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Oh, trigonometry~

• 5 posts
• I never figured out trigonometry enough to solve this little problem I have.

As you may know, when a layer rotates in construct, it rotates with its axis included, so the coordinates move with it, something like this:

<img src="http://img707.imageshack.us/img707/5083/001sng.png" border="0" />

<img src="http://img26.imageshack.us/img26/8682/002bcp.png" border="0" />

Now, I need a formula that, knowing the angle of rotation of the layer, tells me the visual position of a point (in the example, I would want (0,y) instead of (x,y)).

Any help would be greatly appreciated n__n!

Edit: note that the point can move freely. So I need to know the absolute position of the point of its relative position in the rotated layer.

• cx,cy are the coordinates of the center of rotation

xR,yR are the coordinates in a rotated layer at angle layerangle

x,y are the coordinates in an unrotated layer

x = cx + cos(-layerangle) * distance(cx,cy,xR,yR)

y = cy + sin(-layerangle) * distance(cx,cy,xR,yR)

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• It doesn't work because if the layer is not rotated (i.e. layerangle = 0), its sine = 0, so y = cy.

:<

• yeah your right

x = cx + cos(-layerangle+angle(cx,cy,xR,yR)) * distance(cx,cy,xR,yR)

y = cy + sin(-layerangle+angle(cx,cy,xR,yR)) * distance(cx,cy,xR,yR)

• Thank you very much! It works now!

You'll be in the credits of every game for the contest, it appears xD

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