Re: Silly question

I'm still getting awkward values:

Allow me to rephrase:

v=
arcsin(Pz / C) for Px^2 + Py^2 >= A^2.
+pi - arcsin(Pz/C) if Pz >= 0
-pi - arcsin(Pz/C) if Pz < 0

u=
+arccos(Px / (A + Bcos(v))) for Py >= 0
-arccos(Px / (A + Bcos(v))) for Py >= 0

Did I understand correctly?


"Robert Israel" <israel@xxxxxxxxxxx> wrote in message
news:deu91c$ejc$1@xxxxxxxxxxxxxxxxxxxxxxxxx
> In article <deu6pj$bqg$1@xxxxxxxxxxxxxxxxxx>,
> Jaco van Niekerk <sparky@xxxxxxxxxx> wrote:
>>Hello
>>
>>I have a torus, say F(u,v) = X(u,v)i + Y(u,v)j + Z(u,v)k
>>
>>with
>> X(u,v) = Acos(u) + Bcos(u)cos(v)
>> Y(u,v) = Asin(u) + Bsin(u)cos(v)
>> Z(u,v) = Csin(v)
>>
>> A, B and C are constants
>
> ... with, I presume, A > B > 0 and C > 0. Yes, this is a torus
> for, say, -pi <= u <= pi and -pi <= v <= pi.
>
>>Now, for the silly part. I have a point P that definitely lies on the
>>surface. Now I need to find a pair (u,v) for the point P. Now this seems
>>trivial, but on my graphics display (I'm writing a program) it does not
>>work! This is what I've done:
>>
>>Pz = Csin(v)
>>sin(v) = Pz/C
>>v = arcsin(Pz/C)
>
> That gives you a v in [-pi/2, pi/2];
> OK if cos(v) >= 0, which is true if Px^2 + Py^2 >= A^2.
> Otherwise v = pi - arcsin(Pz/C) (if Pz >= 0), or -pi - arcsin(Pz/C)
> (if Pz < 0)
>
>>Px = Acos(u) + Bcos(u)cos(v)
>>Px = cos(u)[A + Bcos(v) ]
>>since I have v, I can calculate u
>>cos(u) = Px / (A + Bcos(v))
>>u = arccos(Px / (A + Bcos(v)))
>
> OK if Py >= 0. Otherwise -that.
>
>>Yet, my u, v pair is not what it should be. What am I doing wrong?
>
> Looking in the wrong intervals.
>
> Robert Israel israel@xxxxxxxxxxx
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada
>


.

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