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Pipe-fitting problem 3D



The Next CEO of Stack OverflowFinding $ csc theta $ given $ cot theta $Functional inverse of $sinthetasqrttantheta$$sin(2arccos(x))$, please help me understand how to do these kind of problems.Proving:$tan(20^circ)cdot tan(30^circ) cdot tan(40^circ)=tan(10^circ)$Compound angle formulasHelp needed verifying a trigonometric identityHow to Simplify $-frac14sinfrac x4 + fracsqrt34cosfrac x4 = 0$?Triangular sides$ frac5cos^2x = 7tan x + 3 $Solution to trig simultaneous equations










2












$begingroup$


I have a 3D pipe-fitting problem for which I was able to write the following equations:



$$
y = tan (a)sqrtx^2 + z^2\
z = tan (b)sqrtx^2 + y^2\
y = sin (a)sqrtx^2 + y^2 + z^2\
z = sin (b)sqrtx^2 + y^2 + z^2
$$



$x > 0$ and known; $y$ and $z ge 0$; $0 < a$, $b < 90º$



  1. what choices for $a$ and $b$ will minimize $y + z$?


  2. if $a$ and $b$ are constrained to $45º$, $22.5º$ or $11.25º$, what choices minimize $y + z$?


I tried rearranging and combining these equations, and I got:
$$
y = fracx sin(a)sqrtcos^2(a) - sin^2(b)\
z = fracx sin(b)sqrtcos^2(b) - sin^2(a)
$$



But these equations behave strangely: as $a$ and $b$ go to $0$, $y$ and $z$ go to $0$; and in my problem, that means the pipes never meet.



Here is the physical problem: I have a vertical pipe, and a pipe aligned on the $z$-axis, separated by a fixed horizontal distance $x$. The length of the pipes can be varied. I need to connect the ends with one pipe and $2$ fittings from $0º$ to $90º$.



There is an easy solution with one $45º$ fitting, and one $90º$ fitting, but I am trying to avoid it.



So, did I model the problem correctly? Were my reduced equations correct? I am also having trouble seeing which variables are independent. Any help would be appreciated.










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I have a 3D pipe-fitting problem for which I was able to write the following equations:



    $$
    y = tan (a)sqrtx^2 + z^2\
    z = tan (b)sqrtx^2 + y^2\
    y = sin (a)sqrtx^2 + y^2 + z^2\
    z = sin (b)sqrtx^2 + y^2 + z^2
    $$



    $x > 0$ and known; $y$ and $z ge 0$; $0 < a$, $b < 90º$



    1. what choices for $a$ and $b$ will minimize $y + z$?


    2. if $a$ and $b$ are constrained to $45º$, $22.5º$ or $11.25º$, what choices minimize $y + z$?


    I tried rearranging and combining these equations, and I got:
    $$
    y = fracx sin(a)sqrtcos^2(a) - sin^2(b)\
    z = fracx sin(b)sqrtcos^2(b) - sin^2(a)
    $$



    But these equations behave strangely: as $a$ and $b$ go to $0$, $y$ and $z$ go to $0$; and in my problem, that means the pipes never meet.



    Here is the physical problem: I have a vertical pipe, and a pipe aligned on the $z$-axis, separated by a fixed horizontal distance $x$. The length of the pipes can be varied. I need to connect the ends with one pipe and $2$ fittings from $0º$ to $90º$.



    There is an easy solution with one $45º$ fitting, and one $90º$ fitting, but I am trying to avoid it.



    So, did I model the problem correctly? Were my reduced equations correct? I am also having trouble seeing which variables are independent. Any help would be appreciated.










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I have a 3D pipe-fitting problem for which I was able to write the following equations:



      $$
      y = tan (a)sqrtx^2 + z^2\
      z = tan (b)sqrtx^2 + y^2\
      y = sin (a)sqrtx^2 + y^2 + z^2\
      z = sin (b)sqrtx^2 + y^2 + z^2
      $$



      $x > 0$ and known; $y$ and $z ge 0$; $0 < a$, $b < 90º$



      1. what choices for $a$ and $b$ will minimize $y + z$?


      2. if $a$ and $b$ are constrained to $45º$, $22.5º$ or $11.25º$, what choices minimize $y + z$?


      I tried rearranging and combining these equations, and I got:
      $$
      y = fracx sin(a)sqrtcos^2(a) - sin^2(b)\
      z = fracx sin(b)sqrtcos^2(b) - sin^2(a)
      $$



      But these equations behave strangely: as $a$ and $b$ go to $0$, $y$ and $z$ go to $0$; and in my problem, that means the pipes never meet.



      Here is the physical problem: I have a vertical pipe, and a pipe aligned on the $z$-axis, separated by a fixed horizontal distance $x$. The length of the pipes can be varied. I need to connect the ends with one pipe and $2$ fittings from $0º$ to $90º$.



      There is an easy solution with one $45º$ fitting, and one $90º$ fitting, but I am trying to avoid it.



      So, did I model the problem correctly? Were my reduced equations correct? I am also having trouble seeing which variables are independent. Any help would be appreciated.










      share|cite|improve this question











      $endgroup$




      I have a 3D pipe-fitting problem for which I was able to write the following equations:



      $$
      y = tan (a)sqrtx^2 + z^2\
      z = tan (b)sqrtx^2 + y^2\
      y = sin (a)sqrtx^2 + y^2 + z^2\
      z = sin (b)sqrtx^2 + y^2 + z^2
      $$



      $x > 0$ and known; $y$ and $z ge 0$; $0 < a$, $b < 90º$



      1. what choices for $a$ and $b$ will minimize $y + z$?


      2. if $a$ and $b$ are constrained to $45º$, $22.5º$ or $11.25º$, what choices minimize $y + z$?


      I tried rearranging and combining these equations, and I got:
      $$
      y = fracx sin(a)sqrtcos^2(a) - sin^2(b)\
      z = fracx sin(b)sqrtcos^2(b) - sin^2(a)
      $$



      But these equations behave strangely: as $a$ and $b$ go to $0$, $y$ and $z$ go to $0$; and in my problem, that means the pipes never meet.



      Here is the physical problem: I have a vertical pipe, and a pipe aligned on the $z$-axis, separated by a fixed horizontal distance $x$. The length of the pipes can be varied. I need to connect the ends with one pipe and $2$ fittings from $0º$ to $90º$.



      There is an easy solution with one $45º$ fitting, and one $90º$ fitting, but I am trying to avoid it.



      So, did I model the problem correctly? Were my reduced equations correct? I am also having trouble seeing which variables are independent. Any help would be appreciated.







      trigonometry 3d






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 12 '14 at 0:57







      Michael Afrides

















      asked Feb 10 '14 at 22:20









      Michael AfridesMichael Afrides

      112




      112




















          1 Answer
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          $begingroup$

          Please look at my blog at forexsg.wordpress.com/2014/06/19/solution-for-3d-pipe-fitting-connection-vertical-wye-tee-and-elbow-or-another-wye-tee/



          I descirbed my solution there. /_a and /_b cannot go to zero at the same time. When a is increasing b is decreasing. They are inverse of each other. There are more variables than the equations can define. Two equations cannot define 5 variables. Even with two know angles you need to define at least one other distance x,y or z . So you have to fix two angles and one of the distances x y or z .






          share|cite|improve this answer









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            active

            oldest

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            oldest

            votes









            0












            $begingroup$

            Please look at my blog at forexsg.wordpress.com/2014/06/19/solution-for-3d-pipe-fitting-connection-vertical-wye-tee-and-elbow-or-another-wye-tee/



            I descirbed my solution there. /_a and /_b cannot go to zero at the same time. When a is increasing b is decreasing. They are inverse of each other. There are more variables than the equations can define. Two equations cannot define 5 variables. Even with two know angles you need to define at least one other distance x,y or z . So you have to fix two angles and one of the distances x y or z .






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              Please look at my blog at forexsg.wordpress.com/2014/06/19/solution-for-3d-pipe-fitting-connection-vertical-wye-tee-and-elbow-or-another-wye-tee/



              I descirbed my solution there. /_a and /_b cannot go to zero at the same time. When a is increasing b is decreasing. They are inverse of each other. There are more variables than the equations can define. Two equations cannot define 5 variables. Even with two know angles you need to define at least one other distance x,y or z . So you have to fix two angles and one of the distances x y or z .






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                Please look at my blog at forexsg.wordpress.com/2014/06/19/solution-for-3d-pipe-fitting-connection-vertical-wye-tee-and-elbow-or-another-wye-tee/



                I descirbed my solution there. /_a and /_b cannot go to zero at the same time. When a is increasing b is decreasing. They are inverse of each other. There are more variables than the equations can define. Two equations cannot define 5 variables. Even with two know angles you need to define at least one other distance x,y or z . So you have to fix two angles and one of the distances x y or z .






                share|cite|improve this answer









                $endgroup$



                Please look at my blog at forexsg.wordpress.com/2014/06/19/solution-for-3d-pipe-fitting-connection-vertical-wye-tee-and-elbow-or-another-wye-tee/



                I descirbed my solution there. /_a and /_b cannot go to zero at the same time. When a is increasing b is decreasing. They are inverse of each other. There are more variables than the equations can define. Two equations cannot define 5 variables. Even with two know angles you need to define at least one other distance x,y or z . So you have to fix two angles and one of the distances x y or z .







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jun 21 '14 at 6:49









                user159320user159320

                1




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