not even notice the transition of map repeating itself.
I have a few ideas how to make it:
1) Keeping total of 3x3 world like these all the time which are exactly the same and updated the same way, just the players exists in only one of them.
2) Another way would be to seperate the map into smaller peaces and add them to required place when asked.
Either way it can be complicated to complete it. I remember that more thatn 10 years ago i played some game like that with soldiers following each other in a repeating wold shooting other AI soldiers.
Mostly waned to hear your thoughts about the idea and how it could be achieved. I'm coding in XNA(C#).
Another alternative is to generate noise using libnoise libraries. The beauty of this is that you can generate noise over a theoretical infinite amount of space.
Take a look at the following:
http://libnoise.sourceforge.net/tutorials/tutorial3.html#tile
There is also an XNA port of the above at: http://bigblackblock.com/tools/libnoisexna
If you end up using the XNA port, you can do something like this:
Perlin perlin = new Perlin();
perlin.Frequency = 0.5f; //height
perlin.Lacunarity = 2f; //frequency increase between octaves
perlin.OctaveCount = 5; //Number of passes
perlin.Persistence = 0.45f; //
perlin.Quality = QualityMode.High;
perlin.Seed = 8;
//Create our 2d map
Noise2D _map = new Noise2D(CHUNKSIZE_WIDTH, CHUNKSIZE_HEIGHT, perlin);
//Get a section
_map.GeneratePlanar(left, right, top, down);
GeneratePlanar is the function to call to get the sections in each direction that will connect seamlessly with the rest of your world.
If the game is tile based I think what you should do is:
Keep only one array for the game area.
Determine the visible area using modulo arithmetics over the size of the game area mod w and h where these are the width and height of the table.
E.g. if the table is 80x100 (0,0) top left coordinates with a width of 80 and height of 100 and the rect of the viewport is at (70,90) with a width of 40 and height of 20 you index with [70-79][0-29] for the x coordinate and [90-99][0-9] for the y. This can be achieved by calculating the index with the following formula:
idx = (n+i)%80 (or%100) where n is the top coordinate(x or y) for the rect and i is in the range for the width/height of the viewport.
This assumes that one step of movement moves the camera with non fractional coordinates.
So this is your second alternative in a little bit more detailed way. If you only want to repeat the terrain, you should separate the contents of the tile. In this case the contents will most likely be generated on the fly since you don't store them.
Hope this helped.