Comments on: Drawing 3D perspective lines in Flash

By: seb

seb — Sun, 27 Apr 2008 16:19:11 +0000

@Evan – thanks!

@Irishman – I’m not sure exactly what you mean but sounds interesting… fire me a mail and I’ll let you know!

cheers

Seb

By: Irishman

Irishman — Sat, 05 Apr 2008 00:15:59 +0000

Hey Seb, as always your examples are inspirational.

Lately I have found a nice workflow between Illustrator -> Blender -> Papervision3D combined with Jim Armstrong’s Singularity library to generate cubic bezier curves.

Basically it would allow one to draw abstract line effects similar to those shown on this site.

http://veerle.duoh.com/blog/comments/using_the_blend_tool_and_distort_zig_zag_effect_in_illustrator/

Let me know if you are interested and I can forward it on. It might make a nice tut.

Cheers, Tim

By: Evan

Evan — Fri, 04 Apr 2008 22:44:40 +0000

Very inspirational Seb! Can’t wait as I see this
line effect all the time on commercials on television and it will be cool to have it in Papervision.

By: Seb Lee-Delisle

Seb Lee-Delisle — Tue, 25 Mar 2008 23:20:54 +0000

hey keith! All that work and you just notice your sliders huh?

But yes should have credited you, sorry! Thank you for your nice easy components… love em!

@nik, yeah I’ve got a couple of sessions planned for FlashBelt, wait and see

cheers
Seb

By: Keith Peters

Keith Peters — Tue, 25 Mar 2008 18:49:35 +0000

Nice sliders.

By: Nik

Nik — Tue, 25 Mar 2008 03:23:43 +0000

Nice work Seb!
Can’t wait to see what else you’ve been up to when you come to Minneapolis this year for Flashbelt!

Nik

By: seb

seb — Mon, 24 Mar 2008 18:43:05 +0000

Hey guys, glad you like it so far – integrating it into PV3D is gonna be a bit tricky though… but I’m on it.. .

Andy! Nice work, your vector maths is clearly a lot better than mine… I pretty much understood your explanation though.

Maybe I’ll post a code snippet later, the only possible problem with my algorithm is that I need to get the distance between the 2 points, and as we all know, the Math.sqrt method is heavy on the CPU.

I don’t mind trig solutions though, as long as there aren’t any sqrts involved! So how would you calculate the angle of r0?

cheers!

Seb

By: Andy Geers

Andy Geers — Mon, 24 Mar 2008 18:20:16 +0000

Been working on a vector based solution… I can get so far but then have to resort to trig to do the final step.
Start with these definitions:
* the vector _x0_ is the centre of one circle
* the vector _x1_ is the centre of the other
* the vector _r0_ is the radius of the first circle, in the direction of the point we want to draw our line from (_t0_)
* the vector _r1_ is the radius of the second circle, in the direction of the point we want to draw our line to (_t1_)

We know that the vector from _t0_ to _t1_ (_t1_ – _t0_) is at 90 degrees to _r0_, and _r0_ is parallel to _r1_ (based upon your diagrams, which I assume are correct).

Therefore (_t1_ – _t0_) * _r0_ = 0, but we know that _t0_ = _x0_ + _r0_ and _t1_ = _x1_ + _r1_, and we can define _r1_ as n * _r0_, where n = |_r1_| / |_r0_|, so we end up with this:

[(_x0_ + n_r0_) - (_x1_ + _r0_)] * _r0_ = 0

Do a bit of rearranging and you end up with this:
_r0_ * (_x0_ – _x1_) = (1 – n) * |_r0_|^2
…where everything is a known quantity except the _r0_ on the far left. This is where I have to resort to trigonometric methods – you could use the definition of the dot product as |a||b|cos(theta) to work out the angle of _r0_ and from there get to the answer. But maybe there’s a neater way to solve that last step.

Hope you followed all that!

Andy
(from barcampbrighton2)

By: Nate Chatellier

Nate Chatellier — Mon, 24 Mar 2008 16:25:01 +0000

Sweet! I can’t wait to start playing with it!

By: noah aronsson-brown

noah aronsson-brown — Mon, 24 Mar 2008 16:12:05 +0000

That’s really elegant and cool! I would love to have true lines in PV3D too, especially to use on my portfolio front page ( http://www.deafmetal.com/portfolio/ ). I can’t wait to see it implemented. Great work!