[How-to] Find the middle point between sides of two distant objects.
[In reference to my attachment]
I would like to find the middle point between object A's right-most segment (in green) and object B's left-most segment (in blue), quickly, effectively and reliably.
Ideally, there's a function i'm not familiar with that exists . I would rather not resort to having to create more objects to "figure out" the middle point manually, so-to speak.
Dandreu,
In a more general case than assumed in the first post, if you wish to have an actual key point, you may:
1) Direct Select each of the path segments in question and Ctrl/Cmd+C+F+X+F,
2) Select either and rotate it by 180 degrees,
3) Object>Blend>Blend Options, with 1 Step, then Object>Blend>Make, then Object>Blend>Expand,
4) Select the line in the middle and Object>Path>Add Anchor Point.
You may also do a similar way where you add an Anchor Point at the middle of the segments and ClickDrag with the Line Segment Tool between them, but that is a bit destructive.
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P Prakashyou could try something similar to this, for each table pair that you want to compare:
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As for your original question, there is no direct way to do it.
Especially not the way you phrased it,
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They have classes and instances.
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Type it in google, and you'll see people are working hard
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phase_differece.vi 66 KByou could try something similar to this, for each table pair that you want to compare:
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How to find the time difference between two times
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KrisThis is a procedure taht do the job
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LD$Date_Deb IN DATE DEFAULT SYSDATE
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,LN$MI OUT PLS_INTEGER
,LN$SS OUT PLS_INTEGER
) RETURN NUMBER
IS
dif NUMBER ;
BEGIN
IF LD$Date_Fin < LD$Date_Deb THEN
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END IF ;
SELECT LD$Date_Fin - LD$Date_Deb INTO dif FROM dual ;
SELECT TRUNC ( LD$Date_Fin - LD$Date_Deb) INTO LN$JJ FROM dual ;
SELECT TRUNC ( (LD$Date_Fin - LD$Date_Deb) * 24) - ( LN$JJ * 24 ) INTO LN$HH FROM dual ;
SELECT TRUNC ( (LD$Date_Fin - LD$Date_Deb) * 1440) - ( (LN$HH * 60) + ( LN$JJ * 1440) ) INTO LN$MI FROM dual ;
SELECT TRUNC ( (LD$Date_Fin - LD$Date_Deb) * 86400) - ( (LN$MI * 60) + (LN$HH * 3600) + ( LN$JJ * 3600 * 24 ) ) INTO LN$SS FROM dual ;
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Francois -
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I've answered your question in the other thread here:
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