Message #1222

From: Roice Nelson <roice3@gmail.com>
Subject: Re: [MC4D] Re: MHT633 v0.1 uploaded
Date: Thu, 28 Oct 2010 10:13:26 -0500

>
> To select position of the ball center use "Area center" slider. On the
> left side it makes the area centered in the camera position (it’s good for
> strong FishEye view and for frequent shift-left-drag movement) and on the
> right side area is centered at the point of view (=rotation center). If you
> have slider in this position, area will not be recalculatied during
> left-drag rotations, because center remains the same.

Thank you for the explanation of this slider. I had not understood what it
was, and the program behavior makes perfect sense now. I find things more
intuitive when the slider is set all the way at the left (on Cam), but
that’s probably because I heavily use left-click drag.

> Second image is the view from inside of the cell. Of cource there you are
> inside of the central sticker, but from inside it’s transparent, so you
> don’t see its faces. I say - why not? It’s the only position from where you
> can see all stickers of the cell (but they hide the rest of space :( ) And
> you can see that face is really infinite :)
>

Thanks for this too, and I completely agree with you now that I understand
exactly where the viewer is in this situation.

> And don’t wait much from sticker shirinking: you will see more of 1C, but
> I’m not sure about stickers on the back side. And probably view with small
> stickers will be a little nonrealistic…

I wondered about how much the 1C would block everything as well. I guess it
will depend on how much it shrinks relative to everything else. One view I
like to play with, and which helps my intuition, is faceShrink = 1.0 and
small sticker shrink values (the so called "continuous cubie" view). This
lets me see the cell boundary locations while still peering through the
space. So I’m still looking forward to sticker shrink, whenever you are
able to support it.

You mentioned the {4,4,3} early on. I figure a difficulty there is the
slicing, since the vertex figure is not a tetrahedron. I am curious if you
were able to make progress with it or not though.

Thanks again for this puzzle. It is highly enjoyable to think about, and
such a fun foothold for gaining a better understanding of hyperbolic space.

All the best,
Roice