I found it, I found it, GOD is so good.
The problem was in the definition of the 'ANG' setq. Why I defined that
I'll never know, I think it had to do with have a triangle upside down.
Well, any way, I started out by defining the roof pitch, asking for
user input for the roof pitch
, then I next did this, I defined a
starting point. (setq sd1 (list 0 0)), I then next input the pitch, so
if the user input, lets say 6.0 for the pitch the the next setq went
like this, (setq sd2 (polar sd1 up pitch)), up is defined as
(setq up (* pi (/ 90.0 180.0))) and remember the pitch was set to 6.0.
Next it took sd2 and moved it right the run
(setq sd3 (polar sd2 rt 12.0)). Then it did some funky stuff, like,
(setq sda1 (angle sd2 sd1) and (setq sda2 (angle sda1 sda3)). Now the
next item is what I use for setting the roof pitch,
(setq ANG (abs (- SDA2 SDA1))), which doesnt make any sence to me why I
did this, but it didn't work. Like I said I think it has to do with
inverting the geometrical triangle. This is why I figure this, sda1 is
the angle dn in radians, right? So it would be 4.71239 in radians which
is the same as (setq dn (* pi (/ 270.0 180.0)) = (setq sda1 (angle sd2
sd1), but the part which I should have used is (setq sda2 (angle sd1
sd3)) = 0.463648 radians which just happens to be 26 Degs 33 min 54 sec
= 26.57 degs or a 6in pitch. This is what I used to corect the statment
and get the heel to draw correctly. But anywho, this statment, (setq
ANG (abs (- SDA2 SDA1))) returns this, SDA1 = 4.71239 radians and
SDA2 = 0.463648 radians, so 0.463648 minus 4.71239 equals -4.248742
radians whaich is a -243.4349848 degs?????????? So what could this be?
--
Nick Haury
AEC Design Services
4959 W. 5th Street
Greeley, CO 80634
Office and Fax (970)356-5511
Cell (970)313-3162
aecdesign@truevine.net
ftp site: http://www.truevine.net/~aecdesign@truevine.net
ECCAD wrote:
> Nick,
> It might help to attach a small drawing, with the
> expected result..and label the points. Also show
> approx. where the PT8 is located when incorrect.
>
> Bob
