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Mathcad 15 , Symbolic Evaluate ?

lvl107
20-Turquoise

Mathcad 15 , Symbolic Evaluate ?

Hello Everyone.
From :

Symbolic_Evaluate.PNG

Mathcad 15 , How to : M , Symbolic Evaluate ---> ?
Thanks in advance for your time and help.
Best Regards.
Loi.

ACCEPTED SOLUTION

Accepted Solutions
ttokoro
20-Turquoise
(To:lvl107)

11 REPLIES 11
ttokoro
20-Turquoise
(To:lvl107)

A=P1,B=P2,C=P3

image.png

lvl107
20-Turquoise
(To:ttokoro)

  I greatly appreciate your time and help. Many, many, many thanks, Ttokoro. 🤔 🙄. I got it :

I.png

   The reason that I need help with this question is the following :

1.PNG

   Best Regards.

     Loi.

Werner_E
25-Diamond I
(To:lvl107)

  • Your "Minimize" command is only minimizing the function f which calculates the distance from O to P. You should not be surprised that the result is the point P (or a point close to it within the set numerical precision)
    If you want the other yellow equations seen as constraints you have to create a solve block by adding "Given" in front of it.
    Furthermore its advisable to decrease the value of CTOL
  • The functions of @ttokoro you try to use to draw the arc are not suitable for real Mathcad. Mathcad does not provide the row selector which ttokoro used (you can't just replace it for the column selector as you did) and Mathcads 3D plot does not accept NaN's in the data values.
    Furthermore in the other thread ttokoro used 101 point for the linePS, you are only creating 11 points. So obviously you can't use a loop from 0 to 101!
    The programs for PSline and Cline have to be rewritten to suit the specifics of Mathcad which are different from Prime. 
    And although the 3D plot in Prime is extremely poor and far inferior to that in Mathcad, the ability to use NaN's and also the simpler data structure to plot a curve is a significant improvement over Mathcad (as is the introduction of the row selector).
    Conclusion: PTC did an incredible amount wrong, but not everything 😉

BTW, as already mentioned in the other thread - ttkoros way to plot the arc will fail if the arc angle is equal to or greater than 180 degree. Furthermore the points created to represent the arc are not equally spaced - points in the middle of the arc are farther apart. Using enough points can hide that effect efficiently, though.

 

Werner_E_0-1693476375424.png

 

Werner_E
25-Diamond I
(To:ttokoro)

@ttokoro 

Unfortunately your "solution" usually returns a wrong (or better: not the desired) result! It works OK in this case because the z-coordinate of the desired center is 0. But your symbolics solution will always use 0 for the z-coordinate of the result.

Reason is that your approach just returns ONE (of the infinite number of) point with the equal distance to all three given points. This usually is NOT a point in the plane of the three points.

You would need to add this coplanarity constraint in some way, too!!

 

Your solution could be written simpler by using an auxiliary function for the square of the distance of two points:

Werner_E_5-1693483065731.png

 

But if we translate all points by, lets say, (5;5;5), the result is not the center we look for (which should be (5;5;5).

Werner_E_6-1693483094556.png

Mathcad knows that there are an infinite number of solutions

Werner_E_8-1693483168496.png

which of course all lie on a straight line:

Werner_E_9-1693483255180.png

What we are looking for is the intersection of this line with the plane P1 P2 P3 ....

Werner_E_10-1693483496454.png

 

Guess that means - back to the drawing board 😉
Good luck!

 

BTW, in the other thread I also had troubles talking the symbolics into providing a solution for the problem and I ended up with a solution which was way more elaborate than I'd had wished it to be:

Werner_E_0-1693484576306.png

 

ttokoro
20-Turquoise
(To:Werner_E)

Werner_E
25-Diamond I
(To:ttokoro)

THATS sure a true solution and its beautiful because it is a direct calculation and does not require the  usage of the weak symbolic "solve".

Could you elaborate in detail on the thinking behind of this?  Never mind - got it!

Werner_E_0-1693490801889.png

 

 

Werner_E
25-Diamond I
(To:Werner_E)

Here is a slightly different function, based on the barycentric coordinates of the circumcircle and their conversion to cartesian coordinates

Werner_E_1-1693498801710.png

 

or with three separate point as input

Werner_E_2-1693499078032.png

 

 

LucMeekes
23-Emerald III
(To:lvl107)

Here's your problem, symbolically solved in Mathcad 11:

LucMeekes_0-1693514208142.png

(The symbolic answer is several pages wide and probably will not show in Mathcad 15)

Now with the given points P1...P3 I get your desired result:

LucMeekes_6-1693516015288.png

 

That's as close as numerical non-zero gets to 0.

But I can fill in arbitrary numbers, such as:

LucMeekes_7-1693516041249.png

or

LucMeekes_8-1693516063492.png

 

Success!
Luc

 

lvl107
20-Turquoise
(To:LucMeekes)

  I save your ***.mcd into ***.xmcd . And I think your MC11-engine is more powerful than my MC15-engine in this case. :

L.PNG 

  Best Regards.

      Loi.

lvl107
20-Turquoise
(To:LucMeekes)

  Many, many thanks again, Luc. 🤔 🙄 🤔 :

LM_3Dcurve.png

 

  Best Regards.

     Loi.

lvl107
20-Turquoise
(To:lvl107)

LM_3Dcurve .png

LM_3Dcurve_  .png

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