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Re: Help with understanding matrices
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Re: Help with understanding matrices


  • Subject: Re: Help with understanding matrices
  • From: Simon Topliss <email@hidden>
  • Date: Tue, 19 Feb 2008 19:06:33 +0000

Thanks for the reply, Deivy. I've got plenty to think about now. I'll start throwing some real jobs at this and see how it goes. Skewing of an image is pretty rare, so this is more of a theoretical problem for now.


On 19 Feb 2008, at 03:54, deivy petrescu wrote:


On Feb 15, 2008, at 9:20, Simon Topliss wrote:


On 14 Feb 2008, at 21:07, Mark J. Reed wrote:
This page has a good overview of matrix transforms - it's about Flash,
but it uses the same Adobe-standard terminology:


http://www.senocular.com/flash/tutorials/transformmatrix/

Yes, I've read that. That's where I came up with the modified handler that uses the format below.


mvalue_a = x scale
mvalue_b = y skew
mvalue_c = x skew
mvalue_d = y scale
mvalue_tx = x translation (position)
mvalue_ty = y translation (position)

Ok, you're mixing up interpretations here. The first four values in
the matrix represent a combination of scaling with *either* skew or
rotation. You can consider either the skew or the rotation as the
primary value and the other one is the result of combining it with the
scale. But it doesn't make sense to try and extract both rotation and
skew from a matrix... you'll get two values that depend on each other.

Probably. I'm pretty poor at math!

I think I understand. But what I'm still not sure about is what if an object has been scaled, rotated and skewed along one axis (e.g horizontally skewed)?

This matrix represents an image that is at 100%, but has been horizontally skewed 12 degrees, then rotated 45 degrees.

{class:matrix, mvalue_a:0.707106769085, mvalue_b:-0.707106769085, mvalue_c:-0.556806564331, mvalue_d:-0.857406973839, mvalue_tx:-1.012733203125E+4, mvalue_ty:2137.982421875}

Rotating and skewing an image isn't common, but I'd like to get the handler nailed-down for all instances.

For example:

set {theta_deg, hscale, vscale, hskew, vskew} to decode_matrix ({mvalue_a:0.707106769085, mvalue_b:-0.707106769085, mvalue_c:-0.556806564331, mvalue_d:-0.857406973839, mvalue_tx:-1.012733203125E+4, mvalue_ty:2137.982421875})

log -- (*-38.21843067341, 0.900018629234, -1.091323521466, -128.21843067341, -38.21843067341*)

on decode_matrix(mat)
	set _a to mvalue_a of mat
	set _b to mvalue_b of mat
	set _c to mvalue_c of mat
	set _d to mvalue_d of mat
	set theta to atan2 {_c, _a}
	set cos_theta to cos (theta)
	set hscale to _a / cos_theta
	set vscale to _d / cos_theta
	set hskew to 180 / pi * theta - 90
	set vskew to 180 / pi * theta
	set theta_deg to theta / pi * 180
	return {theta_deg, hscale, vscale, hskew, vskew}
end decode_matrix

Ideally, what I want the handler to return would be {45.0, 100.0, 100.0, 12.0, 0.0}. So I guess some other calculations are going to be required?

Simon

Simon,
there are some constyraints. As Mark said, there are only so much you can do with the information you have. Besides, I have no idea on how to use "mvalue_ty" or "mvalue_tx". It could help.
Using the handler above and, since I knew what I needed to get :) this is the handler I came up with:


---

decode_matrix({mvalue_a:0.707106769085, mvalue_b:-0.707106769085, mvalue_c:-0.556806564331, mvalue_d:-0.857406973839, mvalue_tx:-1.012733203125E+4, mvalue_ty:2137.982421875})

on decode_matrix(mat)
	set _a to mvalue_a of mat
	set _b to mvalue_b of mat
	set _c to mvalue_c of mat
	set _d to mvalue_d of mat
	set thetax to atan (_b / _a)---x rotation
	set thetay to -1 * (atan (_c / _d))---y rotation
	set cos_thetax to cos (thetax)
	set cos_thetay to (cos (thetay))
	set hscale to _a / cos_thetax
	set vscale to _d / cos_thetay
	set thetax to 180 / pi * thetax
	set thetay to 180 / pi * thetay
	set skew to thetay - thetax
	return {thetax, thetay, hscale, vscale, skew}
end decode_matrix

--->{-45.0, -32.999998357716, 0.999999982886, -1.022340583597, 12.000001642284}

---

Noticed that I get skew not vskew or hskew. The problem is I have no way if you actually rotated 33 degrees then skewed vertically 12 or you rotated 45 degrees and skewed horizontally 12 (to the other side).
Once those 2 other variables are understood you have a better chance.


Deivy
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  • Follow-Ups:
    • Re: Help with understanding matrices
      • From: deivy petrescu <email@hidden>
References: 
 >Help with understanding matrices (From: Simon Topliss <email@hidden>)
 >Re: Help with understanding matrices (From: Simon Topliss <email@hidden>)
 >Re: Help with understanding matrices (From: deivy petrescu <email@hidden>)
 >Re: Help with understanding matrices (From: Simon Topliss <email@hidden>)
 >Re: Help with understanding matrices (From: "Mark J. Reed" <email@hidden>)
 >Re: Help with understanding matrices (From: Simon Topliss <email@hidden>)
 >Re: Help with understanding matrices (From: deivy petrescu <email@hidden>)

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