I've heard that you can tilt a part by a precise amount using the .CFrame property. However, I'm unclear on how to use it. The following code does not work:
Workspace.Part.CFrame = CFrame.new(90,0,45)
It is not rotating the part by 90 degrees and 45 degrees. What am I doing wrong?
First, use the CFrame.fromEulerAnglesXYZ function to create a new CFrame pointing in the direction you wish. Then, use Vector3 math to move the CFrame into the desired position.
EG.
local cframe = CFrame.fromEulerAnglesXYZ(XRADIANS, YRADIANS, ZRADIANS)
cframe = (cframe - cframe.p) + Vector3.new(XPOS,YPOS,ZPOS)
The documentation states that a Coordinate Frame (CFrame) constructor that takes 3 parameters is defining a position offset. Therefore, your example code would move the part 90 along the x-axis and 45 along the z-axis. To perform a rotation as you attempted see the CFrame.fromEulerAnglesXYZ function.
The arguments taken specify position, not rotation
I had this trouble too when I was starting to CFrame. They are RADIANS, not DEGREES. I have written a quick CFraming guide on ROBLOX, here.
If you're struggling with radians, you should look at the ROBLOX wiki page on radians to gain a basic understanding: wiki.roblox.com/index.php/Radians
Thanks!
-pighead10
-- Rotates the part by 90 degrees. If you want to change the axis your
-- rotating it on Use a different placement such as CFrame.Angles(math.rad(90),0,0)
Workspace.Part.CFrame = Workspace.Part.CFrame * CFrame.Angles(0, math.rad(90), 0)
Related
I have a question that I've been wanting to know the answer to for a while. How to get the position and direction the player character is looking? I want to know this because I need to make a system that involves this.
So in ROBLOX, all BaseParts have a property named CFrame which represents the Position and Orientation of that BasePart.
Now if you wanna find the where the character is looking, we could check the direction the character's Head is facing by utilizing its CFrame. (since we can't get the CFrame of a model). To do this, we can reference the character Head and then get the LookVector property of its CFrame. And voila you got the direction where the character is facing. But there is a slight issue, you see-LookVector is not a positional vector but rather a directional vector, as such it will have a Magnitude (length) of 1. So if we want to find the position the character is looking at, we need to multiply this LookVector by a number which will denote the number of studs in the direction of the character we want to look and then add it with the position of the Head.
So based on the above, you can do this:
local Head: BasePart = LocalPlayer.Character.Head.CFrame
local Direction: Vector3 = Head.LookVector
local Distance: number = 3 -- Look 3 studs in the direction of the `Head`
local Target: Vector3 = Head.Position + (Direction * Distance)
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 have a problem to use this bulge arc (dxf parser) function in C++ getArcDataFromBulge().
https://github.com/Embroidermodder/Embroidermodder/blob/master/libembroidery/geom-arc.c
I have my drawArc() function which need 'start angle' and 'sweep angle' parameters from this getArcDataFromBulge() function.
My drawArc() function use OpenGL 2D coordinate system with right side zero angle position and when I get values from getArcDataFromBulge() and recalculate it (0+-, 180+-, 360+-) I have something like unexpected opposite angles as results. It looks like clockwise-counterclockwise problem, but I'm think is not, I'm not sure. Do you have some idea what is going on?
For example:
tempBulge.bulge := 0.70;
arcMidAngle := RadToDeg( atan2(tempBulge.arcMidY - tempBulge.arcCenterY,
tempBulge.arcCenterX - tempBulge.arcMidX) );
After calculaton: arcMidAngle = 179.999
When I add and subtract from this point half of arc chord angle, I get start and end angles of my arc: 90°, 270° but it's not the same arc when I open dxf with some CAD software, it is opposite than origin drawing.
If you have an arc from 0° to 90°, it could be a 1/4 circle or a 3/4 circle.
You need to parse the $ANGDIR and $ANGBASE variables from the HEADER section which tells you in which direction angles are defined ($ANGDIR) and where the 0° angle starts ($ANGBASE) within that specific DXF file:
Variable Group code Description
$ANGBASE 50 Angle 0 direction
$ANGDIR 70 1 = Clockwise angles, 0 = Counterclockwise
For DXF, if $ANGBASE = 0, then 0° is on the right of the center, alike Windows.
Furthermore, in DXF, the positive Y-axis is upwards, in contrast to many Windows API's where the positive Y-axis is downwards.
I have a physics body, and I want it to move forward in the direction that it is facing. I'm only thirteen which I hope explains why I'm so bad at trigonometry. Can anyone tell me how to do this in Corona?
I'm gonna assume you want to push your object with a force. Either way we'll need to get an x and y component of the direction your body is facing. Here's how to get the x and y from the rotation angle:
-- body is your physics body
local angle = math.rad(body.rotation) -- we need angle in radians
local xComp = math.cos(angle) -- the x component
local yComp = -math.sin(angle) -- the y component is negative because
-- "up" the screen is negative
(note: if this doesn't give the facing direction, you may need to add 90, 180, or 270 degrees to your angle, for example: math.rad(body.rotation+90) )
The above code will give you the x and y components of the unit vector in the direction of the rotation. You'll probably also need some multiplier to get the magnitude of force you want.
local forceMag = 0.5 -- change this value to apply more or less force
-- now apply the force
body:applyLinearImpulse(forceMag*xComp, forceMag*yComp, body.x, body.y)
Here's where I got the math: http://www.mathopenref.com/trigprobslantangle.html. Using a unit vector simplifies the math because the hypotenuse is always 1
How about making your own character moving towards an angle before using the confusing physics?
angle = math.rad(Insert the angle you want here)
character.x = character.x - math.sin(angle)
character.y = character.y + math.cos(angle)
Er. You don't need Trigonometry just to move the object.
Add
object:translate(distanceToMoveInXAxis,distanceToMoveInYAxis)
Or if you want to perform a transition,
transition.to(object,{x=object.x + distanceToMoveInXAxis,y=object.y + distanceToMoveInYAxis})
It is stated, that to rotate a line by a certain angle, you multiply its end point coordinates by the matrix ({Cos(a), Sin(a)} {-Sin(a) Cos(a)}), where a is rotation angle. The resulting two numbers in matrix will be x and y coordinates of rotated line's end point. Rotation goes around line's start point.
Simplifying it, new coordinates will be {x*Cos(a) - y*Sin(a)} for x and {x*Sin(a) + y*Cos(a)} for y.
Task is to rotate a triangle, using this method. But the following code that uses this method, is giving out some crap instead of rotated image (twisted form of original triangle, rotated by "random" angle):
x0:=200;
y0:=200;
bx:=StrToInt(Edit1.Text);
by:=StrToInt(Edit2.Text);
cx:=StrToInt(Edit4.Text);
cy:=StrToInt(Edit5.Text);
a:=StrToInt(Edit3.Text);
//Original triangle
Form1.Canvas.Pen.Color:=clBlue;
Form1.Canvas.MoveTo(x0,y0);
Form1.Canvas.LineTo(bx,by);
Form1.Canvas.LineTo(cx,cy);
Form1.Canvas.LineTo(x0,y0);
//New triangle
Form1.Canvas.Pen.Color:=clGreen;
Form1.Canvas.MoveTo(x0,y0);
b1x:=Round(bx*cos(a*pi/180)-by*sin(a*pi/180));
b1y:=Round(bx*sin(a*pi/180)+by*cos(a*pi/180));
c1x:=Round(cx*cos(a*pi/180)-cy*sin(a*pi/180));
c1y:=Round(cx*sin(a*pi/180)+cy*cos(a*pi/180));
Form1.Canvas.LineTo(b1x,b1y);
Form1.Canvas.MoveTo(x0,y0);
Form1.Canvas.LineTo(c1x,c1y);
Form1.Canvas.LineTo(b1x,b1y);
end;
Well, I'm out of ideas. What am I doing wrong?
Thanks for your time.
The formula you are using rotates a point around (0, 0). To achieve the required result change your calculation to:
b1x:=x0 + Round((bx-x0)*cos(a*pi/180)-(by-y0)*sin(a*pi/180));
b1y:=y0 + Round((bx-x0)*sin(a*pi/180)+(by-y0)*cos(a*pi/180));
c1x:=x0 + Round((cx-x0)*cos(a*pi/180)-(cy-y0)*sin(a*pi/180));
c1y:=y0 + Round((cx-x0)*sin(a*pi/180)+(cy-y0)*cos(a*pi/180));
You appear to be rotating each individual line round its initial start point coordinates. So line 1 will get rotated about its start point (x0,y0); then line 2 will get rotated about bx,by; then line 3 will get rotated round cx. This will result in a twisted triangle. Instead you will need to rotate all three lines round the start point of the first line.