Rotation

[Nablo]

What is it you are trying to achieve?

I am trying to rotate a box 90 degrees in turing with its axis of rotation in the center of the box. I am also trying to rotate a few lines continuously (sort of like a car wheel spinning) with its axis of rotation being the center of the oval on which the lines are situated on.


What is the problem you are having?

I do not know how I am to apply sin, cosin, tan, or even if I should use them (I know grade 10 math) to achieve this 3D rotation on the shapes.


Describe what you have tried to solve this problem

I've tried looking through the Turing Index, several submission on compsci as well as any tutorials that were avaliable on compsci but I unfortunately have yet to see the solution to my specific problem.


Post any relevant code (You may choose to attach the file instead of posting the code if it is too long)

I only have a picture of the objects which I am trying to rotate.


EDIT

I figured out how to rotate the rectangular shape 90 degrees. Although, I am still wondering how to do rotate the spikey shape 1 degree at a time, I'm thinking for loop + trig but I don't have enough knowledge to go from there...

Please specify what version of Turing you are using
Turing V. 4.1

[Insectoid]

Look up Pic.Rotate(). It's exactly what you need.

[Nablo]

I already read up the pic rotate module. The thing is that I am trying to rotate a turing generated shape rather than a picture.

[Insectoid]

Oh, then you need to use math! You can't really rotate a rectangle, or any pre-defined shape. If you want to rotate it, you need to draw all the individual lines for each item and draw.fill() the insides. To rotate, you apply rotation math to each vertex in your shape with a function (you'll have to write it yourself). Google probably has information about this, but I warn you it might be a bit advanced.

[Nablo]

What is rotation math, and as you can see, my pictures might be too complex for them to be constructed by lines.

[Dreadnought]

I see your "spikey things" are kinda round. Depending on what you're actually trying to do with them, you might be able to make this work with pic.rotate after all and save yourself a hassle.

Check this out!

Turing:

View.Set ("offscreenonly") var pic1_Array : array 1 .. 36 of int var pic2_Array : array 1 .. 36 of int % Make picture 1 Draw.FillMapleLeaf (110, 110, 190, 190, 12) pic1_Array (36) := Pic.New (100, 100, 200, 200) cls % Make picture 2 Draw.FillStar (110, 110, 190, 190, 12) pic2_Array (36) := Pic.New (100, 100, 200, 200) % Note that when I make a picture, I make it bigger than necessary %      to make sure the edges don't get cut off by pic.rotate % Generate the rotations ofthe picture (here I make a picture for every 10 degrees for i : 1 .. 35     pic1_Array (i) := Pic.Rotate (pic1_Array (36), 10 * i, 50, 50)     pic2_Array (i) := Pic.Rotate (pic2_Array (36), 360 - (10 * i), 50, 50) end for % Now the fun part! for j : 1 .. 15     % This is a coloured background, to show that you can still draw these pictures over other graphics     if j ~= 12 then % my pictures are red (12) so I skip this colour         Draw.FillBox (0, 0, maxx, maxy, j)         View.Update         for i : 1 .. 36             Draw.FillBox (0, 0, maxx, maxy, j)             % Draw the pictures using picMerge (you could use picCopy if your background was plain white)             Pic.Draw (pic1_Array (i), 100, 100, picMerge)             Pic.Draw (pic2_Array (i), 300, 100, picMerge)             View.UpdateArea (100, 100, 200, 200)             View.UpdateArea (300, 100, 400, 200)             Time.DelaySinceLast (40)         end for     end if end for

It could work for a rectangle too, but you might need to make big picture for it to not cut off edges (you can figure out how big using the Pythagorean theorem).

[RandomLetters]

From http://en.wikipedia.org/wiki/Transformation_matrix

"For rotation by an angle θ counter clockwise about the origin, the functional form is x' = xcos θ − ysin θ and y' = xsin θ + ycos θ"

It's easy to see how this works if you look at rotating points on a unit circle.

However, this only rotates point around the origin (0,0).

For objects, you will need to translate all the coordinates so that the axis of rotation is at the origin (i.e. translate them by (-px, -py)). Then, rotate them around the "origin". And then translate them back onto the screen coordinates.

[Raknarg]

Turing:

View.Set ("offscreenonly, graphics:700;700") const xPos := 350 const yPos := 350 var x, y, b : int := 0 % variables for the Mouse.Where var t : real % Theta, or the angle loop     Mouse.Where (x, y, b)     if x = xPos and y > yPos then         t := 270     elsif x < xPos and y = yPos then         t := 0     elsif x = xPos and y < yPos then         t := 90     elsif x > xPos and y = yPos then         t := 180     elsif x = xPos and y = yPos then         t := 0     elsif x > xPos then         t := arctand ((y - yPos) / (x - xPos)) + 180     else         t := arctand ((y - yPos) / (x - xPos))     end if     % This thing here calculates your angle. It'll make sense if you've ever done vectors before. It uses the function tan-1. The ones before it are for mathematical correctness; you cannot divide anything by zero.     Draw.ThickLine (xPos, yPos, xPos - round (40 * cosd (t)), yPos - round (40 * sind (t)), 4, 7) % This is important for drawing shapes according to an angle. You add onto the original postion. In the case of x, its the radius * cos(angle). For y it's radius * sin (angle)     View.Update     cls end loop

It's pretty simple. This is exacrly what randomletters was talking about. You can use this idea to rotate an object. Basically what you are doing is changing the x and y properties according to the angle given. You can imagine it as a right-angled triangle: In this triangle, you have the x (adjacent), the y (opposite) and your hypotenuse (distance a certain line covers). Lets go over some math:

How do you calculate the opposite if you have an angle and a hypotenuse? Like so:
opp (y) = hyp * sin (angle)

How do you calculate the adjacent if you have an angle and a hypotenuse? Like so:
adj (x) = hyp * cos (angle)

Now you have useful information. Instead of the length of the line, you now have the change in x and the change in y for this line, which you can use in drawing a line.

I probably explained this poorly, so if it's still confusing, the internet is a great place to look.

[Velocity]

@ raknarg do you know the formula for cod and sin and tan or something? you some sort of super genius? how do you implement it in a long line that so easily :OOOOO

[Raknarg]

Yeah, it's basic trig:

sin (theta) = opposite/hypotenuse
cos (theta) = adjacent/hypotenuse

Or do you mean the formula to for cos and sin themselves? no.

[smool]

alternatly, if you wanted to rotate not to a set angle but rather rotate it by x degree more than what you already have:

X' = Xcos(theta) - Ysin(theta)
Y' = Xsin(theta) + Ycos(theta)

That rotates around coordinate (0, 0), so you would have to translate everything, rotate, then translate again. You might want to have a procedure. Your code might look like this:

Turing:

View.Set ("offscreenonly") var x : array 1..4 of int := init (100, 100, 300, 300) var y : array 1..4 of int := init (100, 300, 300, 100) var mx, my, mxpast, bnum : int := 0 proc Rotate (var x_, y_ : int, xmid, ymid, theta : int)     var x, y : int     x_ -= xmid     y_ -= ymid     x := round (x_ * cosd (theta) - y_ * sind (theta))     y := round (x_ * sind (theta) + y_ * cosd (theta))     x_ := x + xmid     y_ := y + ymid end Rotate loop     mousewhere (mx, my, bnum)     for a : 1..4         Rotate (x (a), y (a), 200, 200, mx - mxpast)     end for     Draw.FillPolygon (x, y, 4, 42)     View.Update     delay (10)     cls     mxpast := mx end loop

[Raknarg]

Note that in that procedure, the rotation is inverted when your my is above 200. Don't know if thatwas intentional.

[Nablo]

Thanks a lot guys! I really appreciate all of your help! Just another thing, I tried using the transformation Matrix but as one of you guys already said, it rotates the object around the origin rather than a point set by you. Raknarg and smool, I havn't tried using that approach but I will! Promise to send you guys the program once I am done with it! Dreadnough, I checked out your program and it is quite interesting! I'm still trying to understand all of your programs but hopefully I'll get through the mind boggling stage and understand the. Again, thanks a lot for your help!

Ps. What does -= or += mean?

[Aange10]

Turing:

var n : int n += 1 % is the same as n := n + 1 % Likewise n -= 1 % Is the same as n := n - 1

[Insectoid]

In case you're wondering, /=, *=, and I think even div= and mod= are valid (I don't have Turing available to test div/mod though).