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Okay, this is driving me nuts. I have a bunch of 4 x .5 x 1 boxes which
are stacked corner to corner in a stair pattern. Given their layout, I
thought/assumed that they make a 30 degree angle with the ground (a
plane), but when I added another box on the side and rotated it up 30
degrees, it didn't line up with the blocks!
I'm stumped. Barring some kind of bug, I can't imagine why I'm having
this problem; the math isn't particularly complicated here. Can anyone
help me with this? Code and example image follow:
#declare stair_row = union
{
#declare stair_row_members = 1;
#declare stair_row_count = 0;
#while (stair_row_count < stair_row_members)
box { <0, 0, 0> <4, .5, 1> }
#declare stair_row_count = stair_row_count + 1;
#end
}
#declare stair_block = union
{
#declare stair_rows = 10;
#declare stair_row_count = 0;
#while (stair_row_count < stair_rows)
object { stair_row translate <0, stair_row_count/2, stair_row_count> }
#declare stair_row_count = stair_row_count + 1;
#end
}
#declare stair_rail = union
{
box { <0, -in, 0> <in*1, 0, 15> rotate <-30, 0, 0> }
}
I'm using MacMegaPOV 0.7 (I don't care for the present interface of the
official beta) on Mac OS 8.6, if anyone cares.
-Xplo
Post a reply to this message
Attachments:
Download 'mnah2-a.jpg' (13 KB)
Preview of image 'mnah2-a.jpg'
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The angle between the horizontal and the diagonal of a box with width 1 and
height .5 is
arctan(.5/1) = 26.56505117
The *sine* of 30 is 1/2. Not the tangent. If the *diagonal* had a length of
1, and the box had a height of .5, then the angle would be 30 degrees.
- Slime
[ http://www.slimeland.com/ ]
[ http://www.slimeland.com/images/ ]
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Instead of .5 use the tangent of your desired angle (30) for your y value. Since, your
z value is equal to (1) this will work
(unless I just got a giant case of stupid going).
.5773503
Grim
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Oops, you'll have to change your translation of the steps in the y-direction by that
same value times your stair_row_count, in order
for this to work out.
Grim
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Slime wrote:
>
> The angle between the horizontal and the diagonal of a box with width 1 and
> height .5 is
>
> arctan(.5/1) = 26.56505117
>
> The *sine* of 30 is 1/2. Not the tangent. If the *diagonal* had a length of
> 1, and the box had a height of .5, then the angle would be 30 degrees.
Doh!! Stupid trig functions, I always get them mixed up.. could have
sworn it was sine I wanted.
Thanks a lot.
-Xplo
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"Xplo Eristotle" <inq### [at] unforgettable com> wrote in message
news:3CA92AB5.8DAC1279@unforgettable.com...
> Okay, this is driving me nuts. I have a bunch of 4 x .5 x 1 boxes which
> are stacked corner to corner in a stair pattern. Given their layout, I
> thought/assumed that they make a 30 degree angle with the ground (a
> plane), but when I added another box on the side and rotated it up 30
> degrees, it didn't line up with the blocks!
>
> I'm stumped. Barring some kind of bug, I can't imagine why I'm having
> this problem; the math isn't particularly complicated here. Can anyone
> help me with this? Code and example image follow:
>
> #declare stair_row = union
> {
> #declare stair_row_members = 1;
> #declare stair_row_count = 0;
> #while (stair_row_count < stair_row_members)
> box { <0, 0, 0> <4, .5, 1> }
> #declare stair_row_count = stair_row_count + 1;
> #end
> }
>
> #declare stair_block = union
> {
> #declare stair_rows = 10;
> #declare stair_row_count = 0;
> #while (stair_row_count < stair_rows)
> object { stair_row translate <0, stair_row_count/2, stair_row_count> }
> #declare stair_row_count = stair_row_count + 1;
> #end
> }
>
> #declare stair_rail = union
> {
> box { <0, -in, 0> <in*1, 0, 15> rotate <-30, 0, 0> }
> }
>
> I'm using MacMegaPOV 0.7 (I don't care for the present interface of the
> official beta) on Mac OS 8.6, if anyone cares.
>
> -Xplo
Perhaps because the tangent of 30 degrees is about 0.577350269, not 0.5 ?
Robert Chaffe
http://www.donovansweb.com/~chaffe/
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D'oh ! Figures someone else would get the answer to you before I did. Oh, well.
Robert Chaffe
http://www.donovansweb.com/~chaffe/
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There's a pretty good way to remember this. It worked for me, and has for two decades.
sine = opposite/hypotenuse (oh)
cosine = adjacent/hypotenuse (ah)
tangent = opposite/adjacent (oa)
Grim
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"GrimDude" <vos### [at] gulfnet> wrote in message news:3ca9f1ae@news.povray.org...
> There's a pretty good way to remember this. It worked for me, and has for two
decades.
>
> sine = opposite/hypotenuse (oh)
> cosine = adjacent/hypotenuse (ah)
> tangent = opposite/adjacent (oa)
You mean "SOHCAHTOA"? I learnt that at school. I used to be good at trig. with a
tables book... (still got it), wish I kept
it up...
I used to play the trumpet fairly good too, but time winds on.
<sniff>
~Steve~ ;)
>
> Grim
>
>
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25ct wrote:
> You mean "SOHCAHTOA"? I learnt that at school. I used to be good
> at trig. with a tables book... (still got it), wish I kept it up...
I still find it easier to picture, at least, using vectorised versions.
e.g.
x = r*COS(t)
y = r*SIN(t)
t = TAN-1(x/y)
r = SQRT((x^2)+(y^2))
--
Tim Cook
http://empyrean.scifi-fantasy.com
-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------
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